Tensors and Universal Properties

Abstract

We will review the basic properties of the tensor product and use them to illustrate the basic notion of a universal property, which we will see repeatedly.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

We will review the basic properties of the tensor product and use them to illustrate the basic notion of a universal property, which we will see repeatedly.

If R is a commutative ring and M, N, and P are R-modules, then a bilinear map f : M × N → P is a map satisfying

$$\displaystyle{f(r_{1}m_{1} + r_{2}m_{2},n) = r_{1}f(m_{1},n) + r_{2}f(m_{2},n),\quad r_{i} \in R,m_{i} \in M,n \in N,}$$

$$\displaystyle{f(m,r_{1}n_{1} + r_{2}n_{2}) = r_{1}f(m,n_{1}) + r_{2}f(m,n_{2}),\quad r_{i} \in R,n_{i} \in N,m \in M.}$$

More generally, if M ₁,…, M _k are R-modules, the notion of a k-linear map M ₁ × ⋯ × M _k → P is defined similarly: the map must be linear in each variable.

The tensor product $M \otimes _{R}N$ is an R-module together with a bilinear map $\otimes: M \times N\longrightarrow M \otimes _{R}N$ satisfying the following property.

Universal Property of the Tensor Product. If P is any R-module and p: M × N → P is a bilinear map, there exists a unique R-module homomorphism $F: M \otimes N\longrightarrow P$ such that $p = F \circ \otimes$.

Why do we call this a universal property? It says that $\otimes: M \times N\longrightarrow M \otimes N$ is a “universal” bilinear map in the sense that any bilinear map of M × N factors through it. As we will explain, the module $M \otimes _{R}N$ is uniquely determined by the universal property. This is important beyond the immediate example because often objects are described by universal properties. Before we explain this point (which is obvious if one thinks about it correctly), let us make a categorical observation.

If $\mathcal{C}$ is a category, an initial object in $\mathcal{C}$ is an object X ₀ such that, for each object Y, the Hom set $\mathrm{Hom}_{\mathcal{C}}(X_{0},Y )$ consists of a single element. A terminal object is an object X _∞ such that, for each object Y, the Hom set $\mathrm{Hom}_{\mathcal{C}}(Y,X_{\infty })$ consists of a single element. For example, in the category of sets, the empty set is an initial object and a set consisting of one element is a terminal object.

FormalPara Lemma 9.1.

In any category, any two initial objects are isomorphic. Any two terminal objects are isomorphic.

FormalPara Proof.

If X ₀ and X ₁ are initial objects, there exist unique morphisms f : X ₀ → X ₁ (since X ₀ is initial) and g : X ₁ → X ₀ (since X ₁ is initial). Then g ∘ f : X ₀ → X ₀ and $1_{X_{0}}: X_{0}\longrightarrow X_{0}$ must coincide since X ₀ is initial, and similarly $f \circ g = 1_{X_{1}}$. Thus f and g are inverse isomorphisms. Similarly, terminal objects are isomorphic. □

FormalPara Theorem 9.1.

The tensor product $M \otimes _{R}N$ , if it exists, is determined up to isomorphism by the universal property.

FormalPara Proof.

Let $\mathcal{C}$ be the following category. An object in $\mathcal{C}$ is an ordered pair (P, p), where P is an R-module and p : M × N → P is a bilinear map. If X = (P, p) and Y = (Q, q) are objects, then a morphism X → Y consists of an R-module homomorphism f : P → Q such that q = f ∘ p. The universal property of the tensor product means that $\otimes: M \times N\longrightarrow M \otimes N$ is an initial object in this category and therefore determined up to isomorphism. □

Of course, we usually denote $\otimes (m,n)$ as $m \otimes n$ in $M \otimes _{R}N$. We have not proved that $M \otimes _{R}N$ exists. We refer to any text on algebra for this fact, such as Lang [116], Chap. XVI.

In general, by a universal property we mean any characterization of a mathematical object that can be expressed by saying that some associated object is an initial or terminal object in some category. The basic paradigm is that a universal property characterizes an object up to isomorphism.

A typical application of the universal property of the tensor product is to make $M \otimes _{R}N$ into a functor. Specifically, if μ : M → M′ and ν : N → N′ are R-module homomorphisms, then there is a unique R-module homomorphism $\mu \otimes \nu: M \otimes _{R}N\longrightarrow M^{\prime} \otimes _{R}N^{\prime}$ such that $(\mu \otimes \nu )(m \otimes n) = \mu (m) \otimes \nu (n)$. We get this by applying the universal property to the R-bilinear map $M \times N\longrightarrow M^{\prime} \otimes N^{\prime}$ defined by $(m,n)\longmapsto \mu (m) \otimes \nu (n)$.

