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.



1 Answer 1


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 $(ab)c = a(bc)$.

  • $\begingroup$ 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... $\endgroup$
    – Potkin57
    Commented Sep 11, 2015 at 0:09
  • 1
    $\begingroup$ 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 $S_{1} \to S_{1}$). $\endgroup$ Commented Sep 11, 2015 at 18:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .