# Ostensibly different products on Ext-groups

The following is presumably not the greatest generality in which this question makes sense.

1. Given a ring $k$, graded-commutative if it helps, and a Hopf-algebra $A$ over $k$, there is a Yoneda product making $\textrm{Ext}_A^*(k, k)$ into a ring (since $k$ is graded, this actually is a bigraded object, but I suspect the grading on $k$ is immaterial so I have suppressed it). We have from this construction an

2. There is also a product defined the following way $\newcommand{\E}{\textrm{Ext}}\E_A^*(k,k) \otimes_A \E^*_A(k,k) \to \E_{A \otimes_A A}^*(k\otimes_A k, k\otimes_A k) \to \E_A^*(k,k)$ using the external product on $\textrm{Ext}$.

3. One has a development of this idea: $\newcommand{\E}{\textrm{Ext}}\E_A^*(k,k) \otimes_k \E^*_A(k,k) \to \E_{A \otimes_k A}^*(k\otimes_k k, k\otimes_k k) \to \E_A^*(k,k)$ using the coproduct structure $A \to A \otimes_k A$ to change the base-ring. This appears to give a $k$-algebra rather than an $A$-algebra structure on $\textrm{Ext}$.

Should we expect these products to coincide? If they don't always coincide, are there conditions that ensure they do?

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They do coincide, in the examples I know. I am not sure how general this is, I think it is always true, but I don't know a reference or the key word to help you get started on a proof. –  Sean Tilson Mar 18 '11 at 2:52
In 2. the tensor products should be taken over $k$ instead of $A$. That's for two reasons: (i) $Ext_A$ usually doesn't have an $A$-module structure, but is only a module over the center of $A$ and therefore in particular over $k$. (ii) In the definition of a Hopf-algebra the tensor products occuring in the product and coproduct are taken over $k$. –  Ralph Mar 18 '11 at 8:07

Show that each of those products distributes over the others, and use Hilton-Eckmann (over a field, or for $A$ projective; in general, I don't know...) This ends up being then an exercise in using naturality.

It can be done more concretely, too. For example, to show (1) and (2) are the same, one can show that if $E$ and $E'$ are $n$- and $m$-extensions of $k$ by $k$, then $E\circ E'$, the Yoneda composition, is equivalent as an $(n+m)$-extension, to $E\otimes E'$; this also shows (1) is (3), because why I wrote $E\otimes E'$, an iterated extension of $A$-modules, is really the result of restricting scalars along the coproduct of $A$ from the iterated extension $E\otimes E'$ of $A\otimes A$-modules. On the other hand, to show directly that (2) and (3) are the same, show that computation of both in terms of the bar resolution is actually the same.

Later. Let $E$ and $E'$ be $n$- and $m$-extensions of $A$-modules of $k$ by $k$. For example, $E$ is $$0\to k\to E_{n-1}\to E_{n-2}\to\cdots\to E_0\to k$$ I'm going to grade this complex so that $E_0$ is in degree $0$, and write $\bar E$ for the result of simply taking away the rightmost $k$, so that $\bar E$ is a complex with homology $k$, concentrated in degree $0$. Then $\bar E\otimes\bar E'$ is a complex of $A\otimes A$-modules, which, by the Künneth formula, has homology $k\otimes k$ in degree $0$. Since in degree $n+m$ the complex $\bar E\otimes\bar E'$ has $k\otimes k$, we can add the $k\otimes k$ to it in degree $-1$ and obtain an $(n+m)$-extension of $k\otimes k$ by $k\otimes k$ in the category of $A\otimes A$-modules, which is what I denote above $E\otimes E'$. If we restrict scalars along the diagonal map, we get a complex $\Delta^\*(E\otimes E')$, which is the product of $E$ and $E'$ accoding to your product (3).

The other important observation is the following: what I wrote $\bar E\otimes\bar E'$ is a double complex with the shape of a rectangle. By looking at it mildly hard, you can see that if you go from the module in the largest degree (which is $k\otimes k$) towards the one in degree zero by walking along one side and then along another, you get a simple complex which is more or less trivially isomorphic to what now I would write $\overline{E\circ E'}$, the truncation of the Yoneda composition (if you walk along the other two sides of the rectangle, you get $\pm\overline{E'\circ E}$, and pursuing this you get a very concrete proof of graded-commutativity) Finally, you have to notice that there are maps of extensions $E\circ E'\to E\otimes E$ (and $\pm E'\circ E\to E\otimes E'$, of course)

Sorry if this came out rather messy: it is the kind of things that maximally adapts to an explanation face-to-face in front of a blackboard!

That (1) and (2) coincide is also shown here in a general nonsense manner.

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that is always the way isn't it. –  Sean Tilson Mar 18 '11 at 2:58
Thanks for the answer, which is both helpful and promising. I am afraid I don't understand it, being inexperienced with this sort of thing. What is $E \otimes E'$? I think if I understood this I would understand everything. Thanks again. –  Ben Williams Mar 18 '11 at 3:53
The argument in the first paragraph also shows that the product is (graded) commutative, right? It's just like the proof that the fundamental group of a topological group is abelian. –  John Palmieri Mar 18 '11 at 4:08
Thanks especially for the preprint, which I think meets my needs. I have qualms about the concrete construction though, because neither the functor $\otimes_A$ or $\otimes_k$ is exact (I assume $\otimes_k$ is what is meant by $\otimes$ in the explicit calculation). –  Ben Williams Mar 18 '11 at 6:47
I suppose I should really try to understand the exterior product construction as a construction using the derived tensor product in the derived category of $A$-modules. –  Ben Williams Mar 18 '11 at 6:52