## This paper was written for our blog August 21, 2008

Posted by David Speyer in Algebraic Geometry, mathematical physics, Number theory, Paper Advertisement, papers, representation theory.

I’ve recently been reading a paper which ties together a number of this blog’s themes: Canonical Quantization of Symplectic Vector Spaces over Finite Fields by Gurevich and Hadani. I’m going to try to write an introduction to this paper, in order to motivate you all to look at it. It really has something for everyone: symplectic vector spaces, analogies to physics, Fourier transforms, representation theory of finite groups, gauss sums, perverse sheaves and, yes, $\theta$ functions. In a later paper, together with Roger Howe, the authors use these methods to prove the law of quadratic reciprocity and to compute the sign of the Gauss sum. For the experts, Gurevich and Hadani’s result can be summarized as follows: they provide a conceptual explanation of why there is no analgoue of the metaplectic group over a finite field. Not an expert? Keep reading!

Let’s start with some simple “physics”. Let $H$ be the vector space of complex-valued functions on the real line. (We probably want to impose some conditions on these functions, but this is (1) a blog post (2) “physics” and (3) only for motivation. I can think of no better time to ignore analytic issues.) There are two important operators on $H$: the operator $x$, which sends $f(x)$ to $x f(x)$ and the operator $D$ which sends $f(x)$ to $i \partial f/ \partial x$. They obey the relation $D \circ x - x \circ D = i \mathrm{Id}$. Basically all of the math in an introductory quantum mechanics course is playing with these operators.

For our purposes, we would rather exponentiate these operators. The exponential $e^{i a x}$ is the operator which takes $f(x)$ to $e^{i a x} f(x)$ and the exponential $e^{i b D}$ is the operator which takes $f(x)$ to $f(x-b)$. (If you haven’t seen this before, it is worth thinking about.) These exponentiated operators obey the relation

$e^{i b D} \circ e^{i a x} = e^{i ab} \cdot e^{i a x} \circ e^{i b D}.$

It is nice to build an abstract group to capture the formal properties of the above relation. The Heisenberg group is the group of formal symbols

$z e^{i a x + i b D}$

where $z$ is a complex number of norm ${1}$ and $a$ and $b$ are real numbers. Multiplication is determined by the following relations:

For any $z$ in $S^1$, the element $z$ is central.

$e^{i a x} e^{i a' x} = e^{i (a + a') x}$

$e^{i b D} e^{i b' D} = e^{i (b + b') D}$

$e^{i a x} e^{i b D} = e^{(1/2) i a b} e^{i a x+ i b D}$

$e^{i b D} e^{i a x} = e^{-(1/2) i a b} e^{i a x+ i b D}$.

So $H$ is a representation of the Heisenberg group. The Stone-von Neumann Theorem roughly states that $H$ is, up to isomorphism, the only irreducible representation of the Heisenberg group where the symbol $z$ acts by multiplication by the scalar $z$.

Let $V$ be the vector space of formal linear combinations $a x + b D$, with the skew symmetric bilinear form $\omega(a_1 x + b_1 D, a_2 x + b_2 D) = a_1 b_2 - a_2 b_1$. Then we can more concisely define the Heisenberg group by saying that it is the group of symbols $z e^{i v}$, with $z \in S^1$ and $v \in V$, given the relations that $z$ is central and

$e^{i v_1} e^{i v_2} = e^{(i/2) \omega(v_1, v_2)} e^{i (v_1+v_2)}$.

Mathematical physicists like to work with $V$, without choosing a splitting of $V$ into transverse subspaces $\mathbb{R} x$ and $\mathbb{R} D$. Note that the definition of the Heisenberg group respects that aesthetic, but the definition of $H$ does not. For future reference, we set $L_1 = \mathbb{R} x$ and $L_2 = \mathbb{R} D$.

An interruption: all of this works equally well for functions on a vector space of dimension larger than ${1}$. I find that right about now is the transition point where that added generality becomes illuminating, rather than confusing. You might want to go back and rework everything for functions on a $g$ dimensional space. Note that, if you do things correctly, $L_1$ and $L_2$ will naturally be dual vector spaces. I’ll keep writing in the one dimensional case, then switch to the general case when I talk about what is new in Gurevich and Hadani.

