Classification of minimal immersions of conformally flat 3-tori and 4-tori into spheres by the first eigenfunctions

Abstract

The Clifford 2-torus in $\mathbb {S}^3$ and the equlaterial 2-torus in $\mathbb {S}^5$ are known as the only minimal immersions of 2-tori into spheres by the first eigenfunctions (called $\lambda _1$-minimal for short). For $n\ge 3$, the Clifford n-torus in $\mathbb {S}^{2n-1}$ might be the only known example of $\lambda _1$-minimal n-tori in the literature. By discussing the general construction of homogeneous minimal flat n-tori in spheres, we construct several new examples of $\lambda _1$-minimal flat 3-tori and 4-tori. In particular, the existence of 2-parameter family of non-congruent $\lambda _1$-minimal flat 4-tori is shown for the first time. We obtain the complete classification for $\lambda _1$-minimal immersions of conformally flat 3-tori and 4-tori in spheres, by some detailed investigations of shortest vectors in lattices, which could be of independent interests. Using them, we also solve the Berger’s problem (finding the maximal value of the dilation-invariant functional $\lambda _1(g)V(g)^{\frac{2}{n}}$) among all flat 3-tori and 4-tori.

1 Introduction

The study of minimal immersions of Riemannian manifolds into spheres is an interesting topic in differential geometry. It builds a deep link between the spectral theory and minimal submanifold theory. The famous theorem of Takahashi [31] states that the isometric immersion $x:(M^n,g)\rightarrow \mathbb {S}^m$ is minimal, if and only if the coordinate functions of x are eigenfunctions of the Laplacian with respect to the eigenvalue n.

Given a closed Riemannian manifold $(M^n, g)$, we denote the volume and the first eigenvalues of $(M^n, g)$ respectively by V(g) and $\lambda _1(g)$. The dilation-invariant functional

$$\begin{aligned} {\mathcal {L}}_1(g)\triangleq \lambda _1(g)V(g)^{\frac{2}{n}} \end{aligned}$$

on the set of all smooth Riemannian metrics is a basic functional considered in spectral theory. It was shown in [10] (see also [12]) that the critical metric of ${\mathcal {L}}_1(g)$ among all smooth Riemannian metrics on $M^n$ (called $\lambda _1$-critical metric) admits an isometric minimal immersion of $M^n$ in spheres by the first eigenfunctions (called $\lambda _1$-minimal immersion).

Moreover, it was proved by Hersch [15] that on the topological 2-sphere, among all smooth Riemannian metrics, $8\pi$ is the maximal value of ${\mathcal {L}}_1(g)$, which can only be attained by the standard metric. In 1973, Berger [1] obtained the maximal value of ${\mathcal {L}}_1(g)$ on the topological 2-torus among all flat metrics. This value is attained by the $\lambda _1$-minimal immersion of the equilateral 2-torus in $\mathbb {S}^5$. Since then, finding the uniform upper bound of ${\mathcal {L}}_1(g)$ among all smooth metrics is referred as Berger’s problem. By introducing the conformal volume, Li and Yau [21] solved the case of $\mathbb {R}P^2$ in 1982, whose upper bound is attained by the $\lambda _1$-minimal immersion of $\mathbb {R}P^2$ in $\mathbb {S}^4$ (i.e., Veronese surface). Due to also the work of El Soufi and Ilias [9], on any closed manifold of dimension n, Li–Yau’s conformal volume can be used to provide an upper bound for ${\mathcal {L}}_1(g)$ in the conformal class [g]. In the mean time, they also show that for the conformal manifold $(M^n, [g])$ admitting a $\lambda _1$-minimal immersion in spheres, the volume of such $\lambda _1$-minimal immersion is exactly equal to the conformal volume. Therefore it realizes the supremum for ${\mathcal {L}}_1(g)$ in [g]. Furthermore, for a given conformal manifold, Montiel–Ros [23] and El Soufi-Ilias [10] proved there exists at most one $\lambda _1$-minimal metric in the conformal class. Combining this with the existence of $\lambda _1$-maximal metric, Nadirashvili finally solved the Berger’s problem completely for the topological 2-torus in [25]. We point out that for the case of Klein bottle, the Berger’s problem has also been solved due to the work of Nadirashvili [25], Jakobson-Nadirashvili-Polterovich [17] and El Soufi-Giacomini-Jazar [13], where one $\lambda _1$-minimal immersion of the Klein bottle is presented (see [16, 17]). For surfaces of higher genus, we refer to [19, 22, 28] and references therein. In contrast to the dimension 2, Colbois and Dodziuk proved in [4] that there is no uniform bound for the functional ${\mathcal {L}}_1(g)$ on any closed manifold of dimension $n\ge 3$. This implies that one might consider the Berger’s problem restricting to a given conformal class (see [11, 18, 27] and reference therein), for which the investigation of $\lambda _1$-minimal immersions of higher dimensional manifolds into spheres plays an important role.

In the literature, there have been several known classes of $\lambda _1$-minimal submanifolds in spheres. A famous conjecture of Yau states that any closed embedded minimal hypersurface in $\mathbb {S}^{n+1}$ is $\lambda _1$-minimal. Due to the work of Muto–Onita–Urakawa [24] and Tang–Yan [32] on this conjecture (see also [34]), we know all isoparametric hypersurfaces and some ones of their focal submanifolds form a class of $\lambda _1$-minimal submanifolds in spheres. Another class of examples is due to Takahashi [31], who proved that for any positive integer k, up to a dilation of the metric, any compact irreducible homogeneous Riemannian manifold can be immersed minimally into a certain sphere by the kth eigenfunctions (we call it $\lambda _k$-minimal immersion for short). Later, the case of sphere equipped with the constantly curved metric was investigated in detail by Do Carmo and Wallach [6]. It was proved that when $n\ge 3$, the linearly full $\lambda _k$-minimal immersion of n-sphere has rigidity if and only if $k\le 3$. Moreover, they also proved that the immersion will span the full k-eigenspace when $k\le 3$. In this paper, we will show that these two properties do not hold for minimal flat tori of dimension 4, even for the case of $\lambda _1$-minimal immersion. To be precise, we construct a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{11}$, which has rigidity but does not span the whole eigenspace (see Example 4.9). Furthermore, a 2-parameter family of non-congruent $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{23}$ is also constructed, among which there is a 1-parameter family living in $\mathbb {S}^{15}$, neither rigid nor fully-spanning the eigenspace (see Example 1.1, Proposition 6.3 and Remark 6.4).

By Kenmotsu [20] and Bryant [2], all minimal flat 2-tori in spheres are homogeneous. Let $\Lambda _n$ denote a lattice of rank n. In [2], Bryant proved that a flat torus $T^2=\mathbb {R}/\Lambda _2$ admits minimal immersions in spheres, if and only if the Gram matrix of $\Lambda _2$ is rational (i.e., all entries are rational numbers) up to some dilation. This implies there are infinite non-congruent minimal flat 2-tori in spheres. But among them, there are only two $\lambda _1$-minimal ones: the Clifford 2-torus in $\mathbb {S}^3$, and the equilateral 2-torus in $\mathbb {S}^5$. This classification is due to the work of Montiel–Ros [23] and El Soufi-Ilias [10].

In contrast to the plentiful results on dimension 2 in the literature, minimal flat tori of higher dimension haven’t been investigated so much, especially for those $\lambda _1$-minimal ones. As far as we know, the Clifford n-torus in $\mathbb {S}^{2n-1}$ is the only known $\lambda _1$-minimal example (see [26]). In this paper, we construct four non-congruent $\lambda _1$-minimal flat 3-tori, a 2-parameter family and another sixteen non-congruent $\lambda _1$-minimal flat 4-tori in spheres. Among them, the 2-parameter family described as follows is the most interesting.

Example 1.1

Denote by $\{e_i\}$ the standard basis of $\mathbb {R}^4$. The flat 4-torus

$$\begin{aligned} T^4=\mathbb {R}^4/\textrm{Span}_{\mathbb {Z}}\{e_1-e_4, e_2-e_4, e_3-e_4, 2e_4\} \end{aligned}$$

admits a 2-parameter family of non-congruent $\lambda _1$-minimal immersions in $\mathbb {S}^{23}$ given as follows:

$$\begin{aligned} \begin{aligned}&\!\!\Big (a_1e^{i\pi (u_1+u_2+u_3+u_4)},a_1e^{i\pi (u_1+u_2-u_3-u_4)},a_1e^{i\pi (u_1-u_2+u_3-u_4)},a_1e^{i\pi (-u_1+u_2+u_3-u_4)},\\&\quad a_2e^{i\pi (u_1+u_2+u_3-u_4)},a_2e^{i\pi (u_1+u_2-u_3+u_4)},a_2e^{i\pi (u_1-u_2+u_3+u_4)}, a_2e^{i\pi (u_1-u_2-u_3-u_4)},\\&\quad a_3e^{2i\pi u_1},a_3e^{2i\pi u_2},a_3e^{2i\pi u_3},a_3e^{2i\pi u_4}\Big ), \end{aligned} \end{aligned}$$

where $0\le a_1\le a_2\le a_3$ and $a_1^2+a_2^2+a_3^2=\frac{1}{4}$. See Sect. 6 for more details.

It turns out that the examples constructed in Sect. 4 exhaust all non-congruent $\lambda _1$-minimal immersions of conformally flat 3-tori and 4-tori into spheres.

Theorem 1

• (1) Up to congruence, there are five $\lambda _1$-minimal immersions of conformally flat 3-tori in spheres, see Table 1.

• (2) Up to congruence, there are sixteen, as well as a 2-parameter family, $\lambda _1$-minimal immersions of conformally flat 4-tori in spheres, see Table 2.

For the definition of irreducible and reducible appearing in Tables 1 and  2, see Sect. 4.

Table 1 The classification of $\lambda _1$-minimal immersions of confromally flat 3-tori

Table 2 The classification of $\lambda _1$-minimal immersions of confromally flat 4-tori

The construction of these examples is based on the variational characterizations (see Theorems 3.2 and  3.6) we obtained for homogeneous minimal flat tori in spheres. Roughly speaking, the construction of a homogeneous minimal flat n-torus in some sphere is equivalent to finding a 2-tuple $\{Y, Q\}$ (we call it matrix data) satisfying some constrains, where $Q^{-1}$ is a Gram matrix of the lattice corresponding to this torus, and Y is a set of finite integer vectors in $\mathbb {Z}^n$ describing the linear relations between lattice vectors involved in the minimal immersion. Theoretically, all homogeneous minimal flat tori in spheres can be constructed by the approaches we provided (see Sect. 3.1). To construct $\lambda _1$-minimal flat n-tori, we also need to deal with the problem of finding all shortest vectors in a lattice, which is important but difficult in the theory of lattice (or geometry of numbers). Fortunately, a result of Ryshkov [29] proved in Minkowski’s reduction theory can be used in our construction to overcome this obstacle.

To classify all $\lambda _1$-minimal immersions of conformally flat 3-tori and 4-tori in spheres, it follows from the work of El Soufi and Ilias [10] that we only need to classify all $\lambda _1$-minimal immersions of flat 3-tori and 4-tori in spheres. It turns out that they are all homogeneous (in fact, a sufficient condition is given in Proposition 2.7 for general minimal flat tori in spheres to be homogeneous). Note that the moduli space of flat tori (modulo isometry) is $SL(n,\mathbb {Z})\setminus GL(n,\mathbb {R})\,/\,O(n)$ (see [36], or Sect. 2). The action of $SL(n,\mathbb {Z})$ makes the classification highly nontrivial. However, our variational characterization suggests that all we need to do is to find out all the possible integer sets Y, where Q is uniquely determined if it exists. To do this, a coarse classification to lattices of rank no more than 4 is given at first (see Theorem 5.8), from which some necessary constrains on Y can be obtained. Then after introducing some invariants to the set of shortest lattice vectors, we can determine all the possibilities of Y up to the action of $SL(n,\mathbb {Z})$.

The volumes for these $\lambda _1$-minimal flat tori we construct are also calculated (see Sect. 4). It follows from the theory of conformal volume that these $\lambda _1$-minimal metrics maximize the functional $\lambda _1(g)V(g)^{\frac{2}{n}}$ in their respective conformal classes. Among all $\lambda _1$-minimal flat 3-tori, the maximal value of $\lambda _1(g)V(g)^{\frac{2}{n}}$ is $4 \root 3 \of {2} \pi ^2$, and it is $4 \root 2 \of {2} \pi ^2$ among all $\lambda _1$-minimal flat 4-tori (see Tables 1 and 2). In fact, using the investigation about lattices given in Sects. 5 and  6, we can prove the following theorem, which can be seen as a generalization of Berger’s result from flat 2-tori to flat 3 and 4-tori.

Theorem 2

Consider the dilation-invariant functional $\lambda _1(g)V(g)^{\frac{2}{n}}$ on the topological n-torus.

(1) When $n=3$, among all flat metrics,

$$\begin{aligned} \lambda _1(g)V(g)^{\frac{2}{n}}\le 4 \root 3 \of {2} \pi ^2, \end{aligned}$$

and the equality is attained by the $\lambda _1$-minimal flat 3-torus given in Example 4.5.

(2) When $n=4$, among all flat metrics,

$$\begin{aligned} \lambda _1(g)V(g)^{\frac{2}{n}}\le 4 \sqrt{2} \pi ^2, \end{aligned}$$

and the equality is attained by those $\lambda _1$-minimal flat 4-tori given in Example 1.1.

Inspired by this theorem, it is natural to consider the Berger’s problem on conformally flat 3-tori and 4-tori: whether $4 \root 3 \of {2} \pi ^2$ and $4 \sqrt{2} \pi ^2$ are respectively the upper bounds of $\lambda _1(g)V(g)^{\frac{2}{n}}$ on the topological 3-torus and 4-torus among all smooth conformally flat metrics.

In [11], El Soufi and Ilias exhibited a class of flat n-tori for which the endowed flat metric maximizes $\lambda _1(g)V(g)^{\frac{2}{n}}$ on its conformal class (see Corollary 3.1 in their paper). Combining their result with our work (Theorems 1 and  5.8), we can partially solve the above problem.

Theorem 3

Suppose g is a smooth Riemannian metric on the the topological n-torus. If g is conformal equivalent to a flat metric whose first eigenspace is of dimension no less than 2n, then $\lambda _1(g)V(g)^{\frac{2}{n}}\le 4 \root 3 \of {2} \pi ^2$ when $n=3$, and $\lambda _1(g)V(g)^{\frac{2}{n}}\le 4 \sqrt{\pi ^2}$ when $n=4$.

The paper is organized as follows. In Sect. 2, we firstly recall the basic spectral theory of flat tori, and then discuss the homogeneity of minimal flat tori in spheres. Section 3 is devoted to presenting our basic setup on homogeneous minimal flat tori, as well as the variational characterizations obtained for them. New examples of $\lambda _1$-minimal flat 3-tori and 4-tori are constructed in Sect. 4. We devote Sect. 5 to investigate the shortest vectors of lattices, where a coarse classification is given to lattices of rank no more than 4. The classification of $\lambda _1$-minimal immersions of conformally flat 3-tori and 4-tori are obtained in Sect. 6. A class of $\lambda _1$-minimal flat n-tori is presented in Sect. 7 as an application of our construction method in higher dimensions. Finally, Sect. 8 is devoted to discussing Berger’s problem on conformally flat 3-tori and 4-tori, where Theorems 2 and 3 are proved.

2 On isometric minimal immersions of flat tori

In this section, we will firstly recall the basic theory of flat tori. Then a sufficient condition for minimal flat tori in spheres to be homogeneous will be given.

2.1 Flat tori and lattices

It is well known that a flat torus $T^n$ of dimension n can be described as

$$\begin{aligned} T^n=\mathbb {R}^n/{\Lambda _n}, \end{aligned}$$

where $\Lambda _n$ is a lattice of rank n on $\mathbb {R}^n$. Set $L_n$ to be the generator matrix of $\Lambda _n$, which means $\Lambda _n$ can be generated by row vectors of $L_n$. Two tori $T^n=\mathbb {R}^n/{\Lambda _n}$ and $\widetilde{T}^n=\mathbb {R}^n/{\widetilde{\Lambda }_n}$ are isometric if and only if $\Lambda _n$ and $\widetilde{\Lambda }_n$ are isometric, i.e., there exists an orthogonal matrix O and an unimodular matrix $U\in SL(n,\mathbb {Z})$, such that $L_n=U\,\widetilde{L}_n\, O$, where $L_n$ (w.r.t. $\widetilde{L}_n$) is a generator matrix of lattice $\Lambda _n$ (w.r.t. $\widetilde{\Lambda }_n$). It follows that the moduli space of flat n-tori is

$$\begin{aligned} SL(n,\mathbb {Z})\setminus GL(n,\mathbb {R})\,/\,O(n). \end{aligned}$$

The dual lattice of $\Lambda _n$ is defined to be a lattice $\Lambda _n^{*}$, whose generator matrix ${L}_n^*$ satisfies ${L}_n({L}_n^*)^t=I_n.$ The spectrum of $T^n=\mathbb {R}^n/{\Lambda _n}$ is

$$\begin{aligned} \textrm{Spec}(T^n)=\Big \{4\pi ^2|\xi |^2\,\Big |\,\xi \in \Lambda _n^*\Big \}, \end{aligned}$$

and $e^{2\pi \langle \xi ,u \rangle i}$ is an eigenfunction corresponding to the eigenvalue $4\pi ^2|\xi |^2$, where $u=(u_1,u_2,\cdots ,u_n)$ is the coordinates of $\mathbb {R}^n$, such that the flat metric on $\mathbb {R}^n$ ($T^n$) can be written as $du_1^2+d u_2^2+\cdots +d u_n^2$.

2.2 Minimal homogeneous flat tori in spheres

Let $T^n=\mathbb {R}^n/{\Lambda _n}$ be a flat torus. Assume n is an eigenvalue of this torus, whose eigenspace is of dimension 2N. It follows that there are exactly N distinct lattice vectors (up to $\pm 1$) having the length $\frac{\sqrt{n}}{2\pi }$ in the dual lattice $\Lambda _n^*$, which are denoted by

$$\begin{aligned} \xi _1,\xi _2,\ldots ,\xi _N. \end{aligned}$$

By the theorem of Takahashi, any minimal isometric immersion of $T^n$ in spheres can be expressed as follows:

$$\begin{aligned} x=\begin{pmatrix}\Theta _1&\Theta _2&\cdots&\Theta _N\end{pmatrix} A:T^n=\mathbb {R}^n/{\Lambda _n} \longrightarrow \mathbb {S}^{2N-1}, \end{aligned}$$

(1)

where $\Theta _r=\begin{pmatrix}\cos \theta _r&\sin \theta _r\end{pmatrix},$ $\theta _r=2\pi \langle \xi _r,u\rangle$ for $~1\le r\le N$, and A is a $2N\times 2N$ matrix. Write

$$\begin{aligned} AA^t=\begin{pmatrix} A_{11}&{} A_{12}&{}\cdots &{} A_{1N}\\ A_{21}&{} A_{22}&{}\cdots &{} A_{2N}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ A_{N1}&{} A_{N2}&{}\cdots &{} A_{NN}\end{pmatrix},~~~ A_{rs}=\begin{pmatrix} A_{rs}^{11}&{} A_{rs}^{12}\\ A_{rs}^{21}&{} A_{rs}^{22}\end{pmatrix}, \end{aligned}$$

we have the following conclusions.

