User lavender honey - MathOverflow

Answer by Lavender Honey for "Understanding" Gal(\bar Q/Q)

2009-11-01T17:25:36Z

$\newcommand{\bb}{\mathbb}$ This is a response to Charles' remark to JSE's answer, why doesn't $\bar{\bb Q}$ come with a standard algebraic closure inside the complex numbers $\bb C$?

First, if one considers an abstract extension $K/\bb Q$, then $K$ has $d = [K:\bb Q]$ embeddings into the complex numbers, which can not, a priori, be distinguished in any way. (e.g. maps from $K = \bb Q[x]/(x^3-2)$ to $\bb C$ require a "choice" of $2^{1/3}$ in $\bb C$.

Of course, this doesn't answer Charles' question, which, I imagine, is more along the following lines. Why doesn't one simply start with the complex numbers, and then consider the set of algebraic numbers inside $\bb C$? The resulting field is clearly isomorphic to $\bar{\bb Q}$, and, moreover, comes with a canonical embedding into $\bb C$.

The problem arises when one wants to define Frobenius elements. Defining such elements amounts to giving a choice of embedding from $\bar{\bb Q}$ into $\bar{\bb Q}_p$. So there is a choice to be made for every $p$! Thinking of $\bar{\bb Q}$ inside $\bb C$ fixes this choice for "$p=0$" only.

To make this completely explicit, consider the splitting field $K$ of $x^3 - 2$. In the "fields live inside $\bb C$" optic, $K$ is the field $\bb Q(2^{1/3},\sqrt{-3})$ where $2^(1/3)$ is real and the imaginary part of $\sqrt{-3}$ is positive. Clearly $Gal(K/\bb Q) = S_3$, where we can think of $(123)$ as sending $2^{1/3}\mapsto e^{2\pi i/3} 2^{1/3}$. If $p = 7$, then $Frob_p$ has order 3 in $S_3$. We can ask, does $Frob_p = (123)$ or $(132)$? We find that $$ Frob_p = (123) \quad\mbox{if } \sqrt{-3}\equiv 2\mod 7 $$ $$ Frob_p = (132) \quad\mbox{if } \sqrt{-3}\equiv 5\mod 7 $$ and knowing that the imaginary part of $\sqrt{-3}$ is positive does not allow us to determine $Frob_p$ without choosing an embedding of $K$ into $\bar{\bb Q}_7$.

Answer by Lavender Honey for Restricting representations to lattices

2012-09-10T03:08:06Z

The result is essentially the statement of Borel's stability theorem for $\mathrm{SL}_n(\mathbf{R})$, see for example Theorem 4.39 of the following:

http://people.uleth.ca/~dave.morris/books/IntroArithGroups.pdf

Sometimes Borel's stability theorem is phrased in terms of one of the corollaries, which in this case would be that any lattice in $\mathrm{SL}_n(\mathbf{R})$ is Zariski dense. Then to deduce the result one would also have to note that all finite dimensional representations of $\mathrm{SL}_n(\mathbf{R})$ are algebraic.

Answer by Lavender Honey for Existence of points on varieties which avoid a given number field.

2012-06-11T22:52:30Z

Yes, this follows from a Theorem of Moret-Bailly, see for example Corollary 1.5

http://math.stanford.edu/~conrad/vigregroup/vigre05/mb.pdf

Roughly speaking, given a finite set $S$ of primes with $C(K_v)$ is non-empty, this produces a field $L$ with $C(L) \neq \emptyset$ and $L_v = K_v$ for all $v \in S$.

To guarantee that $L \cap K' = K$, one may as well assume that $K'/K$ is Galois with Galois group $G$. Then for every conjugacy class $g \in G = \mathrm{Gal}(K'/K)$, let $v$ be a prime such that $\langle \mathrm{Frob}_v \rangle = \langle g \rangle \in G$ and $C(K_v) \ne \emptyset$. (The existence of such $v$ follows from Cebotarev, the Weil conjectures, and Hensel's Lemma.) If $S$ is the resulting set, then one may find $L$ with $C(L)$ non-empty and $L_v = K_v$ for all $v \in S$, and so (by Cebotarev) that $L \cap K' = K$.

This theorem gets used all the time in "potential modularity" theorems.

Answer by Lavender Honey for The resultant of an arbitrary polynomial and a cyclotomic polynomial

2012-05-28T07:25:36Z

For an algebraic number $\gamma$ which is not a root of unity, Baker's theorem gives a bound (uniform in $n$) of the form $|\gamma^n - 1| > n^{-C}$ for some constant $C$ (only depending on $\gamma$).

