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Stream: learning: questions

Topic: Ext Groups in the Deligne-Kelly Tensor Product?

Chris Grossack (she/they) (Sep 15 2025 at 03:40):

Given two algebras $A$ and $B$, there's this nice fact that the category $(A \otimes B)\text{-mod}$ is equivalent to the category $(A\text{-mod}) \boxtimes (B\text{-mod})$, the Deligne-Kelly tensor product of the module categories.

How does one do homological algebra in this $\boxtimes$-category?

Iirc, every object in $(A\text{-mod}) \boxtimes (B\text{-mod})$ is a colimit of objects of the form $M \boxtimes N$ for $M$ an $A$-module and $N$ a $B$-module, and my intuition is that we can get a projective resolution for one of these by finding an $A$-projective resolution $P_\bullet \to M$ and a $B$-projective resolution $Q_\bullet \to N$ and then applying $\boxtimes$ to these in the sort of "obvious" way, using terms that look like $\bigoplus_{a+b = n} P_a \boxtimes Q_b$...

In particular, if $A$ and $B$ are algebras over a field $k$, so that $\text{Hom}(M \boxtimes N, M' \boxtimes N') \cong \text{Hom}_A(M,M') \otimes_k \text{Hom}_B(N,N')$, this should give us a Künneth formula for the $\text{Ext}^\bullet$-groups as computed in $(A\text{-mod}) \boxtimes (B\text{-mod})$

Is my intuition right? Is this written down somewhere? There might also be a higher powered way to make this "obvious", and I would also be interested in that.

Thanks in advance! ^_^

John Baez (Sep 15 2025 at 11:47):

I like your question. I don't know the answer, but I fear bringing in the Deligne-Kelly tensor product may not increase the number of good answers you get. You could instead ask something like:

Given algebras $A$ and $B$ over a field $k$, with modules $M$ and $N$, $M \otimes_k N$ is naturally a module over $A \otimes B$. Is there a way to compute the cohomology of $M \otimes_k N$ in terms of the cohomology of $M$ and $N$, or obtain a projective resolution of $M \otimes_k N$ in terms of resolutions for $M$ and $N$?

And this sounds like something algebraists will have thought about.

fosco (Sep 15 2025 at 12:35):

I am a bit confused by the fact that the same symbol $\boxtimes$ is used for a monoidal structure on module categories, and for what seems a monoidal structure of objects, that, however, live in different categories.

Can it help the fact that there is a coproduct diagram of $\mathbb Z$-algebras (are A,B algebras over the same base ring?) $A\to A\otimes B \leftarrow B$? This gives functors relating the module categories over A,B and their tensor.

John Baez (Sep 15 2025 at 12:39):

fosco said:

I am a bit confused by the fact that the same symbol $\boxtimes$ is used for a monoidal structure on module categories, and for what seems a monoidal structure of objects, that, however, live in different categories.

Given an object $M$ in $(A\text{-mod})$ and an object $N$ in $(A\text{-mod})$ you can get an object in the Deligne-Kelly tensor product of these categories, $(A\text{-mod}) \boxtimes (B\text{-mod})$, and Chris is following tradition in calling this object $A \boxtimes B$.

John Baez (Sep 15 2025 at 12:43):

That's how the Deligne-Kelly tensor product of finitely cocomplete $k$-linear categories works: given two such categories, say $X$ and $Y$, and objects $M \in X, N \in Y$, you get an object $M \boxtimes N \in X \boxtimes Y$.

Yes, we are overloading the symbol $\boxtimes$ here, but in a way typical of tensor products: given vectors in vector spaces $v \in V, w \in W$, we get a vector $v \otimes w \in V \otimes W$.

John Baez (Sep 15 2025 at 12:45):

But anyway, this is an example of why I recommended Chris not to bring up the Deligne-Kelly tensor product: their actual question here can be asked without mentioning it.

Chris Grossack (she/they) (Sep 15 2025 at 15:49):

I think there's something fancy you can do, which also agrees with my intuitive answer -- Modules $M$ and $N$ over $A$ and $B$ can be thought of as sheaves on $\text{Spec}(A)$ and $\text{Spec}(B)$ (actually the $A$ and $B$ I'm interested in are noncommutative, so we're pushing this geometric intuition even further than normal algebraic geometry does).

With this in mind, the module $M \boxtimes N$ is naturally viewed as a sheaf on $\text{Spec}(A \otimes B) = \text{Spec}(A) \times \text{Spec}(B)$, indeed it's basically $\pi_1^* M \otimes_{A \otimes B} \pi_2^* N$ if you think about it geometrically, and in this case you should really expect a Künneth formula to hold!

