Tensor products

If $V$ and $W$ are finite dimensional vector spaces then the Cartesian product $V \times W$ is naturally a vector space called the direct sum of $V$ and $W$ and denoted $V\oplus W$. The tensor product is a more complicated object. To define it we start by defining for any set $X$ the free vector space over $X$, $F(X)$. This is the set of all maps from $X$ to $\mathbb{R}$ which are zero except at a finite number of points. We define the vector space structure by adding and scalar multiplying maps. Each $x$ gives rise to a function $\Delta(x)$ which is one at $x$ and zero elsewhere. We therefore have a map $\Delta\colon X \to F(X)$. By construction the span of the image of $\Delta$ is all of $F(X)$.

The special property of the free vector space over $X$ is the following.

Proposition D.1   Let $f \colon X \to U$ be any map from $X$ into a vector space $U$ then there is a unique linear map $\hat f F(X) \to U$ such that $\hat f \circ \Delta = f$.

Proof. The general element of $F(X)$ is

$$\sum_{i=1}^n a_i \Delta(x_i)$$

for $a_i \in \mathbb{R}$. We define

$${\hat f}(\sum_{i=1}^n a_i \Delta(x_i)) = \sum_{i=1}^n a_i f(x^i).$$

$\qedsymbol$

Given two vector spaces $V$ and $W$ we can define $F(V \times W)$. This is an infinite dimensional vector space. We shall denote $\Delta((v, w))$ by $\Delta(v, w)$. Consider the subspace $Z$ defined as the span of all elements of the form

$$\Delta(\lambda v +\mu v',w) - \lambda \Delta(v,w)- \mu \Delta(v',w)$$

and

$$\Delta(v, \lambda w +\mu w') - \lambda \Delta(v,w) - \mu \Delta(v,w')$$

for any real numbers $\lambda$ and $\mu$ and vectors $v,v' \in V$ and $w, w' \in W$. Let us denote

$$V\otimes W= F(V\times W)/Z$$

and define a map

$$\otimes \colon V\times W \to V \otimes W$$

by

$$v\otimes w = \Delta(v,w) + Z.$$

We have

Proposition D.2   The map $\otimes \colon V\times W \to V \otimes W$ is bilinear.

Proof. We check the first factor only

$$\begin{split}(\lambda v + \mu v') \otimes w = & \Delta(\lambd... ...,w)+ Z)\\ =& \lambda v \otimes w + \mu v'\otimes w \end{split}$$

$\qedsymbol$

From Proposition D.1 we know that any map $f \colon V \times W \to U$, where $U$ is a vector space extends to a map $\hat f \colon F(V \times W) \to U$. Standard linear algebra tells us that we can take the quotient to get a map $\tilde f\colon V \otimes W \to U$ if $\hat f(Z) = 0$. The map is defined by $v\otimes w \to f(v, w)$. For example if $v^* \in V^*$ and $w^* \in W^*$ then $v \otimes w \to v^*(v)w^*(w)$ defines a linear map from $V \otimes W \to \mathbb{R}$.

Let $\{v^1, \dots, v^n\}$ be a basis of $V$ and $\{ w^1, \dots, w^m\}$ be a basis of $W$. Consider the set of $mn$ vectors $v^i\otimes w^j$ in $V \otimes W$. We wish to show that they form a basis. First we check that they span the space $V \otimes W$. As the elements of $V \otimes W$ are finite linear combinations of elements of the form $v\otimes w$ it suffices to show that these are all in the span of the vectors $v^i\otimes w^j$. But this follows from the bilinearity. If $v = \sum_{i=1}^n a_i v^i$ and $w = \sum_{j=1}^m b_j w^j$ then

$$v\otimes w = \sum_{i=1}^n \sum_{j=1}^m a_ib_j v^i\otimes w^j.$$

To show that they are linearly independent assume that

$$0 = \sum_{i=1}^n \sum_{j=1}^m a_{ij} v^i \otimes w^j.$$

Let $v_i^*$ and $w_j^*$ be the dual bases of $V^*$and $W^*$. That is $v_i^*(v^j) = \Delta_i^j$ and $w_i^*(w^j) = \Delta_i^j$. Then apply the map $V \otimes W \to \mathbb{R}$ defined by $v^*_i$ and $w^*_j$ to this equation to obtain $a_{ij} = 0$. So we have proved.

Proposition D.3   If $V$ and $W$ are finite dimensional vector spaces then

$$\dim(V\otimes W) = \dim(V)\dim(W).$$

We can iterate tensor products. If $V$ and $W$ and $U$ are vector spaces we can form $(V\otimes W) \otimes U$ and $V\otimes (W\otimes U)$. These different vector spaces are in fact isomorphic via the map

$$(v \otimes w)\otimes u \mapsto v \otimes (w \otimes u).$$

We use this map to identify these two spaces and ignore the brackets. We write $V \otimes U \otimes W$ for the triple tensor product. More generally we can form finitely many tensor products.

We also need to know about tensor products of maps. If $X \colon V \to V'$ is linear and $Y \colon W \to W'$ is linear then we can define a map

$$V \times W \to V'\otimes W'$$

by $(v,w) \mapsto X(v)\otimes Y(v)$. This is a bilinear map so factors to a map $V\otimes W \to V'\otimes W'$ which we denote by $X \otimes Y$. It is defined by $(X\otimes Y)(v\otimes w) = X(v)\otimes Y(w).$

We have seen that any bilinear map $V \times W \to \mathbb{R}$ gives rise to a linear map $V \otimes W \to \mathbb{R}$. It is easy to show that this is an isomorphism. More generally if for any collection of vector spaces $V_1, \dots, V_k$ we denote by Mult$(V_1\times \dots\times V_k, \mathbb{R})$ the space of all multilinear maps from $V_1 \times \dots V_k \to \mathbb{R}$ we have

Proposition D.4   If $V_1, \dots, V_k$ are vectors spaces then there is a natural isomorphism

$$Mult(V_1\times\dots\times V_k, \mathbb{R}) \to (V_1\otimes V_2 \otimes \dots \otimes V_k)^*$$

defined by

$$f \mapsto (v_1\otimes \dots \otimes v_k \mapsto f(v_1, \dots, v_k)).$$