Posted by: yanzhang | November 26, 2010

Being Unthankful for Unenlightening Proofs

I’m actually getting a lot of research done these days to post much, but in case you guys are still dropping by, happy holidays. =)

Some personal good news: we finally got that epic (in the “epically long” sense and not the “epically awesome” sense, but I’m still happy it is done) paper about counting invertible matrices over finite fields up on the ArXiv: . However, despite the season, I’m not in the most cheery mood about the paper, because something I wanted to qualitatively understand remains at large:

Take some finite field, and some even n. Consider the number of invertible n \times n skew-symmetric matrices and the number of invertible n-1 \times n-1 symmetric matrices. They equal. Furthermore, they both equal the number of invertible n \times n symmetric matrices with zeroes on the diagonal.

We proved this (it is Propostion 3.6 as of the latest version), but I don’t think we quite understand why (for either of the equivalences). I didn’t feel like the proof we used, a “twisty” bijection, showed us anything (we mention a “high-brow” way, but I don’t know if it is too helpful in its current form). I guess during the editing process I just wanted to get the damn thing finished, but now it is bugging me like all sorts of other things tend to do near holiday midnights. If anyone has ideas, please let me know. Extra points for using the words “Lie Superalgebras” because then I probably owe Steven Sam a beer.




  1. A naive comment: do you expect that the corresponding schemes have the same etale cohomology? Can you compute their singular cohomology over C or something?

  2. I guess this may be the kind of thing I’m looking for. In general I’m distrustful of high-brow methods because I think they tend to hide what is really going on (I define “what is really going on” as the simplest level of abstraction needed to explain the phenomenon at hand), but this may be exactly what I need. The problem is I don’t have the functional understanding to begin such a calculation (only vague knowledge). If you want to teach me how I may start the setup, that would be helpful.

  3. I wouldn’t know where to start, either; I just know that the method exists. I don’t necessarily think it hides “what is really going on”: there is a great example of how it works at this blog post by Jordan Ellenberg.

  4. I guess I should mention something else: to show that the singular cohomology is the same it suffices to show that they are homotopic, and one might be able to do this explicitly and/or in a conceptual way.

  5. Yeah, it would be really nice to see something like this. Thanks for the link, I’ll take a look (though I doubt my computational skills would be useful in me convincing myself, I can try to learn)

    As for your second point – sure, homotopy equivalence is definitely stronger than what I would need. I just feel dumb because my algebraic topology is very non-functional – if you gave me an argument I would be able to follow, but I’d feel rusty actually computing the cohomology groups (and I doubt I’d have any intuition of the homotopy groups).

  6. This paper calculates the (compact) Betti numbers of the complex manifolds of invertible symmetric and invertible skew-symmetric matrices (see Theorem 1.7 and Theorem 1.7′):

    They agree when for nxn skew-symmetric and n-1 x n-1 symmetric when n is even, as expected. My limited understanding of this kind of stuff says that the Grothendieck-Lefschetz trace formula plus étale cohomology versus singular cohomology comparison theorems turn those theorems into the theorems we have about finite field point counting. He mentions that this machine doesn’t apply since these varieties are not projective, but I don’t think that’s necessary? Even if doesn’t apply, he gives alternate proofs (Theorem 1.9 and Theorem 1.8′).

  7. Very nice theorem!

    Here’s how I would describe what’s going on. Following the notations of Jones’s paper above, write Sk(n) for the variety of nxn invertible skew-symmetric matrices, and Sym(n-1) for the variety of (n-1)x(n-1) invertible symmetric matrices. Let’s denote by qDer(n) the variety of nxn symmetric matrices with zero diagonal.

    So you show that all these guys have the same number of F_q-points, and Jones shows that the first two have the same number of F_q-points. I would describe Jones’s argument as “motivic” in the sense that he establishes that Sk(n) and Sym(n-1) represent the same class in the Grothendieck ring K_0(Var). It seems quite possible that you guys have proved that qDer(n) represents the same class as well. These motivic identities are quite often proved by “twisty bijections” and I see no fault in this.

    Jones also shows that the Betti numbers of Sk(n) agree with those of Sym(n-1). This _suggests_ an identity on point-counts but does not immediately imply it, because, as Jones points out, the varieties in question are not projective. This means that it’s not obvious that the etale Betti numbers of Sk(n)/F_p (resp. Sym(n-1), qDer(n)) agree with the topological Betti numbers of the corresponding complex algebraic variety. I mean, they probably do! But something would need to be proved.

    Certainly your theorem suggests that qDer(n) might have the same cohomology as Sk(n), and it would be interesting to prove this. (Note that if your argument can indeed be made motivic it would already show that, e.g., qDer(n) and Sym(n-1) and Sk(n) all had the same topological Euler characteristic.)

  8. @JSE and @qchu:

    Thanks a lot for your suggestions. I can’t pretend to follow everything you say, but I’ve printed out the said paper and a few references and will try my best to understand.

    Much appreciated.

  9. I think no counting arguments are necessary. Only direct exhibition of correspondence.
    Start with upper triangular matrices of size n xn. with 0 diagonal.
    These are nilpotent of degree at most n-1. So consider (I – N ) it has an explicit inverse
    ( I + N + N^2 +… + N^(n-1) ).
    The matrix { I + N ) – ( 1 – N )Transpose} = S is explicitly symmetric with 0 diagonal.
    It is invertible by expanding (I-N) { S } (I-Ntran)inverse.
    get left multiply first = { I – (I-N)inv (I-N)tran)} (i-Ntran)inv =
    (I-Ntran)inv – (I-N)inv [cancel{(i-N)tran) * (i-Ntran)inv } = I ].

    This should complete the skew to symmetric 0 diag correspondence.

    I suspect the n-1 x n-1 symm invertible correspondence is the projection of the
    symmetric 0 diag invertible space onto the space of matricies of form
    ev 0
    0 n-1 symm invertible by an orthognal rotation so symmetric is preserved.
    I think this is just the first step in Gram Schmidt diagonalization.
    I guess the question becomes here the issue of wheather 2 0 diag can land on the same
    n-1 symm for maybe different ev1 and ev2. I am too lazy to argue that unique inverses
    will prevent this even if different rotations are used in the 2 reductions.
    I suppose that maybe here the finite count of the matrices COULD enter but I doubt it.

    If this is nonsense, I apologize but way back then I did try to work in the finite groups of lie type.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: