User freddie manners - MathOverflow

Is there a "right" proof of Riemann's Theta Relation?

2013-04-12T22:14:11Z

Let $\theta$ denote the usual Jacobi Theta function (with auxiliary parameter $\tau = i$, for simplicity), i.e.

$$ \theta(z) = \sum_{n \in \mathbb{Z}} \exp(-\pi (a + n)^2 + 2 \pi i n z) \ . $$

I'm interested in Riemann's quartic theta relation, which is (barring mistakes):

$$ \theta(x_1) \theta(y_1) \theta(u_1) \theta(v_1) = \frac{1}{2} \sum_{\eta \in \lbrace 0, 1/2, i/2, (1+i)/2 \rbrace } c_\eta \theta(x + \eta) \theta(y + \eta) \theta(u+\eta) \theta(v+\eta) $$

where $c_\eta$ are some exponential factors, and $x_1$, $y_1$, $u_1$, $v_1$ are certain linear functions of the variables $x$, $y$, $u$, $v$ as follows:

$$ x_1 = (x + y + u + v) / 2 $$ $$ y_1 = (x + y - u - v) / 2 $$ $$ u_1 = (x - y + u - v) / 2 $$ $$ v_1 = (x - y - u + v) / 2 \ . $$ (This is taken from Mumford's "Tata lectures on theta I", chapter 1, section 5.)

There are many variations. They're all fairly straightforward to prove with bare hands and the above formula; that is precisely the line taken in Mumford's book and every other reference I've seen. However, I've been promised (somewhere) that every fact about special functions should have a "nice" interpretation coming from representation theory, and in particular from their interpretation as matrix coefficients (here, of the Heisenberg group). Is there a clean proof along those lines?

To put it another way, if I only told you some properties of $\theta$ and its variants -- being an eigenfunction of certain operators, periodicity conditions and so on -- but not its formula, is there an enlightening reason to expect something like the Riemann theta formula to hold?

As usual, apologies if I've missed this in a standard reference.

Answer by Freddie Manners for Joint (close to uniform) distribution in finite fields

2013-03-21T14:32:21Z

I think the below is correct; it hasn't been very thoroughly checked. One moral seems to be that this question is nicer in the $L^2$ distance than the $L^1$ distance (because the proof uses Fourier analysis); but I think I can deduce something like your $L^1$ result at the end.

Write $k$ for the cardinality of $\mathbb{F}$. Let $F : \mathbb{F}^2 \rightarrow \mathbb{R}$ be the law of $(A, B)$, minus the uniform distribution; i.e. $F(x, y) = \mathbb{P}(A = x, B = y) - 1/k^2$.

For $a, b \in \mathbb{F}$ define $f_{a b}(r) = \sum_{x, y \in \mathbb{F}} F(x, y) 1\lbrace{a x + b y = r\rbrace}$. Alternatively, this is equivalent to $f_{a b}(r) = \mathbb{P}(a A + b B = r) - 1/k$.

So, as I understand it your hypothesis is that $\|f_{a b}\|_1 \le \varepsilon$ for all $(a, b) \ne (0, 0)$, and you want to conclude a bound on $\|F\|_1$, where I use the counting measure on $\mathbb{F}$ and $\mathbb{F}^2$ to define the $L^1$ norm.

My approach is to apply Fourier analysis to $F$ and $f_{a b}$; so it will be convenient to immediately replace the $L^1$ estimate on $f_{a b}$ with the weaker estimate $\|f_{a b}\|_2 \le \varepsilon$ (as $\|\cdot\|_1 \ge \|\cdot\|_2$ wrt the counting measure).

Fix $\chi$ a non-trivial character of $\mathbb{F}$. Then $\chi_r(x) = \chi(r x)$ ranges over all the characters of $\mathbb{F}$ as $r$ ranges over $\mathbb{F}$, and so the Fourier transform of $f_{a b}$ is

$$\widehat{f_{a b}}(r) = \sum_{s \in \mathbb{F}} f_{a b}(s) \chi_r(-s) = \sum_{x, y} F(x, y) \chi_r(-(a x + b y))$$

The characters on $\mathbb{F}^2$ are $\chi_{u, v}(x,y) = \chi(u x + v y)$, so we deduce

$$\widehat{f_{a b}}(r) = \sum_{x, y} F(x, y) \chi_{a r, b r}(-(x, y)) = \widehat{F}(a r, b r) $$