Lemma 2.1

$\Big \{\theta _r\pm \theta _s\,\Big |\, 1\le r\ne s\le N\Big \}\cap \Big \{0, \pm 2\theta _j\,\Big |\, 1\le j\le N\Big \}=\varnothing .$

Proof

This follows from the fact that $0<|\xi _r\pm \xi _s|<|2\xi _j|$ for $r\ne s$. $\square$

Lemma 2.2

For every r, $1\le r\le N$, we have

$$\begin{aligned} A_{rr}=\begin{pmatrix} a_r&{}\\ {} &{}a_r\end{pmatrix}. \end{aligned}$$

Proof

By definition, we have $A_{rr}^{12}= A_{rr}^{21}$. From $|x|=1$, we can obtain that

$$\begin{aligned} \begin{aligned} 2&=\sum _{1\le r\le N}[( A_{rr}^{11}- A_{rr}^{22})\cos (2\theta _r)+( A_{rr}^{11}+ A_{rr}^{22})+( A_{rr}^{12}+ A_{rr}^{21})\sin (2\theta _r) \\&\quad +\sum _{1\le r\ne s\le N}[( A_{rs}^{11}- A_{rs}^{22})\cos (\theta _r+\theta _s)+( A_{rs}^{11}+ A_{rs}^{22})\cos (\theta _r-\theta _s)\\&\quad +( A_{rs}^{12}+ A_{rs}^{21})\sin (\theta _r+\theta _s) +( A_{rs}^{21}- A_{rs}^{12})\sin (\theta _r-\theta _s)]. \end{aligned} \end{aligned}$$

Note that for all $1\le r\le N$, $\theta _r\not \equiv 0$. So it follows from the Lemma 2.1 that $A_{rr}^{11}- A_{rr}^{22}=0,~ A_{rr}^{12}+ A_{rr}^{21}=0,$ which complete the proof. $\square$

Since x is an isometric immersion, we have

$$\begin{aligned} \langle \frac{\partial x}{\partial u_k},\frac{\partial x}{\partial u_l}\rangle =\delta _{kl},~~~1\le k,l\le n, \end{aligned}$$

(2)

where $u=(u_1, u_2, \cdots , u_n)$ is the coordinates of $\mathbb {R}^n$. Using the expression (1) of x, (2) can be rewritten as below:

$$\begin{aligned}{} & {} \sum _{r}\xi _{rk}\xi _{rl}\Theta _r \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix} A_{rr}\begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix}\nonumber \\{} & {} \quad \Theta _r^t+\sum _{r<s}(\xi _{rk}\xi _{sl}+\xi _{sk}\xi _{rl})\Theta _r \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix} A_{rs}\begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix} \Theta _s^t=\frac{\delta _{kl}}{4\pi ^2}, \end{aligned}$$

(3)

where $\xi _{rk}$ is the k-th coordinates of $\xi _r$, and we have used the fact that

$$\begin{aligned} \Theta _r\begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix} A_{rs}\begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix}\Theta _s^t=\Theta _s\begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix} A_{sr}\begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix}\Theta _r^t, \end{aligned}$$

which can be verified easily. It follows from Lemma 2.2 that the first term in the left hand side of the Eq. (3) is constant. Hence for all $1\le k,l\le n$, we have

$$\begin{aligned} \begin{aligned} 0&=\sum _{r<s}(\xi _{rk}\xi _{sl}+\xi _{sk}\xi _{rl})[( A_{rs}^{11}- A_{rs}^{22})\cos (\theta _r+\theta _s)+( A_{rs}^{11}+ A_{rs}^{22})\cos (\theta _r-\theta _s)\\&\quad + ( A_{rs}^{12}+ A_{rs}^{21})\sin (\theta _r+\theta _s) +( A_{rs}^{21}- A_{rs}^{12})\sin (\theta _r-\theta _s)]. \end{aligned} \end{aligned}$$

(4)

Define ${\mathcal {E}}=\{\xi _r\pm \xi _s,1\le r<s\le n\}$, we call the set of pairs

$$\begin{aligned} \{(\xi _{r_1},\xi _{s_1}), \ldots ,(\xi _{r_p},\xi _{s_p}), (\xi _{r_{p+1}}, -\xi _{s_{p+1}}), \ldots , (\xi _{r_q}, -\xi _{s_q})\} \end{aligned}$$

$\eta$-set if

$$\begin{aligned} \xi _{r_1}+\xi _{s_1}=\cdots =\xi _{r_p}+\xi _{s_p}=\xi _{r_{p+1}}-\xi _{s_{p+1}}=\cdots =\xi _{r_q}-\xi _{s_q}=\eta \in {\mathcal {E}}. \end{aligned}$$

Using $|\xi _{r_j}|=|\xi _{s_j}|$ we have

$$\begin{aligned} \langle \eta , \eta \rangle =|\xi _{r_j}|^2+|\xi _{s_j}|^2\pm 2\langle \xi _{r_j},\xi _{s_j}\rangle =2\langle \xi _{r_j},\eta \rangle >0,\quad 1\le j\le q. \end{aligned}$$

It is straightforward to verify that $\pm \xi _{r_1},\ldots ,\pm \xi _{r_q}, \pm \xi _{s_1},\ldots ,\pm \xi _{s_q}$ are distinct with each other.

Denote by $\xi _{r_{j}}\odot \xi _{s_{j}}$ the symmetric product of $\xi _{r_{j}}$ and $\xi _{s_j}$, we have the following lemma.

Lemma 2.3

Let $x:T^n=\mathbb {R}^n/{\Lambda _n} \longrightarrow \mathbb {S}^{ m}$ be a linearly full minimal flat torus. If for any $\eta \in {\mathcal {E}}$, the $\eta$-set forms a linearly independent set of symmetric products $\{ \xi _{r_1}\odot \xi _{s_1}, \xi _{r_2}\odot \xi _{s_2}, \cdots ,\xi _{r_q}\odot \xi _{s_q}\}$, then m is odd and x is homogeneous.

Proof

Set

$$\begin{aligned} t_{r_{a}s_a}\triangleq A_{r_{a}s_{a}}^{11}- A_{r_{a}s_{a}}^{22},~1\le a\le p,~~~t_{r_{b}s_b}\triangleq A_{r_{b}s_{b}}^{11}+ A_{r_{b}s_{b}}^{22},~p+1\le b\le q. \end{aligned}$$

By the linear independence of trigonometric functions, it follows from (4) that ${t_{r_{1}s_{1}},\ldots ,t_{r_{q}s_{q}}}$ satisfy the following linear equations:

$$\begin{aligned}{} & {} (\xi _{r_{1}j}\xi _{s_{1}k}+\xi _{r_{1}k}\xi _{s_{1}j})t_{r_{1}s_{1}}\nonumber \\{} & {} \quad +(\xi _{r_{2}j}\xi _{s_{2}k}+\xi _{r_{2}k}\xi _{s_{2}j})t_{r_{2}s_{2}}\nonumber \\{} & {} \quad +\cdots +(\xi _{r_{q}j}\xi _{s_{q}k}+\xi _{r_{q}k}\xi _{s_{q}j})t_{r_{q}s_{q}}=0,~~1\le j,k\le n. \end{aligned}$$

(5)

Note that (5) is equivalent to

$$\begin{aligned} t_{r_{1}s_{1}}\xi _{r_{1}}\odot \xi _{s_{1}}+t_{r_{2}s_{2}}\xi _{r_{2}}\odot \xi _{s_{2}}+\cdots +t_{r_{q}s_{q}}\xi _{r_{q}}\odot \xi _{s_{q}}=0. \end{aligned}$$

It implies $t_{r_{j}s_{j}}=0$ for all $1\le j\le q$, since these symmetric products are linearly independent. Similarly, the other coefficients also vanish in (4) and we have $A_{rs}=0~ (r\not = s)$.

By embedding $\mathbb {S}^{ m}$ into $\mathbb {S}^{2N-1}$, we can assume the immersion x has the form as given in (1). From Lemma 2.2 and $A_{rs}=0~ (r\not = s)$, the matrix A in the expression (1) satisfies

$$\begin{aligned} AA^t=\text {diag}\Big \{a_1, a_1, a_2, a_2, \ldots , a_N, a_N\Big \}, \end{aligned}$$

with

$$\begin{aligned} a_1+a_2+\cdots +a_N=1,~~0\le a_1 \le a_2\le \cdots \le a_N,~1\le j\le N. \end{aligned}$$

Consider the QR decomposition of $A^t$, there exists an upper triangular matrix L such that

$$\begin{aligned} L^tL=\text {diag}\Big \{a_1, a_1, a_2, a_2, \ldots , a_N, a_N\Big \}, \end{aligned}$$

which implies L must be diagonal, i.e.,

$$\begin{aligned} L=\text {diag}\Big \{\sqrt{a_1},\sqrt{a_1},\sqrt{a_2},\sqrt{a_2},\ldots \ldots ,\sqrt{a_N},\sqrt{a_N}\,\Big \}. \end{aligned}$$

Hence, up to an orthogonal transformation, x can be expressed as below:

$$\begin{aligned} x=\begin{pmatrix}\Theta _1&\Theta _2&\cdots&\Theta _N\end{pmatrix} \text {diag}\Big \{\sqrt{a_1},\sqrt{a_1},\sqrt{a_2},\sqrt{a_2},\ldots ,\sqrt{a_N},\sqrt{a_N}\,\Big \}. \end{aligned}$$

Since x is linearly full, we can obtain that m is odd, and x is the orbit of a torus group acting at

$$\begin{aligned} (\sqrt{a_1},\sqrt{a_1},\sqrt{a_2},\sqrt{a_2},\ldots ,\sqrt{a_\frac{ m+1}{2}},\sqrt{a_\frac{ m+1}{2}}). \end{aligned}$$

$\square$

Definition 2.4

A finite set V of rank k in $\mathbb {R}^n$ is called unimodular, if all the k-dimensional parallelepipeds spanned in V have a same volume.

Remark 2.5

Suppose V is a finite set of rank k and satisfies the unimodular condition, then for $\{v_1, v_2, \ldots , v_{l}\}$ and $\{w_1, w_2, \ldots , w_{l}\}$, which are two arbitrary collections of l vectors in V with $l\le k$, there must have

$$\begin{aligned} v_1\wedge v_2\wedge \cdots \wedge v_{l}=\pm w_1\wedge w_2\wedge \cdots \wedge w_{l} \end{aligned}$$

when $\textrm{Span}_\mathbb {R}\{v_1, v_2, \ldots , v_{l}\}= \textrm{Span}_\mathbb {R}\{w_1, w_2, \ldots , w_{l}\}$.

Lemma 2.6

If $\{\xi _1,\xi _2, \ldots , \xi _N\}$ satisfies the unimodular condition, then for any $\eta \in {\mathcal {E}}$, the $\eta$-set forms a linearly independent set $\{\xi _{r_1}\odot \xi _{s_1}, \xi _{r_2}\odot \xi _{s_2}, \ldots ,\xi _{r_q}\odot \xi _{s_q}\}$.

Proof

We claim that $\eta , \xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_q}$ are linearly independent. Then by extending $\{\eta , \xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_q}\}$ to a basis of $\mathbb {R}^n$, it is easy to see that

$$\begin{aligned} \xi _{r_1}\odot \xi _{r_1},~~ \xi _{r_2}\odot \xi _{r_2},~ \ldots ,~~\xi _{r_q}\odot \xi _{r_q},~~\xi _{r_1}\odot \eta ,~~\xi _{r_2}\odot \eta ,~ \ldots ,~~\xi _{r_q}\odot \eta \end{aligned}$$

are linearly independent. Combining this with $\xi _{s_j}=\pm (\eta -\xi _{r_j})$, we can derive that

$$\begin{aligned} \xi _{r_1}\odot \xi _{s_1},~~ \xi _{r_2}\odot \xi _{s_2},~ \ldots ,~~\xi _{r_q}\odot \xi _{s_q} \end{aligned}$$

are linearly independent.

Now it suffices to prove the above claim. We prove it by contradictions. First we extend $\eta$ to a maximal linearly independent subset in $\{\eta , \xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_q}\}$, which can be assumed to be $\{\eta , \xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_{t-1}}\}$ with $t<q+1$. Then $\eta , \xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_{t-1}}, \xi _{r_{t}}$ must be linearly dependent. It is left to discuss the following two cases.

The first case is that $\xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_{t-1}}, \xi _{r_t}$ are linearly dependent.

Let

$$\begin{aligned} \xi _{r_t}=c_1\xi _{r_1}+c_2\xi _{r_2}+\cdots \cdots +c_{t-1}\xi _{r_{t-1}}. \end{aligned}$$

(6)

It follows from the unimodular condition that these coefficients all take values in $\{0,\pm 1\}$. Taking inner product of (6) with $\eta$, we can obtain

$$\begin{aligned} \sum _{j=1}^{t-1} c_j=1. \end{aligned}$$

(7)

Note that

$$\begin{aligned} \xi _{s_t}\wedge \xi _{r_1}\wedge \cdots \wedge \xi _{r_{t-1}}=\pm \,\eta \wedge \xi _{r_1}\wedge \cdots \wedge \xi _{r_{t-1}}\ne 0. \end{aligned}$$

For any $1\le j\le t-1$, using $\xi _{s_j}=\pm (\eta -\xi _{r_j})$, we have

$$\begin{aligned} \xi _{s_t}\wedge \xi _{r_1}\wedge \cdots \wedge \xi _{r_{j-1}}\wedge \xi _{s_j}\wedge \xi _{r_{j+1}}\wedge \cdots \wedge \xi _{r_{t-1}}=\pm (c_j-1)\xi _{s_{t}}\wedge \xi _{r_1}\wedge \cdots \wedge \xi _{r_{t-1}}. \end{aligned}$$

From the unimodular condition again we get

$$\begin{aligned} c_j-1\in \{0,\pm 1\},~~1\le j\le t-1, \end{aligned}$$

which implies $c_j\in \{0,1\}$ for all $1\le j\le t-1$. Combining this with (7), we derive that only one $c_j$ is nonzero and equals 1. It follows that $\xi _{r_t}=\xi _{r_j}$, contradicting with the fact that $\pm \xi _{r_1},\ldots ,\pm \xi _{r_q}, \pm \xi _{s_1},\ldots ,\pm \xi _{s_q}$ are distinct with each other.

The second case is that $\xi _{r_1}, \xi _{r_2}, \ldots , \xi _{r_t}$ are linearly independent. Then we can assume

$$\begin{aligned} \eta =c_1\xi _{r_1}+c_2\xi _{r_2}+\cdots \cdots +c_t\xi _{r_t}. \end{aligned}$$

(8)

Taking inner product of (8) with $\eta$, we can obtain

$$\begin{aligned} \sum _{j=1}^t c_j=2. \end{aligned}$$

(9)

On the other hand, using $\xi _{s_j}=\pm (\eta -\xi _{r_j})$, we have

$$\begin{aligned} \xi _{s_j}\!=\!\pm \left( c_1\xi _{r_1}+\cdots +c_{j-1}\xi _{r_{j-1}}+(c_j-1)\xi _{r_j}+c_{j+1}\xi _{r_{j+1}}+\cdots +c_t\xi _{r_t}\right) ,~~1\!\le \! j\!\le \! t. \end{aligned}$$

It follows from the unimodular condition that

$$\begin{aligned} c_j, c_j-1\in \{0,\pm 1\},~~1\le j\le t, \end{aligned}$$

which implies $c_j\in \{0,1\}$ for all $1\le j\le t$. Combining this with (9), we derive that only two $c_j$ are nonzero and equal 1, which can be assumed to be $c_1$ and $c_2$. It follows that $\eta =\xi _{r_1}+\xi _{r_2}$. So we have $\xi _{r_2}=\pm \xi _{s_1}$ also contradicting to the fact that $\pm \xi _{r_1},\ldots ,\pm \xi _{r_q}, \pm \xi _{s_1},\ldots ,\pm \xi _{s_q}$ are distinct. $\square$

By Lemmas 2.3 and  2.6 we immediately obtain the following proposition.

Proposition 2.7

Let $x:T^n=\mathbb {R}^n/{\Lambda _n} \longrightarrow \mathbb {S}^{ m}$ be a linearly full minimal flat torus. If $\{\xi _1,\xi _2, \ldots , \xi _N\}$ satisfies the unimodular condition, then m is odd and x is homogeneous.

Remark 2.8

In Sect. 6, we will show that all $\lambda _1$-minimal flat tori of dimension no more than 4 are homogeneous. In fact all of them satisfy the unimodular condition with only one exception.

3 Variational characterizations of homogeneous minimal flat tori

In this section, we give two variational characterizations for homogeneous minimal flat n-tori in spheres, from which two construction approaches can be derived. The construction of $\lambda _1$-minimal flat n-tori is also discussed.

Let $x: T^n=\mathbb {R}^n/\Lambda _n\rightarrow \mathbb {S}^m$ be a homogeneous minimal isometric immersion, with the metric given by

$$\begin{aligned} \frac{4\pi ^2}{n}(du_1^2+du_2^2+\cdots +du_n^2). \end{aligned}$$

Suppose $\{\xi _j\}_{j=1}^{N}$ are all lattice vectors (up to $\pm 1$) in $\Lambda _n^{*}$ of length 1. From the homogeneity of x, we can assume $m=2N-1$ and write x as follows:

$$\begin{aligned} x=(c_1e^{i\theta _1},c_2e^{i\theta _2},\ldots \ldots ,c_Ne^{i\theta _N}), \end{aligned}$$

(10)

where $\theta _j=2\pi \langle \xi _j, u\rangle ,~1\le j\le N$. Then we have

$$\begin{aligned} \begin{pmatrix} \xi _1^t&\xi _2^t&\cdots&\xi _N^t \end{pmatrix} \begin{pmatrix} c_1^2&{}&{}&{}\\ &{}c_2^2&{}&{}\\ &{}&{}\ddots &{}\\ &{}&{}&{}c_N^2 \end{pmatrix} \begin{pmatrix} \xi _1\\ \xi _2\\ \vdots \\ \xi _N \end{pmatrix}= \frac{1}{n}I_n. \end{aligned}$$

(11)

Assume $\{\eta _1,\eta _2,\ldots ,\eta _n\}$ is a generator of $\Lambda _n^{*}$. Then there exist integers $a_{j_k}$ such that

$$\begin{aligned} \xi _j=a_{j_1}\eta _1+a_{j_2}\eta _2+\cdots \cdots +a_{j_n}\eta _n,\quad 1\le j\le N. \end{aligned}$$

We denote

$$\begin{aligned} A_j=(a_{j_1},a_{j_2},\ldots \ldots ,a_{j_n}),\quad Y^t=(A_1^t,\cdots ,A_N^t), \quad 1\le j\le N, \end{aligned}$$

and use Y also representing the set of its row vectors if no confusion caused.

