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Stream: learning: questions

Topic: decomposing 2d TQFTs

Joe Moeller (Feb 26 2021 at 20:48):

@Ethan Kowalenko, @Sadaf Kadir, and I are studying 2d TQFTs, but we are not experts, and we could use some help. We are considering the TQFTs associated to the group algebras of $\mathbb Z/4$ and $\mathbb Z/2 \oplus \mathbb Z/2$ over a field of characteristic 0, say $\mathbb Q$ or $\mathbb C$. By this paper by Sawin we know that 2d TQFTs should decompose into a direct sum of simples. We want to compute this decomposition for the TQFTs associated to these group algebras. By theorem 3 in this paper by Perlis and Walker we know the group algebras are not isomorphic as algebras. So they're definitely not isomorphic as Frobenius algebras. Can anybody help us compute these decompositions?

John Baez (Feb 26 2021 at 21:05):

Some preliminary remarks:

Theorem 3 in that paper by Perlis and Walker doesn't say anything about group algebras over $\mathbb{C}$, does it?

I'm a bit scared to work over $\mathbb{Q}$ because I'm a physicist: we study TQFTs over $\mathbb{C}$, so that's what I mainly know about. But I can try thinking about $\mathbb{Q}$.

Over $\mathbb{Q}$ I think it's easy to see that the group algebras of $\mathbb{Z}/4$ and $\mathbb{Z}/2 \oplus \mathbb{Z}/2$ are nonisomorphic. $\mathbb{Q}[\mathbb{Z}/4]$ contains an obvious fourth root of unity, coming from a generator of $\mathbb{Z}/4$. But I'm pretty sure $\mathbb{Q}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$ doesn't contain a fourth root of unity. At least I don't see one!

But this sort of argument fails over $\mathbb{C}$, since over $\mathbb{C}$ every group algebra contains an nth root of unity for all n. Are you claiming that

$\mathbb{C}[\mathbb{Z}/4] \ncong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

or not?

Joe Moeller (Feb 26 2021 at 21:09):

I don't actually want to claim anything.

John Baez (Feb 26 2021 at 21:09):

Oh, darn.

John Baez (Feb 26 2021 at 21:10):

Math is so much more fun when people claim stuff.

Joe Moeller (Feb 26 2021 at 21:10):

Well then sure, I'll claim that.

John Baez (Feb 26 2021 at 21:10):

Okay, good. Then I claim that as algebras,

$\mathbb{C}[\mathbb{Z}/4] \cong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

Joe Moeller (Feb 26 2021 at 21:11):

Great, what's the isomorphism?

Joe Moeller (Feb 26 2021 at 21:11):

By the way, Todd gave an argument for this last week to me, but reading it now I'm a little confused. I don't know if he's interested in joining in.

John Baez (Feb 26 2021 at 21:12):

In fact I claim that if $G$ is any finite abelian group with n elements,

$\mathbb{C}[G] \cong \mathbb{C} \oplus \cdots \oplus \mathbb{C}$

where the right side is a direct sum of n copies of $\mathbb{C}$. I'm just claiming an isomorphism of algebras here, not Frobenius algebras - one step at a time!

John Baez (Feb 26 2021 at 21:12):

This claim would imply my previous claim - right?

Jason Erbele (Feb 26 2021 at 21:18):

Since isomorphism is transitive and $|\mathbb{Z}/4| = 4 = |\mathbb{Z}/2 \oplus \mathbb{Z}/2|$, I would make the claim that
$\mathbb{C}[G] \cong \mathbb{C} \oplus \dotsb \oplus \mathbb{C}$ for finite groups $G$ does indeed imply $\mathbb{C}[\mathbb{Z}/4] \cong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$. :upside_down:

John Baez (Feb 26 2021 at 21:19):

Indeed!

John Baez (Feb 26 2021 at 21:20):

So, next: does it seem correct that to prove this:

if $G$ is an abelian group with n elements, there is an algebra isomorphism

$\mathbb{C}[G] \cong \mathbb{C} \oplus \cdots \oplus \mathbb{C}$

where the right side is a direct sum of n copies of $\mathbb{C}$.

it's enough to prove it for cyclic groups?

Joe Moeller (Feb 26 2021 at 21:20):

Right, this is what Todd was saying too. That's making more sense now.

John Baez (Feb 26 2021 at 21:21):

Does the "right" mean you agree it's enough to prove the result for cyclic groups?

John Baez (Feb 26 2021 at 21:21):

If so, why?

Joe Moeller (Feb 26 2021 at 21:23):

No I wrote the "right" before I saw that part.

