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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.

See also

Note iconIncomplete This article is incomplete. Many more examples needed, demonstrating different ways in which tensor products are used. And more general discussion of the different uses is needed too.

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

Comments

Can the notation be

Can the notation be explained? Is K[x] different from K(x) different from K((x))?

Clarification

Anonymous:
Usually, all those are different algebraic structures:

K[x] represents the ring of polynomials on one variable (x) with coefficients on K, i.e., $K[x]=\{\sum_{i=0}^n k_i