In particular, if $s_n:=\prod |\alpha^n_i - 1|$, then Baker's method gives the following estimate uniform in $n$ (for $\alpha_i$ not a root of unity): $$n \log \mathcal{M}(\alpha) + A \ge \log|s_n| \ge n \log \mathcal{M}(\alpha) - A - B \log|n|,$$ where $\mathcal{M}(\alpha)$ is the Mahler measure of $\alpha$. Proof: for $|\alpha_i| > 1$, one has $\log|\alpha^n_i - 1| \sim n \log|\alpha_i|$ up to $O(1)$, which gives rise to the term $n \log \mathcal{M}(\alpha)$; the term $\log|\alpha^n_i - 1|$ for $|\alpha_i| \le 1$ is trivial to bound from above and can be bound from below by Baker's Theorem. The logarithm of the Mahler measure is the sum of $\log|\alpha_i|$ for the roots $|\alpha_i|> 1$. By a theorem of Kronecker this sum is positive if $\alpha$ is not a root of unity.

OTOH, if $\Phi_n(x)$ is the $n$th cyclotomic polynomial and $t_n:=\prod |\Phi_n(\alpha_i)|$, then we deduce that $\sum_{d|n} \log(t_n) = \log(s_n)$, and hence $$\log(t_n) = \sum_{d|n} \log(s_{n/d}) \mu(d) \ge \varphi(n) \log \mathcal{M}(\alpha) - d(n)(A + B \log(n)),$$ where $d(n)$ is the number of divisors of $n$. The bounds $\phi(n) \gg n^{1-\epsilon}$ and $d(n) = n^{\epsilon}$ easily give the asymptotic relation $$\log(t_n) \sim \varphi(n) \log \mathcal{M}(\alpha) \gg 1$$ as $n$ goes to infinity.

FWIW, the bounds of Baker (and the other bounds used above) are effective, so for any particular $\alpha$ one could in principle find all $n$ with $t_n = 1$. (n.b. Gelfond's estimate would give $\epsilon \cdot n$ instead of $B \log|n|$ as an error term, which is enough to show that $s_n \rightarrow \infty$ but not $t_n$.)

Answer by Lavender Honey for On the Hasse-Weil L-function of $P^n$

2012-05-21T07:02:41Z

$\newcommand\GL{\mathrm{GL}}$ $\newcommand\SL{\mathrm{SL}}$ $\newcommand\R{\mathbf{R}}$

There is, of course, an autormorphic representation (Hecke character) $\chi$ for $\GL(1)$ whose $p$-adic avatar is the cyclotomic character. From this, one thus has the isobaric sum (following Langlands and Jacquet-Shalika) $$\pi = 1 \boxplus \chi \boxplus \ldots \boxplus \chi^{n-1}$$ which is an automorphic representation for $\GL(n)$ with $L(\pi,s) = L(\mathbf{P}^n,s)$.

But all of this is somewhat irrelevant to your action question, to which the answer is not really. The reason $q$-expansions arise for $\GL(2)$ has to do with the fact that $\SL_2(\R)/\mathrm{SO}_2(\R)$ is the upper half plane, and $\SL_2(\mathbf{Z})$ contains the element $z \mapsto z+1$ for which $q = e^{2 \pi i z}$ is invariant. $\SL_n(\R)/\mathrm{SO}_n(\R)$ is quite a different beast.

Edit You don't seem satisfied with my answer, but I think you seem to be missing a basic principle: if you write down something at random, there's no reason it should be interesting. The theory of automorphic forms, however labyrinthian, has an incredibly precise structure. If you take an automorphic representation $\pi$ corresponding to a classical modular form, then the representation theory of $\mathrm{GL}_2(\mathbf{R})$ will tell you that a lowest weight vector for the discrete series will be annihilated by some differential operator which one can compute to be the Cauchy-Riemann equations, and hence the corresponding function on the upper half plane will be holomorphic: this is a very specific reason why holomorphic functions might be associated to automorphic forms. If, instead, one works with an automorphic form for $\mathrm{GL}(3)$, then the representation theory will (under appropriate conditions) produce an expansion in terms of Whittaker functions which will have a completely different flavour.

It might also be worth remarking that already since Langlands time people have thought hard about the problem of transfer, that is, for example, starting with an automorphic representation for $\GL(2)$ and producing one for $\GL(3)$. The methods used to prove these results are almost exclusively via the trace formula - in particular, they proceed via harmonic analysis and representation theory, rather than explicit manipulations with functions (holomorphic or otherwise). Thus (addressing your comment) the hope that one might explicitly decompose the symmetric space of $\GL(3)$ in some way is a little too optimistic.

Finally, you can (of course) recover $L(\pi,s)$ from $f(q)$, so $f(q)$ does carry (in some sense) all the information of $L(\pi,s)$. Moreover, one can recover $f(q)$ as the inverse Mellin transform of $L(\pi,s)$ adjusted by appropriate Gamma factors. However, you will find that when $n > 2$ the relevant Gamma factors involved will not play well with respect to the functional equation, and thus there will be no reason to expect that $f(q)$ will not have any nice properties (such as a nice functional equation). Even for $n = 1$ you have to cheat a little (replacing $q^n$ by $q^{n^2}$), and moreover the theta function is (automorphically) more naturally thought of as associated to the metaplectic group rather than $\GL(2)$.

Answer by Lavender Honey for Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

2012-02-24T18:44:12Z

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n), $$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

Answer by Lavender Honey for Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

2012-02-24T00:58:17Z

Here is a partial answer, which shows that there are no zeros for $z = s + i t$ with $|t| \ge 4$ .

Let $\psi(z):= \Gamma'(z)/\Gamma(z)$ be the digamma function. If $z = s + i t$, then $$\frac{d}{ds} |\Gamma(z)|^2 = \frac{d}{ds} \Gamma(z) \Gamma(\overline{z}) = |\Gamma(z)|^2 \left(\psi(z) + \psi(\overline{z})\right).$$ (Both $\Gamma(z)$ and $\psi(z)$ are real for real $z$, and so satisfy the Schwartz reflection principle.) The product formula for the Gamma function implies that there is an identity $$\psi(z) = - \ \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{z + n} \right) = 1 - \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{z + n} \right),$$ and hence $$\psi(z) + \psi(\overline{z}) = 2(1 - \gamma) + \sum_{n=1}^{\infty} \left(\frac{2}{n + 1} - \frac{1}{z + n} - \frac{1}{\overline{z} + n} \right).$$ Suppose that $z = s + i t$, and that $s \in [0,1]$. Then $$ \frac{2}{n + 1} - \frac{1}{s + i t + n} - \frac{1}{s - i t + n} = \frac{2(s^2 + t^2 + n s - s - n)}{(1+n)(n^2 + 2 n s + s^2 + t^2)} \ge \frac{-2}{(n^2 + t^2)}.$$ (The last inequality comes from ignoring all the positive terms in the numerator, and then setting $s = 0$ in the denominator.) It follows that $$\psi(z) + \psi(\overline{z}) \ge 2(1 - \gamma) - \sum_{n=1}^{\infty} \frac{2}{n^2 + t^2},$$ which is positive for $t$ big enough, e.g. $|t| \ge 4$. On the other hand, $$\psi(z + 1) + \psi(\overline{z} + 1) = \psi(z) + \psi(\overline{z}) + \frac{1}{z} + \frac{1}{\overline{z}} = \psi(z) + \psi(\overline{z}) + \frac{2s}{|z|^2}.$$ In particular, if $\psi(z) + \psi(\overline{z})$ is positive for $s \in [0,1]$ for some particular $t$, it is positive for all $s$ and that particular $t$. It follows that, if $|t| > 4$, that $|\Gamma(s + it)|^2$ is increasing as a function of $s$. In particular, if $|t| > 4$, then any equality $$|\Gamma(s + i t)| = |\Gamma(1 - (s + i t))| = |\Gamma(1 - s + i t)|$$ implies that $s = 1/2$.

Since this method applies equally well to $\Gamma(z) + \theta \cdot \Gamma(1 - z)$ for any $|\theta| = 1$, it is not sufficient to answer the question .

(NDE's comment seem to suggest one can reduce to the case of $z$ with real part in $[0,1]$ which is handled by this method, but I don't understand the remark. I made this communitity wiki if someone wants to complete the argument.)

Answer by Lavender Honey for The graph of congruences between modular forms

2012-01-10T00:22:53Z

Suppose that $f$ has level $N$, and suppose that $N$ is divisible by $p$. Then it is well known that $f$ is congruent modulo (some prime above) $p$ to a form $g$ of level $M$ dividing $N$ (and high weight), where $M$ is prime to $p$. In particular, by induction, all forms $f$ are connected to a form $g$ of level $1$ in at most $d$ steps, where $d$ is the number of distinct prime divisors of $N$. Yet all level one forms are congruent to $\Delta$ modulo $2$. This is actually related to ideas behind the proof of Serre's conjecture:

http://en.wikipedia.org/wiki/Serre%27s_modularity_conjecture

Answer by Lavender Honey for irreducible polynomial with repect to prime number

2012-01-02T20:39:07Z

The answer is yes to both questions: it is a theorem of Brillhart, Filaseta, and Odlyzko (see corollary 2, p.1058):

http://cms.math.ca/10.4153/CJM-1981-080-0

Answer by Lavender Honey for gcd of three numbers

2011-12-31T06:39:34Z

The answer is yes, there exists $c$ and $d$, even with $c = 1$, and $d \ll (\log(n))^{O(1)}$. This follows from a result of Iwaniec.

It suffices to assume that $(a,b) = 1$. Suppose that $(n,b) = e$, which implies that $(a,e) = 1$.

Since $b$ is divisible by $e$, it follow that $(a+bk,e) = 1$ for all integers $k$. Let $m$ denote the largest factor of $n$ such that $(m,e) = 1$, it clearly satisfies $(m,b) = 1$.

If $(a+bk,m) = 1$, then because $(a+bk,e) = 1$, one also has $(a+bk,n) = 1$.

Hence the problem becomes: given an integer $m$, and an arithmetic progression $$a, a + b, a + 2b, a+3b, a+4b, \ldots $$ with common difference $b$ prime to $m$, can one find a small integer $d$ such that $a+db$ is prime to $m$? Equivalently, let $g \in \mathbf{Z}$ be any multiplicative inverse to $b$ mod $m$, then does there exist a small $d$ such that $ag + dbg$ is prime to $m$? Equivalently, does there exist a small $d$ such that $ag + d$ is prime to $m$?

Recall the definition of the Jacobsthal function: $j(m)$ is the smallest integer such that any arithmetic progression of length $j(m)$ (with common difference one) contains an element which is co-prime to $m$.

If $m|n$, then $j(m) \le j(n)$. In particular, there exists a $d \le j(m) \le j(n)$ with $c = 1$ such that $ac+bd$ is prime to $n$.

Finally (the hard part), by a result of Iwaniec (Demonstratio Math. 11 (1978), 225-231 (MR0499895)), if $n$ has $r$ distinct prime factors, then $j(n) \ll r^2 \log^2 r$, which implies that $$j(n) \ll \log^{2}(n) = \log(n)^{O(1)}.$$

Answer by Lavender Honey for Free subquotient of Galois representations coming from Hida theory

2011-11-30T06:45:44Z

$\newcommand\T{\mathbf{T}_{\mathfrak{m}}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\m{\mathfrak{m}}$ $\newcommand\F{\mathbf{F}}$ $\newcommand\Frob{\mathrm{Frob}}$ $\newcommand\rhobar{\overline{\rho}}$ $\newcommand\eps{\epsilon}$

First, as Professor Emerton mentions, the construction of $L^{+}$ you gave is not necessarily free over $\T$. Thus, I will interpret your question as asking the following: does there exist an exact sequence:

$$0 \rightarrow L^{+} \rightarrow (\T)^2 \rightarrow L^{-} \rightarrow 0$$ of $\T[G_{\Q_p}]$-modules where $L^{+}$ and $L^{-}$ are free $\T$-modules of rank one.

( Edit Perhaps this extra remark might be useful. Suppose that $L = (\T)^2$ admits a free rank one quotient $L^{-}$. Since $L^{-}$ is free, it admits a section $L^{-} \rightarrow L$, and hence the kernel $L^{+}$ of $L \rightarrow L^{-}$ is also free. Thus the existence of a free rank one quotient asked for in the question is equivalent to the existence of the exact sequence above.)

The answer to this question, in general, is no. The following argument is implicitly contained in papers of Wiese on the failure of multiplicity one and weight one forms.

The action of $G_{\Q_p}$ on $L^{+}$ is unramified and so acts via $G_{\mathbf{F}_p}$. Thus $\Frob_p$ acts on a basis vector as multiplication by some element of $\T$. Since $\T$ is determined by its action on classical eigenforms, one may identify this element with the Hecke operator $U$. In particular, $U \in \T$ (it wasn't clear whether your $\T$ included $U$ or not).

The exact sequence remains exact after tensoring with $\T/\m$, for dimension reasons. It follows that the sequence is split as a sequence of $\T$-modules. Hence it remains exact after quotienting out by any ideal of $\T$.

Suppose that $\rhobar: G_{\Q} \rightarrow \mathrm{GL}_2(\F_p)$ is irreducible and modular (mod-$p$) of weight $1$. Suppose, moreover, that $\rhobar(\Frob_p)$ acts by a scalar $\lambda$. Associated to $\rhobar$ is a mod-$p$ weight $1$ form $f = \sum a_n q^n \in \F_p[[q]]$. If $A$ is the Hasse invariant, then then $Af$ and $f^p$ are both mod-$p$ modular forms of weight $p$. One can check that all elements of the $\F_p$-vector space $\{Af,f^p\}$ are eigenvalues for all the Hecke operators $T_l$ for $(l,p) = 1$, but the operator $T$ (and so $U$, which is the same as $T$ in weight $> 1$) satisfies $(U - \lambda)^2 = 0$ but does not act by a scalar. Since $U$ acts invertibly on this vector space, it gives rise to a surjective map: $$\T \rightarrow \F_p[\eps]/\eps^2,$$ where the image of $T_l$ lands in $\F_p$ for all $(l,p) = 1$, but $U$ does not act by a scalar. Let $I$ be the kernel.

