What Is the Tensor Product of Vector Spaces?

[eastside00_99]

You have a real rigorous side to you Fredrik. I like that, but if I may, I will outline all that we have done:

(1) Create a definition for an object which may or may not exist
(2) Proved abstractly that if it does exist, then it is unique up to isomorphism
(3) As you pointed out, that if it does exist, then it tells us something about all bilinear maps from VxW to any vector space Z

We began this discussion with the construction of the tensor product and I suggested we back up and give the abstract definition first. We have given the abstract definition so that now if we look at the construction we will have proved existence and we will automatically have the conclusions of 2 and 3. This should be enough to convince you that the definition will work if we prove existence.

 

[Hurkyl]

I would say that that's what I'm doing already. I'm trying to use the definition of "tensor product" that Eastside posted to prove that it's unique up to isomorphisms, but that's probably not what you meant.

The existence of the map you seek is his definition of the tensor product... (at least, the one he gave in post #2)

 

[George Jones]

First a recap.

The tensor product of vector spaces $V$ and $W$ is a vector space $V \otimes W$ together with a bilinear mapping $\otimes : V \times W \rightarrow V \otimes W$, such that if $X$ is a vector space and $\tau$ is a bilinear mapping from $V \times W$ into $X$, then there exists a unique linear mapping $f$ from $V \otimes W$ into $X$ such that $\tau = f \circ \otimes$.

Note that, as eastside00_99 has said, the existence of such a unique $f$ is part of the definition of tensor product.

Let $V \otimes W$ together with a bilinear mapping $\otimes : V \times W \rightarrow V \otimes W$ be a tensor product.

Let $V \square W$ together with a bilinear mapping $\square : V \times W \rightarrow V \square W$ also be a tensor product.

Now, to prove uniqueness, use this definition of tensor product four times.

First time: take the tensor product to be $V \otimes W$, $X$ to be $V \square W$ with $\tau = \square$. The definition says that there is unique linear mapping $f$ from $V \otimes W$ into $V \square W$ such that $\square = f \circ \otimes$.

Second time: take the tensor product to be $V \square W$, $X$ to be $V \otimes W$ with $\tau = \otimes$. The definition says that there is unique linear mapping $g$ from $V \square W$ into $V \otimes W$ such that $\otimes = g \circ \square$.

Combining these gives $\otimes = g \circ f \circ \otimes$ and $\square = f \circ g \circ \square$.

Third time: take the tensor product to be $V \otimes W$, $X$ to be $V \otimes W$ with $\tau = \otimes$. The definition says that there is unique linear mapping $h$ from $V \otimes W$ into $V \otimes W$ such that $\otimes = h \circ \otimes$. $h = I_{V \otimes W}$ (the identity on $V \otimes W$) cleary works. But, since (as above), $\otimes = g \circ f \circ \otimes$, $h = g \circ f$ also works. Because, by definition, $h$ is unique, $I_{V \otimes W} = g \circ f$.

Fourth time: take the tensor product to be $V \square W$, $X$ to be $V \square W$ with $\tau = \square$. Use the definition to show $I_{V \square W} = f \circ g$.

Therefore, $f : V \otimes W \rightarrow V \square W$ and $g : V \square W \rightarrow V \otimes W$ are both isomorphisms, and $g = f^{-1}$.

 

[Fredrik]

Thank you George. That cleared up a lot. Now I understand everything except a detail in the construction, but I'll give that some more thought later today.

Actually all that anyone would have needed to say to get me past my main concern is what you said here:

take [...] $X$ to be $V \square W$

I had a feeling I was overlooking something very simple, but for some reason I couldn't see that all I needed to do was to let the "alternative" tensor product be the arbitrary bilinear function mentioned in the definition of the tensor product.

I also appreciate that your post explained why we assume that the functions you called $\tau$ (the ones eastside called $f_*$) are unique. (Because it allows us to identify $g\circ f$ and $f\circ g$ with the appropriate identity maps).

I also realized that no bilinear map can be injective, since we want things like B(ax,y)=B(x,ay) to hold.