This is points in John's direction of not worrying to much about the tensor product of categories

Chris Grossack (she/they) (Sep 15 2025 at 15:55):

Actually, yeah, this idea shouldn't be so hard to formalize algebraically. Thinking about these "pullbacks of sheaves" (read: induction of scalars along the inclusions $A,B \to A \otimes B$) was useful, thank you! That should work, formally the same, in the noncommutative setting too. Then we're working with a usual tensor product of $(A \otimes B)$-modules and the result should fall out, rather than needing to think abou this weird "external" tensor product

Mike Shulman (Sep 15 2025 at 19:14):

Is it really always true that $\hom_{A\otimes B}(M\otimes_k N, M'\otimes_k N') \cong \hom_A(M,M') \otimes_k \hom_B(N,N')$? If so, I feel like I should have known it, but I don't remember learning it.

Chris Grossack (she/they) (Sep 16 2025 at 21:31):

I guess I haven't checked myself, but if you believe that $(A \otimes B)\text{-mod} \simeq (A\text{-mod}) \boxtimes (B\text{-mod})$ (which is mentioned in many places, here's the first I found) then this follows from the explicit descriptions of the $\boxtimes$ product

Chris Grossack (she/they) (Sep 16 2025 at 21:33):

It's also fairly believable, since if you have a map $M \boxtimes N \to M' \boxtimes N'$ which is $A \otimes B$-linear, note that $B$ acts trivially on the $M$ piece but (presumably) nontrivially on the $N'$ piece, so it feels difficult to come up with an $A \otimes B$-linear map that somehow "mixes" the factors. That's not a proof, obviously, but it's some evidence

Mike Shulman (Sep 16 2025 at 22:36):

Is it obvious that it's true even when $A=B=k$? When the vector spaces are finite-dimensional I can count dimensions and see that they're the same, and so "obviously" the canonical map must be an isomorphism, but it's not obvious to me how that generalizes to infinite dimensions. Or is it only claimed in the finite-dimensional case?

Chris Grossack (she/they) (Sep 17 2025 at 00:08):

Yeah, you probably want $A$ and $B$ to be finite dimensional as vector spaces over $k$. I don't even know how to make sense of the $\boxtimes$-product (computationally) in other settings. See, for instance, the relevant nlab article, which also cites a paper by Etingof and Ostrik where they explicitly ask for a finite dimensional assumption. Iirc ENGO's book on tensor algebras has a version for coalgebras, though, which I guess makes sense since I remember @John Baez telling me at one point that coalgebras somehow behave more finitely than algebras, for reasons I still haven't internalized

John Baez (Sep 17 2025 at 08:59):

The ultimate reason is that when we have a coalgebra $A$ and we apply the comultiplication $\Delta : A \to A \otimes A$ to any $a \in A$, the result is a finite linear combination of elements $a_1 \otimes a_2$. Somehow one can leverage this into a proof of the Fundamental Theorem of Coalgebras, which says that any element of a coalgebra lies in some finite-dimensional sub-coalgebra. Using the fact that the internal direct sum of two sub-coalgebras is a sub-coalgebra, it follows that every coalgebra is a filtered colimit of finite-dimensional sub-coalgebras.

I'm trying to figure out the proof of this Fundamental Theorem. I've looked at it, and my eyes rolled over the Sweedler notation and it seemed sort of obvious, but it seems a lot less obvious as I'm trying to mentally re-create it. How do you get the finite-dimensional sub-coalgebra containing $a \in A$? You form

$\Delta(a) = \sum a_i \otimes b_i$

and you take the linear combinations of $a$ and all the $a_i$ and $b_i$. Maybe that's the desired sub-coalgebra. Or maybe you have to repeat this process and comultiply the $a_i$ and $b_i$ and also take the linear combinations of all the elements you get that way. Or....

The question is why this process terminates. Coassociativity must help.

Mike Shulman (Sep 17 2025 at 15:09):

What about the modules $M,N,M',N'$ -- do they have to be finite dimensional too?

Mike Shulman (Sep 17 2025 at 15:10):

The nLab also claims that there is a version "without the finiteness constraints and using the tensor product of categories with finite colimits".

Joe Moeller (Sep 17 2025 at 23:53):

Here's the link to the proof of the Fundamental Theorem of Coalgebras: https://planetmath.org/FundamentalTheoremOfCoalgebras

The Sweedler notation is not so bad. The proof does hinge on coassociativity. I'll try summarizing with minimal notation.

We're going to define the sub-coalgebra of an element x by double comultiplying. Comultiply x, then comultiply the first slot of the result. Each comultiplication results in a sum over tensors, so our thing is a sum over two indices, the first term over j, the third over i, and the middle term over both.

Now we define the subspace that will be our subcoalgebra D to be the span of all the terms in the middle slot. x is in this subspace because x is equal to applying the counit to the first and third slots of our double comuliplication. This is just using (co)unitality of a (co)monoid. But the counit of the first and third slots gives the coefficients of the linear combination that shows x is in the span.