The Galois representation on $(\T)^2/I \simeq (\F_p[\eps]/\eps^2)^2$ is equal to $\rhobar \oplus \rhobar$. This follows from a result of Boston-Lenstra-Ribet, since $T_l$ is acting by a scalar for each $(l,p) = 1$. It follows, by assumption, that the action of $G_{\Q_p}$ on $L^{+}/I L^{+} \simeq \T/I$ must also be trivial, because this is a sub-representation of $\rhobar \oplus \rhobar$. On the other hand, as we have seen, the action of Frobenius on $L^{+}$ and thus $L^{+}/I L^{+} = \T/I$ is given by $U$, which is acting non-trivially $\T/I$ by the construction of $I$. This is a contradiction.

Such representations $\rhobar$ exist (for example, with $p = 2$, and level $\Gamma_0(431)$) as mentioned in Professor Emerton's answer.

Answer by Lavender Honey for Infinite exponential representation of real numbers

2011-11-06T17:25:27Z

First, even though I think this is a fun question, it's not really research mathematics and I'm not sure it belongs on mathoverflow. (You know that some really smart people answer questions on math.stackexchange, right?) As was noted in Robert's answer, one is investigating the sequence $x_{n+1} = | \log(x_n)|$, which makes sense for all $x_0$ outside some countable set. Moreover, the periodic points will (surely) be transcendental, and it will be impossible to prove this fact (except for periods of length $1$). So what should one expect for the number of $+$ and $-$ signs? Here is a heuristic description of what happens, with the caveat that I have made no attempt to be rigorous (to do so, I would compare this with the probability theory of the Gauss–Kuzmin distribution).

First, choose $x_0$ in $[0,\infty]$ according to some probability measure $f_0(x)$, with cumulative probability distribution $F_0(x)$. Let $f_n(x)$ be the distribution of $x_n$. By definition, $f_n(x)$ must satisfy the following equation:

$$\int^{b}_{a} f_n(x) dx = \int_{e^a}^{e^b} f_{n-1}(x) dx + \int_{e^{-b}}^{e^{-a}} f_{n-1}(x) dx$$ for all $b \ge a \ge 0$. If $F_n(x) = \int^{x}_{0}f_n(t)dt$ is the cumulative distribution function of $x_n$, then this equation becomes: $$F_n(b) - F_n(a) = F_{n-1}(e^b) - F_{n-1}(e^a) + F_{n-1}(e^{-a}) - F_{n-1}(e^{-b}).$$ Letting $a = 0$, one obtains: $$F_n(z) = F_{n-1}(e^z) - F_{n-1}(e^{-z}).$$ Given some basic assumption on $F_0(z)$, the sequence of functions $F_n(z)$ converges to the unique increasing function $F(z)$ such that $$F(z) = F(e^z) - F(e^{-z}).$$ The independence of $F(z)$ on $F_0(z)$ implies that this function should describe the cumulative distribution function of $x_n$ for $n$ sufficiently large for almost all initial values $x_0$. By choosing random functions $F_0(z)$, one can estimate that $$F(1) \simeq 0.6518\ldots$$ Since the sign in the exponential is determined by whether $x_n > 1$ or not, this implies that the ratio of $-$ signs to $+$ signs (for almost all initial values, which presumably includes $x_0 = 2$) is roughly $1.872$ to $1$. This seems to confirm what you observed experimentally. Moreover, since the function $F(z)$ is strictly increasing, it follows by Kolmogorov's zero-one law that, for almost all initial values $x_0$, that there are arbitrarily long runs of $+$ signs, $-$ signs, etc.

Edit: To make this completely rigorous, one can make $\mathrm{R}^{+}$ a compact measure space given by the measure specified by $\mu([a,b]) = F(b) - F(a)$. The function $T:=|\log(x)|$ (modified so that $T(1) = 1$) is then measure preserving and (as is relatively easy to check) ergodic. The claims then follow from the Birkhoff Ergodic theorem, for suitable choices of test function $f$ (like the step function which is zero for $x < 1$ and one for $x > 1$). (BTW, I may have added a few more decimal digits above than was really justified.)

Answer by Lavender Honey for Continuous extensions reals and to p-adic numbers

2011-08-26T21:41:02Z

The answer is no, and one can essentially use the same construction as in the answer: http://mathoverflow.net/questions/42460

Specifically, enumerate the non-zero rationals $\{r_1,r_2, \ldots\}$ in some way. Now consider the function: $$f(x) = \sum_{n=1}^{\infty} c_n x^{n^2} \prod_{i=1}^{n} (x - r_i).$$ If $c_n \in \mathbf{Q}$, then this is a well defined function from rationals to rationals. On the other hand, $f(x)$ converges to an analytic function in $\mathbf{Q}_v$ if and only if the coefficients of this power series converge to zero fast enough. Since the coefficients of the power series in the range $k = n^2$ to $k < (n+1)^2$ are simply the cofficients of $c_n x^{n^2} \prod_{i=1}^{n} (x - r_i)$, this can be ensured by forcing these coefficients to be very highly divisible by the first $n$ primes, and small in the archimedean sense (by including a very very large prime in the denominator).

