## Tensors #4

First off, hello there. As it currently stands, I’ve been up for more than 20 hours, in an attempt to straighten out my sleep schedule once and for all (this is probably futile, but I can only hope not). I’m bored, have no energy to move, and certainly am in no state to learn, but require something to keep me from sleeping, so as a consequence, you get the present post, in which I will mindlessly regurgitate the procedure for constructing the tensor product of two vector spaces (this is therefore the much-awaited “existence proof” which I procrastinated due to its finicky nature and pronouncèd lack of aesthetic appeal). I apologize in advance if I manage to botch anything, since I probably won’t read over it until I’ve had a good sleep or something. Anyways, here goes nothing.

§5. Construction of the tensor product (as a quotient of a free vector space)

Suppose as usual that we fix a field $\mathbb{F}$ and we are given two $\mathbb{F}$-vector spaces $\mathsf{V}$ and $\mathsf{W}$. In order to “construct a tensor product”, we must do two things: first of all, construct the space $\mathsf{T} := \mathsf{V} \otimes \mathsf{W}$ itself, exhibit a suitable bilinear mapping $\otimes : \mathsf{V} \times \mathsf{W} \to \mathsf{T}$, and prove that the universal property is satisfied.

The construction will go as follows: we will take the free vector space on $\mathsf{V} \times \mathsf{W}$, denoted $C(\mathsf{V} \times \mathsf{W})$, define a certain “special” subspace $N$, and then quotient it out. The mapping $\otimes$ will then be given simply by $\pi$, the canonical projection (“reduction mod $N$“).

So let’s do this! Let $N$ be the subspace of $C(\mathsf{V} \times \mathsf{W})$ generated by the vectors $(\lambda x_1 + \mu x_2, y) - \lambda(x_1,y) - \mu(x_2, y)$ and $(x, \lambda y_1 + \mu y_2) - \lambda (x,y_1) - \mu (x,y_2)$, where $x, x_1, x_2 \in \mathsf{V}$, $y, y_1, y_2 \in \mathsf{W}$, and $\lambda, \mu \in \mathbb{F}$. Now then, let us define $\mathsf{T}$ to be the quotient $C(\mathsf{V} \times \mathsf{W}) / N$. Defining the map $\otimes : \mathsf{V} \times \mathsf{W} \to \mathsf{T}$ by putting $v \otimes w := \pi(v,w)$ where $\pi : C(\mathsf{V} \times \mathsf{W}) \to \mathsf{T}$ is the canonical projection (or “quotient map”), we can observe bilinearity quite easily. Since every vector of the form $(x, \lambda y_1 + \mu y_2)$ differs from $\lambda(x, y_1) + \mu(x, y_2)$ only by an element of $N$, we observe that

$x \otimes (\lambda y_1 + \mu y_2) = \pi(x, \lambda y_1 + \mu y_2) = \lambda \pi(x, y_1) + \mu \pi(x, y_2) = \lambda( x \otimes y_1 ) + \mu( x \otimes y_2)$

thereby demonstrating that $\otimes$ is linear in its second argument. In an identical fashion it is shown that $\otimes$ is linear in its first argument. Therefore $\otimes$ is a bilinear map. It now remains to check that (T1) and (T2) are satisfied.

To verify (T1), we need simply show that the vectors $v \otimes w$ generate $\mathsf{T}$. Let $z \in \mathsf{T}$ be arbitrary. We know this element must be the equivalence class of some element of $C(\mathsf{V} \times \mathsf{W})$ (note the direct analogue between subspaces and equivalence relations when we talk about quotient spaces). Hence (due to the general form of such an element) we see that

$\displaystyle z = \pi \sum_{\alpha, \beta} \lambda_{\alpha \beta} (v_\alpha, w_\beta) = \sum_{\alpha, \beta} \lambda_{\alpha \beta} \pi(v_\alpha, w_\beta) = \sum_{\alpha, \beta} \lambda_{\alpha \beta} (v_\alpha \otimes w_\beta).$

To verify (T2), suppose we are given a bilinear map $f : \mathsf{V} \times \mathsf{W} \to \mathsf{Z}$, where $\mathsf{Z}$ is some vector space. Note that the set of all pairs $(x,y)$ for $x \in \mathsf{V}$ and $y \in \mathsf{W}$ form a basis for the space $C(\mathsf{V} \times \mathsf{W})$, therefore this is enough to uniquely determine a linear map $\psi : C(\mathsf{V} \times \mathsf{W}) \to \mathsf{Z}$ such that $\psi(x,y) = f(x,y)$. In order to see that this map induces our desired map $\varphi : \mathsf{T} \to \mathsf{Z}$, we must show that the subspace $N$ is sitting inside the kernel of $\psi$. For example, suppose we have some vector of the form

$z = (\lambda x_1 + \mu x_2, y) - \lambda(x_1, y) - \mu(x_2, y).$

Then we observe that by the definition of $\psi$ and the bilinearity of $f$,

$\psi(z) = \psi(\lambda x_1 + \mu x_2, y) - \lambda \psi(x_1, y) - \mu \psi(x_2, y) = f(0, y) = 0.$

A similar phenomenon occurs if $z = (x, \lambda y_1 + \mu y_2) - \lambda(x,y_1) - \mu(x,y_2)$. Therefore indeed $N \subseteq \ker \psi$, and we therefore know by the universal property of quotient spaces that $\psi$ induces a linear map $\varphi$ from the quotient space (tensor product) $\mathsf{T}$, into $\mathsf{Z}$, such that $\varphi \circ \pi = \psi$ and hence $\varphi \circ \otimes = f$ thereby proving (T2). This completes the construction.