Ethan Kowalenko (Feb 26 2021 at 21:23):

Hopefully due to the fact that the group algebra of the product of finite abelian groups is isomorphic to the tensor product of the group algebras of the factors

John Baez (Feb 26 2021 at 21:23):

Ethan Kowalenko said:

Hopefully due to the fact that the group algebra of the product of finite abelian groups is isomorphic to the tensor product of the group algebras of the factors

Alas, that fact is false...

John Baez (Feb 26 2021 at 21:23):

... but it's on the right track in a way.

Ethan Kowalenko (Feb 26 2021 at 21:24):

Oh no!

John Baez (Feb 26 2021 at 21:28):

Maybe you can spot your mistake and fix it... it's not a big mistake in terms of how many words you need to change.

Ethan Kowalenko (Feb 26 2021 at 21:31):

Must be that the words "tensor product" should become "direct sum"

Joe Moeller (Feb 26 2021 at 21:31):

Product of abelian groups is also coproduct, and I think taking group algebra is a left adjoint. So I think he should have said "the group algebra of the product of finite abelian groups is isomorphic to the direct sum of the group algebras of the factors".

Ethan Kowalenko (Feb 26 2021 at 21:33):

Or in my more hands-on brain: the basis of the direct sum will be pairs of elements from each factor with their own relations, and multiplication is component wise

Ethan Kowalenko (Feb 26 2021 at 21:36):

Wait, that can't be right... Then $\mathbb{C}[\mathbb{Z}/2\times\mathbb{Z}/3]$ would be 5 dimensional and 6 dimensional as a vector space

Joe Moeller (Feb 26 2021 at 21:38):

Right, good point.

David Michael Roberts (Feb 26 2021 at 21:50):

Something something conjugacy classes?

Ethan Kowalenko (Feb 26 2021 at 21:51):

So I'm back to believing that $\mathbb{C}[\mathbb{Z}/n\times\mathbb{Z}/m]\simeq\mathbb{C}[\mathbb{Z}/n]\otimes\mathbb{C}[\mathbb{Z}/m]$, since the multiplication is tensor component-wise and it should work for the natural bases by sending $e_{(a,b)}\mapsto e_a\otimes e_b$

John Baez (Feb 26 2021 at 22:27):

Whoops, I screwed up! Ethan was right in the first place. Sorry!

Joe Moeller (Feb 26 2021 at 22:30):

I just realized that my argument was right, but I ended up in the wrong place. Tensor is coproduct for commutative algebras.

John Baez (Feb 26 2021 at 22:32):

So yeah: since every finite abelian group is a product of finite cyclic groups, and since taking the group algebra sends products of groups to tensor products of algebras, to prove my claim it's enough to prove it for finite cyclic groups.

John Baez (Feb 26 2021 at 22:36):

So does anyone have a guess about why

$\mathbb{C}[\mathbb{Z}/n]$

is isomorphic, as an algebra, to a direct sum of n copies of $\mathbb{C}$?

Todd Trimble (Feb 26 2021 at 22:59):

Did you still want me to join in?

Todd Trimble (Feb 26 2021 at 23:00):

Because that last question is easy to answer (and I already answered elsewhere).

Todd Trimble (Feb 26 2021 at 23:04):

Well, just @ me if you do. Otherwise, I can let you guys puzzle it out (but I already gave Joe the argument -- if he doesn't believe me, he can query me).

Ethan Kowalenko (Feb 26 2021 at 23:42):

Joe told me a bit about @Todd Trimble 's argument, but there's one piece that seems like a leap to me. Basically, the reasoning seemed to be that $\mathbb{C}[\mathbb{Z}/n]\simeq \mathbb{C}[x]/(x^n-1)$ as algebras, and the right hand side, as $\mathbb{C}[x]$-modules, is isomorphic to a direct sum of the quotients $\mathbb{C}[x]/(x-\eta)$ where $\eta$ ranges over all $n$-th roots of 1. Each of these is one-dimensional, so we're done if we can make sure that i can upgrade my $\mathbb{C}[x]$-isomorphisms to an isomorphisms of $\mathbb{C}$-algebras

Ethan Kowalenko (Feb 26 2021 at 23:42):

The last part is what I'm not seeing

Todd Trimble (Feb 26 2021 at 23:53):

That's not actually what I said: I was working with algebra isomorphisms. I quoted something called the Chinese remainder theorem: if $R$ is a ring and if $a, b \in R$ are relatively prime elements, then the evident ring homomorphism $R/(ab) \to R/(a) \times R/(b)$ is an isomorphism. The idea of proof is that from $ua + vb = 1$, then $ua$ and $vb$ are idempotents modulo $(ab)$, and these map to the idempotents $(0, 1) \in R/a \times R/b$ and $(1, 0) \in R/a \times R/b$ respectively. From there, it's easy to check that $R/ab \to R/a \times R/b$ is surjective, and that the kernel is trivial.

Now apply this idea plus induction to the factorization $x^n - 1 = \prod_{i=1}^n (x - \zeta^i)$ where $\zeta$ is a primitive root. The factors $x-\zeta^i$ are mutually relative prime. This gives an algebra isomorphism

$\mathbb{C}[x]/(x^n-1) \cong \prod_{i=1}^n \mathbb{C}[x]/(x-\zeta^i)$

and each of the factors on the right is isomorphic to $\mathbb{C}$.

John Baez (Feb 27 2021 at 00:01):

This is slick. There's another way to think about this that doesn't require so much knowledge of algebra. You can explicitly find n idempotents in the group algebra of $\mathbb{Z}/n$, say $p_j$, with $p_j p_k = 0$ when $j \ne k$. This instantly implies the group algebra is isomorphic to the direct sum of n copies of $\mathbb{C}$.

Todd Trimble (Feb 27 2021 at 00:02):

Gee, isn't that pretty much what I said?

John Baez (Feb 27 2021 at 00:03):

All roads lead to Rome and many are homotopic.

Todd Trimble (Feb 27 2021 at 00:03):

Well, we're both using idempotents in more or less identical ways. Anyhoo...

John Baez (Feb 27 2021 at 00:04):

But my road doesn't require knowing about ideals, the Chinese remainder theorem, etc.

Todd Trimble (Feb 27 2021 at 00:04):

Just putting things into broad context, then.

John Baez (Feb 27 2021 at 01:22):

Yes! The advantage of the proof I sketched is that 1) you barely need anything more than the definition of algebra and isomorphism to check that it works, and it 2) gives you an explicit algebra isomorphism

$\mathbb{C}[\mathbb{Z}/n] \cong \bigoplus_{j \in \mathbb{Z}/n} \; \mathbb{C}$

The big disadvantage is that 1) you have to guess the formula for the idempotents $p_j$, or have someone hand it to you (which you'll notice I didn't do) and 2) then you have to check that they're idempotents and obey $p_j p_k = 0$ when $j \ne k$. All this can look very mysterious if one doesn't have any context.

And your proof provides that context.

John Baez (Feb 27 2021 at 01:24):

I guess I'd call your proof a "ring theory" proof or "commutative algebra" proof.

Todd Trimble (Feb 27 2021 at 01:25):

Yeah, I don't think we disagree. One can follow the ideas I was indicating to get explicit idempotents (although I don't really feel like doing that at the moment).

Fawzi Hreiki (Feb 27 2021 at 01:26):

I guess you could kill a fly with a canon and invoke the representation theory stuff

John Baez (Feb 27 2021 at 01:27):

I probably find the "Fourier transform" proof closest to my heart: here we use the fact that the group algebra of a finite abelian group $A$ is isomorphic to the algebra of functions on the dual abelian group $A^\ast$, together with the fact that $(\mathbb{Z}/n)^\ast \cong \mathbb{Z}$. Here the explicit isomorphism

$\mathbb{C}[\mathbb{Z}/n] \cong \bigoplus_{j \in \mathbb{Z}/n} \; \mathbb{C}$

turns out to be the Fourier transform (or "Pontryagin duality" if one wants to show off).

John Baez (Feb 27 2021 at 01:30):

But I also like the "group representation theory" proof, which is really the "Fourier transform" proof in a more general context. If you're gonna study Frobenius algebras, you'd better get to like group algebras $\mathbb{C}[G]$ of general finite groups $G$, not just abelian ones. So soon you'll know that the minimal central idempotents in $\mathbb{C}[G]$ correspond to (isomorphism classes) of irreducible representations of $G$. And for $G = A$ abelian, these irreducible representations are just the elements of $A^\ast$.

John Baez (Feb 27 2021 at 01:31):

So, $\mathbb{C}[\mathbb{Z}/n]$ has minimal central idempotents coming from elements of $(\mathbb{Z}/n)^\ast \cong \mathbb{Z}/n$, and these are the guys $p_j$ that I was alluding to.

John Baez (Feb 27 2021 at 01:32):

There are probably even more ways to tackle this, since group algebras of finite abelian groups are such fundamental things.

John Baez (Feb 27 2021 at 01:33):

But I'd never even seen the "ring theory proof".

Todd Trimble (Feb 27 2021 at 01:36):

The Chinese remainder theorem is actually quite a marvelous thing. It's at the heart of partial fraction decompositions, Lagrange interpolation, and much else.

John Baez (Feb 27 2021 at 01:40):

Somehow I've never fallen in love with it. My favorite proofs, the "Fourier analysis" proof and "group representation theory" proof, display my residual love affair with analysis... I think of them as tiny baby versions of harmonic analysis on compact Lie groups, or compact Hausdorff topological groups.