Here is a definition of the Heisenberg group which uses the subspace $L_2$ but not the subspace $L_1$. Let $S^1 e^{i L_2}$ denote the abelian subgroup of the Heisenberg group consisting of expressions $z e^{i \ell}$ where $\ell \in L_2$. The quotient of the Heisenberg group by $S^1 e^{i L_2}$ is canonically isomorphic to $L_1$. Rather than realize $H$ as functions on $L_1$, we will realize it as functions on the quotient of the Heisenberg group by $S^1 e^{i L_2}$. The language of induced representations lets us do this in a very concise way: $H$ is (canonically isomorphic to) the induction from $S^1 e^{i L_2}$ of the obvious one dimensional representation of $z e^{i \ell}$. By “the obvious one dimensional representation”, I mean $z e^{i \ell} \mapsto z$.

For any one dimensional subspace $L$ of $V$, we can define the group $S^1 e^{i L}$. Let $H(L)$ be the induction to the Heisenberg group of the character $z e^{i \ell} \mapsto z$ of $S^1 e^{i L}$. (If you are doing the $g$-dimensional generalization, then $L$ should be $g$ dimensional and $\omega|_L$ should be zero, i.e., $L$ should be Lagrangian.) All of the representations $H(L)$ are isomorphic, but not canonically so.

To get some intuition, let’s look at $H(L_2)$ and $H(L_1)$. To find an isomorphism, we must find a way to translate functions on $L_1$ to functions on the dual space $L_2$. The Fourier transform,

$\hat{f}(y) = C \int_{x \in M_1} e^{i \omega(y,x)} f(x) \ dx$

is such an isomorphism, and the standard fact that Fourier transform switches $x$ and $D$ turns into the fact that this is a map of representations. That $C$ is the constant I always forget, you know, the one that is something like $1/\sqrt{2 \pi}$. The issue of this constant is actually very important for us.

For almost any two subspaces $M_1$ and $M_2$, we can similarly define a map $H(M_2)$ to $H(M_1)$ by

$\hat{f}(y) = C(M_1, M_2) \int_{x \in M_1} e^{i \omega(y,x)} f(x) dx$

where we use the canonical identifications between $H(M_i)$ and functions on $M_{3-i}$. (Why “almost any”? Stay tuned!) We hit a technical point: what do I mean by $dx$? I need to choose a volume form on $M_1$ in order to take the integral. So, properly, we should work not with subspaces $M_1$ and $M_2$ but with subspaces equipped with a nondegenerate volume form.

But now something strange happens! Even once the issue about volume forms is straightened out, there is no compatible continuous way to choose the constants $C(M_1, M_2)$. The space of one dimensional spaces of $V$, equipped with volume forms, is a Mobius strip. Requiring that form to be nondegenerate cuts the strip down the center, leaving a cylinder. If you travel around that cylinder and come back to where you started, you’ll find you’ve picked up a sign error!

I’m sure there is a lot to say here about connections, topology and so forth. But I want to explain what happens in Gurevich and Hadani’s work.

Instead of working with $\mathbb{R}$, they work over the finite field $\mathbb{F} := \mathbb{Z}/p$ for $p$ an odd prime. Fix a nontrivial $p^{\textrm{th}}$ root of unity $\zeta$. Let $V$ be a $\mathbb{F}$-vector space of dimension $2g$, equipped with a perfect skew symmetric form $\omega$. Define the discrete Heinsenberg group to be the group of formal symbols $\zeta^k e^{v}$, with $v \in V$, subject to the relations that $\zeta$ is central and

$e^{v_1} e^{v_1} = \zeta^{(1/2) \omega(v_1, v_2)} e^{v_1 + v_2}$.

For any Lagrangian subspace $L$ of $V$, we can define $H(L)$ as before. If $L'$ is any lagrangian subspace transverse to $L$, we can identify $H(L)$ with complex-valued functions on $L'$. If $M_1$ and $M_2$ are transverse, there is an isomorphism $H(M_2) \to H(M_1)$ by

$\hat{f}(y) = C(M_1, M_2) \sum_{x \in M_1} \zeta^{\omega(y,x)} f(x).$

(If $M_1$ and $M_2$ are not transverse, we can’t identify $H(M_1)$ with $\mathbb{C}^{M_2}$. This is why I said “almost any” two subspaces above. We’ll talk more about this issue later.)

