## Tensors #2

Welcome back to the beginning of my journey into the world of multilinear algebra. In this post, I want to discuss the definition I previously gave of the tensor product of two vector spaces. There are many ways to define it, and the definition I gave was through a universal property. One thing I have found particularly enthralling is the way we can prove things about the tensor product by simply making use of this universal property. I will show an example of this shortly.

The main point of this post is to discuss in detail two different equivalent formulations of the universal property. I realize I still have not proved existence or uniqueness of the tensor product, but don’t worry. I will do this soon.

§3. Discussion of the universal property of the tensor product

The universal property of the tensor product. (Note that the labels used here are not consistent with those in my post).

I’ll omit the preamble and assume that from the last post, you know the general setup that we’ve put into place. Recall that we say that a pair $(\mathsf{T}, \otimes)$ consisting of a vector space $\mathsf{T}$ and a bilinear map $\otimes : \mathsf{V} \times \mathsf{W} \to \mathsf{T}$ has the universal property if

• (T) If $\mathsf{Z}$ is some other vector space and $f : \mathsf{V} \times \mathsf{W} \to \mathsf{Z}$ is bilinear, then there exists a unique linear map $\varphi : \mathsf{T} \to \mathsf{Z}$ such that $f = \varphi \circ \otimes$.

Condition (T) above turns out to be equivalent to the following two conditions holding:

• (T1) The span of $\mathrm{Img}(\otimes) := \{ v \otimes w : v \in V, w \in W \}$ is all of $\mathsf{T}$.
• (T2) If $\mathsf{Z}$ is some other vector space and $f : \mathsf{V} \times \mathsf{W} \to \mathsf{Z}$ is bilinear, then there exists a linear map $\varphi : \mathsf{T} \to \mathsf{Z}$ such that $f = \varphi \circ \otimes$.

At first glance you may have thought (T) and (T2) are the same. However, there is a subtle difference. (T) tells us that the map $\varphi$ for some fixed $f$ is unique. (T2) tells us that such a map $\varphi$ exists, but says nothing about uniqueness.

Let’s get right into it, then. Why are these equivalent? Well, let’s take care of the easy direction first: suppose (T1) and (T2) hold. Let an arbitrary $f$ (as described above) be given, and suppose that $\varphi_1, \varphi_2 : \mathsf{T} \to \mathsf{Z}$ are such that $f = \varphi_1 \circ \otimes = \varphi_2 \circ \otimes$. Our goal is to prove $\varphi_1 = \varphi_2$. However, this is quite easy. Note that by the above, we see that $f(v,w) = \varphi_1( v \otimes w ) = \varphi_2( v \otimes w ).$ However, by (T1), these vectors $v \otimes w$ generate all of $\mathsf{T}$! So therefore $\varphi_1$ and $\varphi_2$ agree on a generating set for $\mathsf{T}$, whereby immediately we obtain $\varphi_1 = \varphi_2$. Wonderful.

Now for the slightly more difficult direction. Suppose (T) holds. It is clear that (T2) holds, but (T1) is not so obvious. When I was thinking about this, I realized I didn’t really understand the proof of this fact given in the book I’m reading (Greub), because it made use of some odd notation I’m not familiar with. So here’s my own proof, which I hope will be clearer. How is uniqueness of the map related to whether or not these vectors $v \otimes w$ generate $\mathsf{T}$? Well, here’s a trick. Suppose to the contrary that the vectors $v \otimes w$ do not generate $\mathsf{T}$. Let $\mathsf{Z} = \mathsf{T}$ and pick our map $f$ to be nothing more than $\otimes$. Then (T) tells us there should be a unique $\varphi$ such that $\otimes = \varphi \circ \otimes$. You’re probably thinking already “OK, so we can just choose $\varphi$ to be the identity map”. That is correct, but it’s not the only choice. What we do know is that $\varphi$ restricted to the span of $\mathrm{Img}(\otimes)$ must be the identity map (why?). However, the key idea is that if $\mathrm{Img}(\otimes)$ does not generate all of $\mathsf{T}$, then there will be some “arbitrary choice” to be made in coming up with the map $\varphi$. As I previously stated, it is valid to choose $\varphi$ to be the identity, but we can also define $\varphi$ to be the identity on the subspace generated by $\mathrm{Img}(\otimes)$, and then map everything outside the subspace into the subspace (clearly, a map so obtained will not be the identity map on $\mathsf{T})$.

You can turn this into a formal proof if you want, but that’s the idea. In the next post, I will man up and prove existence and uniqueness. Cheers.

just another guy trying to make the diagrams commute.
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### 3 Responses to Tensors #2

1. Saifuddin Syed says:

For the first direction, I don’t understand why T1 is needed to show \varphi_1 = \varphi_2 .

2. dx7hymcxnpq says:

The invocation of (T1) is crucial. Just because $\varphi_1( v \otimes w) = \varphi_2( v \otimes w)$ for all $v \in \mathsf{V}, w \in \mathsf{W}$ does not mean they are the same map on $\mathsf{T}$. All it means is that their restrictions to the subspace generated by the vectors $v \otimes w$ must be the same. If this subspace turns out to be contained properly within $\mathsf{T}$ then you can’t conclude $\varphi_1 = \varphi_2$ because you don’t know anything about how each map behaves outside of this subspace.

3. dx7hymcxnpq says:

Thing about the tensor product: it’s a space which is large enough to properly capture the possibilities for multilinear (well, in this case bilinear) maps, but it’s also small/tight enough to provide you with a unique means of decomposition (I’m referring to the maps $\varphi$ here). In that sense it’s very special.