What is a Tensor Product?

Question

If you were to explain the concept of a tensor product to an undergraduate(post linear algebra), how would you do so?

I would like to hear your definition, your take, on the definition of a tensor product. Explain what exact it means to tensor modules over a ring. And, if you feel it necessary, explain the significance of multiple tensor products, $\bigotimes$.

Answer 1

In linear algebra, we deal with vector spaces and linear maps, and we are happy. But we also want to talk about inner products, bilinear forms and such things, and these are not linear maps. Instead of developing a completelu new theory, we do what we do best: reduce the problem to one we know how to solve. The tensor product allows us to turn bilinear maps $V\times W\to U$ into linear maps $V\otimes W\to U$. And this makes us happy again. This is important, because we like being happy.

Once we have established that, I would explain the actual construction and basic properties of the tensor product to an undergraduate by pointing him to the library. People with a reasonable background of linear algebra are perfectly prepared, in my experience, to read the construction and so on.

Answer 2

https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html

Timothy Gowers posted this on his website. He has a very good explanation of tensor product, it's definition, and why we use it. I found it very useful when I was beginning my study of them.

Answer 3

Let ${ V, W, X}$ be ${ k - }$vector spaces.

Let

$${ V \times W \overset{f}{\longrightarrow} X }$$

be a bilinear map.

Can we factor the bilinear map

$${ V \times W \overset{f, \, \, \text{bilin}}{\longrightarrow} X }$$

as

$${ V \times W \overset{\pi, \, \, \text{bilin}}{\longrightarrow} V \otimes W \overset{\tilde{f}, \, \, \text{lin}}{\longrightarrow} X }$$

where ${ \pi }$ is a canonical bilinear map?

We want to have

• ${ V \otimes W }$ is a vector space.
• ${ (v, w) \mapsto \pi(v, w) }$ is bilinear.
• For any bilinear map ${ V \times W \overset{f}{\longrightarrow} X }$ there is a unique linear map ${ V \otimes W \overset{\tilde{f}}{\longrightarrow} X }$ such that ${ f(v,w) = \tilde{f}(\pi(v, w)) . }$

To ensure the first property, we can consider the free vector space ${ \mathcal{F}(V \times W) . }$

We will pick a quotient ${ \mathcal{F}(V \times W) / U }$ such that

$${ {\begin{aligned} &\, V \times W \overset{\pi}{\longrightarrow} \mathcal{F}(V \times W) / U, \quad \\ &\, \pi(v, w) = (v, w) + U \end{aligned}} }$$

is a valid candidate for ${ \pi . }$

Note that the first condition is satisfied. Note that to satisfy the second condition we want

$${ {\begin{aligned} &\, \text{Want: } \\ &\, (v _1 + v _2, w) + U = (v _1 , w) + (v _2, w) + U , \\ &\, (\alpha v, w) + U = \alpha (v, w) + U, \\ &\, (v, w _1 + w _2) + U = (v , w _1) + (v, w _2) + U, \\ &\, (v, \alpha w) + U = \alpha (v, w) + U \end{aligned}} }$$

that is

$${ {\begin{aligned} &\, \text{Want: } \\ &\, (v _1 + v _2, w) - (v _1 , w) - (v _2, w) \in U , \\ &\, (\alpha v, w) - \alpha (v, w) \in U, \\ &\, (v, w _1 + w _2) - (v , w _1) - (v, w _2) \in U, \\ &\, (v, \alpha w) - \alpha (v, w) \in U . \end{aligned}} }$$

Hence we can define ${ U }$ to be the subspace generated by the elements

$${ U := {\begin{aligned} &\, \text{Subspace generated by the elements} \\ &\, (v _1 + v _2, w) - (v _1 , w) - (v _2, w) , \\ &\, (\alpha v, w) - \alpha (v, w), \\ &\, (v, w _1 + w _2) - (v , w _1) - (v, w _2), \\ &\, (v, \alpha w) - \alpha (v, w) . \end{aligned}} }$$

This fixes ${ \pi . }$ It suffices to verify the third condition is satisfied.

Let ${ V \times W \overset{f}{\longrightarrow} X }$ be a bilinear map. Consider the map

$${ {\begin{aligned} &\, \mathcal{F}(V \times W) / U \overset{\tilde{f}}{\longrightarrow} X , \\ &\, \tilde{f}\left( \sum _{\text{finite}} \alpha _i (v _i, w _i) + U \right) := \sum _{\text{finite}} \alpha _i f \left(v _i, w _i \right) . \end{aligned}} }$$

Is ${ \tilde{f} }$ well defined? Is ${ \tilde{f} }$ a linear map?

Yes.

The linear map

$${ {\begin{aligned} &\, \mathcal{F}(V \times W) \longrightarrow X, \\ &\, \left( \sum _{\text{finite}} \alpha _i (v _i, w _i) \right) \longmapsto \sum _{\text{finite}} \alpha _i f(v _i, w _i) \end{aligned}} }$$

vanishes on the subspace ${ U . }$ This induces the linear map ${ \tilde{f} }$ above.

Hence ${ \tilde{f} }$ is a well defined linear map.

Is ${ f = \tilde{f} \circ \pi }$?

Yes.

Note that

$${ f(v, w) = \tilde{f}(\pi(v, w)) }$$

that is ${ f = \tilde{f} \circ \pi . }$

Is ${ \tilde{f} }$ the unique linear map ${ \mathcal{F}(V \times W) / U \longrightarrow X }$ such that ${ f = \tilde{f} \circ \pi }$?

Yes.

Suppose there were another linear map ${ \tilde{f} ^{’} : \mathcal{F}(V \times W) / U \longrightarrow X }$ with the property ${ f = \tilde{f} ^{’} \circ \pi . }$ Note that ${ \tilde{f} ^{’} ((v, w) + U) = f(v, w). }$ Hence by linearity ${ \tilde{f} ^{’} \left( \sum _{\text{finite}} \alpha _i (v _i, w _i) + U \right) = \sum _{\text{finite}} \alpha _i f(v _i, w _i) . }$ Hence ${ \tilde{f} ^{’} = \tilde{f} . }$

Hence we have the required factorisation result.

We denote ${ \pi(v, w) }$ by ${ \otimes (v, w) = v \otimes w . }$

We can also show (for eg, as in Knapp’s Algebra book) that any bilinear map ${ \pi }$ with the above three properties is essentially unique (i.e. unique upto composing by isomorphisms).