As another example of an object that can be defined by a universal property, let V be a vector space over a field F. Let us ask for an F-algebra $\bigotimes V$ together with an F-linear map $i: V \longrightarrow \bigotimes V$ satisfying the following condition.

Universal Property of the Tensor Algebra. If A is any F-algebra and ϕ : V →A is an F-linear map then there exists a unique F-algebra homomorphism $\varPhi:\bigotimes V \longrightarrow A$ such that r = ρ ∘ i.

It should be clear from the previous discussion that this universal property characterizes the tensor algebra up to isomorphism. To prove existence, we can construct a ring with this exact property as follows. Let unadorned $\otimes$ mean $\otimes _{F}$ in what follows. By ${\otimes }^{k}V$ we mean the k-fold tensor product $V \otimes \cdots \otimes V$ (k times); if k = 0, then it is natural to take ${\otimes }^{0}V = F$ while ${\otimes }^{1}V = V$. If V has finite dimension d, then ${\otimes }^{k}V$ has dimension d ^k. Let

$$\displaystyle{\bigotimes V =\bigoplus _{ k=0}^{\infty }\left ({\otimes }^{k}V \right ).}$$

Then $\bigotimes V$ has the natural structure of a graded F-algebra in which the multiplication ${\otimes }^{k}V \times {\otimes }^{l}V \longrightarrow {\otimes }^{k+l}V$ sends

$$\displaystyle{(v_{1} \otimes \cdots \otimes v_{k},u_{1} \otimes \cdots \otimes u_{l})\longrightarrow v_{1} \otimes \cdots \otimes v_{k} \otimes u_{1} \otimes \cdots \otimes u_{l}.}$$

We regard V as a subset of $\bigotimes V$ embedded onto ${\otimes }^{1}V = V$.

FormalPara Proposition 9.1.

The universal property of the tensor algebra is satisfied.

FormalPara Proof.

If ϕ : V → A is any linear map of V into an F-algebra, define a map $\varPhi:\bigotimes V \longrightarrow A$ by $\varPhi (v_{1} \otimes \cdots \otimes v_{k}) = \phi (v_{1})\cdots \phi (v_{k})$ on ${\otimes }^{k}V$. It is easy to see that Φ is a ring homomorphism. It is unique since V generates $\bigotimes V$ as an F-algebra. □

A graded algebra over the field F is an F-algebra A with a direct sum decomposition

$$\displaystyle{A =\bigoplus _{ k=0}^{\infty }A_{ k}}$$

such that $A_{k}A_{l} \subseteq A_{k+l}$. In most examples we will have A ₀ = F. Elements of A _k are called homogeneous of degree k. The tensor algebra is a graded algebra, with ${\otimes }^{k}V$ being the homogeneous part of degree k.

Next we define the symmetric and exterior powers of a vector space V over the field F. Let V ^k denote V × ⋯ × V (k times). A k-linear map f : V ^k → U into another vector space is called symmetric if for any σ ∈ S _k it satisfies $f(v_{\sigma (1)},\ldots,v_{\sigma (k)}) = f(v_{1},\ldots,v_{k})$ and alternating if $f\big(v_{\sigma (1)},\ldots,v_{\sigma (k)}) =\varepsilon (\sigma )\,f(v_{1},\ldots,v_{k})$, where $\varepsilon: S_{k}\longrightarrow \{ \pm 1\}$ is the alternating (sign) character. The kth symmetric and exterior powers of V, denoted ${\vee }^{k}V$ and ${\wedge }^{k}V$, are F-vector spaces, together with k-linear maps $\vee: {V }^{k}\longrightarrow {\vee }^{k}V$ and $\wedge: {V }^{k}\longrightarrow {\wedge }^{k}V$. The map $\vee$ is symmetric, and the map $\wedge$ is alternating. We normally denote $\vee (v_{1},\ldots,v_{k}) = v_{1} \vee \cdots \vee v_{k}$ and similarly for $\wedge$. The following universal properties are required.

Universal Properties of the Symmetric and Exterior Powers: Let f : V ^k →U be any symmetric (resp. alternating) k-linear map. Then there exists a unique F-linear map ϕ : ∨ ^k V →U (resp. ${\wedge }^{k}V \longrightarrow U$ ) such that $f = \phi \circ \vee$ (resp. $f = \phi \circ \wedge$ ).