Answer by Lavender Honey for super-transitive group action

2011-08-02T18:59:44Z

Let $A$ denote the set of all integers except $i$. Since $G_A$ is transitive, it follows that there exists an element $\sigma \in G$ that fixes $i$ and sends any fixed $j \ne i$ to some $k \ne i$. From this one quickly deduces that $G$ is $2$-transitive, and hence primitive.

Choose a prime $p$ such that $n-3 \ge p > n/2$ (by a modification of Chebyshev's proof of Bertrand's postulate, this is true whenever $n \ge 8$). Let $B = \{1,2, \ldots, p\}$. Since the stabilizer $G_B \subset G$ of $B$ acts transitively on $B$, by the orbit-stabilizer theorem, $G_B$ (and hence $G$) has an element of order $p$. Since $n-p < p$, it follows that $G$ contains a $p$-cycle. But a well known theorem of Jordan says that any primitive permutation group containing a $p$ cycle for $p \le n-3$ is the symmetric or alternating group.

This leaves $n \le 7$. Note that since $G_A$ acts transitively for any $A$, it follows that $|G|$ is divisible by $|A|$.

For $n = 1$, $2$, and $3$, $G$ has to be $S_n$ (easy). For $n = 4$ and $5$, $|G|$ must be divisible by $12$ and $60$ respectively. So $G = A_n$ or $S_n$ in these cases. Yet both $A_n$ and $S_n$ satisfy the defining condition when $n \ge 4$. For $n = 7$, $|G|$ must be divisible by $420$. The only primitive subgroups of $S_7$ with order divisible by $420$ are $A_7$ and $S_7$. This leaves $n = 6$, where $|G|$ is divisible by $60$. It follows that $G$ could possibly be $A_5$, $S_5$, $A_6$, or $S_6$. The latter two groups work. The group $S_5$ with its natural embedding (stabilizing one point) does not work, because it does not act transitively on the full set $\{1,2,\ldots,6\}$. The same remark applies to the standard embedding of $A_5$. It suffices, then, to check the exotic actions of $A_5$ and $S_5$. Here is a "computation free" argument.

This exotic action of $S_5$ on $6$ points is realized by the action of $\mathrm{PGL}_2(\mathbf{F}_5)$ on the projective line over $\mathbf{F}_5$. The action of $\mathrm{PGL}_2(\mathbf{F}_5)$ is sharply $3$-transitive, and the action of $A_5 = \mathrm{PSL}_2(\mathbf{F}_5)$ is $2$-transitive. It follows that both groups satisfy the conditions for $|A| = 1$ and $|A| = 2$. The first group satisfies the condition for $|A| = 3$ since it is $3$-transitive. Suppose that $|A| = 5$. Then we may assume by transitivity that $A = \mathbf{F}_5 = \mathbf{P}^1(\mathbf{F}_5) \setminus \{\infty\}$. Both groups act transitively on this set via the element $z \mapsto z + 1$. Finally, let $|A| = 4$. Since both groups are $2$-transitive, we may assume that $A = \mathbf{F}^{\times}_5 = \mathbf{P}^1(\mathbf{F}_5) \setminus \{0,\infty\}$. The subgroup of elements of $\mathrm{PGL}_2(\mathbf{F}_5)$ which fix $0$ and $\infty$ are of the form $z \mapsto a z$ for $a \in \mathbf{F}^{\times}_5$, which acts transitively. The set of elements of $\mathrm{PGL}_2(\mathbf{F}_5)$ which send $0$ to $\infty$ and $\infty$ to $0$ are of the form $z \mapsto a/z$ for $a \in \mathbf{F}^{\times}_5$. Yet the corresponding elements in $\mathrm{PSL}_2(\mathbf{F}_5)$ are of the form $z \mapsto a^2 z$ and $z \mapsto a^2/z$, which does not act transitively since it only sends quadratic residues to quadratic residues. Thus the complete list of such groups is:

(i) $G = S_n$ for $n \le 3$.

(ii) $G = A_n$, $S_n$ for $n \ge 4$.

(iii) $G = \mathrm{PGL}_2(\mathbf{F}_5) = S_5$ when $n = 6$.

Answer by Lavender Honey for The parity conjecture

2011-07-29T21:21:03Z

For convenience, restrict to elliptic curves over $\mathbf{Q}$ (there are more general results/conjectures over number fields). There are three possible parities one could consider:

(i) The parity of the rank of $E(\mathbf{Q})$.