Set

$$\begin{aligned} Q=\begin{pmatrix} \eta _1\\ \eta _2\\ \vdots \\ \eta _n \end{pmatrix} \begin{pmatrix} \eta _1^t&\eta _2^t&\cdots&\eta _n^t \end{pmatrix}. \end{aligned}$$

It is a positive-definite matrix, called Gram matrix of $\Lambda ^*_n$. It is well-known that the volume of x equals $\frac{2^n\pi ^{n}}{\sqrt{n^n\det (Q)}}.$

Remark 3.1

The minimal immersion x given in (10) is uniquely determined by the following data set

$$\begin{aligned} \{Y,~ Q,~ (c_1^2, c_2^2, \cdots , c_N^2)\}, \end{aligned}$$

which is called the matrix data of x, and will be used to present examples in Sect. 4.

It is obvious that the condition $|\xi _j|=1$ is equivalent to

$$\begin{aligned} A_jQA_j^t=1, \end{aligned}$$

(12)

i.e., $A_j$ lies on the hyper-ellipsoid ${\mathcal {Q}}$ determined by

$$\begin{aligned} \begin{pmatrix} x_1&x_2&\cdots&x_n \end{pmatrix} Q\begin{pmatrix} x_1\\ x_2\\ \vdots \\ x_n \end{pmatrix}=1. \end{aligned}$$

Furthermore, it is straightforward to verify that the flat condition (11) is equivalent to

$$\begin{aligned} c_1^2 A_1^tA_1+c_2^2A_2^tA_2+\cdots \cdots +c_N^2A_N^tA_N=\frac{1}{n}Q^{-1}. \end{aligned}$$

(13)

3.1 Two variational characterizations

Consider the space S(n) of $n\times n$ symmetric matrices over $\mathbb {R}$, which is a $\frac{n(n+1)}{2}$-dimensional Euclidean space endowed with the inner product:

$$\begin{aligned} \langle S_1, S_2\rangle = {\text {tr}}(S_1S_2),\quad S_1, S_2\in S(n). \end{aligned}$$

For any given vector $v\in \mathbb {R}^n$, $\Pi _a(v):=\{M|\langle v^tv, M\rangle =vMv^t=a,a\in \mathbb {R}\}$ is an affine hyperplane dividing S(n) into two half spaces:

$$\begin{aligned} S^+_a(v)=\{M|vMv^t\ge a\}, ~~~S^-_a(v)=\{M|vMv^t\le a\}. \end{aligned}$$

Let $\Sigma$ be the set of semi-positive definite matrices. Then it is well known that $\Sigma =\cap _{v\in \mathbb {R}^n} S^+_0(v)$ is a convex cone in S(n) with the set of positive definite matrices as its interior, which is denoted by $\Sigma _+$. By our definition, we have $Q^{-1}\in \Sigma _+$, and $A_j^t A_j\in \Sigma , 1\le j\le N$.

Given a subset $X\subset \mathbb {Z}^n$, let $C_X$ be the convex hull spanned by $A^tA$ for all $A\in X$, and $V_X$ be the affine subspace $\cap _{A\in X}\Pi _{{1}}(A)\subset S(n)$. Moreover, we also consider the smooth linear submanifold $W_X= V_X\cap \Sigma _+$ (see Fig. 1), whose geometric meaning is the set of all hyper-ellipsoids passing through X.

Fig. 1

$W_X$ in $\Sigma _+$

With respect to these notations, the condition (12) is equivalent to $Q\in W_Y$, and (13) is equivalent to $\frac{Q^{-1}}{n}\in C_Y$.

Theorem 3.2

Let $x: T^n=\mathbb {R}^n/\Lambda _n\rightarrow \mathbb {S}^{2N-1}$ be an isometric homogeneous immersion. If x is minimal, then Q is a critical point of the determinant function restricted on $W_Y$, where Y is the set of integer vectors determined by x, and Q the Gram matrix of $\Lambda _n^*$ under some chosen generator.

Conversely, given a set X of integer vectors, if Q is a critical point of the determinant function restricted on $W_X$ and $\frac{Q^{-1}}{n}$ lies in the convex hull $C_X$, then the torus $\mathbb {R}^n/\Lambda _n$ determined by $Q^{-1}$ (as the Gram matrix of $\Lambda _n$) admits an isometric minimal immersion in $\mathbb {S}^{2N-1}$.

Proof

Let $\gamma (t)$ be a smooth curve in $W_Y$ passing through Q at $t=0$. Set $f(t)\triangleq \det {\gamma (t)}$, it is easy to verify that

$$\begin{aligned} f^{'}(0)=f(0)\langle Q^{-1}, \gamma '(0)\rangle . \end{aligned}$$

As an isometric homogeneous immersion, x is minimal if and only if $Q\in W_Y$, $Q^{-1}\in C_Y$. The conclusion follows from that $\textrm{Span}\{ A^tA ~|~A\in Y\}$ is the normal space of $W_Y$ at $\gamma (0)=Q$. $\square$

Remark 3.3

We point out that if Q is a critical point of the determinant function restricted on $W_X$, then $\frac{Q^{-1}}{n}$ lies in $\textrm{Span}\{A^tA\, |\, A\in X\}$ automatically. In fact let $Q+tS$ be an arbitrary segment in $W_X$, then it follows from $\langle Q^{-1},S\rangle$=0 that

$$\begin{aligned} \langle \frac{Q^{-1}}{n},Q+tS\rangle =1, \end{aligned}$$

which implies $\frac{Q^{-1}}{n}\in \textrm{Span}\{A^tA\, |\, A\in X\}$.

As a result, if all the vectors $A_i$ in X can be arranged as row vectors to form a block diagonal matrix, then the critical point Q is also block diagonal.

Although the following conclusion is well known, we give a proof here for completeness.

Lemma 3.4

The function $\ln \circ \det$ restricted on $\Sigma _+$ is strictly concave.

Proof

Given a positive definite matrix $P\in \Sigma _+$, Let $\gamma (t)$ be a smooth curve in $\Sigma _+$ with $\gamma '(0)=S\in S(n)\backslash \{0\}$. Let $f(t)=\ln \circ \det (\gamma (t))$, then we have $f'(t)={\text {tr}}\left( \gamma (t)^{-1}S\right)$ and

$$\begin{aligned} f''(0) =-{\text {tr}}\left( P^{-1}SP^{-1}S\right) =-{\text {tr}}\left( HSHHSH\right) =-|HSH|^2< 0, \end{aligned}$$

where H is a square root of the positive definite matrix $P^{-1}$ (i.e., $H^2=P^{-1}$), and we have used the fact that $S\in S(n)$. $\square$

Therefore, for any given subset $X\subset \mathbb {Z}^n$, if $W_X\ne \varnothing$, the determinant function $\det$ has only one critical point in $W_X$, which is the maximal point. This leads to the following corollary about uniqueness.

Corollary 3.5

Let $x: T^n=\mathbb {R}^n/\Lambda _n\rightarrow \mathbb {S}^{2N-1}$ and ${\tilde{x}}: \widetilde{T}^n=\mathbb {R}^n/\widetilde{\Lambda }_n\rightarrow \mathbb {S}^{2N-1}$ be two isometric homogeneous minimal immersions, if after choosing suitable generators respectively, x and $\tilde{x}$ share the same subset $Y\subset \mathbb {Z}^n$, then $T^n$ and $\widetilde{T}^n$ are isometric. Moreover, x is congruent to ${\tilde{x}}$ if $\{A^tA\,|\, A\in Y\}$ are linearly independent.

It seems that we can use the following approach to construct a homogeneous minimal immersion of flat torus. Choose a subset $X\subset \mathbb {Z}^n$, determine whether $W_X$ is not empty, and then discriminate whether the maximal point Q of $\det$ restricted on $W_X$ lies in the the convex hull $C_X$.

Note that in general the dimension of $V_X$ is $n(n+1)/2-\sharp (X)$, which implies $W_X$ could be empty if X involves too many vectors. Even if $V_X$ exists, it also could have no intersection with $\Sigma _+$. For example, we will get a degenerated matrix when

$$\begin{aligned} X=\begin{pmatrix} 1&{}&{}&{}1&{}&{}1\\ &{}1&{}&{}1&{}1&{}\\ &{}&{}1&{}&{}1&{}1 \end{pmatrix},~~ \text { for which }~~ Q=\begin{pmatrix} 1&{}-1/2&{}-1/2\\ -1/2&{}1&{}-1/2\\ -1/2&{}-1/2&{}1 \end{pmatrix}. \end{aligned}$$

It seems to be a challenge to obtain a general method by which one can efficiently construct the desirable integer set X.

Next we give an alternative method to determine homogeneous minimal immersions of flat tori, by use of a variational characterization of $Q^{-1}$ (comparing Theorem 3.2).

Theorem 3.6

Let $x: T^n=\mathbb {R}^n/\Lambda _n\rightarrow \mathbb {S}^{2N-1}$ be a homogeneous minimal flat torus, where N is the half dimension of the eigenspace of $T^n$ corresponding to n. If x is minimal and linearly full, then $\frac{Q^{-1}}{n}$ lies in the interior of $C_Y$, and is a critical point of the determinant function restricted on $C_Y$.

Conversely, given a finite set X of integer vectors such that $C_X\cap \Sigma _+\ne \varnothing$, if $P\in \overset{\circ }{C_X}$ is a critical point of the determinant function restricted on $C_X$, then the torus $\mathbb {R}^n/\Lambda _n$ determined by nP (as the Gram matrix of $\Lambda _n$) admits an isometric minimal immersion in some $\mathbb {S}^{2N-1}$.

Proof

For the first part of this theorem, $\frac{Q^{-1}}{n}\in C_Y$ just follows from the flat condition (13). To prove that $\frac{Q^{-1}}{n}$ is a critical point of $\det$ restricted on $C_Y$, we only need to check whether the derivatives of $\det |_{C_Y}$ at this point equal zero, which can be computed directly and are omitted here.

For the converse part, suppose $X=\{A_1, A_2, \cdots , A_k\}$, and $P=\sum _{j=1}^k y_j A_j^tA_j$. It follows from $P\in \overset{\circ }{C_X}$ that $y_j> 0$ for all $1\le j\le k$. We only need to prove that $\frac{P^{-1}}{n}\in W_X$, i.e.,

$$\begin{aligned} \langle \frac{P^{-1}}{n}, A_j^tA_j\rangle =1,~~~1\le j\le k. \end{aligned}$$

(14)

For any given $2\le j\le k$, consider the line segment

$$\begin{aligned} \gamma _j(t)=P+t(A_j^tA_j-A_1^tA_1). \end{aligned}$$

It is obvious that for sufficient small t, $\gamma _j(t)$ lies in $C_X$. So P is a critical point of $\det (\gamma _j(t))$, which implies that $\langle P^{-1}, \gamma _j'(0)\rangle =0$. Therefore we have

$$\begin{aligned} \langle P^{-1}, A_j^tA_j\rangle =\langle P^{-1}, A_1^tA_1\rangle ,~~~2\le j\le k. \end{aligned}$$

It follows from $\sum _{j=1}^k y_j=1$ that

$$\begin{aligned} n=\langle P^{-1},P \rangle =\sum _{j=1}^k y_j\langle P^{-1}, A_j^tA_j\rangle =\big (\sum _{j=1}^k y_j\big )\langle P^{-1}, A_1^tA_1\rangle =\langle P^{-1}, A_1^tA_1\rangle , \end{aligned}$$

which completes the proof. $\square$

Remark 3.7

The above theorem provides another approach to construct minimal flat tori. One can choose a finite set X of integer vectors such that $C_X\cap \Sigma _+\ne \varnothing$ (see Fig. 2), and calculate the the maximal point P of $\det$ on $C_X$. If $P\in \overset{\circ }{C_X}$, then the matrix data $\{\frac{P^{-1}}{n}, X\}$ provides a minimal flat n-torus. Otherwise, let $C_{X'}$ be the face of $C_X$ so that $P\in \overset{\circ }{C}_{X'}$ (such face could have high codimension and the existance is due to the compactness of $C_X$), then $\{\frac{P^{-1}}{n}, X'\}$ provides a minimal flat n-torus.

This remark can be seen as a generalization of Bryant’s characterization to minimal flat 2-tori in spheres, see Proposition 3.3 in [2].

Fig. 2

$C_X$ and its faces

3.2 Homogeneous $\lambda _1$-minimal flat tori

To determine homogeneous $\lambda _1$-minimal immersions of flat tori, one has to solve the following problem.

Problem

Firstly how to choose a subset $X\subset \mathbb {Z}^n$, such that in $W_X$ there is a maximal point Q of $\det$, and $Q^{-1}/n\in \overset{\circ }{C_X}$. Then how to discriminate in the interior of the hyper-ellipsoid determined by Q, whether there exist other nonzero integer vectors, i.e., to discriminate whether $vQv^t\ge 1$ hold for all $v\in \mathbb {Z}^n\backslash \{0\}$.

In $\Sigma _{+}$, the infinite constraints

$$\begin{aligned} vMv^t\ge 1~ \text {for~ all~} v\in \mathbb {Z}^n\backslash \{0\} \end{aligned}$$

(15)

define a non-compact convex domain $\Omega$.

So X can determine a homogeneous $\lambda _1$-minimal immersion of some flat torus if and only if $Q\in \partial \Omega$. Given a positive definite matrix Q, it seems that to verify whether $Q\in \partial \Omega$, i.e., satisfying (15), infers an infinite process. However, due to Minkowski’s reduction theory of lattices (see [30] for reference), only finite inequalities need to be checked. This can also be seen from the following simple theorem.

Theorem 3.8

For a given $n\times n$ positive definite matrix Q with diagonal entries 1, there exist n integers $a_k(Q)>0$ $(1\le k\le n)$ such that $vQv^t\ge 1$ for all $v\in \mathbb {Z}^n\backslash \{0\}$ if and only if it holds for all integer vectors in

$$\begin{aligned} S\triangleq \big \{v=(v_1,\ldots ,v_n)\in \mathbb {Z}^n \, \big |\, |v_k|< a_k(Q), 1\le k\le n\big \}. \end{aligned}$$

Proof

Let $\{\xi _i\}$ be a basis of $\mathbb {R}^n$ with Gram matrix Q, $\Pi _k$ the $(n-1)$-dimensional subspace spanned by $\xi _1,\ldots ,\xi _{k-1},\xi _{k+1},\ldots ,\xi _n$. So the distance of $\xi _k$ to $\Pi _k$ is a fixed value $d_k>0$, which can be determined from the entries $a_{ij}$ of Q by using the least square method, such as

$$\begin{aligned} d_n^2=|\xi _n|^2-|\xi _n^\top |^2=1-|\xi _n^\top |^2=1-(a_{1n},\ldots ,a_{(n-1)n})(a_{ij})^{-1}_{1\le i,j\le n-1}(a_{1n},\ldots ,a_{(n-1)n})^t, \end{aligned}$$

(16)

where $\xi _n^\top$ is the orthogonal projection of $\xi _n$ into $\Pi _n$.

Suppose $a_k(Q)$ is the integer such that $a_k(Q)-1 \le d_k^{-1} < a_k(Q)$, then we will get $|\alpha |\ge |c_k|d_k>1$ for any $\alpha =c_1\xi _1+\cdots +c_n\xi _n$ when some $|c_k|\ge a_k(Q)$. $\square$

Although the criterion given in the above theorem only infers finite steps, it still requests a lot of computations. For $n\le 6$, we give an alternative criterion which comes from Minkowski’s reduction theory (see [29]). For the case $n>6$, so far we do not find such an explicit description in the literature. Applied to our case, the criterion is stated as below.

Theorem 3.9

Suppose Q is an $n\times n$ positive definite matrix with diagonal entries 1, and $n\le 6$. Then $vQv^t\ge 1$ holds for all $v\in \mathbb {Z}^n\backslash \{0\}$, if and only if, it holds for those integer vectors v, whose entries take values from the first n integers in each rows below with arbitrary sign (“$+$” or “−”):

$$\begin{aligned} \begin{array}{llllll} 1&{}1&{}0&{}0&{}0&{}0\\ 1&{}1&{}1&{}0&{}0&{}0\\ 1&{}1&{}1&{}1&{}0&{}0\\ 1&{}1&{}1&{}1&{}1&{}0\\ 1&{}1&{}1&{}1&{}2&{}0\\ 1&{}1&{}1&{}1&{}1&{}1\\ 1&{}1&{}1&{}1&{}1&{}2\\ 1&{}1&{}1&{}1&{}2&{}2\\ 1&{}1&{}1&{}1&{}2&{}3\,.\\ \end{array} \end{aligned}$$

4 Examples of $\lambda _1$-minimal flat tori of dimension 3 and 4

In this section, we present a series of $\lambda _1$-minimal immersions for flat tori $T^3=\mathbb {R}^3/\Lambda _3$ and $T^4=\mathbb {R}^4/\Lambda _4$, which are all homogeneous. Instead of giving the explicit petty immersions, we use the matrix data

$$\begin{aligned} \{Y,~ Q,~ (c_1^2, c_2^2, \cdots , c_N^2)\} \end{aligned}$$

introduced in the last section (see Remark 3.1). Recall that Q is a Gram matrix of the lattice $\Lambda _n^*$ under some chosen generator, which can determine a flat torus $T^n=\mathbb {R}^n/\Lambda _n$; N denotes the number of distinct shortest lattice vectors of $\Lambda ^*_n$ up to $\pm 1$, i.e., the half dimension of the first eigenspace of flat torus $T^n=\mathbb {R}^n/\Lambda _n$; Y is an $n\times N$ integer matrix describing the linear relations of those shortest lattice vectors; $c_i^2$ involves the information of immersion.

First of all, given two $\lambda _1$-minimal tori $f_i:T^{n_i}\rightarrow \mathbb {S}^{m_i} (i=1,2)$, one can construct a new $\lambda _1$-minimal $(n_1+n_2)$-torus by the following direct product (see [3, 33, 35])

$$\begin{aligned} f=\left( \sqrt{\frac{n_1}{n_1+n_2}}f_1,\sqrt{\frac{n_2}{n_1+n_2}}f_2\right) :T^{n_1}\times T^{n_2}\rightarrow \mathbb {S}^{m_1+m_2+1}. \end{aligned}$$

(17)

In the sequel, $\lambda _1$-minimal torus constructed in such way is called reducible, and those non-product ones are called irreducible.