By Parseval's identity, we get that $\frac{1}{k} \sum_r |\widehat{f_{a b}}(r)|^2 \le \varepsilon^2$ for every $(a, b) \ne (0, 0)$. We remark that $\widehat{F}(0, 0) = \sum_{x, y} F(x, y) = 0$. Summing over all possible $a, b$ and double-counting, we get that

$$ \sum_{a, b} \sum_r |\widehat{F}(a r, b r)|^2 = (k - 1) \sum_{u, v} |\widehat{F}(u, v)|^2 \le k^3 \varepsilon^2$$

as each non-zero $(u, v)$ is counted once for each non-zero $r$, and we can ignore the zero terms. By another application of Parseval,

$$ \sum_{x, y} |F(x, y)|^2 = \frac{1}{k^2} \sum_{u, v} |\widehat{F}(u, v)|^2 \le \frac{k \varepsilon^2}{k - 1} $$

So, $\|F\|_2 \le \varepsilon \sqrt{\frac{k}{k-1}}$ : this is the $L^2$ result. Applying Cauchy-Schwarz we get something like

$$ \|F\|_1 \le k \|F\|_2 \le k \varepsilon \sqrt{\frac{k}{k - 1}} $$

which (even after Kevin Costello's correction below) means this result is stronger than you asked for by a factor $\sqrt{k (k - 1)}$, meaning I'm still slightly suspicious of the proof.

Answer by Freddie Manners for Question about the Hardy-Littlewood method (quite basic)

2013-01-26T23:16:47Z

I'll split this into two questions:

Why can $f$ be large near rationals with small denominator?
Why is $f$ small away from these points (i.e. on the minor arcs)?

For (1), I should first clarify that $f$ isn't always large at these rationals; it could well be zero. But it is large some of the time.

Let's look at the example of $f(a/8)$ for $a \in \{1, 3, \dots, 7\}$ , when (say) $k=2$. We have:

\[f(a/8) = \sum_{n=1}^N e(a\, n^2 / 8) \]

Since $e(an/8)$ only detects the residue of $n \bmod{8}$, we have:

\[f(a/8) = \frac{N}{8} \sum_{n \in \mathbb{Z} / 8 \mathbb{Z}} e(a\, n^2 / 8) + O(1) \]

But $n^2$ is congruent to $0 \bmod{8}$ a quarter of the time, $4 \bmod{8}$ a quarter of the time and $1 \bmod{8}$ half the time, so:

\[ f(a/8) = N \left(\frac{1}{4} + \frac{1}{4} e(a/2) + \frac{1}{2} e(a/8) \right) + O(1) = \frac{N}{2} e(a/8) + O(1) \]

(recalling $a$ was odd). That's going to be pretty large (absolute value about $N/2$). The reason is essentially that $f(a/8)$ is detecting a huge bias in the squares, namely that many more than average are $1 \bmod 8$.

The same's going to happen for other denominators: only about half of all residues modulo $q$ get to be squares (for $q$ a prime power), and this bias causes large values of $f(a/q)$.

Another way of saying exactly the same is that $(n^2 / 8) \pmod{1}$ has a very non-uniform distribution, with a lot of the mass bunched at $1/8$.

This brings us to (2). The question is: might the squares have similar bias " $(\bmod{\sqrt{2}})$ "? I have no idea where such a bias might plausibly come from, and fortunately in any case the answer is no. That is, if you were to draw the set:

\[ \left\{ n^2 / \sqrt{2} \pmod{1}\,:\, n \in \{1 \dots N\} \right\} \]

as a subset of $[0, 1)$, you'd find they were smeared out all over the place. More formally, the proportion of these points contained in any interval $[a, b]$ is roughly $(b-a)$, and we say the set is equidistributed modulo 1. As $N$ becomes large, this equidistribution becomes quantitatively better.

Now, by Weyl's Equidistribution Criterion (many good references on the web), my previous statement is essentially equivalent to $f(a / b \sqrt{2})$ being small, whenever $a, b \in \mathbb{Z}$ are smallish. (Intuitively, you get a lot of cancellation in the definition of $f$.) In practice, it's usual to do the converse to this, i.e. deduce equidistribution from bounds on $f(a / b \sqrt{2})$; but I think equidistribution shows more intuitively what is going on in the minor arc bounds.

How do you prove it? Well, there's some work to be done at some point. Proving Weyl's Inequality gives quite a general bound of this form. Another (essentially equivalent) statement is Van der Corput's lemma (see e.g. the first half of Terry Tao's post on this):

If $a_n$ is a real sequence such that $(a_{n+h} - a_n)$ equidistributes $\bmod{1}$ for every non-zero $h \in \mathbb{Z}$, then $a_n$ equidistributes $\bmod{1}$.

In our case, $((n + h)^2 - n^2) / \sqrt{2} = (2 h n + h^2) / \sqrt{2}$ is reasonably easily shown to equidistribute $\bmod{1}$, for fixed $h$.

Hope some of that helps.

The space of lattices and modular forms of weight 1/2

2012-06-18T11:05:50Z

Suppose my favorite way of thinking about modular forms is as functions on the space of (real, 2D) lattices. One can identify this space with $SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R})$, i.e. bases for the lattice up to reparameterization.

A function $f : SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R}) \rightarrow \mathbb{C}$ is a modular form of weight $k$ if it satisfies the "scaling relation":

$f(a\ R_\theta\ T) = a^{-k} e^{- i k \theta} f(T)$

where $R_\theta$ denotes the appropriate rotation matrix, and $a$ a positive real.

[More precisely $f$ only has to be defined on one connected component of the space.]

(I think) the standard definition is equivalent to this, by considering the value of $f$ on canonical lattices $\langle 1,\ z \rangle$ for $z$ in the complex upper half-plane. Note it's entirely clear in this language that e.g. $G_4$ is a modular form of weight 4.

So, my question is: is it possible to make sense of modular forms of half-integral weight (for concreteness, say $\vartheta$) in a similar way?

I'm aware that a necessary step is to pass to some kind of double cover such as $Mp_2$, to make the scaling relation make sense for $k=1/2$; but I am having trouble making this sufficient. In particular, as sub-questions:

What group plays the role of $SL_2(\mathbb{Z})$?
While I could extend $\vartheta$ by "brute force" to a function on e.g. $Mp_2$ (by callously applying the scaling relation), is there a natural way to define $\vartheta$ on the larger space, similar to the "obvious" definition:
$G_4(\Lambda) = \sum_{w \in \Lambda \setminus 0} w^{-4}$
?

Apologies in advance if this is standard -- I've been unable to locate a satisfactory answer in the literature.

Thanks,
Freddie

Is it known that $(F_p^{\times} \ltimes F_p, F_p)$ is not a relative expander family?

2012-03-19T19:19:34Z

Shalom [edit: originally M. Burger] showed that the pair $(\mathrm{SL}_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$ has Relative Property (T) with respect to standard generating sets.

(The action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{Z}^2$ is the usual one, i.e. the semidirect product can be thought of as a group of affine transformations $x \mapsto A x + b$ where $A \in \mathrm{SL}_2(\mathbb{Z})$ and $b \in \mathbb{Z}^2$.

If we reduce $\mathrm{mod}\ p$, we can think of this as giving an "efficient" way of generating the translations $x \mapsto x + b$ for $b \in F_p^2$.)

A one-dimensional variant in the finite case is whether there exist bounded size subsets $S_p \subset F_p^{\times} \ltimes F_p$ and $\delta > 0$ such that the relative Kazhdan constant:

$\kappa\ (F_p^{\times} \ltimes F_p, F_p, S_p) \ge \delta$

i.e. whether the pairs $(F_p^{\times} \ltimes F_p, F_p)$ can form a relative expander family.

An equivalent formulation: do there exist bounded size sets $S_p$ of affine transformations on $F_p$, such that no non-empty subset $U \subset F_p$, $|U| \leq p/2$ is almost invariant with respect to all of them, i.e.

$\neg \exists U: \forall s \in S: |s(U) \cap U| > \frac{99}{100} |U|$

I believe the answer is no if one uses standard "generating" sets (they needn't actually generate) such as $x \mapsto x + 1,\ x \mapsto ax$, even if $a$ is allowed to vary with $p$. This is very slightly surprising, as these do generate all translations "efficiently" in the weaker sense of logarithmic diameter.