(ii) The parity of the $p$-Selmer rank of $E$ for a prime $p$.

(iii) The parity of the order of vanishing of the $L$-function of $E$ (determined by the root number).

Conjecturally these parities are all the same, and this follows from the BSD conjecture. The parity conjecture usually refers to the claim that (i) and (iii) are the same. The claim that (i) and (ii) are the same (your question) follows from the conjecture that Sha(E) is finite (it is almost equivalent to this conjecture), and this is only known in the cases where one can (essentially) prove the entire BSD for $E$ (for example, by work of Kolyvagin). The conjecture that (ii) and (iii) are the same is also sometimes called the parity conjecture. There has been progress on this question in recent years by Nekovar, and most recently by the brothers Dokchister. The current state of the art is that (ii) and (iii) are now known to be the same for all elliptic curves over $\mathbf{Q}$. A survey article by Tim Dokchitser can be found here:

http://arxiv.org/pdf/1009.5389v1

Answer by Lavender Honey for Limits of p-adic Representations

2011-06-29T05:51:32Z

Both $2$ and $3$ are immediately false by considering limits of ($p$-adic) powers of the cyclotomic character.

Answer by Lavender Honey for Action of PGL(2) on Projective Space

2011-06-05T04:57:22Z

A KT-field $(F,+,\times,\sigma)$ consists of a neardomain $(F,+,\times)$ together with an involutionary automorphism $\sigma$ satisfying $$\sigma(1 + \sigma(x)) = 1 - \sigma(1 + x)$$ for all $x \in F \setminus {0,1}$. (My impression is that neardomains are quite weak entities, e.g. $F^{\times}$ is required to be a group but it may not be commutative, $(F,+)$ is not even necessarily a group. Industrious MO reader adds the definition of a neardomain to this answer if they wish.) Sharply $3$-transitive groups are determined up to isomorphism as permutation groups on $\mathbf{P}^1(F) = F \cup { \infty }$ consisting of maps of the form:

(i): $x \mapsto a + m x, \quad \infty \mapsto \infty$

(ii): $x \mapsto a + \sigma(b + m x), \quad \infty \mapsto a, \quad - m^{-1} b \mapsto \infty$,

where $a,b \in F$ and $m \in F^{\times}$.

Consider the set of elements $\gamma \in G$ such that $\gamma(0) = \infty$ and $\gamma(\infty) = 0$. They are given exactly by mappings of the form: $$\gamma: x \mapsto \sigma( \lambda x)$$ for any $\lambda \in F^{\times}$. If all such $\gamma$ have order two, then $$\sigma(\lambda \sigma(\lambda x)) = x$$ for all $x, \lambda \in F^{\times}$. Setting $x = \lambda^{-1}$, it follows that $\sigma(\lambda) = \lambda^{-1}$ for all $\lambda \in F^{\times}$. Since $\sigma$ is an automorphism, it follows that $F^{\times}$ is commutative. From a Theorem of Kirby (see below), it follows that $(F,+,\times)$ is actually a commutative field, and $G = \mathrm{PGL}_2(F)$.

All the results and definitions of this answer can be gleamed from the math review: MR0997066 (91b:20004a) of a paper by William Kerby. The paper is only $3$-pages long, so I assume that is is relatively elementary - although I can't access it myself, and it may refer to previous results. (Full disclosure, all I did was type "sharply 3-transitive" into mathscinet, I don't actually know what a neardomain actually is.) In case your actual purpose is to generalize this result to $(\infty,\pi)$-whatzit categories with creamy rice pudding centres, you might want to take a glance at the actual paper.

Arithmetic progressions modulo $p$ under the squaring map

2011-02-10T01:53:02Z

I feel that the following problem should be known, but I'm not sure where to look for it.

Fix a real constant $\frac{1}{2} \ge \epsilon > 0$. For varying primes $p$, Let $A_p$ denote the set of residue classes coming from the first $\lfloor p \epsilon \rfloor$ integers. Let $B_p$ denote the squares (modulo $p$) of the elements of $A_p$. Then one might ask whether $$\lim_{p \rightarrow \infty} \frac{|A_p \cap B_p|}{|A_p|} =^{?} \epsilon.$$ It's true for $\epsilon = \frac{1}{2}$, but that's a degenerate case where $B_p$ can essentially be replaced by $\mathbf{F}^{\times 2}_p$, in which case the answer follows from any non-trivial upper bound on character sums (say the Polya-Vinagradov inequality). Is it true more generally?

Answer by Lavender Honey for How to show modularity of an elliptic curve?