Remark 4.1

A reducible $\lambda _1$-minimal flat torus constructed from two homogeneous and $\lambda _1$-minimal tori is also homogeneous. For such torus, by definition, suitable generator of the corresponding dual lattice can be chosen such that the matrix Y is block diagonal. Then it follows from Remark 3.3 that the Gram matrix Q is also block diagonal. Conversely, a $\lambda _1$-minimal flat torus given by diagonal data set is a reducible $\lambda _1$-minimal flat torus.

Example 4.2

As stated in the introduction, the Clifford torus and equilateral torus are two (only two) $\lambda _1$-minimal flat 2-torus, whose matrix data are

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} 1 \end{array} \right) ,~~~ Y=\left( \begin{array}{ccc} 1 &{} 0 \\ 0 &{} 1 \end{array} \right) ,~~~ (c_1^2,\, c_2^2)=\left( \frac{1}{2},\, \frac{1}{2}\right) ,\\{} & {} \quad Q=\left( \begin{array}{ccc} 1 &{} -\frac{1}{2} \\ -\frac{1}{2} &{} 1 \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccc} 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 \end{array} \right) ,~~~ (c_1^2,\, c_2^2,\, c_3^2)=\left( \frac{1}{3},\, \frac{1}{3},\,\frac{1}{3}\right) . \end{aligned}$$

Taking one from these two tori, considering the direct product (as in (17)) of it with the unite circle in $\mathbb {R}^2$, we can obtain two examples of $\lambda _1$-minimal flat 3-torus in $\mathbb {S}^5$ and $\mathbb {S}^7$, whose volume is respectively $\frac{8\sqrt{3}}{9}\pi ^3$ and $\frac{16}{9}\pi ^3$. Their matrix data can be stated as follows:

$$\begin{aligned} Q=\left( \begin{array}{ccc} 1 &{} 0 &{}0 \\ 0 &{} 1 &{}0 \\ 0 &{}0 &{} 1 \\ \end{array} \right) ,~~~ Y=\left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) ,~~~ (c_1^2,\, c_2^2,\, c_3^2)=\left( \frac{1}{3},\, \frac{1}{3},\, \frac{1}{3}\right) ,\\ Q=\left( \begin{array}{ccc} 1 &{} -\frac{1}{2} &{} 0 \\ -\frac{1}{2} &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccc} 1 &{} 0 &{} 1&{}0 \\ 0 &{} 1 &{} 1&{}0 \\ 0 &{} 0 &{} 0&{}1 \\ \end{array} \right) ,~~~ (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2)=\left( \frac{2}{9},\, \frac{2}{9},\,\frac{2}{9},\, \frac{1}{3}\right) . \end{aligned}$$

Next we give some data with neither Q nor Y being block diagonal, which are obtained originally by applying our variational characterization, where some tedious but routine computation is involved, and we omit it here. Alternatively, one can verify directly that all these data fit (12) and (13), and satisfy the criterion given in Theorem 3.9, hence provide irreducible examples of $\lambda _1$-minimal flat tori.

Example 4.3

N=4. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{ccc} 1 &{} -\frac{1}{3} &{} -\frac{1}{3} \\ -\frac{1}{3} &{} 1 &{} -\frac{1}{3} \\ -\frac{1}{3} &{} -\frac{1}{3} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccc} 1 &{} 0 &{} 0&{}1 \\ 0 &{} 1 &{} 0 &{}1\\ 0 &{} 0 &{} 1 &{}1\\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2)= \left( \frac{1}{6},\, \frac{1}{6},\, \frac{5}{48},\, \frac{3}{16},\, \frac{3}{16},\, \frac{3}{16}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 3-torus in $\mathbb {S}^7$, which is irreducible, linearly full and has volume $2\pi ^3$.

Example 4.4

N=5. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{ccc} 1 &{} -\frac{1}{2} &{} -\frac{1}{4} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{4} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccc} 1 &{} 0 &{} 1&{}0&{}1 \\ 0 &{} 1 &{} 1&{}0&{}1\\ 0 &{} 0 &{} 0 &{}1&{}1\\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2)\\{} & {} \quad =\left( \frac{2}{9},\, \frac{2}{9},\,\frac{2}{9},\, \frac{1}{9},\,\frac{2}{9}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 3-torus in $\mathbb {S}^9$, which is irreducible, linearly full and has volume $\frac{32\sqrt{3}}{27}\pi ^3$.

Example 4.5

N=6. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{ccc} 1 &{} -\frac{1}{2} &{} 0 \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{2} \\ 0 &{} -\frac{1}{2} &{} 1 \\ \end{array} \right) ,~ Y^t=\left( \begin{array}{cccccc} 1 &{} 0 &{} 1&{}0&{}0&{}1 \\ 0 &{} 1 &{} 1&{}0&{}1&{}1\\ 0 &{} 0 &{} 0 &{}1&{}1&{}1\\ \end{array} \right) \\{} & {} \quad ,(c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2)\\{} & {} \quad =\left( \frac{1}{6},\, \frac{1}{6},\,\frac{1}{6},\, \frac{1}{6},\,\frac{1}{6},\,\frac{1}{6}\right) . \end{aligned}$$

gives a $\lambda _1$-minimal flat 3-torus in $\mathbb {S}^{11}$, which is irreducible, linearly full and has volume $\frac{8\sqrt{6}}{9}\pi ^3$.

We will prove in the next section that Examples 4.2 to  4.5 enumerate all examples of $\lambda _1$-minimal flat 3-tori in spheres.

Example 4.6

Taking the direct products (as in (17)) of two $\lambda _1$-minimal flat 2-tori, one can obtain three reducible $\lambda _1$-minimal flat 4-tori in $\mathbb {S}^7$, $\mathbb {S}^9$ and $\mathbb {S}^{11}$, they are all linearly full with volumes $\pi ^4$, $\frac{3\sqrt{3}}{4}\pi ^4$ and $\frac{4}{3}\pi ^4$, respectively.

Taking the direct product (as in (17)) of one $\lambda _1$-minimal flat 3-torus given in Examples 4.3 to  4.5 with the unite circle in $\mathbb {R}^2$, we can obtain another three reducible $\lambda _1$-minimal flat 4-tori in $\mathbb {S}^9$, $\mathbb {S}^{11}$ and $\mathbb {S}^{13}$, they are all linearly full with volumes $\frac{2\sqrt{3}}{3}\pi ^4$, $\frac{4}{3}\pi ^4$ and $\sqrt{2}\pi ^4$, respectively.

For brevity, we omit the data set of these 6 reducible $\lambda _1$-minimal flat 4-tori.

Example 4.7

$N=5$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{4} &{} -\frac{1}{4} &{} -\frac{1}{4} \\ -\frac{1}{4} &{} 1 &{} -\frac{1}{4} &{} -\frac{1}{4} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} 1 &{} -\frac{1}{4} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} -\frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccc} 1 &{} 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\,c_5^2)\\{} & {} \quad =\left( \frac{1}{5},\, \frac{1}{5},\,\frac{1}{5},\, \frac{1}{5},\,\frac{1}{5}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{9}$, which is irreducible, linearly full and has volume $\frac{16\sqrt{5}}{25}\pi ^4$.

Example 4.8

$N=6$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} -\frac{1}{6} &{} -\frac{1}{6} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{6} &{} -\frac{1}{6} \\ -\frac{1}{6} &{} -\frac{1}{6} &{} 1 &{} -\frac{1}{3} \\ -\frac{1}{6} &{} -\frac{1}{6} &{} -\frac{1}{3} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccccc} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2)=\left( \frac{1}{6},\, \frac{1}{6},\,\frac{5}{48},\,\frac{3}{16},\,\frac{3}{16},\,\frac{3}{16}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{11}$, which is irreducible, linearly full and has volume $\frac{3}{2}\pi ^4$.

Example 4.9

$N=7$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} -\frac{1}{4} &{} -\frac{1}{4} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{4} &{} -\frac{1}{4} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} 1 &{} \frac{1}{4} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} \frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2,\, c_7^2)=\left( \frac{1}{6},\, \frac{1}{6},\, 0,\,\frac{1}{6},\,\frac{1}{6},\,\frac{1}{6},\,\frac{1}{6}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{13}$, which is irreducible, and has volume $\frac{8\sqrt{3}}{9}\pi ^4$. Note that it is not linearly full in $\mathbb {S}^{13}$ but in $\mathbb {S}^{11}$. As far as we know, this is the first explicit example of $\lambda _1$-minimal immersion in the literature that does not span the whole first eigenspace. Note that branched minimal surfaces with extra eigenfunctions have been studied by Ejiri and Kotani in [7, 8].

Example 4.10

$N=7$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} -\frac{1}{4} &{} -\frac{1}{8} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{4} &{} -\frac{1}{8} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} 1 &{} -\frac{1}{4} \\ -\frac{1}{8} &{} -\frac{1}{8} &{} -\frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2,\, c_7^2)=\left( \frac{1}{6},\, \frac{1}{6},\,\frac{1}{6},\, \frac{1}{12},\,\frac{1}{12},\,\frac{1}{6},\,\frac{1}{6}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{13}$, which is irreducible, linearly full and has volume $\frac{8\sqrt{3}}{9}\pi ^4$.

Example 4.11

$N=7$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array} {cccc} 1 &{}\frac{\sqrt{13} - 7}{12} &{}\frac{\sqrt{13} - 7}{12} &{}\frac{4-\sqrt{13} }{6} \\ \frac{\sqrt{13} - 7}{12} &{} 1 &{}\frac{1 - \sqrt{13}}{6} &{}\frac{\sqrt{13} - 7}{12} \\ \frac{\sqrt{13} - 7}{12} &{}\frac{1 - \sqrt{13}}{6} &{} 1 &{}\frac{\sqrt{13} - 7}{12} \\ \frac{\sqrt{13} - 4}{6} &{}\frac{\sqrt{13} - 7}{12} &{}\frac{\sqrt{13} - 7}{12} &{} 1 \\ \end{array} \right) ,~~~ Y^t= \left( \begin{array}{ccccccc} 1 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2,\, c_7^2)\\{} & {} \quad =\left( \frac{\sqrt{13}-2}{12},\,\frac{5-\sqrt{13}}{8},\,\frac{5-\sqrt{13}}{8},\,\frac{\sqrt{13}-2}{12},\, \frac{\sqrt{13}-2}{12},\,\frac{\sqrt{13}-2}{12},\,\frac{5-\sqrt{13}}{12}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{13}$, which is irreducible, linearly full and has volume $\frac{\sqrt{26\sqrt{13}-70}}{3}\pi ^4$.

As far as we know, this is the first example of $\lambda _1$-minimal immersion in the literature whose Gram matrix are not rational, comparing to that minimal 2-tori always have rational Gram matrices up to a rescaling(see [2]).

Example 4.12

$N=8$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} 0 &{} -\frac{1}{4} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{2} &{} 0 \\ 0 &{} -\frac{1}{2} &{} 1 &{} -\frac{1}{4} \\ -\frac{1}{4} &{} 0 &{} -\frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 \\ \end{array} \right) , \\{} & {} \quad (c_1^2,\, c_2^2,\, c_3^2,\, c_4^2,\, c_5^2,\, c_6^2,\, c_7^2,\, c_8^2)= \left( \frac{1}{8},\, \frac{1}{8},\, \frac{1}{8},\, \frac{1}{8},\, \frac{1}{8},\, \frac{1}{24},\, \frac{1}{6},\, \frac{1}{6}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{15}$, which is irreducible, linearly full and has volume $\frac{2\sqrt{6}}{3}\pi ^4$.

Example 4.13

$N=8$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} \frac{\sqrt{3}}{2}-1 &{} 1-\frac{\sqrt{3}}{2} \\ -\frac{1}{2} &{} 1 &{} \frac{1}{2}-\frac{\sqrt{3}}{2} &{} \sqrt{3}-2 \\ \frac{\sqrt{3}}{2}-1 &{} \frac{1}{2}-\frac{\sqrt{3}}{2} &{} 1 &{} \frac{1}{2}-\frac{\sqrt{3}}{2} \\ 1-\frac{\sqrt{3}}{2} &{} \sqrt{3}-2 &{} \frac{1}{2}-\frac{\sqrt{3}}{2} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\; c_2^2,\; c_3^2,\; c_4^2,\; c_5^2,\; c_6^2,\; c_7^2,\; c_8^2)\\{} & {} \quad = \left( \frac{3- \sqrt{3}}{12},\; \frac{\sqrt{3}}{12}, \; \frac{3-\sqrt{3}}{12},\; \frac{\sqrt{3}}{12}, \;\frac{3-\sqrt{3}}{12},\; \frac{\sqrt{3}}{12}, \;\frac{\sqrt{3}}{12},\;\frac{3- \sqrt{3}}{12}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{15}$, which is irreducible, linearly full and has volume $\frac{\sqrt{12+8 \sqrt{3}}}{3}\pi ^4$. This is another example of $\lambda _1$-minimal immersion whose Gram matrix are not rational.

Example 4.14

$N=9$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} 0 &{} -\frac{1}{6} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{2} &{} \frac{1}{6} \\ 0 &{} -\frac{1}{2} &{} 1 &{} -\frac{1}{2} \\ -\frac{1}{6} &{} \frac{1}{6} &{} -\frac{1}{2} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\; c_2^2,\; c_3^2,\; c_4^2,\; c_5^2,\; c_6^2,\; c_7^2,\; c_8^2,\; c_9^2)\\{} & {} \quad =\left( \frac{1}{8},\; \frac{1}{8},\;\frac{1}{12},\; \frac{1}{12},\;\frac{1}{8},\;\frac{1}{12},\;\frac{1}{8},\;\frac{1}{8},\;\frac{1}{8}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{17}$, which is irreducible, linearly full and has volume $\frac{16}{9}\pi ^4$.

Example 4.15

$N=9$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} -\frac{1}{4} &{} \frac{1}{4} \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{4} &{} -\frac{1}{2} \\ -\frac{1}{4} &{} -\frac{1}{4} &{} 1 &{} -\frac{1}{4} \\ \frac{1}{4} &{} -\frac{1}{2} &{} -\frac{1}{4} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{ccccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 &{}0\\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 0 &{} 1 &{} 1 &{}1\\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 &{}0\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 &{}1\\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\; c_2^2,\; c_3^2,\; c_4^2,\; c_5^2,\; c_6^2,\; c_7^2,\; c_8^2,\; c_9^2)\\{} & {} \quad =\left( \frac{1}{9},\; \frac{1}{9},\;\frac{1}{9},\; \frac{1}{9},\;\frac{1}{9},\;\frac{1}{9},\;\frac{1}{9},\;\frac{1}{9},\;\frac{1}{9}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{17}$, which is irreducible, linearly full and has volume $\sqrt{3}\pi ^4$.

Example 4.16

$N=10$. The following matrix data

$$\begin{aligned}{} & {} Q=\left( \begin{array}{cccc} 1 &{} -\frac{1}{2} &{} 0 &{} 0 \\ -\frac{1}{2} &{} 1 &{} -\frac{1}{2} &{} 0 \\ 0 &{} -\frac{1}{2} &{} 1 &{} -\frac{1}{2} \\ 0 &{} 0 &{} -\frac{1}{2} &{} 1 \\ \end{array} \right) ,~~~ Y^t=\left( \begin{array}{cccccccccc} 1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 &{} 0 &{} 0 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 &{} 0 &{} 1 &{} 1 &{} 1 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 1 &{} 1 &{} 1 \\ \end{array} \right) ,\\{} & {} \quad (c_1^2,\; c_2^2,\; c_3^2,\; c_4^2,\; c_5^2,\; c_6^2,\; c_7^2,\; c_8^2,\; c_9^2,\;c_{10}^2)\\{} & {} \quad =\left( \frac{1}{10},\; \frac{1}{10},\;\frac{1}{10},\; \frac{1}{10},\;\frac{1}{10},\;\frac{1}{10},\;\frac{1}{10},\;\frac{1}{10},\;\frac{1}{10},\;\frac{1}{10}\right) \end{aligned}$$

gives a $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{19}$, which is irreducible, linearly full and has volume $\frac{4}{\sqrt{5}}\pi ^4$.

We will prove in the next section that Examples 4.6 to  4.16, and Example 1.1 enumerate all examples of $\lambda _1$-minimal flat 4-tori in spheres.

5 Shortest vectors in lattices

As mentioned in the introduction, the classification of $\lambda _1$-minimal tori of dimension 3 and 4 in spheres relies on deep investigation of shortest lattice vectors for lattices of rank 3 and 4. This section is devoted to the discussion of some related properties of lattices, which are of independent interest.

Let $\Lambda ^*_n$ be a lattice of rank n, N be the number of distinct shortest lattice vectors of $\Lambda ^*_n$ up to $\pm 1$, and $\Xi$ be the set of these N shortest vectors, which are denoted by

$$\begin{aligned} \xi _1, \xi _2, \ldots ,\xi _N. \end{aligned}$$

For simplicity, in the sequel, these shortest vectors are assumed to be of unit length.

Definition 5.1

A set of $k+1$ vectors in $\mathbb {R}^n$ is called a generic $(k+1)$-tuple if it is of rank k and any k vectors in it are linearly independent.

Definition 5.2

A lattice $\Lambda _n^*$ with $N\ge n$ is called s-reducible if there is a non-trivial decomposition $\mathbb {R}^n=V_1\oplus V_2$ such that

$$\begin{aligned} \Xi =(\Xi \cap V_1)\cup (\Xi \cap V_2). \end{aligned}$$

It will be called s-irreducible otherwise.

It is easy to see that if $N=n$, or $N=n+1$ and $\Xi$ is not a generic $(n+1)$-tuple, or $\text {rank}(\Xi )<n$, then $\Lambda _n^*$ is s-reducible.

Lemma 5.3

For an s-irreducible lattice $\Lambda _n^*$ of rank $n\ge 3$, there always exists a generic k-tuple in $\Xi$ for some $k\ge 4$.

Proof

Let $\{\xi _1,\ldots ,\xi _n\}\subset \Xi$ be of rank n. Suppose the opposite that there are only generic 3-tuples in $\Xi$. Then for any $n+1\le i\le N$, when we write $\xi _i$ as a linear combination of these generator vectors, there are exactly two non-zero coefficients. From the s-irreducible assumption we know $N\ge n+2$. We can assume $\xi _{n+1}=a_1\xi _1+a_2\xi _2$ with $a_1a_2\not =0$. Consider two subspaces $V_1\triangleq \textrm{Span}\{\xi _3,\ldots ,\xi _n\}$ and $V_2\triangleq \textrm{Span}\{\xi _1,\xi _2\}$, it follows that there exists at least one vector, say $\xi _{n+2}$, belonging to neither $V_1$ nor $V_2$.