As usual, the symmetric and exterior algebras are characterized up to isomorphism by the universal property. We may construct ${\vee }^{k}V$ as a quotient of ${\otimes }^{k}V$, dividing by the subspace W generated by elements of the form $v_{1} \otimes \cdots \otimes v_{k} - v_{\sigma (1)} \otimes \cdots \otimes v_{\sigma (k)}$, with a similar construction for ${\wedge }^{k}$. The universal property of ${\vee }^{k}V$ then follows from the universal property of the tensor product. Indeed, if f : V ^k → U is any symmetric k-linear map, then there is induced a linear map $\psi: {\otimes }^{k}V \longrightarrow U$ such that $f = \psi \circ \otimes$. Since f is symmetric, $\psi$ vanishes on W, so $\psi$ induces a map ${\vee }^{k}V = {\otimes }^{k}V/W\longrightarrow U$ and the universal property follows.

If V has dimension d, then ${\vee }^{k}V$ has dimension $\binom{d + k - 1}{k}$, for if x ₁,…, x _d is a basis of V, then $\left \{x_{i_{1}} \vee \cdots \vee x_{i_{k}}\,\vert \,1 \leqslant i_{1} \leqslant i_{2} \leqslant \cdots \leqslant i_{k} \leqslant d\right \}$ is a basis for ${\vee }^{k}V$. On the other hand, the exterior power vanishes unless $k \leqslant d$, in which case it has dimension $\binom{d}{k}$. A basis consists of $\{x_{i_{1}} \wedge \cdots \wedge x_{i_{k}}\,\vert \,1 \leqslant i_{1} < i_{2} < \cdots < i_{k} \leqslant d\}$. The vector spaces ${\vee }^{k}V$ may be collected together to make a commutative graded algebra:

$$\displaystyle{\bigvee V =\bigoplus _{ k=0}^{\infty }{\vee }^{k}V.}$$

This is the symmetric algebra. The exterior algebra $\bigwedge V =\bigoplus _{k} {\wedge }^{k}V$ is constructed similarly. The spaces ${\vee }^{0}V$ and ${\wedge }^{0}V$ are one-dimensional and it is natural to take ${\vee }^{0}V = {\wedge }^{0}V = F$.

Exercises

FormalPara Exercise 9.1.

Let V be a finite-dimensional vector space over a field F that may be assumed to be infinite. Let $\mathcal{P}(V )$ be the ring of polynomial functions on V. Note that an element of the dual space V ^∗ is a function on V, so regarding this function as a polynomial gives an injection ${V }^{{\ast}}\longrightarrow \mathcal{P}(V )$. Show that this linear map extends to a ring isomorphism $\bigvee {V }^{{\ast}}\longrightarrow \mathcal{P}(V )$.

FormalPara Exercise 9.2.

Prove that if V is a vector space, then $V \otimes V \mathop{\cong}(V \wedge V ) \oplus (V \vee V )$.

FormalPara Exercise 9.3.

Use the universal properties of the symmetric and exterior power to show that if V and W are vector spaces, then there are maps ${\vee }^{k}f: {\vee }^{k}V \longrightarrow {\vee }^{k}W$ and ${\wedge }^{k}f: {\wedge }^{k}V \longrightarrow {\wedge }^{k}W$ such that

$$\displaystyle{{\vee }^{k}f(v_{ 1} \vee \cdots \vee v_{k}) = f(v_{1}) \vee \cdots \vee f(v_{k}),{\wedge }^{k}f(v_{ 1} \wedge \cdots \wedge v_{k}) = f(v_{1}) \wedge \cdots \wedge f(v_{k}).}$$

FormalPara Exercise 9.4.

Suppose that V = F ⁴. Let $f: V \longrightarrow V$ be the linear transformation with eigenvalues a, b, c, d. Compute the traces of the linear transformations ${\vee }^{2}f$ and ${\wedge }^{2}f$ on ${\vee }^{2}V$ and ${\wedge }^{2}V$ as polynomials in a, b, c, d.

FormalPara Exercise 9.5.

Let A and B be algebras over the field F. Then $A \otimes B$ is also an algebra, with multiplication $(a \otimes b)(a^{\prime} \otimes b^{\prime}) = aa^{\prime} \otimes bb^{\prime}$. Show that there are ring homomorphisms $i: A \rightarrow A \otimes B$ and $j: B \rightarrow A \otimes B$ such that if f : A → C and g : B → C are ring homomorphisms into a ring C satisfying f(a) g(b) = g(b) f(a) for a ∈ A and b ∈ B, then there exists a unique ring homomorphism $\phi: A \otimes B \rightarrow C$ such that ϕ ∘ i = f and ϕ ∘ j = g.

FormalPara Exercise 9.6.

Show that if U and V are finite-dimensional vector spaces over F then show that

$$\displaystyle{\bigvee (U \oplus V )\mathop{\cong}\left (\bigvee U\right ) \otimes \left (\bigvee U\right )}$$

and

$$\displaystyle{\bigwedge (U \oplus V )\mathop{\cong}\left (\bigwedge U\right ) \otimes \left (\bigwedge U\right ).}$$