# Understanding tensor-products in the commutative diagram of a k-Algebra

I'm having trouble getting to grips with the commutative diagram for an algebra over a field $k$.

The main problem is that my understanding of the tensor product is weak.

I have seen $V \otimes W$ defined as the set of $k$-valued bilinear functions on $V^* \times W^*$.

I have also seen $V \otimes W$ defined as as the dual space of the space of bilinear forms on $V \times W$.

These don't seem obviously the same to me, but I assume that they are equivalent.

In general I understand a dual space $V^*$, given vector space $V$ over $k$, as the set of all linear maps $V \rightarrow k$. I guess in the above definitions we're looking at bilinear maps.

My problem with the diagram is that I can't understand the use of the $\otimes$ along the arrows (the arrows I understand as similar to the morphisms I studied in group theory). I understand the elements of $k \otimes A$, or $A \otimes A$, as maps onto $k$, on the basis of what I said above. But where do id$\otimes i$ and id$\otimes \mu$ "belong"? These aren't part of the algebra per se? Perhaps the simplest way I can put it is: what is happening along these arrows?

I've included the diagram below. I hope what I've said makes sense. I feel like I'm halfway to understanding, but any help would be appreciated.

The definitions of tensor product you're using aren't the best. The proper way to define the tensor product of 2 vector spaces $V$ and $W$ is via the universal property.

First, a map $B: V \times W \to U$ (for $k$-vector spaces $V, W, U$) is bilinear if it is linear in each entry. Then the tensor product $V \otimes W$ is another $k$-vector space, equipped with a bilinear map $i: V \times W \to V\otimes W$, which satisfies the following universal property: Given any bilinear map $B: V \times W \to U$, there is a unique linear map $\tilde{B} : V \otimes W \to U$ such that $\tilde{B} \circ i = B$. Or in other words, every bilinear map out of $V \times W$ factors uniquely through $i$ and a linear map out of $V \otimes W$.

It is an exercise to show that the tensor product $V \otimes W$ (along with the map $i$) is uniquely determined up to a unique isomorphism (just take two of them and use the universal properties to give isomorphisms between them).

Another consequence of the universal property is that we have an isomorphism between $Bil(V \times W, U)$ (the space of bilinear maps from $V \times W$ to $U$) and $Hom(V\otimes W, U)$ (the space of linear maps from $V \otimes W$ to $U$). Applying this to the case of $U = k$, we get

$Bil(V \times W, k) \cong Hom(V \otimes W, k) = (V \otimes W)^* \cong V^* \otimes W^*,$

and hence $V^* \otimes W^*$ can be identified with what you are calling the bilinear forms on $V \times W$. Or switching things around, $V \otimes W$ can be identified with the space of bilinear forms on $V^* \times W^*$. This is the first definition you wrote down above. We can also recover the second definition you wrote down as follows:

$V \otimes W \cong {(V \otimes W)^{*}}^* \cong (V^* \otimes W^*)^* \cong Bil(V \times W, k)^*.$

Now elements of $V \otimes W$ can be written as sums of what elementary tensors, i.e. tensors of the form $v \otimes w$, for $v \in V$ and $w \in W$. So a general tensor in $V \otimes W$ has the form $\sum_{i} (v_{i} \otimes w_{i})$. And the linear map $\tilde{B}$ acts on such a tensor as follows:

$\tilde{B}(\sum_{i} (v_{i} \otimes w_{i})) = \sum_{i} B(v_{i},w_{i})$.

(Again, you can see this by staring at the universal property). It turns out that if $\{e_{i}\}$ is a basis of $V$ and $\{d_{j}\}$ is a basis of $W$, then $\{e_{i} \otimes d_{j}\}$ is a basis of $V \otimes W$.

Now as a final thing, lets see what it means to take the tensor product of 2 linear maps $\phi: V \to V'$ and $\psi : W \to W'$. Well, the first thing we can do is to take the product of these maps, and post compose with $i' : V' \times W' \to V' \otimes W'$. This gives the map

$i' \circ (\phi \times \psi) : V \times W \to V' \otimes W', \ \ (v,w) \mapsto \phi(v) \otimes \psi(w).$

As you can check, this is bilinear, and therefore factors through the tensor product of $V$ and $W$. Therefore, what we end up with is a map

$\phi \otimes \psi : V \otimes W \to V' \otimes W', \ \ v \otimes w \mapsto \phi(v) \otimes \psi(w)$.

In the definition of a $k$-algebra, the map $\mu: A \otimes A \to A, a \otimes b \mapsto ab$ is meant to encode the multiplication. This is a bilinear map from $A \times A \to A$, and hence it makes sense to use the tensor product. As you can check, the first two diagrams are just encoding the identities $1a = a, a1 = a$, and the last diagram is encoding associativity $(ab)c = a(bc)$.

• This has been incredibly helpful, thank you very much. In paragraph 3, should $V \otimes U$ read $V \otimes W$? And the exercise you suggested, I just need to consider two possible tensor products and mappings to arbitrary vector spaces $U$. But I can replace my "mapped to" vector space, $U$, with one of my tensor product vector spaces? I think I can get an identity map out of that... Commented Sep 11, 2015 at 0:09
• Yeah thats the idea. If $S_{1}$ and $S_{2}$ are tensor products, then you get unique maps $S_{1} \to S_{2}$ and $S_{2} \to S_{1}$. You need to show that these maps are inverses (Hint: what is the unique map $S_{1} \to S_{1}$). Commented Sep 11, 2015 at 18:12