This is one reason I panicked for a second when Joe suggested looking at Frobenius algebras over $\mathbb{Q}$. How do you do Fourier transforms over a field with practically no roots of unity? :fear:

Todd Trimble (Feb 27 2021 at 01:45):

Well, yes. It gets a bit more Galois-theoretic, or something.

Todd Trimble (Feb 27 2021 at 01:50):

Meaning for example that in the case of e.g. $\mathbb{Q}[\mathbb{Z}/n]$, you factor $x^n-1$ into irreducible polynomials $p(x)$, and then $\mathbb{Q}[x]/(p(x))$ gets you into number field extensions.

Ethan Kowalenko (Feb 27 2021 at 19:02):

Todd Trimble said:

That's not actually what I said: I was working with algebra isomorphisms. I quoted something called the Chinese remainder theorem: if $R$ is a ring and if $a, b \in R$ are relatively prime elements, then the evident ring homomorphism $R/(ab) \to R/(a) \times R/(b)$ is an isomorphism. The idea of proof is that from $ua + vb = 1$, then $ua$ and $vb$ are idempotents modulo $(ab)$, and these map to the idempotents $(0, 1) \in R/a \times R/b$ and $(1, 0) \in R/a \times R/b$ respectively. From there, it's easy to check that $R/ab \to R/a \times R/b$ is surjective, and that the kernel is trivial.

Now apply this idea plus induction to the factorization $x^n - 1 = \prod_{i=1}^n (x - \zeta^i)$ where $\zeta$ is a primitive root. The factors $x-\zeta^i$ are mutually relative prime. This gives an algebra isomorphism

$\mathbb{C}[x]/(x^n-1) \cong \prod_{i=1}^n \mathbb{C}[x]/(x-\zeta^i)$

and each of the factors on the right is isomorphic to $\mathbb{C}$.

Ah okay, I think Joe actually said the right argument but I must have spaced on what the remainder theorem actually said.... This is great!

Todd Trimble (Feb 27 2021 at 19:33):

I guess then I should have read Joe's "Well then sure, I'll claim that [there isn't such an isomorphism]" (received at 4:10 PM on my clock) merely as a flippant response to John and not necessarily as calling into question my explanation to him before.

John Baez (Feb 27 2021 at 20:12):

I sort of bullied him into claiming it because he seemed to be saying it, and I find myself much more motivated to prove something if someone just claimed the opposite. James Dolan and I used to work this way: we called it our "good cop, bad cop routine".

In fact that paper he cited just claimed

$\mathbb{Q}[\mathbb{Z}/4] \ncong \mathbb{Q}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

as algebras, not

$\mathbb{C}[\mathbb{Z}/4] \ncong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

which is false.

I still haven't gotten around to thinking about the original question, which is whether

$\mathbb{C}[\mathbb{Z}/4] \cong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

as Frobenius algebras, using the natural Frobenius structure on the group algebra of a finite group.

In fact each of these two algebras has a 4-parameter family of Frobenius algebra structures, since $\mathbb{C}$ has a 1-parameter family of Frobenius algebra structures. So, you can certainly find ways to choose isomorphic Frobenius algebra structures for these two algebras... but Joe's question was presumably about the "natural" Frobenius algebra structures.

By the way, unless I'm really confused, I believe we have

$\mathbb{C}[\mathbb{Z}/4] \ncong \mathbb{C}[\mathbb{Z}/2 \oplus \mathbb{Z}/2]$

as Hopf algebras, using the natural Hopf algebra structure on a group algebra. The Hopf algebra should know everything about the group.

Todd Trimble (Feb 27 2021 at 20:26):

Well, that paper made a correct claim because the right side is (as an algebra) a direct product of four copies of $\mathbb{Q}$, whereas the left side is a product of the form $\mathbb{Q}[i] \times \mathbb{Q} \times \mathbb{Q}$.

I haven't thought about the Frobenius algebra stuff yet. Maybe I'll have a chance this evening.

Finally, you're not really confused. :-)

Jalex Stark (Apr 09 2021 at 22:11):

I'm not sure exactly what a hopf algebra is but I think I've heard they only know the character table of the group?

John Baez (Apr 10 2021 at 02:14):

You can recover a finite group from its Hopf algebra by looking at the "grouplike elements" of the Hopf algebra: the elements $x$ such that $\Delta(x) = x \otimes x$, where $\Delta$ is the "comultiplication" of the Hopf algebra.

John Baez (Apr 10 2021 at 02:15):

Then the multiplication of these grouplike elements in the Hopf algebra is exactly the multiplication in the group!

John Baez (Apr 10 2021 at 02:16):

I should be a bit careful about infinite groups, but it seems to me that nothing I said requires the group to be finite, either: any group $G$ gives a Hopf algebra $k[G]$ whose elements are finite $k$-linear combinations of elements of $G$. (In other words, the free vector space on $G$ becomes a Hopf algebra.)