Totally Geodesic Surfaces of Riemannian Symmetric Spaces

Abstract

A submanidfold S of a Riemannian manifold is called a totally geodesic submanifold if every geodesic of S is also a geodesic of M. Totally geodesic submanifolds of Riemannian symmetric spaces have long been studied by many mathematicians. We give a classification of non-flat totally geodesic surfaces of the Riemannian symmetric space of type AI, AIII and BDI.

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1 Introduction

Let G be a compact simple Lie group and θ be an involutive automorphism of G. We denote by $\mathfrak{g}$ the Lie algebra of G and denote also by θ the differential of θ. Let $\mathfrak{k}$ be the set of all θ-invariant elements of $\mathfrak{g}$ and K be a Lie subgroup of G of which Lie algebra coincides with $\mathfrak{k}$.

Let $\langle,\rangle$ be an Ad(G)-invariant inner product on $\mathfrak{g}$ and $\mathfrak{p}$ be the orthogonal complement of $\mathfrak{k}$. We extend the restriction of $\langle,\rangle$ on $\mathfrak{p}$ to the G-invariant Riemannian metric on G∕K and denote it also by $\langle,\rangle$.

A subspace $\mathfrak{s}$ of $\mathfrak{p}$ is called a Lie triple system if it satisfies $[[\mathfrak{s},\mathfrak{s}]\mathfrak{s}] \subset \mathfrak{s}$. There exits a one-to-one correspondence between the set of totally geodesic submanifold of M through the origin $o = eK$ and the set of Lie triple systems in $\mathfrak{p}$ [1].

Important constructions and classification results of totally geodesic submanifolds in Riemannian symmetric spaces are summarized in an expository article by S. Klein [2].

In [3] the author classified non-flat totally geodesic surfaces in irreducible Riemannian symmetric spaces where G is SU(n), Sp(n) or SO(n). The main tool used in [3] is the representation theory of SU(2). The aim of this article is to introduce the outline of the contents of [3].

2 Irreducible Representation of SU(2)

In this section, we review real and complex irreducible representations of SU(2).

Let H, X, Y be a basis of the complexification of the Lie algebra $\mathfrak{s}\mathfrak{u}(2)$ of SU(2) satisfying

$$\displaystyle{ [H,X] = 2X,\quad [H,Y ] = -2Y,\quad [X,Y ] = H. }$$

(1)

2.1 Complex Irreducible Representations

If we denote by V _d the set of polynomial functions on $\mathbb{C}^{2}$ and by ρ _d the contragradient action of SU(2) on V _d , then $(V _{d},\rho _{d})$ is a complex irreducible representation of SU(2). On the other hand, every finite dimensional complex irreducible representation of SU(2) is equivalent to $(V _{d},\rho _{d})$ for some positive integer d.

The next proposition plays an important role in our classification.

Proposition 1.

Let (V,ρ) be a (d + 1)-dimensional complex irreducible representation of SU(2) and $\left < \,,\,\right >$ be an SU(2)-invariant Hermitian inner product on V. If we put λ the largest eigenvalue of ρ(H) and v ₀ ∈ V be a corresponding eigen vector, then we have λ = d and $\rho (Y )^{i}(v_{0})$ is an eigen vector of ρ(H) corresponding to the eigenvalue (λ − 2i).

Let $\varepsilon _{i}$  (0 ≤ i ≤ d) be arbitrary complex numbers with $\vert \varepsilon _{i}\vert = 1$ , and put $v_{i} = \frac{\varepsilon _{i}} {\vert \rho (Y )^{i}v_{0}\vert }\ \rho (Y )^{i}v_{ 0}$  (0 ≤ i ≤ d). Then v ₀ , v ₁ , ⋯ , v _d is an orthonormal basis of V _d and the matrix representations of ρ(H), ρ(X), ρ(Y ) with respect to $v_{0},\cdots \,,v_{d}$ are as follows

$$\displaystyle\begin{array}{rcl} & & \rho (H) = \left [\begin{array}{cccc} d& 0 &\cdots & 0\\ 0 &d - 2 &\cdots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 &\cdots & - d \end{array} \right ],\quad \rho (X) = \left [\begin{array}{ccccc} 0 & 0 &\cdots & 0 &0\\ c_{ 1} & 0 &\cdots & 0 &0\\ 0 &c_{ 2} & \cdots & 0 &0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 &\cdots &c_{ d}&0 \end{array} \right ], {}\\ & & \rho (Y ) = \left [\begin{array}{cccccc} 0&c_{1}' & 0 &\cdots & 0 \\ 0& 0 &c_{2}' & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0& 0 & 0 &\cdots &c_{d}' \\ 0& 0 & 0 &\cdots & 0 \end{array} \right ],\qquad \begin{array}{ll} \mathrm{where}\quad c_{i}' = \overline{c_{i}} \\ \qquad \vert c_{i}\vert = \sqrt{i(d - i + 1)}. \end{array} {}\\ \end{array}$$

2.2 Real Irreducible Representations

Let (V, ρ) be a complex representation of SU(2) and v ₁, ⋯ , v _N be a basis of V. We denote by $\overline{V }$ the complex vector space, which is V itself as an additive group and the scalar multiplication is defined by $c {\ast} x = \overline{c}\,x$ $(c \in \mathbb{C},\ x \in V )$. Define the action $\overline{\rho }$ of SU(2) on $\overline{V }$ so that

$$\displaystyle{\overline{\rho }\left (\sum z_{i} {\ast} v_{i}\right ) =\sum z_{i} {\ast}\rho (v_{i}).}$$

The representation $(\overline{V },\overline{\rho })$ is called the conjugate representation of (V, ρ). 

A complex irreducible representation (V, ρ) of G is said to be a self-conjugate representation if there exists a conjugate-linear automorphism $\hat{j}: V \rightarrow V$ which commute with ρ(g) for any g ∈ SU(2). A conjugate-linear automorphism commuting with ρ is called a structure map of (V, ρ).

Let (V, ρ) be a self-conjugate representation and $\hat{j}$ be a structure map. By Schur’s lemma, $\hat{j}^{2} = c$ for some constant. It is known that the constant c is a real number and (V, ρ) is said to be of index 1 (resp. − 1) if c > 0 (resp. c < 0).

Each complex irreducible representation $(V _{d},\rho _{d})$ of SU(2) is a self-conjugate representation and its index is equal to (−1)^d. If d is an even integer, the subspace of V _d invariant under the structure map $\hat{j}$ is a real irreducible representation of SU(2). If d is an odd integer, V _d (viewed as a real representation by restricting the coefficient field from $\mathbb{C}$ to $\mathbb{R}$) is also a Real irreducible representation and V _d admits a structure of vector space over the field of quaternions.

3 Classification

The standard orthonormal basis of $\mathbb{R}^{N}$ or $\mathbb{C}^{N}$ will be denote by e ₁,⋯ ,e _N . We denote by G _ij  (i ≠ j) the skew-symmetric endomorphism satisfying

$$\displaystyle{G_{ij}(e_{j}) = e_{i},\quad G_{ij}(e_{i}) = -e_{j},\quad G_{ij}(e_{k}) = 0\quad (k\neq i,j),}$$

and by S _ij the symmetric endomorphism

$$\displaystyle{S_{ij}(e_{j}) = e_{i},\quad S_{ij}(e_{i}) = e_{j},\quad S_{ij}(e_{k}) = 0\quad (k\neq i,j).}$$

3.1 AI:  SU(n)∕SO(n)

We denote by τ the conjugation on $\mathbb{C}^{N}$ with respect to $\mathbb{R}^{N}$ and denote by θ the involutive automorphism on SU(N) defined by θ(g) = τ ∘ g ∘τ (g ∈ SU(n)).

Theorem 1.

Let M be a non-flat totally geodesic surface of SU(n)∕SO(n) and U be the set of all elements in SU(n) leaving M invariant.

• (i) There exists an orthogonal direct sum decomposition of $\mathbb{C}^{n}$ by τ-invariant and U-invariant subspaces.

• (ii) Let X ₂ , X ₃ be a basis of the Lie triple system corresponding to M with

  $$\displaystyle{[[X_{2},X_{3}],X_{2}] = 4\,X_{3},\quad [[X_{2},X_{3}],X_{3}] = -4\,X_{2}.}$$

  Assume that $\mathbb{C}^{n}$ is U-invariant. There exists an element $g = [u_{1},\cdots \,,u_{n}] \in SO(n)$ such that

  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{2}& =& \sqrt{-1}\,\sum _{i=1}^{n}\,(n - 2\,i)\,E_{ i,i} {}\end{array}$$

  (2)
  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{3}& =& -\sqrt{-1}\,\left [\sum _{i=1}^{n-2}\,\sqrt{i(n - i)}S_{ i,i+1} +\,\varepsilon \, \sqrt{n - 1}S_{n-1,n}\right ] {}\end{array}$$

  (3)

  where

  $$\displaystyle{\varepsilon = \left \{\begin{array}{@{}l@{\quad }l@{}} 1 \quad &\mathrm{if\ }n \equiv 1\ (\bmod 2),\\ \pm 1\quad &\mathrm{if\ }n \equiv 0\ (\bmod 2). \\ \quad \end{array} \right.}$$

Proof.