We assume $\xi _{n+2}=b_1\xi _1+b_2\xi _3$ with $b_1b_2\not =0$. Then it is straightforward to verify that $\{\xi _2,\xi _3,\xi _{n+1},\xi _{n+2}\}$ forms a generic 4-tuple, which gives us a contradiction. $\square$

Definition 5.4

A lattice $\Lambda _n^*$ is called prime if this lattice can be generated by any n linearly independent vectors in $\Xi$.

Remark 5.5

By definition, for a prime lattice, $\Xi$ always satisfies the unimodular condition defined in Definition 2.4.

The following conclusion is easy to obtain.

Lemma 5.6

Let $\Lambda _n^*$ be a prime lattice generated by $\{\alpha _1,\ldots ,\alpha _n\}\subset \Xi$. Then for any $\xi \in \Xi$, we have

$$\begin{aligned} \xi =a_1\alpha _1+\cdots +a_n\alpha _n,\quad a_i\in \{0,\pm 1\}. \end{aligned}$$

For the lattices of rank no more than 4, we obtain a sufficient condition to discriminate whether it is prime. The covering radius in lattice theory will be used, which is defined for a lattice $\Lambda _n^*$ as follows:

$$\begin{aligned} \mu (\Lambda _n^*)\triangleq \inf \Big \{r\,\Big |\,\mathbb {R}^n= \cup _{p\in \Lambda _n^*} B^n(p,r)\Big \}, \end{aligned}$$

where $B^n(p,r)$ is the ball centered at p with radius r. Any ball with radius larger than $\mu (\Lambda ^*_ n)$ must contain a point of $\Lambda _n^*$. The following well-known estimate (see [14]) will be used to prove our main theorem in this section.

Lemma 5.7

Let $\Lambda _n^*$ be a lattice of rank n, if $\textrm{rank}(\Xi )=n$ and the shortest vectors of $\Lambda _n^*$ are of unit length,

then

$\mu (\Lambda _n^*)\le \frac{{\sqrt{n}}}{{2}}.$

Theorem 5.8

Suppose $\Lambda _n^*$ is a lattice of rank n with $\textrm{rank}(\Xi )$=n. If $n\le 4$, then either $\Lambda _n^*$ is prime, or it can be generated by the row vectors of

$$\begin{aligned} \begin{pmatrix} 1&{}0&{}0&{}0\\ 0&{}1&{}0&{}0\\ 0&{}0&{}1&{}0\\ \frac{1}{2}&{}\frac{1}{2}&{}\frac{1}{2}&{}\frac{1}{2} \end{pmatrix}. \end{aligned}$$

(18)

Proof

It is obviously true for $n=1$. Suppose it is true for $n=k-1$ ($\le 3$), we will show that in the case of $n=k$ the conclusion is also true. Let $\{\xi _1, \xi _2,\cdots ,\xi _k\}$ be any given linearly independent shortest vectors in $\Lambda _k^*$. Let $\Lambda _{k-1}^*$ be the sublattice of rank $k-1$ containing $\{\xi _2,\cdots ,\xi _k\}$. Then it follows from the inductive hypothesis that there exists a generator $\{\eta _1,\xi _2,\cdots ,\xi _k\}$ of $\Lambda _k^*$ such that

$$\begin{aligned} \begin{pmatrix} \xi _1\\ \xi _2\\ \vdots \\ \xi _k \end{pmatrix}=\begin{pmatrix} a_{1}&{}a_{2}&{}\cdots &{}a_{k}\\ &{}1&{}&{}\\ &{}&{}\ddots &{}\\ &{}&{}&{}1 \end{pmatrix}\begin{pmatrix} \eta _1\\ \xi _2\\ \vdots \\ \xi _k \end{pmatrix}, \end{aligned}$$

where $a_1, a_2, \ldots , a_k\in \mathbb {Z}$, we can assume they are all non-negative and $a_1>0$ by changing directions of these vectors. Since ${\eta _1+c\xi _2,\xi _2,\ldots ,\xi _k}$ $(c\in \mathbb {Z})$ is still a generator of $\Lambda _k^*$, we get

$$\begin{aligned} \xi _1=a_1(\eta _1+c\xi _2)+(a_2-ca_1)\xi _2+\cdots +a_k\xi _k. \end{aligned}$$

So suitable value c will make $a_1>a_2-ca_1\ge 0$. Therefore, we may assume $a_1>a_i\ge 0$ for $2\le i\le k$.

If $a_1=1$ then $a_i=0$ for $2\le i\le k$ and we derive that ${\xi _1, \xi _2,\ldots ,\xi _k}$ is a generator of $\Lambda _k^*$. If $a_1\ge 2$, let $\eta _1=\eta _1^{\perp }+\eta _1^{\top }$, where $\top$ (w.r.t. $\perp$) denotes the orthogonal projection onto $\Pi _0\triangleq \textrm{Span}_\mathbb {R}\{\xi _2,\ldots ,\xi _k\}$ (w.r.t. the normal space of $\Pi _0$ ). So $\xi _1=a_1\eta _1^{\perp }+\xi _1^{\top }$ and

$$\begin{aligned} 0< |\eta _1^{\perp }|^2=\frac{|\xi _1|^2-|\xi _1^{\top }|^2}{a_1^2}=\frac{1-|\xi _1^{\top }|^2}{a_1^2}\le \frac{1}{4}. \end{aligned}$$

It follows that the intersection of k-ball $B^k(0,1)$ with the hyperplane $\Pi \triangleq \eta _1^\perp +\Pi _0$ is a $(k-1)$-ball $B^{k-1}(\eta _1^\perp , r)$ with

$$\begin{aligned} r^2=\frac{a_1^2-1}{a_1^2}+\frac{|\xi _1^\top |^2}{a_1^2}\ge \frac{3}{4}. \end{aligned}$$

When $k<4$, we can see r is larger than the covering radius of $\Lambda _{k-1}^*$. Therefore, in $B^{k-1}(\eta _1^\perp , r)\subset B^k(0,1)$, there exists at least one point of $\Lambda _k^*\cap \Pi$. This contradicts with our assumption that 1 is the shortest length of $\Lambda _k^*$. Similarly, in the case of $k=4$, if $r^2>\frac{3}{4}$, we can also obtain a contradiction.

Therefore, when $k=4$, we have $r^2=\frac{3}{4}$. This yields $a_1=2$ and $\xi _1^\top =0$. Moreover, $a_2=a_3=a_4=1$. Otherwise, say $a_4=0$, then

$$\begin{aligned} \begin{pmatrix} \xi _1\\ \xi _2\\ \xi _3 \end{pmatrix}=\begin{pmatrix} 2&{}1&{}1\\ &{}1&{}\\ &{}&{}1 \end{pmatrix}\begin{pmatrix} \eta _1\\ \xi _2\\ \xi _3 \end{pmatrix}, \end{aligned}$$

which implies in the sublattice $\textrm{Span}_\mathbb {Z}\{\eta _1, \xi _2, \xi _3\}$, the shortest vectors $\xi _1,\xi _2,\xi _3$ can not form a generator, which contradicts with the induction hypothesis.

Note that $\xi _1^\top =0$ means $\xi _1$ is orthogonal to other $\xi _i$. It follows from $2\eta _1=\xi _1-\xi _2-\xi _3-\xi _4$ and $|\eta _1|\ge 1$ that $|\xi _2+\xi _3+\xi _4|\ge \sqrt{3}$. Moreover, using

$$\begin{aligned} \left| \frac{\xi _1+\xi _2-\xi _3-\xi _4}{2}\right| =|\eta _1+\xi _2|\ge 1, \end{aligned}$$

we have $|\xi _2-\xi _3-\xi _4|\ge \sqrt{3}$. Similarly, $|\xi _2-\xi _3+\xi _4|\ge \sqrt{3}$ and $|\xi _2+\xi _3-\xi _4|\ge \sqrt{3}$. Combining these with the following identities,

$$\begin{aligned}{} & {} |\xi _2+\xi _3+\xi _4|^2+|\xi _2-\xi _3-\xi _4|^2+|\xi _2-\xi _3+\xi _4|^2+|\xi _2+\xi _3-\xi _4|^2\\{} & {} \quad =4(|\xi _2|^2+|\xi _3|^2+|\xi _4|^2)=12, \end{aligned}$$

we can derive that

$$\begin{aligned} |\xi _2+\xi _3+\xi _4|^2=|\xi _2-\xi _3-\xi _4|^2=|\xi _2-\xi _3+\xi _4|^2=|\xi _2+\xi _3-\xi _4|^2=3 \end{aligned}$$

which implies $\{\xi _1, \xi _2, \xi _3, \xi _4\}$ forms an orthonormal basis of $\mathbb {R}^4$ and we complete the proof of this theorem. $\square$

Given a lattice $\Lambda _n^*$ of rank n, to investigate the set $\Xi$ constituted by all shortest lattice vectors up to $\pm 1$, we can consider the intersections of it with all sublattices of rank $n-1$. Let

$$\begin{aligned} m(\Xi ) \triangleq \max \Big \{\sharp (\Xi \cap {\tilde{\Lambda }})\,\Big |\,{\tilde{\Lambda }}\subset \Lambda _n^* \text {~is a sublattice of rank}~ n-1\Big \}, \end{aligned}$$

(19)

and $\Xi '$ be one of the intersection attaining $m(\Xi )$. Note that there may be more than one such intersections.

In the rest part of this section, we always assume $\Lambda _n^*$ is a prime lattice of rank n. For such lattice, we have $m(\Xi )\ge n-1$. Let $\Xi '$ be a chosen intersection attaining $m(\Xi )$, we assume that

$$\begin{aligned} \Xi '=\{\xi _1,\xi _2,\ldots ,\xi _{m(\Xi )}\}, \end{aligned}$$

and $\Lambda _n^{*}$ is generated by $\{\xi _1, \xi _2,\ldots ,\xi _{n-1}, \xi _{m(\Xi )+1}\}$. Then there exist N integer vectors

$$\begin{aligned} \{A_i=(a_{i1}, a_{i2}, \ldots ,a_{in})\}_{i=1}^N\subset \mathbb {Z}^n, \end{aligned}$$

such that

$$\begin{aligned} \begin{pmatrix} \xi _1\\ \xi _2\\ \vdots \\ \xi _N \end{pmatrix}=\begin{pmatrix} A_1\\ A_2\\ \vdots \\ A_N \end{pmatrix} \begin{pmatrix} \xi _1\\ \xi _2\\ \vdots \\ \xi _{n-1}\\ \xi _{m(\Xi )+1} \end{pmatrix}. \end{aligned}$$

(20)

Obviously, $\{A_1,A_2,\ldots ,A_{n-1},A_{m(\Xi )+1}\}$ is the standard basis of $\mathbb {R}^n$. It follows from the prime assumption and Lemma 5.6 that

$$\begin{aligned} a_{ji}\in \{0,\pm 1\},~~~1\le j\le N, 1\le i\le n. \end{aligned}$$

Moreover, by changing the direction of some vectors in $\Xi$ if necessary, we can assume the last coordinate of $A_i$ equals 0 for $1\le i\le m(\Xi )$, and 1 for $m(\Xi )+1\le i\le N$.

We will still abuse the notation Y to denote either the set constituted by $A_i$, or the matrix constructed by $A_i$ as in (20). Note that the prime assumption on $\Xi \subset \Lambda _n^*$ is equivalent to saying that any n linearly independent vectors in Y form a generator of $\mathbb {Z}^n$. For applications in the next section, we conclude some further properties of prime lattices in the following four lemmas.

Lemma 5.9

For a prime lattice, in terms of matrix, all minors of $Y^t$ can only take values in $\{0, \pm 1\}$.

Proof

Suppose the opposite that there is a nonzero minor involving the $i_1\text {th}, i_2\text {th}, \ldots , i_k$th rows and $j_1\text {th}, j_2\text {th}, \ldots , j_k$th columns that is not equal to $\pm 1$, then $\mathbb {Z}^n$ can not be generated by $\{A_{j_1}, A_{j_2}, \ldots ,A_{j_k}\}$ and $\{A_{i_{k+1}},A_{i_{k+2}}, \ldots , A_{i_{n}}\}$, since

$$\begin{aligned} |A_{j_1}\wedge \cdots \wedge A_{j_k}\wedge A_{i_{k+1}}\wedge \cdots \wedge A_{i_n}|>1, \end{aligned}$$

where $\{i_{k+1}, \ldots ,i_n\}$ is the complementary set of $\{i_1, \ldots , i_k\}$ in $\{1,2,\ldots ,n-1,m(\Xi )+1\}$. This gives a contradiction with our prime assumption. $\square$

Lemma 5.10

For any given $1\le i\le n$ and $m(\Xi )+1\le j,k \le N$, we have $a_{ji}a_{ki}\ge 0$, where $a_{ji}$ is the ith coordinate of $A_j$.

Proof

Assume $a_{j i}a_{k i}<0$. Then from $a_{jn}=a_{kn}=1$ we get the minor $\begin{vmatrix} a_{j i}&a_{k i}\\ 1&1 \end{vmatrix}=\pm 2,$ which is a contradiction with our prime assumption. $\square$

For a subset $I\subset \{1,2,\ldots ,n\}$, we say a subset $X\subset \mathbb {Z}^n$ has a partial order according to I, if for any given two vectors $\xi , \eta \in X$, their ith coordinates $\xi _i,\eta _i$ and jth coordinates $\xi _j,\eta _j$ satisfy

$$\begin{aligned} (\xi _i-\eta _i)(\xi _j-\eta _j)\ge 0,\quad i,j\in I. \end{aligned}$$

Lemma 5.11

If there is a vector $A_j \in Y$ such that three coordinates $\{a_{j i_{1}}, a_{j i_{2}}, a_{j i_{3}}\}$ of $A_j$ satisfy

$$\begin{aligned} a_{j i_{1}} a_{j i_{2}}=1,~~~ a_{j i_{3}}=0, \end{aligned}$$

then for any $1\le r \le N$, we have

$$\begin{aligned} a_{r i_1} a_{r i_2}\ge 0, \end{aligned}$$

and the subset $Y_{i_3} \triangleq \{ A_k\in Y\, | \, a_{k i_3}=1\}$ has a partial order according to $\{i_1, i_2\}$, so does the subset $\widehat{Y}_{i_3} \triangleq \{ A_k\in Y\, | \, a_{k i_3}=-1\}$.

Proof

Similarly as in the proof of Lemma 5.10, by considering the minor $\begin{vmatrix} a_{j i_1}&a_{r i_1}\\ a_{j i_2}&a_{r i_2} \end{vmatrix}$, we can derive the first conclusion.

Given two arbitrary vectors $A_k, A_l\in Y_{i_{3}}$, we have $a_{k i_{3}}=a_{l i_{3}}=1$, which implies $a_{k i_{1}}a_{l i_{1}}\ge 0$ and $a_{k i_{2}}a_{l i_{2}}\ge 0$ by considering the minors $\begin{vmatrix} a_{k i_1}&a_{l i_1}\\ 1&1 \end{vmatrix}$ and $\begin{vmatrix} a_{k i_2}&a_{l i_2}\\ 1&1 \end{vmatrix}$, respectively. Therefore, $a_{k i_{1}}-a_{l i_{1}}$ and $a_{k i_{2}}-a_{l i_{2}}$ all take values in $\{0,\pm 1\}$. Consider the minor

$$\begin{aligned} \begin{vmatrix} a_{j i_{1}}&a_{k i_{1}}&a_{l i_{1}}\\ a_{j i_{2}}&a_{k i_{2}}&a_{l i_{2}}\\ 0&1&1 \end{vmatrix}=\pm [(a_{k i_{1}}-a_{l i_{1}})-(a_{k i_{2}}-a_{l i_{2}})]. \end{aligned}$$

It follows from the prime assumption that $(a_{k i_{1}}-a_{l i_{1}})(a_{k i_{2}}-a_{l i_{2}})\ge 0$. Similar discussion can be applied to $\widehat{Y}_{i_{3}}$. $\square$

Lemma 5.12

If $A_1+A_2+\cdots +A_{n-1}+A_{m(\Xi )+1}$ belongs to Y, then all the coordinates of any $A_j\in Y$ must be either $\ge 0$ or $\le 0$. Especially, after changing the directions of some vectors, all entries of Y take values in $\{0, 1\}$.

Proof

Let $A_0=A_1+A_2+\cdots +A_{n-1}+A_{m(\Xi )+1}$. If there is a vector $A_j\in Y$ which doesn’t satisfy the conclusion, then one can find a 2-minor in $A_0$ and $A_j$ taking values other than $\{0, \pm 1\}$. This gives a contradiction with our prime assumption. $\square$

The following definition will also be used in the next section.

Definition 5.13

A vector in $\mathbb {Z}^n$ is called k-null, if it has exactly $n-k$ nonzero coordinates. Under some fixed generator of $\Lambda _n^*$, a lattice vector in $\Lambda _n^*$ is called k-null if its coordinate vector is k-null.

6 Classification of conformally flat and $\lambda _1$-minimal 3-tori and 4-tori

It follows from the works of Montiel–Ros [23] and El Soufi–Ilias [10] that for each conformal structure on compact manifold $(M^n,[g_0])$, there exists at most one metric $g\in [g_0]$ so that $(M^n,g)$ can be minimally immersed into a sphere by the first eigenfunctions. Moreover, if $(M^n,g_0)$ is homogeneous, then such $\lambda _1$-minimal metric must be $g_0$ itself (up to a constant dilation). Note that the flat torus $T^n$ is homogeneous, so we only need to classify all non-congruent $\lambda _1$-minimal immersions of falt 3-tori and 4-tori in spheres.

Let $x: T^n=\mathbb {R}^n/\Lambda _n\rightarrow \mathbb {S}^{2N-1}$ be a $\lambda _1$-minimal flat torus. Here we do not assume it is linearly full, and denote by N the number of distinct shortest lattice vectors of $\Lambda ^*_n$ up to $\pm 1$, i.e., the half dimension of the first eigenspace of flat torus $T^n=\mathbb {R}^n/\Lambda _n$. Without loss of generality, we assume this shortest length is 1. It is well known that $N\ge n$ (see Corollary 3.4 in [5]). Let $\Xi$ be the set of these N shortest vectors, which are denoted by $\xi _1, \xi _2, \ldots ,\xi _N$. Then according to Remark 4.1, x is reducible if and only if $\Lambda ^*_n$ is s-reducible defined as in Sect. 5.

6.1 Classification of conformally flat 3-tori

Theorem 6.1

Up to congruence, there are five distinct $\lambda _1$-minimal immersions of conformally flat 3-tori in spheres. Two of them are reducible ones given in Example 4.2, others are irreducible ones given in Examples 4.3 to 4.5. They are all listed in the Table 1.

Proof

For $\lambda _1$-minimal flat 3-torus, it follows from Theorem 5.8 that the dual lattice $\Lambda _3^*$ is always prime. Hence by Remark 5.5 and Proposition 2.7, the $\lambda _1$-minimal immersion x is homogeneous. Then it follows from Corollary 3.5 that we only need to prove the corresponding integer set Y of x is exactly that given in Examples 4.3 to 4.5, for which $\{A^tA\,|\, A\in Y\}$ is of rank N can be easily checked.