Showing this subspace is a subcoalgebra is where coassociativity comes in. We're going to do this by showing that comultiplying any of the terms from the middle slot lands in $D \otimes D$. The gist of the argument is that you apply a third comultiplication, then coassociate it around and use an assumption of linear independence to peel off the other terms so you end up concluding the comultiplication of the middle term is in first $C \otimes D$, then in $D \otimes C$, and thus $D \otimes D$.

John Baez (Sep 18 2025 at 11:14):

Ah, thanks Joe, that's great! I don't mind Sweedler notation, but it's easy for me, when reading a purely computational proof using this or any clever notation, to say "yeah yeah, that probably works" without thinking about it. And then I can't remember the trick on my own!

So now I will remember this: take an element $c$ of a coalgebra $C$, comultiply it twice to get a linear combination of elements like $c_1 \otimes c_2 \otimes c_3$, and take the space spanned by all the middle terms $c_2$. This is our finite-dimensional sub-coalgebra of $C$ containing $c$. To show it's a sub-coalgebra, comultiply $c$ again and use coassociativity.

John Baez (Sep 18 2025 at 11:15):

This is like a dehydrated version of the proof, and I'm supposed to be smart enough that I can "add water" - do some calculations - whenever I want to reconstitute the whole proof.

John Baez (Sep 18 2025 at 11:16):

It's interesting that the proof requires comultiplying $c$ three times.

Mike Shulman (Sep 18 2025 at 15:16):

Possibly some of the above messages should be moved to another thread about coalgebra?

John Baez (Sep 18 2025 at 17:49):

Sometimes people here are chopping conversations in half, moving part to a new thread, making it harder to follow the overall conversation. For example the stuff about finite-dimensionality and coalgebras here arose from people talking about finite-dimensionality assumptions for results on (external) tensor products of modules of algebras.

To exhibit the inner unity of this conversation: not only is it true that every element of a coalgebra lies in a finite-dimensional sub-coalgebra, every element of a comodule of a coalgebra lies in a finite-dimensional sub-comodule of that coalgebra. The proof is analogous. And this is probably why some results on tensor products of comodules of coalgebras work better than the analogous results on tensor products of modules of algebras, as Chris was saying:

Chris Grossack (she/they) said:

See, for instance, the relevant nlab article, which also cites a paper by Etingof and Ostrik where they explicitly ask for a finite dimensional[ity] assumption. Iirc ENGO's book on tensor algebras has a version for coalgebras, though, which I guess makes sense since I remember John Baez telling me at one point that coalgebras somehow behave more finitely than algebras, for reasons I still haven't internalized

I hope @Chris Grossack (she/they) can now internalize those reasons, thanks to what @Joe Moeller said.

Mike Shulman (Sep 18 2025 at 18:11):

My feeling was that the conversation about algebras got interrupted by an essentially unrelated conversation about coalgebras, and now I have to scroll way up and skip over the coalgebra posts to find what we were saying about algebras and get back to it. I don't see any real connection between the algebra and coalgebra conversations, in fact the point of the latter was that coalgebras are different than algebras. But if you want to keep them all together, okay.

If we're done with coalgebras, the question about algebras that I still want to know about is whether $\hom(M\otimes N, M'\otimes N') \cong \hom(M,M') \otimes \hom(N,N')$ for infinite-dimensional vector spaces over a field.

John Baez (Sep 18 2025 at 20:07):

There's a natural map

$\alpha : \hom(M,M') \otimes \hom(N,N') \to \hom(M\otimes N, M'\otimes N')$

and we're wondering if it's onto. Let me try to argue that it's not.

Let $m_i, n_i, m'_i, n'_i$ be bases of countable-dimensional vector spaces $M,N,M',N'$, respectively. Consider the linear maps $M \otimes N \to M' \otimes N'$ that sends

$m_1 \otimes n_1$ to $m'_1 \otimes n'_1$
$m_2 \otimes n_2$ to $m'_1 \otimes n'_1 + m'_2 \otimes n'_2$
$m_3 \otimes n_3$ to $m'_1 \otimes n'_1 + m'_2 \otimes n'_2 + m'_3 \otimes n'_3$

and so, while sending all other basis vectors to zero. Is this linear map in the image of $\alpha$? I'm hoping not.

Mike Shulman (Sep 18 2025 at 21:41):

I sort of see what you're getting at. The image of a pure tensor $\alpha(f\otimes g)$ can't behave like that on $m_i\otimes n_i$ for any $i>1$. We can "correct" that for any particular $i$ by taking a linear combination of pure tensors, but it feels like since any linear combination is finite we can only correct finitely many of them. But I'm not sure how to make that rigorous.