We omit the proof of (i) and assume that the action of U on $\mathbb{C}^{n}$ is irreducible.

Note that $\mathfrak{k} =\{ X:\theta (X) = X\} = \mathrm{Skew}(n; \mathbb{R})$ and $\mathfrak{p} =\{X:\theta (X) = -X\} = \sqrt{-1}\,\mathrm{Sym}(n; \mathbb{R})$.

If we put $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$ the set of eigenvalues of $H = -\sqrt{-1}\,X_{2} \in \mathrm{Sym}(n; \mathbb{R})$, then by the action of Ad(SO(n)) we may assume that $H = \mathrm{Diag}(a_{1},a_{2},\ \cdots \,,a_{n})$.

If we put

$$\displaystyle{H = [X_{2},X_{3}],\quad X = \frac{1} {2}(\sqrt{-1}\,X_{3} + X_{1}),\quad Y = \frac{1} {2}(\sqrt{-1}\,X_{3} - X_{1}),\quad }$$

we have

$$\displaystyle{[H,X] = 2X,\quad [H,Y ] = -2Y,\quad [X,Y ] = H.}$$

Since a _i are weights of the complex irreducible representation of U we have

$$\displaystyle{a_{1} - a_{2} = a_{2} - a_{3} = \cdots = a_{n-1} - a_{n} = 2.}$$

Put $n = d + 1$ and v ₀ = e ₁. Since each eigenspace (the weight space) of H is one-dimensional there exists $\varepsilon _{i}$ (1 ≤ i ≤ d) such that $e_{i} = \frac{\varepsilon _{i}} {\vert H^{i}v_{0}\vert }H^{i}v_{ 0}$. Thus the matrix H, X and Y are of the form given in the Proposition 1. We can choose unit complex numbers $\varepsilon '_{i}$ (0 ≤ i ≤ d) such that by a change of basis $\{e_{i}\} \rightarrow \{\varepsilon '_{i}e_{i}\}$ all the components of X, Y in the Proposition 1 are changed to real numbers. We omit further detail. □ 

3.2 AIII: SU(p + q)∕S(U(p) × U(q))

We denote by I _n the unit matrix of order n and put $I_{p,q} = \left [\begin{array}{cc} \mbox{ $I_{p}$}& \mbox{ $O$} \\ \mbox{ $O$} &\mbox{ $ - I_{q}$}\\ \end{array} \right ].$

Theorem 2.

Let M be a non-flat totally geodesic surface of $SU(p + q)/S(U(p) \times U(q))$ and U be the set of all elements of SU(p + q) which leave M invariant.

• (i) There exists an orthogonal direct sum decomposition of $\mathbb{C}^{p+q}$ by I _p,q -invariant, U-irreducible subspaces.

• (ii) If V is an I _p,q -invariant, U-irreducible subspace of $\mathbb{C}^{p+q}$ , then we have

  $$\displaystyle{\left \vert \dim \{v \in V: I_{p,q}(v) = v\} -\dim \{ v \in V: I_{p,q}(v) = -v\}\right \vert \leq 1.}$$

• (iii) Assume that the action of SU(2) on $\mathbb{C}^{p+q}$ is irreducible. Let X ₂ , X ₃ be a basis of the Lie triple system corresponding to M with

  $$\displaystyle{[[X_{2},X_{3}],X_{2}] = 4\,X_{3},\quad [[X_{2},X_{3}],X_{3}] = -4\,X_{2}.}$$

  There exists an element $g = [u_{1},\cdots \,,u_{p+q}] \in S(U(p) \times U(q))$ such that

  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{2}& =& \sum _{i=1}^{q}\sqrt{(2i - 1)(p + q + 1 - 2i)}\,G_{ i,p+i} \\ & & \qquad +\sum _{ i=1}^{p-1}\sqrt{2i(p + q - 2i)}\,G_{ p+i,i+1} {}\end{array}$$

  (4)
  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{3}& =& \sqrt{-1}\,\left [\sum _{i=1}^{q}\sqrt{(2i - 1)(p + q + 1 - 2i)}\,S_{ p+i,i}\right. \\ & & \qquad \left.+\sum _{i=1}^{p-1}\sqrt{2i(p + q - 2i)}\,S_{ i+1,p+i}\right ] {}\end{array}$$

  (5)

Proof.