Suppose x is irreducible. It follows from Lemma 5.3 that there exists a generic 4-tuple in $\Xi$, which is assumed to be $\{\xi _1, \xi _2, \xi _3, \xi _4\}$. Moreover, we choose $\{\xi _1, \xi _2, \xi _3\}$ to be a generator of $\Lambda ^*_3$ so that $\xi _4=\xi _1+\xi _2+\xi _3$. Then Lemma 5.12 implies that all coordinates of lattice vectors in $\Xi$ can be assumed to take values in $\{0,1\}$. Therefore $\Xi \setminus \{\xi _1, \xi _2, \xi _3, \xi _4\}$ is constituted by 1-null lattice vectors, the number of which is no more than 3. On the other hand, from

$$\begin{aligned} (\xi _1+\xi _2)\wedge (\xi _2+\xi _3)\wedge (\xi _1+\xi _3)=2\xi _1\wedge \xi _2\wedge \xi _3, \end{aligned}$$

we know this number can not be 3, which implies $N\le 6$. If $N=4$, we obtain Example 4.3. If $N=5$, we obtain Example 4.4 by making a permutation to $\{\xi _1, \xi _2, \xi _3\}$ such that the only 1-null lattice vector in $\Xi$ is $\xi _1+\xi _2$. Similarly, we arrive at Example 4.5 for $N=6$. $\square$

6.2 Classification of conformally flat 4-tori

For $\lambda _1$-minimal flat 4-torus, it follows from Theorem 5.8 that the dual lattice $\Lambda _4^*$ is prime with only one exception, which is generated by the row vectors of (18). We will firstly discuss the $\lambda _1$-minimal immersion of such exceptional torus.

Lemma 6.2

The $\lambda _1$-minimal isometric immersion of such exceptional torus is homogeneous.

Proof

It follows from (18) that the shortest vectors of $\Lambda _4^*$ are composed of the following column vectors $\xi _i$ and $-\xi _i$,

$$\begin{aligned} \left( \begin{array}{rrrrrrrrrrrr} 1&{}0&{}0&{}0&{}{1}/{2}&{}{1}/{2}&{}{1}/{2}&{}{-1}/{2}&{}1/2&{}1/2&{}1/2&{}1/2 \\ 0&{}1&{}0&{}0&{}{1}/{2}&{}{1}/{2}&{}{-1}/{2}&{}{1}/{2}&{}1/2&{}1/2&{}-1/2&{}-1/2\\ 0&{}0&{}1&{}0&{}{1}/{2}&{}{-1}/{2}&{}{1}/{2}&{}{1}/{2}&{}1/2&{}-1/2&{}1/2&{}-1/2\\ 0&{}0&{}0&{}1&{}{1}/{2}&{}{-1}/{2}&{}{-1}/{2}&{}{-1}/{2}&{}-1/2&{}1/2&{}1/2&{}-1/2 \end{array} \right) , \end{aligned}$$

(21)

which can be divided into three blocks: $I_4$, S, $S^t$. It is obvious that S is an orthogonal matrix and $S^3=I_4$.

It suffices to show that the symmetric products from any $\eta$-set are linearly independent, where the conclusion arises according to Lemma 2.7.

Suppose $\eta =\xi _{r_1}\pm \xi _{s_1}$. Either $\{\xi _{r_1},\xi _{s_1}\}$ comes from the same block which can be assumed $I_4$ and thus $|\eta |^2=2$, or from different blocks which can be assumed $I_4$ and S so that $|\eta |^2=1$ or 3. Here we have used the symmetry induced by S.

When $|\eta |^2=3$, we may assume the first coordinate of $\eta$ being 3/2. Then it is easy to verify that there is no other possibilities for $\{\pm \xi _{r_i},\pm \xi _{s_j}\}$ (note that these vectors have to be distinct). Clearly, $\xi _{r_1}\odot \xi _{s_1}$ is linearly independent.

When $|\eta |^2=2$, the other pairs $\{\xi _{r_i},\xi _{s_i}\}$ must also come from a same block, for $\xi _{r_i}$ and $\xi _{s_i}$ have to be orthogonal. We may assume $\eta =(1,1,0,0)$ such that all pairs $\{\xi _{r_i},\xi _{s_i}\}$ are given as follows,

$$\begin{aligned} \eta =\xi _1+\xi _2=\xi _5+\xi _6=\xi _9+\xi _{10}. \end{aligned}$$

Then by direct computation, we obtain that $\xi _1\odot \xi _2$, $\xi _5\odot \xi _6$ and $\xi _9\odot \xi _{10}$ are linearly independent.

When $|\eta |^2=1$, $\eta$ is one of the shortest vectors. We may assume $\eta =\xi _5$ such that all pairs $\{\xi _{r_i},\xi _{s_i}\}$ are given as follows,

$$\begin{aligned} \eta =\xi _1-\xi _{12}=\xi _2+\xi _{11}=\xi _3+\xi _{10}=\xi _4+\xi _9. \end{aligned}$$

Then it is straightforward to verify that $\xi _1\odot \xi _{12}$, $\xi _2\odot \xi _{11}$, $\xi _3\odot \xi _{10}$ and $\xi _4\odot \xi _{9}$ are linearly independent. $\square$

Proposition 6.3

For such exceptional torus, there is altogether a 2-parameter family of $\lambda _1$-minimal isometric immersions in $\mathbb {S}^{23}$ up to congruence, given as follows (see also Example 1.1):

$$\begin{aligned} \begin{aligned}&\!\!(a_1e^{i\pi (u_1+u_2+u_3+u_4)},a_1e^{i\pi (u_1+u_2-u_3-u_4)},a_1e^{i\pi (u_1-u_2+u_3-u_4)},a_1e^{i\pi (-u_1+u_2+u_3-u_4)},\\&\quad a_2e^{i\pi (u_1+u_2+u_3-u_4)},a_2e^{i\pi (u_1+u_2-u_3+u_4)},a_2e^{i\pi (u_1-u_2+u_3+u_4)}, a_2e^{i\pi (u_1-u_2-u_3-u_4)},\\&\quad a_3e^{2i\pi u_1},a_3e^{2i\pi u_2},a_3e^{2i\pi u_3},a_3e^{2i\pi u_4}), \end{aligned} \end{aligned}$$

(22)

where $0\le a_1\le a_2\le a_3$ and $a_1^2+a_2^2+a_3^2=\frac{1}{4}$.

Proof

From (21) one can see that $\Lambda _4^*$ can be generated by $-\xi _1,-\xi _2,-\xi _3,\xi _5$ whose Gram matrix is

$$\begin{aligned} \left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} \frac{1}{2} \\ 0 &{} 1 &{} 0 &{} \frac{1}{2} \\ 0 &{} 0 &{} 1 &{} \frac{1}{2} \\ \frac{1}{2} &{} \frac{1}{2} &{} \frac{1}{2} &{} 1 \\ \end{array} \right) . \end{aligned}$$

(23)

By a straightforward computation, we can derive that the matrix Y characterizing all shortest lattice vectors up to $\pm 1$ is given by

$$\begin{aligned} Y=\left( \begin{array}{cccccccccccc} 1&{}1&{}0&{}0&{} 1&{}0&{}0&{} 1&{}1&{}0&{}0&{}1\\ 1&{}0&{}1&{}0&{} 0&{}1&{}0&{} 1&{}0&{}1&{}0&{}1\\ 0&{}1&{}1&{}0&{} 0&{}0&{}1&{} 1&{}0&{}0&{}1&{}1\\ 1&{}1&{}1&{}1&{} 1&{}1&{}1&{} 1&{}0&{}0&{}0&{}2\\ \end{array} \right) ^t. \end{aligned}$$

We point out that the order of $A_i$ in Y does not coincide with that of $\xi$ in (21). Moreover, for any $c_1^2+c_2^2\le 1/4$,

$$\begin{aligned} \Big (c_2^2,\;c_2^2,\;c_2^2,\;c_2^2,\;\frac{1}{4}-c_1^2-c_2^2,\;\frac{1}{4}-c_1^2-c_2^2,\;\frac{1}{4}-c_1^2-c_2^2,\;\frac{1}{4}-c_1^2-c_2^2,\; c_1^2\;c_1^2,\;c_1^2,\;c_1^2\Big ) \end{aligned}$$

(24)

defines a $\lambda _1$-minimal isometric immersion. In fact, these enumerate all possibilities of the $\lambda _1$-minimal isometric immersion, since the rank of $\{A^tA\,|\, A\in Y\}$ is 10, which can be verified directly. Define $a_1=\sqrt{c_2^2},\; a_2=\sqrt{\frac{1}{4}-c_1^2-c_2^2},\,a_3=\sqrt{c_1^2}$, then these immersions can be written down explicitly as (22).

Next, we discuss the congruence of these immersions. Note that an ambient congruence induces an isometry on the flat torus $T^n=\mathbb {R}^n/\Lambda _n$. It is well known that there are two kinds of isometries on $T^n$. One is produced by the translations on $\mathbb {R}^n$. It is obvious that the ambient congruence corresponding to such isometry can not transform one immersion of the form (22) to another. The other is induced from the orthogonal transformations on $\mathbb {R}^n$ which preserve the lattice $\Lambda _n$, and then $\Lambda _n^*$. Since such orthogonal transformations preserve the lengths and angles of lattice vectors, they induce perturbations between the sets $\{\pm I_4\}$, $\{\pm S\}$ and $\{\pm S^2\}$, which further induce perturbations on $\{a_1, a_2, a_3\}$. $\square$

Remark 6.4

The immersion given in (21) can be seen as a twist product:

$$\begin{aligned} x(\textbf{u})\triangleq \Big (a_1f(\textbf{u}S),\;a_2f(\textbf{u}S^2),\;a_3f(\textbf{u})\Big )\in \mathbb {S}^7(2a_1)\times \mathbb {S}^7(2a_2)\times \mathbb {S}^7(2a_3)\subset \mathbb {S}^{23}, \end{aligned}$$

where $\textbf{u}=(u_1,u_2,u_3,u_4)$, $f(\textbf{u})= (e^{2i\pi u_1},e^{2i\pi u_2},e^{2i\pi u_3},e^{2i\pi u_4})$, and S is the following orthogonal matrix of order 3:

$$\begin{aligned} \left( \begin{array}{rrrr} {1}/{2}&{}{1}/{2}&{}{1}/{2}&{}{-1}/{2} \\ {1}/{2}&{}{1}/{2}&{}{-1}/{2}&{}{1}/{2}\\ {1}/{2}&{}{-1}/{2}&{}{1}/{2}&{}{1}/{2}\\ {1}/{2}&{}{-1}/{2}&{}{-1}/{2}&{}{-1}/{2} \end{array} \right) . \end{aligned}$$

The flat torus involved in $x(\textbf{u})$ is

$$\begin{aligned} \mathbb {R}^4/\textrm{Span}_{\mathbb {Z}}\{e_1-e_4, e_2-e_4, e_3-e_4, 2e_4\}, \end{aligned}$$

with the volume $2\pi ^4$. Note that when $a_3=\frac{1}{2}$, $x(\textbf{u})$ reduces to the double covering of the Clifford 4-torus with the underlying flat torus

$$\begin{aligned} \mathbb {R}^4/\textrm{Span}_{\mathbb {Z}}\{e_1, e_2, e_3, e_4\}. \end{aligned}$$

Theorem 6.5

Up to congruence, Examples 4.6 to  4.16, and Proposition 6.3 exhaust all $\lambda _1$-minimal immersions of conformally flat 4-tori in spheres. They are all listed in the Table 2.

Proof

The exceptional case has been discussed in Proposition 6.3. Next, we assume $\Lambda _4^*$ is prime. It follows from Remark 5.5 and Proposition 2.7 that the $\lambda _1$-minimal immersion x is homogeneous. Combining this with Corollary 3.5, we only need to prove the corresponding integer set Y of x is exactly that given in Example 4.6$\sim$ Example 4.16, for which $\{A^tA\,|\, A\in Y\}$ is of rank N can be easily checked.

Note that all reducible ones have be given in Example 4.6. We will complete the classification of irreducible ones through the following lemmas and propositions involving prime lattices of rank 4. $\square$

In the rest discussion, when some generator of $\Lambda _4^*$ is chosen, we will identify the lattice vectors with their coordinates with respect to the given generator such that all vectors belong to $\mathbb {Z}^4$.

Lemma 6.6

Suppose $\Lambda _4^*$ is an s-irreducible prime lattice of rank 4, if there is no generic 5-tuple in $\Xi$, then $N\le 7$ and Y takes the form as given in Example 4.9.

Proof

It follows from Lemma 5.3 that there always exists a generic 4-tuple in $\Xi$. Set $\{\eta _1, \eta _2, \eta _3, \eta _4\}$ to be a generator of $\Lambda _4^*$ such that the generic 4-tuple is given by

$$\begin{aligned} \{\eta _1, \eta _2, \eta _3, \eta _5\triangleq \eta _1+\eta _2+\eta _3\}. \end{aligned}$$

Using Lemma 5.12, we can assume that all the coordinates of lattice vectors in $\Xi \cap \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ take values in $\{0,1\}$.

Since $\Lambda _4^*$ is s-irreducible, there is at least one vector in $\Xi {\setminus } \textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2, \eta _3\}$ other than $\eta _4$. Let $\eta$ be such vector which can not be 2-null. Otherwise, $\eta =\eta _j\pm \eta _4$ for some $1\le j \le 3$ and then $\{\eta _1, \eta _2, \eta _3,\eta _4, \eta _5,\eta \}{\setminus } \{\eta _j\}$ is a generic 5-tuple. So we can assume $\eta$ is $(a_1, a_2, 0, 1)$ by the symmetry of $\eta _1,\eta _2,\eta _3$. It follows from Lemma 5.11 that $a_1a_2>0$. Therefore, after changing the direction of $\eta$ and $\eta _4$ if necessary we can assume $a_1=a_2=1$. Moreover, using Lemma 5.10 and the partial order according to $\{1,2,3\}$ (see Lemma 5.11), we know there exists no other 1-null vectors in $\Xi {\setminus } \textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2, \eta _3\}$, which means $\Xi {\setminus } \textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2, \eta _3\}=\{\eta _4, \eta \}.$

If $\eta _1+\eta _3$ (resp. $\eta _2+\eta _3$) belongs to $\Xi$, then $\{\eta _4, \eta , \eta _5\}$ together with $\{\eta _1+\eta _3, \eta _1\}$ (resp. $\{\eta _2+\eta _3, \eta _2\}$) constitute a generic 5-tuple. Therefore, besides $\{\eta _1, \eta _2, \eta _3,\eta _5\}$, $\eta _1+\eta _2$ is the only vector may appear in $\Xi$ from $\textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$, which completes the proof of this lemma. $\square$

Lemma 6.7

Suppose $\Xi$ contains a generic 5-tuple X. Then we can choose $\Xi '$ such that $\sharp (\Xi '\cap X)= 3$.

Proof

We assume $X=\{\eta _1, \eta _2, \eta _3, \eta _4, \eta _5\}$ and $\eta _5=\eta _1+\eta _2+\eta _3+\eta _4$. Choose $\{\eta _1, \eta _2, \eta _3, \eta _4\}$ to be a generator of $\Lambda _4^*$, then by Lemma 5.12 we have all coordinates of vectors in $\Xi$ taking values in $\{0,1\}$. Since $\textrm{rank}(\Xi ')=3$, it suffices to prove that we can choose $\Xi '$ such that $\sharp (\Xi '\cap X)\ge 3$.

If $\Xi '$ contains only 0-null or 3-null vectors (i.e. $\eta _i$) then $\Xi =\{\eta _1, \eta _2, \eta _3, \eta _4, \eta _5\}$ and $m(\Xi )=3$. For $m(\Xi )>3$, $\Xi '$ contains at least a 1-null or 2-null vector. Note that by rechoosing a generator of $\Lambda _4^*$ in X, a given 2-null vector (such as $\eta _1+\eta _2$) can be transformed to a 1-null vector (such as $\eta _1+\eta _2+\eta _3$ by choosing $\{\eta _5, -\eta _3, -\eta _4, -\eta _1\}$ as a new generator). Therefore, without loss of generality, we assume there is a 1-null vector $\eta \triangleq \eta _1+\eta _2+\eta _3\in \Xi '$.

It is easy to see that if $\sharp (\Xi '\cap \{\eta _1, \eta _2,\eta _3\})=2$ then $\Xi '=\Xi \cap \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$.

If $\sharp (\Xi '\cap \{\eta _1, \eta _2,\eta _3\})=1$, we can assume it is $\eta _1$. Since $\sharp (\Xi \cap \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\})\ge 4$ and $\eta _1,\eta \in \Xi '$, we know $\sharp (\Xi '{\setminus }\textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\})\ge 2$ in which all the vectors have coordinates 1 with respect to $\eta _4$. It follows from Lemma 5.11 that the existence of $\eta =\eta _1+\eta _2+\eta _3$ implies the vectors in $\Xi '\setminus \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ obey the partial order according to $\{1,2,3\}$, which means there is at most one k-null vector in $\Xi '\setminus \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ for every $0\le k\le 3$. When $\Xi '\setminus \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ is composed of exactly two vectors, we get $\Xi \cap \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ also attaining $m(\Xi )$ so that it can be chosen as the $\Xi '$ we desired. Or else, $\Xi '$ must contain $\eta _4$ or $\eta _5$. Since $\eta _5-\eta _4=\eta \in \Xi '$, we have $\Xi '$ must contain both $\eta _4$ and $\eta _5$ simultaneously, which implies $\eta _1,\eta _4,\eta _5\in \Xi '$. The third one in $\Xi '\setminus \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ must be $\eta _1+\eta _4$ if it’s 2-null or $\eta _2+\eta _3+\eta _4$ if it’s 1-null. Otherwise, $\eta _2$ or $\eta _3$ will be contained in $\Xi '$. Moreover, only one of these two vectors could appear in $\Xi '$ for they violate the partial order according to $\{1,2,3\}$. This is obviously a case of $N=7$.

If $\Xi '\cap \{\eta _1, \eta _2,\eta _3\}=\emptyset$ then $\Xi '\cap \textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2,\eta _3\}=\{\eta \}$. Similarly, we have $\sharp (\Xi '{\setminus }\textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\})\ge 3$ and all the vectors in $\Xi '\setminus \textrm{Span}_{\mathbb {Z}} \{\eta _1, \eta _2,\eta _3\}$ admit at most one k-null vector for every $0\le k\le 3$. Therefore, $\Xi '\cap \{\eta _4,\eta _5\}\not =\emptyset$ and thus $\eta _4,\eta _5\in \Xi '$. Note that there are still others left in $\Xi '\setminus \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$. However, any 1-null (resp. 2-null) minus $\eta _5$ (resp. $\eta _4$) will yield $\Xi '\cap \{\eta _1, \eta _2,\eta _3\}\not =\emptyset$. Hence we finish the proof by this contradiction. $\square$

Remark 6.8

From the proof of above two lemmas one can see that for $N\ge 8$, we can always choose a generic 5-tuple and a generator of $\Lambda _4^*$ as in Lemma 6.7 such that $\Xi '=\Xi \cap \textrm{Span}_\mathbb {Z} \{\eta _1, \eta _2,\eta _3\}$ and $\eta \in \Xi '$.

Proposition 6.9

If $N\le 7$ and there exists a generic 5-tuple in $\Xi$, then we can find a generator of $\Lambda _4^*$ such that Y takes the form as given in Examples 4.7,  4.8,  4.10 and  4.11.

Proof

Suppose $\{\eta _1, \eta _2, \eta _3, \eta _4, \eta _5\}$ is a generic 5-tuple in $\Xi$, with $\eta _5=\eta _1+\eta _2+\eta _3+\eta _4$. When $N=5$, it is easy to see that after choosing $\{\eta _1, \eta _2, \eta _3, \eta _4\}$ to be a generator of $\Lambda ^*_4$, we obtain Y as given in Example 4.7.

When $N\ge 6$, similarly as discussed in the proof of Lemma 6.7, we can assume

$$\begin{aligned} \eta _1, \eta _2, \eta _3, \eta _1+\eta _2+\eta _3, \eta _4, \eta _5\in \Xi . \end{aligned}$$

For the case of $N=6$, by choosing $\{\eta _1, \eta _2, \eta _3, \eta _4\}$ to be a generator of $\Lambda ^*_4$, we arrive at Example 4.8. For the case of $N=7$, if the remainder lattice vector lies on $\textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2, \eta _3\}$, then we obtain Y as given in Example 4.10 after a permutation in $\{\eta _1, \eta _2, \eta _3\}$. Now we assume the remainder lies on $\Xi {\setminus }\textrm{Span}_{\mathbb {Z}}\{\eta _1, \eta _2, \eta _3\}$. Either it is 1-null, which can be assumed $\eta _2+\eta _3+\eta _4$ without loss of generality, so that we arrive at Example 4.11. Or it is 2-null, which can be assumed $\eta _3+\eta _4$ without loss of generality. Then choosing $\{-(\eta _1+\eta _2+\eta _3), \eta _5, -(\eta _3+\eta _4), -\eta _2\}$ as a generator of $\Lambda _4^*$, we also obtain Y as in Example 4.11. $\square$

Proposition 6.10

If $N\ge 8$, then there exist a generator of $\Lambda _4^*$ such that Y takes the form as given in Examples 4.12 to  4.16.

Proof

It follows from Remark 6.8 that there exist a generic 5-tuple $\{\eta _1, \eta _2, \eta _3, \eta _4,\eta _5\}$ and $\Xi '$ such that

$$\begin{aligned} \{\eta _1, \eta _2, \eta _3, \eta \triangleq \eta _1+\eta _2+\eta _3\}\in \Xi ',\quad \eta _5=\eta _1+\eta _2+\eta _3+\eta _4. \end{aligned}$$

By choosing $\{\eta _1, \eta _2, \eta _3, \eta _4\}$ as a generator of $\Lambda _4^*$, it follows from Lemma 5.12 that all coordinates of vectors in Y can only take values in $\{0,1\}$.

If $m(\Xi )=4$, the vectors in $\Xi$ other than $\eta _1, \eta _2, \eta _3, \eta , \eta _4, \eta _5$ don’t lie on $\textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$. As in the proof of Lemma 6.7, they can be assumed $\eta _3+\eta _4$ and $\eta _2+\eta _3+\eta _4$ due to the partial order according to $\{1,2,3\}$ (see Lemma 5.11). It means $\sharp (\Xi \cap \textrm{Span}_{\mathbb {Z}}\{\eta _2,\eta _3,\eta _4\})=5$ against $m(\Xi )=4$. So $m(\Xi )\ge 5$.

When $m(\Xi )=5$, we can assume $\eta _1+\eta _2\in \Xi '$ after making a permutation to $\{\eta _1, \eta _2, \eta _3\}$. Note that $\eta ,\eta _5\in \textrm{Span}_{\mathbb {Z}}\{\eta _1+\eta _2,\eta _3,\eta _4\}$, neither $\eta _3+\eta _4$ nor $\eta _1+\eta _2+\eta _4$ appears in $\Xi$ which will make $\sharp (\Xi \cap \textrm{Span}_{\mathbb {Z}}\{\eta _1+\eta _2,\eta _3,\eta _4\})\ge 6>m(\Xi )$. Due to the partial order according to $\{1,2,3\}$ and the symmetry of $\eta _1$ and $\eta _2$, there is at most one 1-null vector in $\Xi \setminus \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$ which can be assumed $\eta _2+\eta _3+\eta _4$, and at most one 2-null vector in $\Xi \setminus \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$ which can be assumed $\eta _2+\eta _4$. So $N\le 9$. If $N=9$, both $\eta _2+\eta _3+\eta _4$ and $\eta _2+\eta _4$ exist and Y takes the form as given in Example 4.15. If $N=8$ and only $\eta _2+\eta _3+\eta _4$ exists, then we arrive at Example 4.13. If $N=8$ and only $\eta _2+\eta _4$ exists, then after choosing $\{-\eta _2, -\eta _1, -\eta _3, \eta _5\}$ as a new generator, we can also obtain Y as given in Example 4.13.

When $m(\Xi )=6$, up to a permutation of $\{\eta _1, \eta _2, \eta _3\}$, we can assume $\eta _1+\eta _2, \eta _2+\eta _3 \in \Xi '$. From

$$\begin{aligned} \begin{aligned}&(\eta _1+\eta _3)\wedge (\eta _1+\eta _2)\wedge (\eta _2+\eta _3)\wedge \eta _4=(\eta _1+\eta _3+\eta _4)\wedge (\eta _1+\eta _2)\\&\qquad \wedge (\eta _2+\eta _3)\wedge \eta _4\\&\quad =(\eta _2+\eta _4)\wedge (\eta _2+\eta _3)\wedge (\eta _1+\eta _2)\wedge \eta _5 =2\eta _1\wedge \eta _2\wedge \eta _3\wedge \eta _4, \end{aligned} \end{aligned}$$

we know none of $\eta _1+\eta _3$, $\eta _1+\eta _3+\eta _4$ or $\eta _2+\eta _4$ can be contained in $\Xi$. Combining this with the partial order according to $\{1,2,3\}$ and the symmetry of $\eta _1$ and $\eta _3$, there is at most one 1-null vector in $\Xi \setminus \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$ which can be assumed $\eta _2+\eta _3+\eta _4$, and at most one 2-null vector in $\Xi \setminus \textrm{Span}_{\mathbb {Z}}\{\eta _1,\eta _2,\eta _3\}$ which can be assumed $\eta _3+\eta _4$. So $N\le 10$. If $N=8$ then Y takes the form as given in Example 4.12. If $N=10$, both $\eta _2+\eta _3+\eta _4$ and $\eta _3+\eta _4$ exist, which leads to Example 4.16. If $N=9$ and only $\eta _3+\eta _4$ exists, then we obtain Y as given in Example 4.14. If $N=9$ and only $\eta _2+\eta _3+\eta _4$ exists, then after choosing $\{-\eta _3, -\eta _2, -\eta _1, \eta _5\}$ as a new generator, we have also Y taking the form as given in Example 4.14.

The non-existence of $\eta _1+\eta _3$ implies $m(\Xi )<7$ and we finish the proof. $\square$

Remark 6.11

Through the discussion in this section, we can obtain that in a prime lattice of rank $n\le 4$, there exists at most $\frac{n(n+1)}{2}$ distinct lattice vectors of shortest length up to $\pm 1$.

7 On $\lambda _1$-minimal flat tori of higher dimension

Similarly as in Sect. 4, we can construct many examples of higher dimensional $\lambda _1$-minimal flat tori in spheres. For simplicity, we choose a certain class to introduce in this section.

Consider the set $X_n\subset \mathbb {Z}^n$ given by the column vectors of

$$\begin{aligned} X^t_n=\begin{pmatrix} 1 &{} &{} 1 &{} &{} &{} 1 &{} &{} &{} &{} 1\\ &{}1&{} 1 &{} &{}1&{} 1 &{} &{} &{} &{} 1\\ &{} &{} &{}1&{}1&{} 1 &{} &{} &{} &{} 1\\ &{} &{} &{} &{} &{} &{}\ddots &{} &{} &{} \vdots \\ &{} &{} &{} &{} &{} &{} &{} &{}1\cdots &{}1\\ &{} &{} &{} &{} &{} &{} &{}1&{}1\cdots &{} 1\\ \end{pmatrix}_{n\times \frac{n(n+1)}{2}}. \end{aligned}$$

We call $X_n$ the ladder set. It is straightforward to verify that $W_{X_n}$ is a singleton set, only contains

$$\begin{aligned} Q_n=\begin{pmatrix} 1 &{}-\frac{1}{2}&{} &{} &{} &{} \\ -\frac{1}{2}&{} 1 &{}-\frac{1}{2}&{}&{}&{}\\ &{}-\frac{1}{2}&{}1&{}&{}&{}\\ &{}&{}\ddots &{}\ddots &{}\ddots &{}\\ &{}&{}&{}&{}1&{}-\frac{1}{2}\\ &{}&{}&{}&{}-\frac{1}{2}&{}1\\ \end{pmatrix}. \end{aligned}$$

By direct computation, we can obtain that

$$\begin{aligned} Q^{-1}_n=\frac{2}{n+1}\begin{pmatrix} n&{}n-1&{}n-2&{}\cdots &{}3&{}2&{}1\\ n-1&{}(n-1)2&{}(n-2)2&{}\cdots &{}3\cdot 2&{}2\cdot 2&{}2\\ n-2&{}(n-2)2&{}(n-3)3&{}\cdots &{}3\cdot 3&{}3\cdot 2&{}3\\ \vdots &{}\vdots &{}\vdots &{}&{}\vdots &{}\vdots &{}\vdots \\ 3&{}3\cdot 2&{}3\cdot 3&{}\cdots &{}(n-3)3&{}(n-3)2&{}n-2\\ 2&{}2\cdot 2&{}3\cdot 2&{}\cdots &{}(n-2)2&{}(n-1)2&{}n-1\\ 1&{}2&{}3&{}\cdots &{}n-2&{}n-1&{}n\\ \end{pmatrix}, \end{aligned}$$

(25)

and

$$\begin{aligned} Q_n^{-1}=\frac{2}{n+1}\left( A_1^t A_1+A_2^t A_2+\cdots +A_{\frac{n(n+1)}{2}}^tA_{\frac{n(n+1)}{2}}\right) . \end{aligned}$$

(26)

Moreover, for any $(a_1, a_2, \cdots , a_n)\in \mathbb {Z}^n\backslash 0$, we have

$$\begin{aligned}{} & {} (a_1, a_2, \cdots , a_n)Q_n(a_1, a_2, \cdots , a_n)^t=\sum _{i=1}^n a_i^2-\sum _{i=1}^{n-1} a_{i}a_{i+1}\nonumber \\{} & {} \quad =\sum _{i=1}^{n-1} \frac{(a_{i}-a_{i+1})^2}{2}+\frac{a_1^2+a_n^2}{2}\ge 1, \end{aligned}$$

(27)

with the equality holding if and only if $(a_1, a_2, \cdots , a_n)$ comes from $X_n$.

Proposition 7.1

The matrix data set $\{X_n,\, Q_n,\, \frac{2}{n(n+1)}(1,1,\cdots ,1)\}$ provides a $\lambda _1$-minimal flat torus in $\mathbb {S}^{n(n+1)-1}$, which is linearly full and has volume $\frac{(2\sqrt{2})^n}{\sqrt{n+1}(\sqrt{n})^n}\pi ^n$.

One can see that when we take $n=2, 3, 4$, we obtain the equilateral torus, Examples 4.5, and  4.16, respectively.

Next, for a given $1\le k\le n$, we consider the integer set $X_{n,k}$ obtained by removing the last $n-k$ rows from the ladder set $X_n$, i.e.,

$$\begin{aligned} X^t_{n,k}=\begin{pmatrix} 1 &{} &{} 1 &{} &{} &{} 1 &{} &{} &{} &{} \\ &{}1&{} 1 &{} &{}1&{} 1 &{} &{} &{} &{} \\ &{} &{} &{}1&{}1&{} 1 &{} &{} &{} &{} 1\\ &{} &{} &{} &{} &{} &{}\ddots &{} &{} &{} \vdots \\ &{} &{} &{} &{} &{} &{} &{} &{}1\cdots &{}1\\ &{} &{} &{} &{} &{} &{} &{}1&{}1\cdots &{} 1\\ \end{pmatrix}_{n\times (\frac{n(n-1)}{2}+k)}. \end{aligned}$$

(28)

Remark 7.2

Note that when $k=n$, $X_{n,k}=X_n$, which can automatically determine a $\lambda _1$-minimal flat n-torus as discussed above. When $k=1$, $X_{n,k}$ is a block diagonal matrix, from which a reducible $\lambda _1$-minimal flat n-torus can be obtained.

Next let us consider the case of $2\le k\le n-1$. Such kind of $X_{n,k}$ is called the faulted ladder set.

Lemma 7.3

Define

$$\begin{aligned} Q_{n,k}\triangleq Q_n+\frac{1}{2k}(E_{n,n-k}+E_{n-k,n}), \quad \end{aligned}$$

where $E_{i,j}=e_i^t e_j$ and $e_i$ is the i-th row of $I_n$, then $Q_{n,k}\in W_{X_{n,k}}$, and

$$\begin{aligned} Q_{n,k}^{-1}=R_n+S_k-T_k, \end{aligned}$$

with

$$\begin{aligned} R_n=\begin{pmatrix} Q^{-1}_{n-1}&{}\\ {} &{}0 \end{pmatrix},\quad S_k=\begin{pmatrix} O&{}\\ {} &{}Q^{-1}_k \end{pmatrix},\quad T_k=\begin{pmatrix} O&{}&{}\\ {} &{}Q^{-1}_{k-1}&{}\\ {} &{}&{}0 \end{pmatrix}. \end{aligned}$$

Proof

The first conclusion is easy to be verified, we only prove the second one.

Write

$$\begin{aligned} Q_n^{-1}=\frac{2}{n+1}\left( \sum _{1\le i<j\le n}i(n+1-j)\left( E_{i,j}+E_{j,i}\right) +\sum _{1\le i \le n}i(n+1-i)E_{i,i}\right) , \end{aligned}$$

It follows that

$$\begin{aligned} R_n&=\frac{2}{n}\left( \sum _{1\le i<j\le n-1}i(n-j)\left( E_{i,j}+E_{j,i}\right) +\sum _{1\le i \le n-1}i(n-i)E_{i,i}\right) , \end{aligned}$$

(29)

$$\begin{aligned} S_k&=\frac{2}{k+1}\!\!\left( \!\!\sum _{1\le i<j\le k}i(k+1-i)\left( E_{n-k+i,n-k+j}+E_{n-k+j,n-k+i}\right) \right. \nonumber \\&\quad \left. +\!\!\!\sum _{1\le i \le k}i(k+1-i)E_{n-k+i,n-k+i}\right) ,\end{aligned}$$

(30)

$$\begin{aligned} T_k&=\frac{2}{k}\left( \sum _{1\le i<j\le k-1}i(k-i)\left( E_{n-k+i,n-k+j}+E_{n-k+j,n-k+i}\right) \right. \nonumber \\&\quad \left. +\sum _{1\le i \le k-1}i(k-i)E_{n-k+i,n-k+i}\right) . \end{aligned}$$

(31)

Note that we can also express $Q_{n,k}$ as follows:

$$\begin{aligned} \begin{aligned} Q_{n,k}&= \begin{pmatrix} Q_{n-1}&{}\\ {} &{}1 \end{pmatrix}-\frac{1}{2}(E_{n,n-1}+E_{n-1,n})+\frac{1}{2k}(E_{n,n-k}+E_{n-k,n}), \\&=\begin{pmatrix} Q_{n-k}&{}\\ {} &{}Q_k \end{pmatrix}-\frac{1}{2}(E_{n-k+1,n-k}+E_{n-k,n-k+1})+\frac{1}{2k}(E_{n,n-k}+E_{n-k,n})\\&=\!\!\begin{pmatrix} Q_{n-k}&{}&{}\\ {} &{}Q_{k-1}&{}\\ {} &{}&{}1 \end{pmatrix}-\frac{1}{2}(E_{n,n-1}+E_{n-1,n})\!-\!\frac{1}{2}(E_{n-k+1,n-k}\\&\quad +E_{n-k,n-k+1})\!+\!\frac{1}{2k}(E_{n,n-k}+E_{n-k,n}). \end{aligned} \end{aligned}$$

(32)

Combining the fact that

$$\begin{aligned} E_{i,j}E_{k,l}=\delta _{j,k}E_{i,l},~~~1\le i, j, k, l\le n \end{aligned}$$

with (29) $\sim$ (32), we can obtain that

$$\begin{aligned}&Q_{n,k}R_n= \begin{pmatrix} I_{n-1}&{}\\ {} &{}0 \end{pmatrix}+\sum _{n-k< i\le n-1}\frac{n-k-i}{k}E_{n,i}, \\&Q_{n,k}S_k=\begin{pmatrix} 0&{}\\ {} &{}I_k \end{pmatrix}-\sum _{1\le i\le k}\frac{k-i}{k}E_{n-k,n-k+i},\\&Q_{n,k}T_k=\begin{pmatrix} 0&{}&{}\\ {} &{}I_{k-1}&{}\\ {} &{}&{}0 \end{pmatrix}-\sum _{1\le i\le k-1}\frac{i}{k}E_{n,n-k+i}-\sum _{1\le i\le k-1}\frac{k-i}{k}E_{n-k,n-k+i}, \end{aligned}$$

from which the conclusion follows. $\square$

It follows from (26) that

$$\begin{aligned}{} & {} R_n=\frac{2}{n}\left( A_1^tA_1+\cdots +A_{\frac{n(n-1)}{2}}^tA_{\frac{n(n-1)}{2}}\right) =\frac{2}{n}\sum _{j=1}^{n-1}\sum _{i=1}^j A^t_{\frac{j(j+1)}{2}+i}A_{\frac{j(j+1)}{2}+i},\quad \\{} & {} S_k=\frac{2}{k+1}\sum _{j=n-k}^{n-1}\sum _{i=1}^{j+1+k-n} A^t_{\frac{j(j+1)}{2}+i}A_{\frac{j(j+1)}{2}+i},~~~ T_k=\frac{2}{k}\sum _{j=n-k}^{n-2}\sum _{i=1}^{j+1+k-n} A^t_{\frac{j(j+1)}{2}+i}A_{\frac{j(j+1)}{2}+i}, \end{aligned}$$

where $A_i^t$ is the i-th column vector of $X^t_{n,k}$. Note that in $Q_{n,k}^{-1}$, the coefficient of $A^t_{\frac{j(j+1)}{2}+i}A_{\frac{j(j+1)}{2}+i}$ is

$$\begin{aligned} \frac{2}{n}+\frac{2}{k+1}-\frac{2}{k}=\frac{2(k^2+k-n)}{nk(k+1)} \end{aligned}$$

for $n-k\le j\le n-2$ and $1\le i\le j+1+k-n$; the other coefficients are $\frac{2}{n}>0$ or $\frac{2}{k+1}>0$.