Is there a good argument as to why this should fail in general? Or might there be cunning sets $S_p$ such that relative expansion occurs?

Comment by Freddie Manners

2013-04-13T11:05:12Z

@J. Martel: I have to confess I started to get a bit out of my algebraic depth with the later (adelic) stuff in Tata III, so haven't yet managed to get a working understanding of Thm 7.4, but will persevere. Possibly I should just try to figure out what that result is saying in the Euclidean case. @Jeff: That would be great! @Everyone: the equation for $x_1$ in Mumford looks like it must be a typo; corrected in the above.

Comment by Freddie Manners

2013-04-09T17:07:13Z

So, the strategy when A = H is a subgroup generalizes to A being your favorite set with $|A A^{-1}| = O(|A|)$, by a similar packing argument (if you accept a constant factor at the end). The aim is then to find such a set with $|A| \approx \sqrt{N}$. This gives rise to the cyclic example above, and should do what Colin Reid wants for nilpotent groups (see above). I don't know Whether you could use something like this to avoid the use of CFSG in the general case.

Comment by Freddie Manners

2013-03-22T12:15:47Z

My guess is that (1) and (2) are doomed not to expand because they are "too abelian", or "too nilpotent". So for (1), a set like $\{\pm g^{n} : 0 \le n \le N \}$ for $N$ large (but small compared to $p$) is likely to contradict expansion. Something similar should work in (2), considering a set of bounded length words on both your generators. I think this is closely related to [this MO question](mathoverflow.net/questions/91657/…). (3) has a very different flavour. It might be related to expansion in $SL_2$.

Comment by Freddie Manners

2013-03-21T19:31:28Z

I think that's right -- third edit lucky. I think the $L^2$ result should give $\varepsilon$ up to a (basically) constant factor like $k / (k - 1)$, meaning the $L^1$ result should go as $k \varepsilon$ times some constant factor; i.e. this hasn't made too much difference to the conclusion.

Comment by Freddie Manners

2013-03-21T19:13:11Z

Kevin -- thanks! You're absolutely right, and that means there's also an $\varepsilon^2$ that needs tracing through the argument. I'll fix this now and see what I get.

Comment by Freddie Manners

2012-06-20T18:08:03Z

@David Loeffler: fixed (I think), thanks.

Comment by Freddie Manners

2012-06-18T12:25:30Z

Thanks, I will work through the answers to that question. I'm concerned though that I'm looking at the double cover rather than the universal (infinite cyclic) cover. Does the action of $B_3$ on $AdS_3$ project down nicely to an action of $B_3$ on the double cover $Mp_2$, or do we have to first pass to a subgroup of $B_3$?

Comment by Freddie Manners

2012-03-20T11:56:28Z

Well, maybe "distance at most 100^100" or so.

Comment by Freddie Manners

2012-03-20T11:49:53Z

So, this nicely encapsulates my approach in special cases. E.g. if our generators are $x \mapsto x + 1$ and $x \mapsto 2 x$, you can say something like "take a bunch of elements in $F_p$ with the same low bits when written in base 2", since it still vaguely makes sense to talk about elements of $F_p$ in base 2; but this fails if you switch 2 for $a \approx p^{1/10}$. So, the generalization here is "take elements of $F_p$ which have distance at most 100 from 0 on the edges: $x \mapsto x + a^r$, for $0 \leq r < 100$" which -- in hindsight -- is clearly the correct generalization.

Comment by Freddie Manners

2012-03-20T11:44:26Z

Thanks! Putting that together with your blog post, I think "set of all group elements $t \in K$" should read ".. $t \in H$" in para 3, proof of Proposition 5.

Comment by Freddie Manners

2012-03-19T23:37:21Z

@Jesse: right, he then says some things about finite covolume lattices, which I thought might have given the result for $\mathbb{Z}$ implicitly, but it's possible it says something subtly different. Looking again, Shalom cites Burger's 1991 paper (Kazhdan constants for $\mathrm{SL}_3(\mathbb{Z})$), who doesn't appear to cite anyone else, so I think I'll give the credit to him.

Comment by Freddie Manners

2012-03-19T21:30:47Z

Thanks! Since I'm not that familiar with the language that paper is couched in, could you maybe point out which result there is equivalent to relative (T) for $(SL_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$?