User freddie manners - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:28:27Z http://mathoverflow.net/feeds/user/22253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127421/is-there-a-right-proof-of-riemanns-theta-relation Is there a "right" proof of Riemann's Theta Relation? Freddie Manners 2013-04-12T22:14:11Z 2013-05-02T01:51:03Z <p>Let $\theta$ denote the usual Jacobi Theta function (with auxiliary parameter $\tau = i$, for simplicity), i.e.</p> <p>$$\theta(z) = \sum_{n \in \mathbb{Z}} \exp(-\pi (a + n)^2 + 2 \pi i n z) \ .$$</p> <p>I'm interested in Riemann's quartic theta relation, which is (barring mistakes):</p> <p>$$\theta(x_1) \theta(y_1) \theta(u_1) \theta(v_1) = \frac{1}{2} \sum_{\eta \in \lbrace 0, 1/2, i/2, (1+i)/2 \rbrace } c_\eta \theta(x + \eta) \theta(y + \eta) \theta(u+\eta) \theta(v+\eta)$$</p> <p>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:</p> <p>$$x_1 = (x + y + u + v) / 2$$ $$y_1 = (x + y - u - v) / 2$$ $$u_1 = (x - y + u - v) / 2$$ $$v_1 = (x - y - u + v) / 2 \ .$$ (This is taken from Mumford's "Tata lectures on theta I", chapter 1, section 5.)</p> <p>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?</p> <p>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?</p> <p>As usual, apologies if I've missed this in a standard reference.</p> http://mathoverflow.net/questions/124812/joint-close-to-uniform-distribution-in-finite-fields/125173#125173 Answer by Freddie Manners for Joint (close to uniform) distribution in finite fields Freddie Manners 2013-03-21T14:32:21Z 2013-03-21T19:28:24Z <p>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.</p> <p>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. $F(x, y) = \mathbb{P}(A = x, B = y) - 1/k^2$.</p> <p>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 $f_{a b}(r) = \mathbb{P}(a A + b B = r) - 1/k$.</p> <p>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.</p> <p>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).</p> <p>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</p> <p>$$\widehat{f_{a b}}(r) = \sum_{s \in \mathbb{F}} f_{a b}(s) \chi_r(-s) = \sum_{x, y} F(x, y) \chi_r(-(a x + b y))$$</p> <p>The characters on $\mathbb{F}^2$ are $\chi_{u, v}(x,y) = \chi(u x + v y)$, so we deduce</p> <p>$$\widehat{f_{a b}}(r) = \sum_{x, y} F(x, y) \chi_{a r, b r}(-(x, y)) = \widehat{F}(a r, b r)$$</p> <p>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</p> <p>$$\sum_{a, b} \sum_r |\widehat{F}(a r, b r)|^2 = (k - 1) \sum_{u, v} |\widehat{F}(u, v)|^2 \le k^3 \varepsilon^2$$</p> <p>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,</p> <p>$$\sum_{x, y} |F(x, y)|^2 = \frac{1}{k^2} \sum_{u, v} |\widehat{F}(u, v)|^2 \le \frac{k \varepsilon^2}{k - 1}$$</p> <p>So, $\|F\|_2 \le \varepsilon \sqrt{\frac{k}{k-1}}$ : this is the $L^2$ result. Applying Cauchy-Schwarz we get something like</p> <p>$$\|F\|_1 \le k \|F\|_2 \le k \varepsilon \sqrt{\frac{k}{k - 1}}$$</p> <p>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.</p> http://mathoverflow.net/questions/119937/question-about-the-hardy-littlewood-method-quite-basic/119971#119971 Answer by Freddie Manners for Question about the Hardy-Littlewood method (quite basic) Freddie Manners 2013-01-26T23:16:47Z 2013-01-26T23:23:47Z <p>I'll split this into two questions:</p> <ol> <li>Why can $f$ be large near rationals with small denominator?</li> <li>Why is $f$ small away from these points (i.e. on the minor arcs)?</li> </ol> <p>For (1), I should first clarify that $f$ isn't <em>always</em> large at these rationals; it could well be zero. But it is large some of the time.</p> <p>Let's look at the example of <code>$f(a/8)$</code> for <code>$a \in \{1, 3, \dots, 7\}$</code>, when (say) $k=2$. We have:</p> <p><code>$f(a/8) = \sum_{n=1}^N e(a\, n^2 / 8)$</code></p> <p>Since $e(an/8)$ only detects the residue of $n \bmod{8}$, we have:</p> <p><code>$f(a/8) = \frac{N}{8} \sum_{n \in \mathbb{Z} / 8 \mathbb{Z}} e(a\, n^2 / 8) + O(1)$</code></p> <p>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:</p> <p><code>$f(a/8) = N \left(\frac{1}{4} + \frac{1}{4} e(a/2) + \frac{1}{2} e(a/8) \right) + O(1) = \frac{N}{2} e(a/8) + O(1)$</code></p> <p>(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 $1 \bmod 8$.</p> <p>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 $f(a/q)$.</p> <p>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 $1/8$.</p> <p>This brings us to (2). The question is: might the squares have similar bias "<code>$(\bmod{\sqrt{2}})$</code>"? 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:</p> <p><code>$\left\{ n^2 / \sqrt{2} \pmod{1}\,:\, n \in \{1 \dots N\} \right\}$</code></p> <p>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 <em>equidistributed</em> modulo 1. As $N$ becomes large, this equidistribution becomes quantitatively better.</p> <p>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.</p> <p>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 <a href="http://terrytao.wordpress.com/2008/06/14/the-van-der-corputs-trick-and-equidistribution-on-nilmanifolds/" rel="nofollow">Terry Tao's post on this</a>):</p> <blockquote> <p>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 $\bmod{1}$.</p> </blockquote> <p>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 $h$.</p> <p>Hope some of that helps.