Proposition 7.4

Suppose $2\le k\le n-1$.

(1) When $k^2+k>n$, the faulted ladder set $X^t_{n,k}$ given in (28) can determine a linearly full and $\lambda _1$-minimal flat n-torus in $\mathbb {S}^{n(n-1)+2k-1}$.

(2) When $k^2+k=n$, the faulted ladder set $X^t_{n,k}$ given in (28) can determine a linearly full and $\lambda _1$-minimal flat torus in a sphere of dimension $n(n-1)-k^2+3k-1$, with ${k(k-1)}$ dimensional eigenfunctions redundant.

Proof

We assume $\Lambda _n^*$ is the lattice determined by $Q_{n,k}~(1\le k\le n)$, with $\{\xi _1,\cdots ,\xi _n\}$ being a generator. Let $\Lambda _n$ be the dual lattice. Define $\left( c_1^2, c_2^2, \cdots \cdots , c^2_{\frac{n(n+1)}{2}+k}\right)$ as follows:

$$\begin{aligned} c^2_{\frac{j(j+1)}{2}+i}={\left\{ \begin{array}{ll} \dfrac{2(k^2+k-n)}{nk(k+1)},&{}n-k\le j\le n-2 \text { and } 1\le i\le j+1+k-n;\\ \dfrac{2}{n},&{}j<n-k \text { and } 1\le i\le j;\\ \dfrac{2}{k+1},&{}\text {others}. \end{array}\right. } \end{aligned}$$

Since the matrix data set $\Big \{X_{n,k}, Q_{n,k}, \big (c_1^2, c_2^2, \cdots \cdots , c^2_{\frac{n(n+1)}{2}+k}\big )\Big \}$ satisfy (12) and (13), they can determine an isometric minimal immersion of $T^n=\mathbb {R}^n/\Lambda _n$ in $\mathbb {S}^{n(n-1)+2k-1}$.

We are left to show 1 is the shortest length in $\Lambda _n^*$, and all the lattice vectors of this length having coordinates vectors as given in $X_{n,k}$ up to the sign.

Consider the sublattice generated by $\{\xi _1,\cdots ,\xi _{n-1}\}$. It is obvious that it takes $Q_{n-1}$ as the Gram matrix. Using (27), we know 1 is the shortest length in this sublattice. Next, we will use Theorem 3.8 to show this is also true for $\Lambda _n^*$.

Firstly, we calculate the distance from $\xi _n$ to the hyperplane $\textrm{Span}_{\mathbb {R}}\{\xi _1,\ldots ,\xi _{n-1}\}$. Define $v\triangleq (v_1,\ldots ,v_{n-1})$, where $v_{n-k}=1/(2k)$, $v_{n-1}=-1/2$ and the others are 0. By (16) we get

$$\begin{aligned} d_n^2=1-vQ_{n-1}^{-1}v^t=1-\frac{k-1}{2k}=\frac{k+1}{2k}>\frac{1}{4}, \end{aligned}$$

which implies for any vector $\xi =a_1\xi _1+\cdots +a_n\xi _n\in \Lambda _n^*$, if $|a_n|> 1$ then $|\xi |> 1$. When $a_n=0$, as a lattice vector in the sublattice, it is obvious that $|\xi |\ge 1$, with equality holding if and only if $\pm (a_1. a_2, \ldots , a_{n-1})$ belongs to $X_{n-1}$, which can be embedded into $X_{n,k}$. So we only need to discuss the case of $a_n=\pm 1$. By considering $-\xi$ if necessary, we can assume $a_n=1$. Then we have

$$\begin{aligned} |\xi |^2-1&=a_1^2+\cdots +a_{n-1}^2-a_1a_2-\cdots -a_{n-2}a_{n-1}-a_{n-1}+\frac{1}{k}a_{n-k}\\&=\sum _{i=1}^{n-1}\frac{1}{2}(a_i-a_{i+1})^2+\frac{1}{2}\left( a_1^2-1\right) +\frac{1}{k}a_{n-k}, \end{aligned}$$

which is greater than 0 when $a_{n-k}>0$. For the case of $a_{n-k}=0$, it is obvious that $|\xi |^2-1\ge 0$, and the equality holds if and only if $a_1=\cdots =a_{p-1}=0$ and $a_p=\cdots =a_n=1$ for some $p>n-k$, which exactly corresponds to a certain row of $X_{n,k}$. When $a_{n-k}<0$, using Cauchy inequality, we have

$$\begin{aligned} \sum _{i=n-k}^{n-1}(a_i-a_{i+1})^2\ge \frac{1}{k}\left( \sum _{i=n-k}^{n-1}(a_i-a_{i+1})\right) ^2=\frac{1}{k}\left( 1-a_{n-k}\right) ^2>-\frac{2}{k}a_{n-k}, \end{aligned}$$

from which it follows that $|\xi |^2-1> 0$, and we finish the proof. $\square$

Remark 7.5

The $\lambda _1$-minimal flat torus in this section has volume $\frac{\sqrt{k}(2\sqrt{2})^n}{\sqrt{k+1}(\sqrt{n})^{n+1}}\pi ^n$ $(1\le k\le n)$. It comes from the fact that $\det Q_{n,k}=\frac{n(k+1)}{2^nk}$ which again indicates $Q_{n,k}$ is positive-definite. One can easily check that Example 4.4, 4.12 and 4.14 exactly correspond to the case $(n,k)=(3,2)$, (4, 2) and (4, 3), respectively.

8 Berger’s problem on conformally flat 3-tori and 4-tori

As recalled in the introduction, on n-tori ($n\ge 3$), there is no solution to the Berger’s problem, i.e., one can not expect a uniform upper bound for ${\mathcal {L}}(g)$ among all smooth Riemannian metrics. However, if restricted to the flat metric, or a certain class of some conformally flat metrics, we can solve the Berger’s problem for $n\le 4$. That is, we will prove Theorems 2 and  3 given in the introduction.

Note that finding the upper bound for $\lambda _1(g)V(g)^{\frac{2}{n}}$ among all flat n-tori, is equivalent to find the upper bound for volumes of all flat n-tori with $4\pi ^2$ as the first eigenvalue, which can be done by calculating the minimum of the determinant of those lattices (of rank n) with 1 as the shortest length. Since a shortest lattice vector can always be extended to be the first vector of a generator, in terms of the Gram matrix, we only need to calculate the minimum of $\det$ on the following set:

$$\begin{aligned} \Omega _1\triangleq \big \{Q\in \Sigma _{+}\, \big |\, Q_{11}=1, vQv^t\ge 1~{\textrm{for all}}~v\in \mathbb {Z}^n\backslash \{0\}\big \}, \end{aligned}$$

where $Q_{ij}$ is the entry of Q. We still use the notation $\Xi$ to denote the set of shortest lattice vectors up to $\pm 1$ in a given lattice $\Lambda _n^*$.

Lemma 8.1

For the lattice $\Lambda _n^*$ determined by $Q\in \Omega _1$, if $\textrm{rank}(\Xi )<n$, then Q is not a minimal point of $\det$ on $\Omega _1$.

Proof

We assume $\Xi$ is contained in the $(n-1)$-dimensional sublattice $\Lambda _{n-1}$ with a generator $\{\xi _1,\cdots ,\xi _{n-1}\}$ and $\alpha$ is a vector in $\Lambda _n^*\backslash \Lambda _{n-1}$ such that $\{\xi _1,\cdots ,\xi _{n-1},\alpha \}$ is a generator of $\Lambda _{n}^*$.

Set $\alpha =\alpha ^\top +\alpha ^\perp$, in which $\alpha ^\top$ is the orthogonal projection of $\alpha$ onto $\textrm{Span}\{\xi _1,\cdots ,\xi _{n-1}\}$. Let ${\Lambda }^*_{n,t}$ be a new lattice generated by $\{\xi _1,\cdots ,\xi _{n-1},\alpha -t\alpha ^\perp \}$ for $0\le t<1$, $\Pi _{k,t}$ be the affine hyperplane defined by $\sum _{i=1}^{n-1}c_i\xi _i+k(\alpha -t\alpha ^\perp )$ $(c_i\in \mathbb {R})$ for any $k\in \mathbb {Z}$.

Note that there exists an integer $K>0$ such that $\Pi _{k,0}\cap \overline{B(0,1)}=\emptyset$ if and only if $|k|>K$. By continuity, there is $0<t_0<1$ such that for any $0\le t\le t_0$, $\Pi _{k,t}\cap \overline{B(0,1)}=\emptyset$ if and only if $|k|>K$.

As for $0<|k|\le K$, it is easy to see that for any vector of $\Lambda _{n,t}^*\cap \Pi _{k,t}$, its orthogonal projection onto $\textrm{Span}\{\xi _1,\cdots ,\xi _{n-1}\}$ does not depend on t, while $\Pi _{k,t}\cap \overline{B(0,1)}$ is a family of co-centered $(n-1)$-dimensional closed balls expanding as t goes from 0 to $t_0$. Given the fact that all the vectors of $\Lambda _{n,0}^*\cap \Pi _{k,0}$ are located outside of $\Pi _{k,0}\cap \overline{B(0,1)}$, there exists $t_k>0$ such that all the vectors of $\Lambda _{n,t}^*\cap \Pi _{k,t}$ are still located outside of $\Pi _{k,t}\cap \overline{B(0,1)}$ for any $0\le t\le t_k$.

Set $\delta =\min \{t_0,\cdots ,t_K\}$. Then the lattice $\Lambda ^*_{n,\delta }$ admits the same shortest vectors as $\Lambda _n^*$ while the Gram matrix $Q_\delta$ satisfying $\det Q_\delta =(\det (\xi _1,\cdots ,\xi _{n-1},\alpha -\delta \alpha ^\perp ))^2=(1-\delta )^2\det Q$. $\square$

Due to Theorem 5.8 and (23), for a lattice $\Lambda _n^*$ with $\textrm{rank}(\Xi )=n\le 4$, we can always choose a generator of $\Lambda _n^*$ such that the Gram matrix takes 1 as its diagonal entries. Set

$$\begin{aligned} \Omega _2\triangleq \{Q\in \Sigma _{+}\, \big |\, Q_{ii}=1\text { for all }1\le i\le n, \text { and }vQv^t\ge 1 \text { for all } v\in \mathbb {Z}^n\backslash \{0\}\}. \end{aligned}$$

(33)

Then according to Lemma 8.1, we have the minimum of $\det$ on $\Omega _1$ can only be attained on $\Omega _2$.

Lemma 8.2

$\Omega _2$ is convex and compact.

Proof

The convexity of $\Omega _2$ is obvious. To prove the compactness, we only need to prove that $\Omega _2$ is both bounded and closed.

Suppose $Q\in \Omega _2$. It follows from $\langle Q,Q\rangle =\textrm{tr}Q^2<(\textrm{tr}Q)^2=n^2$ that $\Omega _2$ is bounded.

Next, we prove that $\Omega _2$ is closed. Suppose $\{Q_n\}$ is a convergent sequence in $\Omega _2$, whose limit is denoted by $Q_0$. We denote by $D_{nk}$ (resp. $D_{0k}$) the leading principal minor of order k for $Q_n$ (resp. $Q_0$). Note that for all $1\le k\le n$, the submatrix corresponding to $D_{nk}$ is also positive definite. It can determine a sublattice of rank k, which also takes 1 as the shortest length. It follows from the Theorem 13 in [30] (a corollary of the Minkowski’s first theorem) that

$$\begin{aligned} \sqrt{D_{nk}}\ge \frac{V(\mathbb {S}^k)}{2^k}, \end{aligned}$$

where $V(\mathbb {S}^k)$ is the volume of the standard round k-sphere. By continuity, we have $\sqrt{D_{0k}}\ge \frac{V(\mathbb {S}^k)}{2^k}>0$ for all $1\le k\le n$, which implies $Q_0\in \Sigma _{+}$. In the mean time, it is obvious that the diagonal entries of $Q_0$ are all 1 and $v Q_0 v^t\ge 1$ for all $v\in \mathbb {Z}^n\backslash \{0\}$. So we get $Q_0\in \Omega _2$, and thus $\Omega _2$ is closed. $\square$

Lemma 8.3

$\Omega _2$ is a convex polytope.

Proof

We denote by $\{0,\pm 1\}^n$ the set of integer vectors whose coordinates take values only in $\{0,\pm 1\}$. Define

$$\begin{aligned} \begin{aligned} \Omega _3\triangleq \bigg \{Q\in S(n)\, \bigg |\, Q_{ii}=1 \text { for all }1\le i\le n,~ \text { and }vQv^t\ge 1 \text { for all nonzero } v\in \{ 0,\pm 1\}^n\bigg \}. \end{aligned} \end{aligned}$$

(34)

For simplicity, we only prove the case of $n\le 4$. To prove the higher dimensional case, one only need to replace the integer set $\{0,\pm 1\}^n$ in $\Omega _3$ by the finite set of integer vectors appearing in Minkowski’s reduction theory (see [30]). It follows from Theorem 3.9 that $\Omega _2=\Omega _3\cap \Sigma _+$.

We claim that $\Omega _2=\Omega _3$. For this, let us consider the topology on $\Omega _3$ induced from the ambient space S(n). It is easy to show that $\Omega _3$ is convex, which implies $\Omega _3$ is connected. Seeing $\Omega _2$ as a subset in $\Omega _3$, we can firstly derive that it is closed by Lemma 8.2. On the other hand, note that $\Sigma _{+}$ is open in S(n), it follows from $\Omega _2=\Omega _3\cap \Sigma _+$ that $\Omega _2$ is also open in $\Omega _3$. Then the claim follows from the fact that $\Omega _2\ne \emptyset$.

It is not hard to see that the constraints $vQv^t\ge 1$ for all nonzero $v\in \{ 0,\pm 1\}^n$ define a polyhedron in S(n) by the intersection of finite half-spaces, and $\Omega _3$ is one of its facets. Combining this and Lemma 8.2, we derive that $\Omega _2=\Omega _3$ is a convex polytope. $\square$

According to the concavity of $\ln \circ \det$ on $\Sigma _{+}$ introduced in Lemma 3.4, one can easily obtain the next conclusion.

Lemma 8.4

The minimum of $\det$ on $\Omega _2$ is attained at some vertex of $\Omega _2$.

Lemma 8.5

Suppose Q is a vertex of $\Omega _2$, then $\sharp (\Xi )\ge \frac{n(n+1)}{2}$.

Proof

Since Q is a vertex and $\dim {S(n)}=\frac{n(n+1)}{2}$, it satisfies the constraints given in (34), among which there should be a system of linear equations with rank $\frac{n(n+1)}{2}$. As a result, there should be at least $\frac{n(n+1)}{2}$ integer vectors such that $vQv^t= 1$, from which the conclusion follows. $\square$

Proposition 8.6

Suppose $n\le 4$, then every vertex of $\Omega _2$ can determine a $\lambda _1$-minimal flat n-torus in some sphere. Up to congruence, these $\lambda _1$-minimal flat n-tori are

(1) the equilaterial 2-torus in $\mathbb {S}^5$ given in Example 4.2 for $n=2$;

(2) the $\lambda _1$-minimal flat 3-torus in $\mathbb {S}^{11}$ given in Example 4.5 for $n=3$;

(3) the $\lambda _1$-minimal flat 4-torus in $\mathbb {S}^{19}$ given in Example 4.16, or those $\lambda _1$-minimal flat 4-tori given in Example 1.1 for $n=4$.

Proof

Firstly, for these examples, using their matrix data, one can check directly that the Gram matrix Q belongs to $\Omega _2$, and the rank of $\{A^tA | A\in Y\}$ is exactly $\frac{n(n+1)}{2}$. Therefore they all correspond to the vertices of $\Omega _2$.

Suppose Q is a vertex of $\Omega _2$. If the lattice determined by Q is not prime, then we arrive at the only exceptional torus, and the conclusion follows from Proposition 6.3.

Fig. 3

$\Omega _2$ and the distribution of all $\lambda _1$-minimal flat 3-tori

Next, we assume the lattice determined by Q is prime. Let X be the set of integer vectors corresponding to $\Xi$. Combining Remark 6.11 and Lemma 8.5, we can derive that $\sharp (\Xi )=\frac{n(n+1)}{2}$. Furthermore, it follows from the discussion in Sect. 6 that X is exactly the ladder set $X_n$ up to a unimodular transformation in $SL(n,\mathbb {Z})$. Note that $W_{X_n}$ is a singleton set. So Q is congruent to $Q_n$ by a unimodular transformation in $SL(n,\mathbb {Z})$, which completes the proof of this proposition. $\square$

Remark 8.7

It follows from Theorem 3.2, Remark 3.3 that for every facets of $\Omega _2$, in general one can determine a unique flat torus which admitting the $\lambda _1$-minimal immersion in spheres. Conversely, when $n\le 4$, every $\lambda _1$-minimal flat torus can determined a Gram matrix Q in $\Omega _2$.

For instance, when $n=3$, $\Omega _2$ is constituted by matrices $\begin{pmatrix} 1&{} u&{}v\\ u&{}1&{}w\\ v&{}w&{}1 \end{pmatrix}$, with u, v, w all taking values in $[-\frac{1}{2},\frac{1}{2}]$, and satisfying

$$\begin{aligned} -1\le & {} u+v+w\le 1,~~-1\le u+v-w\le 1,~~-1\\\le & {} u-v+w\le 1~~-1\le u-v-w\le 1. \end{aligned}$$

It is straightforward to verify that there is a correspondence between the $\lambda _1$-minimal flat 3-tori and the barycenters of facets of $\Omega _2$. In Fig. 3, it shows the two reducible ones described in Example 4.2 respectively correspond to the gray point (center of the body) and the green point (center of the square), and those points painted by blue (center of the triangle), red (center of the edge) and pink (the vertex) corresponds to Examples 4.3 to  4.5, respectively.

Combining Lemma 8.4 with Proposition 8.6, we can directly calculate the minimum of $\det$ on $\Omega _2$ (hence on $\Omega _1$), from which Theorem 2 follows.

For a given flat metric $g_0$ on n-torus $T^n$, it was proved by El Soufi and Ilias in [11] that $g_0$ maximizes ${\mathcal {L}}(g)$ in $[g_0]$, if the first eigenspace of $g_0$ is of dimension no less than 2n. Combining this with Theorem 2, we can obtain Theorem 3.