We omit the proof of (i).

Assume that the action of U on $\mathbb{C}^{p+q}$ is irreducible.

Take a basis X ₁, X ₂, X ₃ of the Lie algebra $\mathfrak{u}$ of U which satisfy

$$\displaystyle\begin{array}{rcl} & & I_{p,q} \circ X_{1} = X_{1} \circ I_{p,q},\quad I_{p,q} \circ X_{i} = -X_{i} \circ I_{p,q}\quad (i = 2,3), {}\\ & & [X_{1},X_{2}] = 2X_{3},\quad [X_{2},X_{3}] = 2X_{1},\quad [X_{3},X_{1}] = 2X_{2}, {}\\ \end{array}$$

and put

$$\displaystyle{H = -\sqrt{-1}X_{1},\ X = \frac{1} {2}(X_{2} -\sqrt{-1}X_{3}),\ Y = -\frac{1} {2}(X_{2} + \sqrt{-1}X_{3}) = ^{t}\overline{X}.}$$

Since H is a Hermitian matrix, there exists an element g ∈ S(U(p) × U(q)) such that

$$\displaystyle{\mathrm{Ad}(g)H = \mathrm{diag}(a_{1},\cdots \,,a_{p};b_{1},\cdots \,,b_{q})}$$

where a ₁ > ⋯ > a _p and $b_{1} > \cdots > b_{q}$ holds. We denote by $\xi _{i}$ the i-th column vector of g. The set of eigenvalues of H coincides with the set of weights of the (p + q)-dimensional complex irreducible representation of SU(p + q)), namely we have

$$\displaystyle{\{a_{1},\cdots \,,a_{p},b_{1},\cdots \,,b_{q}\} =\{ p + q - 1,\ p + q - 2,\ \cdots \,,\ 1 - p - q\}.}$$

We assume that $a_{1} > b_{1}$ holds.

• We have $a_{1} = p + q - 1$ and $I_{p,q}\xi _{1} =\xi _{1}$, $H \cdot \xi _{1} = (p + q - 1)\,\xi _{1}$ hold.

• From $I_{p,q} \circ Y = -Y \circ I_{p,q}$, we have $I_{p,q}(Y \cdot \xi _{1}) = -Y \cdot \xi _{1}$ and from $[H,Y ] = -2Y$ we have $H(Y \cdot \xi _{1}) = (p + q - 3)\,Y \cdot \xi _{1}$. Thus we have $b_{1} = p + q - 3$ and there exists a complex number γ _i with

  $$\displaystyle{Y \cdot \xi _{1} =\gamma _{1}\,\xi _{p+1},\quad \vert \gamma _{1}\vert = \sqrt{p + q - 1}.}$$

• Similarly we have

  $$\displaystyle{Y \cdot \xi _{p+1} =\gamma _{2}\,\xi _{2},\quad \vert \gamma _{2}\vert = \sqrt{2(p + q - 2)}}$$

  etc.

Finally we have $p - q = 0,1$ and the matrix representation of Y with respect to the basis ξ ₁, ⋯ , ξ _p , ξ _p+1, ⋯ , ξ _p+q is

$$\displaystyle{\mathrm{Ad}(g)Y =\sum _{ i=1}^{q}\gamma _{ 2i-1}\,E_{p+i,i} +\sum _{ i=1}^{p-1}\gamma _{ 2i}\,E_{i+1,p+i}.}$$

Let $\varepsilon _{i}$ (1 ≤ i ≤ p + q) be unit complex numbers and put $g = (\varepsilon _{1}\,\xi _{1},\cdots \,,\varepsilon _{p+q}\,\xi _{p+q})$. We can choose $\varepsilon _{i}$ so that the all of the coefficients γ _2i and γ _2i−1 in the representation of Ad(g)Y above are positive real numbers. From

$$\displaystyle{X_{2} = ^{t}\overline{Y } - Y,\quad X_{ 3} = \sqrt{-1}\,\left (^{t}\overline{Y } + Y \right )}$$

we obtain (4) and (5). □ 

3.3 BDI:  SO(p + q)∕S(O(p) × O(q))

Let θ be the involutive automorphism on $G = SO(p + q)$ defined by

$$\displaystyle{\theta (g) = I_{p,q} \circ g \circ I_{p,q}}$$

and put

$$\displaystyle{K = \left \{g \in SO(p + q):\theta (g) = g\right \} = S(O(p) \times O(q)).}$$