Now, once again, we need to think carefully about volume forms. (And this time, it is less obvious why, because there is a standard counting measure on $M_1$.) However, once that issue is straightened out, a miracle happens: this time, we can choose the constants $C(M_1, M_2)$ so that everything is compatible. In particular, the group $Sp_{2g}(\mathbb{F})$ of automorphisms of $V$ preserving $\omega$ acts on $H$, even though the analgous statement for $Sp_{2g}(\mathbb{R})$ is false.

My understanding is that the above was known before Gurevich and Hadani, but their explanation, and in particular an explicit recipe for choosing the constants $C(M_1, M_2)$, was not. One of the innovations of their paper is that, using perverse sheaves, they extend the Fourier transform to the case where $M_1$ and $M_2$ are not transverse. Also, while I do not follow their proof yet, it seems to involve a number of identities between Gauss sums, so the number theorists should like this too.

A few questions that came to mind reading their paper: (1) Does this result say that the space of lagrangian subspaces of $\mathbb{F}_p^{2g}$ is simply connected? (A statement which is false for $\mathbb{R}^{2g}$.) If not, is there some similar statement about characteristic $p$ topology which is true and give a conceptual explanation for this?

(2) Gurevich and Hadani work very hard to handle the nontransverse case. Is this necessary? Explicitly, consider this simplicial complex whose vertices are lagrangian subspaces equipped with a non-degnerate volume form and whose faces are pairwise transverse collections of lagrangians. Is this complex connected and simply connected? If so, then we could give an alternate definition of $H(M_2) \to H(M_1)$ by composing the definition for the transverse case. Once we checked that going around a triangle in both ways gave the same result, we would know that our map was well defined. This certainly works when $g=1$!

1. Scott Carnahan - August 21, 2008

Shamgar has been a postdoc at Berkeley for the past couple years, and gave several talks about his work with Hadani. It was apparently inspired by a 1982 letter from Deligne to Kazhdan that did not receive much attention for a while.

There is some similar work by Lysenko (not the “biologist”) and V. Lafforgue, related to geometric Langlands and sheafy theta correspondence.

2. Scott Carnahan - August 21, 2008

For question number one, there is a rather unenlightening answer. Symplectic groups over finite fields don’t have nontrivial central extensions, except for some small exceptions (I’m afraid I’ve never seen a proof). The space of Lagrangians is a homogeneous space with connected stabilizer (I think), making the quotient simply connected.

This is notably false for nonarchimedean local fields. Our own Cal Moore did some work on central extensions of these groups in the middle of the last century, and there is exactly one nontrivial isomorphism type in each case.

3. Kea - August 21, 2008

Wow, thanks! So cool to learn about Deligne’s ideas for the quantum (or discrete if you like) Fourier transform and its ‘phase space’ or ‘quantum torus’ (for the physicists who don’t like the word symplectic). You know, such Hilbert spaces are well studied in quantum computation and the field known as Quantum Foundations, but I haven’t come across such a mathematical connection before. Very nice, and very close to some interesting physics.

4. Kea - August 21, 2008

…and I suspect that number theorists will be more intrigued by their remark that the proof will be close in spirit to the procedure of analytic continuation … coupled with the name of Weil.

5. Andy P. - August 24, 2008

Scott — The result you are quoting is due to Steinberg, and is contained in his paper “Générateurs, relations et revêtements de groupes algébriques”. It is also (I think) contained in his “Lectures on Chevalley groups”.

6. Ben Wieland - August 24, 2008

The space of Lagrangians is a homogeneous space with connected stabilizer

That is a hard statement to make sense of for finite fields. The usual way of making sense of it really involves passing to algebraically closed fields, but then the argument would apply just as well to R.

7. Emmanuel Kowalski - August 24, 2008

One has to be careful with things like “simply connected” for algebraic groups: the notion bearing this name in the category of algebraic groups is not the same as the corresponding one when other natural topologies are involved (e.g., SL(n) is simply connected as an algebraic group, but SL(2,R) is not with the usual topology — one reason the metaplectic groups, which are not algebraic groups, come up).

(In a probably different direction, having to do with Comment 6, it may be interesting to note that Steinberg, among others, discusses the philosophy that, over finite fields, a notion analogue to “connectedness” is “generated by unipotents” — I don’t remember the precise reference however).

8. Scott Carnahan - August 25, 2008

I’m not sure what I was thinking when I wrote the bit about “simply connected,” but it only seems to bear a superficial resemblance to actual mathematics. If I were pressed, I’d probably wave my hands and mumble something about spherical buildings, but I don’t know if they’re really useful here.

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