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Stream: deprecated: mathematics

Topic: Dieudonné determinant

John Baez (Feb 10 2021 at 06:15):

I'm teaching a class on Lie groups where as a small side-topic I mention the determinant of a quaternionic matrix - needed to describe the group $\mathrm{SL}(n,\mathbb{H})$. This topic is a bit of a rabbit-hole, so I tried to explain it efficiently on the nLab:

• Dieudonné determinant.

By the way, if you're going to study god-given mathematical structures it's nice to be named Dieudonné.

John van de Wetering (Feb 10 2021 at 11:43):

I'm a bit confused as to why the determinant should be real-valued. What does this determinant correspond to if you restrict to the subset of complex (invertible) matrices? It can't be the regular determinant can it?

Oscar Cunningham (Feb 10 2021 at 12:35):

John Baez said:

Dieudonné showed that for $K$ a division ring, there is a group isomorphism of the form

$\alpha: GL(n,K)/[GL(n,K), GL(n,K)]\rightarrow K^\times/[K^\times, K^\times]$

where

• $GL(n,K)$ is the group of $n \times n$ invertible matrices with entries in $K$,

• $K^\times = GL(1,K)$ is the group of units of $K$,

• and $G/[G,G]$ is the abelianization of the group $G$.

This isn't quite true. When $K=\mathbb{F}_2$ and $n=2$ we get $GL(n,K)/[GL(n,K), GL(n,K)] = \mathbb{Z}/2\mathbb{Z}$, but $K^\times/[K^\times, K^\times] = 1$. Which is a great shame, since otherwise this would be a lovely way of motivating the definition of the determinant.

John Baez (Feb 10 2021 at 17:15):

Oh, interesting! Thanks! I guess $\alpha$ is just a homomorphism. I think I read somewhere that it was an isomorphism. Now where was that? --- or did I just imagine it?

Todd Trimble (Feb 10 2021 at 17:16):

Well, if that's the only case, maybe other hypotheses ruled it out?

Oscar Cunningham (Feb 10 2021 at 17:21):

Right, I think $\mathbb{F}_2^2$ is the only case (although I can't find a reference for this). It's an exceptional object!
So if you made the statement refer to strictly noncommutative division algebras then it would be correct.

John Baez (Feb 10 2021 at 17:22):

By the way, how did you compute $GL(n,K)/[GL(n,K),GL(n,K)]$ in your example, or at least show it's nontrivial? One nice way might be to figure out a way to define an "enhanced determinant" in this example, taking values in $\mathbb{Z}/2\mathbb{Z}$, and actually taking both values.

John Baez (Feb 10 2021 at 17:24):

Right, I think $\mathbb{F}_2^2$ is the only case (although I can't find a reference for this). It's an exceptional object!
So if you made the statement refer to strictly noncommutative division algebras then it would be correct.

Wow! So maybe I read somewhere that it's an isomorphism for skew fields, and someone meant skew fields that are truly skew.

But now I'm really wanting a reference on this amazing claim that $K = \mathbb{F}$, $n = 2$ is the only counterexample!

John Baez (Feb 10 2021 at 17:38):

John van de Wetering said:

I'm a bit confused as to why the determinant should be real-valued.

There's no quaternion-valued determinant that obeys the usual properties one wants from a determinant.

You can define a complex-valued determinant by arbitrarily choosing a copy of the complex numbers inside the quaternions: this is called the Study determinant, and I explained it here. It turns out to be independent of which copy of $\mathbb{C}$ you choose sitting inside $\mathbb{H}$... but it also turns out to be real-valued!

John Baez (Feb 10 2021 at 17:41):

What does this determinant correspond to if you restrict to the subset of complex (invertible) matrices? It can't be the regular determinant can it?

No, it's not. But notice that there is not just one copy of $\mathbb{C}$ sitting inside $\mathbb{H}$. So your demand would be quite strong if we wanted it to work for every copy of $\mathbb{C}$ inside $\mathbb{H}$, and strangely arbitrary if we wanted it to work for just one.

John Baez (Feb 10 2021 at 17:57):

But these are "philosophical" objections; I'm pretty sure that there's just no complex-valued determinant of a quaternionic square matrix that has lots of the properties of determinants one wants. This is the best tour of the subject that I know:

• Helmer Aslaksen, Quaternionic determinants.

Oscar Cunningham (Feb 10 2021 at 18:25):

John Baez said:

By the way, how did you compute $GL(n,K)/[GL(n,K),GL(n,K)]$ in your example, or at least show it's nontrivial? One nice way might be to figure out a way to define an "enhanced determinant" in this example, taking values in $\mathbb{Z}/2\mathbb{Z}$, and actually taking both values.

I used a boring bare-hands proof. If a $2\times 2$ $\mathbb{F}_2$-matrix is invertible it can't have zeros on both diagonals, since then it would have determinant $0$ by the usual formula. And it can't be the all ones matrix because this isn't invertible either. This only leaves six elements, and you can quickly check that all of them are invertible but not order 6, so we must be looking at $S_3$. Then it's well known that the abelianization is $\mathbb{Z}/2\mathbb{Z}$.

John Baez said:

But now I'm really wanting a reference on this amazing claim that $K = \mathbb{F}$, $n = 2$ is the only counterexample!

The results in this short paper https://www.ams.org/journals/proc/1955-006-03/S0002-9939-1955-0068541-X/S0002-9939-1955-0068541-X.pdf are enough.

Oscar Cunningham (Feb 10 2021 at 18:26):

It occurs to me that we have no real reason to be having this discussion about groups. The number $0$ is a perfectly valid determinant. We could abelianize the monoids $\mathrm{End}(V)$ just as easily.

John Baez (Feb 10 2021 at 18:32):

Thanks for the reference and the hands-on proof!

In the nLab stuff I wrote, I lead up to treating the Dieudonné determinant as a monoid homomorphism, but I start with what seems to be the standard approach using groups.

John Baez (Feb 10 2021 at 18:33):

It might ultimately be better to use monoids all along, but I don't have time now to rework the theory.

John Baez (Feb 10 2021 at 18:42):

@Oscar Cunningham - Could you say a bit more about how that short paper does the job? I'm slower at algebra than you. Theorem 7 of that short paper says the commutator subgroup of $GL(n,K)$ is the group generated by transvections (adding a multiple of one row to another) except when where $n = 2$ and $K$ is any ring where 2 is not invertible. But I don't see how that does the job, for two reasons:

1) I don't see why that'd imply the commutator is $K^\times/[K^\times,K^\times]$

2) What about all the division rings (e.g. fields) other than $\mathbb{F}_2$ where 2 is not invertible?

Oscar Cunningham (Feb 10 2021 at 19:22):

I was only thinking of proving that the determinant was isomorphic to what you get when you quotient by the commutator subgroup. So it sufficed to show that the subgroup of determinant 1 was the same as the commutator subgroup. So this doesn't answer (1). I think your question (2) is answered by Theorem 3, which says that $R$ modulo its radical must be the field with 2 elements.

John Baez (Feb 10 2021 at 19:40):

1) I see. So, there are some questions left, which I sadly have no time to think about.

John Baez (Feb 10 2021 at 19:43):

2) In Theorem 3, $R$ is assumed to be a valuation ring, and they show the transvection subgroup of $GL(n,R)$ is contained in the commutator group unless $n = 2$ and 1+1 = 0 in $R$ . Does this somehow cover or imply the case where $R$ I'm sorry, I'm really bad at commutative algebra.

Oscar Cunningham (Feb 10 2021 at 19:46):

The only fact it uses is that there's some $\lambda$ such that both $\lambda$ and $\lambda - 1$ are units. This is fulfilled in any division ring that's not $\mathbb{F}_2$ by taking $\lambda$ to be anything not $0$ or $1$.

John Baez (Feb 11 2021 at 03:01):

Thanks!