We can classify totally geodesic surfaces of $SO(p + q)/S(O(p) \times O(q))$ by similar argument to that on $SU(p + q)/S(U(p) \times U(q))$. But, since there are two types of real irreducible representations of SU(2), the classification result is divided into two cases; (iii) and (iv) in the following theorem. Since it is troublesome to give the representation matrix of the action of $\mathfrak{s}\mathfrak{u}(2)$ on the odd-dimensional real irreducible representation ((iii) in the following theorem), we give only the result without proof.

Theorem 3.

Let M be a non-flat totally geodesic surface of $SO(p + q)/S(O(p) \times O(q))$ and U be the set of all elements in SO(p + q) leaving M invariant.

• (i) There exists an orthogonal direct sum decomposition of $\mathbb{R}^{p+q}$ by I _p,q -invariant and U-irreducible subspaces.

• (ii) For each I _p,q -invariant, U-irreducible subspace V of $\mathbb{R}^{p+q}$ , we have

  $$\displaystyle{\left \vert \dim \{v \in V: I_{p,q}(v) = v\} -\dim \{ v \in V: I_{p,q}(v) = -v\}\right \vert \leq 1.}$$

• (iii) Assume that the action of U on $\mathbb{R}^{p+q}$ is irreducible and $p = q + 1 \geq 3$.

  We denote by p′ the integer part of p∕2 and by q′ the integer part of q∕2. There exists an element g ∈ S(O(p) × O(q)) such that

  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{2}& =& -\sum _{i=1}^{q'}\sqrt{(2i - 1)(p + q + 1 - 2i)}\left (G_{ p+2i-1,2i-1} + G_{p+2i,2i}\right ) \\ & & +\sum _{i=1}^{p'-1}\sqrt{2i(p + q - 2i)}\left (G_{ p+2i-1,2i+1} + G_{p+2i,2i+2}\right ) \\ & & +\left \{\begin{array}{@{}l@{\quad }l@{}} (-\sqrt{2})\sqrt{p\,q}G_{p+q,q}\quad &\quad (\mathrm{if}\ p = 0\mod 2) \\ \sqrt{2}\sqrt{p\,q}G_{p+q-1,p} \quad &\quad (\mathrm{if}\ p = 1\mod 2) \end{array} \right. {}\end{array}$$

  (6)
  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{3}& =& \sum _{i=1}^{q'}\sqrt{(2i - 1)(p + q + 1 - 2i)}\left (G_{ p+2i,2i-1} - G_{p+2i-1,2i}\right ) \\ & & +\sum _{i=1}^{p'-1}\sqrt{2i(p + q - 2i)}\left (G_{ p+2i,2i+1} - G_{p+2i-1,2i+2}\right ) \\ & & +\left \{\begin{array}{@{}l@{\quad }l@{}} (-\sqrt{2})\sqrt{p\,q}G_{p+q,p}\quad &\quad (\mathrm{if}\ p = 0\mod 2) \\ \sqrt{2}\sqrt{p\,q}G_{p+q,p} \quad &\quad (\mathrm{if}\ p = 1\mod 2) \end{array} \right. {}\end{array}$$

  (7)

• (iv) Assume that the action of U on $\mathbb{R}^{p+q}$ is irreducible and p = q. Then p is an even integer, say p = 2p′, and there exists an element g ∈ S(O(p) × O(q)) such that

  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{2}& =& \sum _{i=1}^{p'-1}\sqrt{2i(p - 2i)}\left (G_{ p+i,i+1} + G_{p+p'+i,p'+i+1}\right ) \\ & & -\sum _{i=1}^{p'}\sqrt{(2i - 1)(p + 1 - 2i)}\left (G_{ p+i,i} + G_{p+p'+i,p'+i}\right ) {}\end{array}$$

  (8)
  $$\displaystyle\begin{array}{rcl} \mathrm{Ad}(g)X_{3}& =& \sum _{i=1}^{p'-1}\sqrt{2i(p - 2i)}\left (G_{ p+p'+i,i+1} - G_{p+i,p'+i+1}\right ) \\ & & +\sum _{i=1}^{p'}\sqrt{(2i - 1)(p + 1 - 2i)}\left (G_{ p+p'+i,i} - G_{p+i,p'+i}\right ) {}\end{array}$$

  (9)