How to use tensor products

Quick description

The tensor product is a way to encapsulate the notion of bilinearity, and can be thought of as a multiplication of two vector spaces.

Prerequisites

Linear algebra.

General discussion

The dimension of a tensor product of two vector spaces is precisely the product of their dimensions, so when one wishes to show that a certain vector space is finite dimensional, one can try to show that it is a subspace of a tensor product (or an image of a tensor product) of two finite dimensional vector spaces.

Example 1

Fix a field $K$ . Some notation: $K[x]$ is the polynomial ring in one variable, $K(x)$ is the field of rational functions, $K[ [ x ] ]$ is the ring of formal power series, and $K((x))$ is the field of formal Laurent series.

A power series $u \in K[ [ x ] ]$ is said to be D-finite if it satisfies a linear differential equation $p_d(x)u^{(d)} + \cdots + p_1(x)u' + p_0(x)u = 0$ for some polynomials $p_i(x) \in K[x]$ with $p_d \ne 0$ , and $u^{(j)} = d^j u/ dx^j$ . Let $V_u$ denote the $K(x)$ -subspace of $K((x))$ spanned by the derivatives of $u$ . Then the property of being D-finite can be seen to be equivalent to requiring that the subspace $V_u$ is finite dimensional over $K(x)$ . From this, it is easy to see that the sum of two D-finite generating functions is also D-finite since $V_{u+w} \subseteq V_u + V_w$ . But what about the product of two D-finite generating functions?

We can define a map $V_u \times V_w \to K((x))$ by multiplication: the pair $(a,b)$ simply goes to $ab$ . The subspace spanned by the image of this will contain $V_{uw}$ by the Leibniz rule for taking the derivative of a product. But this map is not linear, so we cannot say much about the dimension of this span. However, it is bilinear, and hence we have an associated linear map $V_u \otimes_{K(x)} V_w \to K((x))$ whose image is precisely the span of the image of the bilinear map, and we see then that $\dim V_{uw} \le \dim(V_u \otimes V_w) = \dim(V_u) \cdot \dim(V_w)$ , so $uw$ is also D-finite.

Example 2

This finite dimensionality argument is used when proving a basic result about affine algebraic groups over fields, namely that they admit a faithful linear representation (and thus are rightfully called linear algebraic groups).

An affine algebraic group $G$ is of the form $Spec(A)$ where $A$ is a $k-$ algebra of finite type endowed with a comultiplication $A\rightarrow A\otimes A$ . Constructing a faithful linear representation of $G$ boils down to finding a surjection $k[x_{ij}]\rightarrow A$ where the polynomial ring $k[x_{ij}]$ is endowed with the usual Hopf algebra structure. This is done by choosing carefully a finite set of generators of the $k-$ algebra $A$ in such a way that the spanned finite dimensional $k-$ vector space $V$ satisfies $\mu(V)\subset V\otimes A$ and writing down the coefficients. See e.g Borel's book Linear Algebraic Book, sections I.1.9 and I.1.10