</p> http://mathoverflow.net/questions/99886/the-space-of-lattices-and-modular-forms-of-weight-1-2 The space of lattices and modular forms of weight 1/2 Freddie Manners 2012-06-18T11:05:50Z 2012-06-20T18:07:37Z <p>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.</p> <p>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":</p> <blockquote> <p>$f(a\ R_\theta\ T) = a^{-k} e^{- i k \theta} f(T)$</p> </blockquote> <p>where $R_\theta$ denotes the appropriate rotation matrix, and $a$ a positive real.</p> <p>[More precisely $f$ only has to be defined on one connected component of the space.]</p> <p>(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.</p> <hr> <p>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?</p> <p>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:</p> <ul> <li>What group plays the role of $SL_2(\mathbb{Z})$?</li> <li>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:<br> $G_4(\Lambda) = \sum_{w \in \Lambda \setminus 0} w^{-4}$<br> ?</li> </ul> <hr> <p>Apologies in advance if this is standard -- I've been unable to locate a satisfactory answer in the literature.</p> <p>Thanks,<br> Freddie</p> http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam Is it known that $(F_p^{\times} \ltimes F_p, F_p)$ is not a relative expander family? Freddie Manners 2012-03-19T19:19:34Z 2012-03-20T00:40:18Z <p>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.</p> <p>(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 $b \in \mathbb{Z}^2$.</p> <p>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 $b \in F_p^2$.)</p> <p>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:</p> <p>$\kappa\ (F_p^{\times} \ltimes F_p, F_p, S_p) \ge \delta$</p> <p>i.e. whether the pairs $(F_p^{\times} \ltimes F_p, F_p)$ can form a relative expander family.</p> <p>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 <em>almost invariant</em> with respect to all of them, i.e.</p> <p>$\neg \exists U: \forall s \in S: |s(U) \cap U| > \frac{99}{100} |U|$</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/127421/is-there-a-right-proof-of-riemanns-theta-relation Comment by Freddie Manners Freddie Manners 2013-04-13T11:05:12Z 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. http://mathoverflow.net/questions/126944/minimal-size-of-subsets-a-b-in-a-finite-group-g-such-that-abg Comment by Freddie Manners Freddie Manners 2013-04-09T17:07:13Z 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. http://mathoverflow.net/questions/125251/more-expanders Comment by Freddie Manners Freddie Manners 2013-03-22T12:15:47Z 2013-03-22T12:15:47Z My guess is that (1) and (2) are doomed not to expand because they are &quot;too abelian&quot;, or &quot;too nilpotent&quot;. 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](<a href="http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam" rel="nofollow" title="is it known that f p times ltimes f p f p is not a relative expander fam">mathoverflow.net/questions/91657/&hellip;</a>). (3) has a very different flavour. It might be related to expansion in $SL_2$. http://mathoverflow.net/questions/124812/joint-close-to-uniform-distribution-in-finite-fields/125173#125173 Comment by Freddie Manners Freddie Manners 2013-03-21T19:31:28Z 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. http://mathoverflow.net/questions/124812/joint-close-to-uniform-distribution-in-finite-fields/125173#125173 Comment by Freddie Manners Freddie Manners 2013-03-21T19:13:11Z 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. http://mathoverflow.net/questions/99886/the-space-of-lattices-and-modular-forms-of-weight-1-2 Comment by Freddie Manners Freddie Manners 2012-06-20T18:08:03Z 2012-06-20T18:08:03Z @David Loeffler: fixed (I think), thanks. http://mathoverflow.net/questions/99886/the-space-of-lattices-and-modular-forms-of-weight-1-2 Comment by Freddie Manners Freddie Manners 2012-06-18T12:25:30Z 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 $B_3$? http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam/91675#91675 Comment by Freddie Manners Freddie Manners 2012-03-20T11:56:28Z 2012-03-20T11:56:28Z Well, maybe &quot;distance at most 100^100&quot; or so. http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam/91675#91675 Comment by Freddie Manners Freddie Manners 2012-03-20T11:49:53Z 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 &quot;take a bunch of elements in $F_p$ with the same low bits when written in base 2&quot;, 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 &quot;take elements of $F_p$ which have distance at most 100 from 0 on the edges: $x \mapsto x + a^r$, for $0 \leq r &lt; 100$&quot; which -- in hindsight -- is clearly the correct generalization. http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam/91675#91675 Comment by Freddie Manners Freddie Manners 2012-03-20T11:44:26Z 2012-03-20T11:44:26Z Thanks! Putting that together with your blog post, I think &quot;set of all group elements $t \in K$&quot; should read &quot;.. $t \in H$&quot; in para 3, proof of Proposition 5. http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam Comment by Freddie Manners Freddie Manners 2012-03-19T23:37:21Z 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. http://mathoverflow.net/questions/91657/is-it-known-that-f-p-times-ltimes-f-p-f-p-is-not-a-relative-expander-fam Comment by Freddie Manners Freddie Manners 2012-03-19T21:30:47Z 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 $(SL_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$?