2011-02-13T06:04:16Z

They explicitly computed quotients of $X_0(N)$ and identified them with elliptic curves. Suppose you can compute the space $S_2(\Gamma_0(N),\mathbf{C})$ of modular forms. An (isogeny class of) elliptic curves of conductor $N$ corresponds (by modularity) to a normalized new Hecke eigenform with coefficients in $\mathbf{Z}$. Given such an $f$, one can compute the periods of $f$, which allows one to write down a Weierstrass equation for $f$. If one can do this in such a way to guarantee that the coefficients of the Weierstrass equation are integers, this allows one to computationally determine exactly the modular elliptic curves of conductor $N$. All this is very well explained in Cremona's book, which is available free online:

http://www.warwick.ac.uk/~masgaj/book/amec.html

The particular method referred to in the paper of [BGZ] was presumably the "method of graphs", see for example here:

http://modular.math.washington.edu/msri06/refs/mestre-method-of-graphs/mestre-en.pdf

The Faltings-Serre method is slightly different; it allows you to determine, given two Galois representations $\rho_1$ and $\rho_2$ from $G_{K}$ to (say) $\mathrm{GL}_n(\mathbf{Z}_p)$ such that:

$\overline{\rho}_1 \simeq \overline{\rho}_2$,
$\rho_1$ and $\rho_2$ are both unramified outside some finite (given) set of places $S$ of $K$.

whether $\rho_1 \simeq \rho_2$. You could use it to prove that an elliptic curve $E/\mathbf{Q}$ is modular by comparing the Galois representation attached to the $p$-adic Tate module of $E$ and the Galois representation attached to the conjecturally corresponding eigenform of level $N$ (EDIT: this works because the Galois representation determines the isogeny class of $E$ by Faltings' proof of the Tate conjecture for abelian varieties). However, this wouldn't be so efficient. However, there are other situations (for example, elliptic curves over other number fields) where the corresponding automophic form is not enough to compute the periods. In this case, the Faltings-Serre method can be used to prove modularity. Taylor uses this to prove that a certain elliptic curve over $\mathbf{Q}(\sqrt{-3})$ is modular.

Answer by Lavender Honey for Construction of abelian varieties from Hilbert modular forms?

2011-02-06T22:23:57Z

There is no problem with constructing an abelian variety $A$ for most Hilbert modular forms of parallel weight $2$, the issue is finding such a variety for all $\pi$. In particular, when $d = [K:\mathbf{Q}]$ is even, there is a local obstruction to the existence of a corresponding Shimura curve which realizes the Galois representation associated to $\pi$. In particular, if $\pi$ has "level one", then no such Shimura curve exists. To construct the Galois representation in this case one has to use congruences; this was done by Taylor in the late 80's. This issue is also discussed here:

http://mathoverflow.net/questions/39485/are-there-motives-which-do-not-or-should-not-show-up-in-the-cohomology-of-any-s

Explicit $p$-adic local Langlands for $p = 2$.

2011-02-04T06:23:04Z

$\newcommand\Q{\mathbf{Q}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Qbar{\overline{\Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\ss{\mathrm{ss}}$ $\newcommand\F{\mathbf{F}}$

Fix an integer $k \ge 2$, let $G = \Gal(\Qbar_2/\Q_2)$, and let $\chi: G \rightarrow \Qbar^{\times}_2$ be the $2$-adic cyclotomic character.

There are exactly two $2$-dimensional $2$-adic representations $\rho: G \rightarrow \mathrm{GL}(V)$ such that:

$\mathrm{det}(\rho) = \chi^{k-1}$,
$V$ is crystalline with Hodge--Tate weights $[0,k-1]$,
Frobenius does not act semisimply on $D_{\mathrm{cris}}(V)$.

(they differ by an unramified quadratic twist.) Choose either of these representations. Let $H$ denote the kernel of $\rho$. Then what is $(G/H)^{ab}$ in terms of $k$? More precisely, what is $(G/H)^{ab}$ as a quotient of $G^{ab} \simeq \widehat{\Q^{\times}_2}$ in terms of $k$? (The answer may depend on the twist, but it is easy to pass from the answer for one to the other.)

A related but slightly different question is to determine $(\overline{V})^{\ss}$; I am interested in both questions.

In principle, I could imagine trying to guess the answer in any particular case by fixing $k$ and assuming that $(G/H)^{ab}$ was locally constant as a function of the trace of Frobenius $a_2$, and then trying to find classical modular forms which were nearby. However, in practice, the valuations of $a_p$ are never that large. This suggests fixing $a_p$ and varying the weight. This leads to my second question: is $(G/H)^{ab}$ locally constant as a function of sufficiently large positive integers $k$ (with the topology coming from weight space)? Does Berger's argument (proving the analogous statement for $(\overline{V})^{\ss}$ when $a_p \ne 0$ and some other mild conditions) apply in this case?

When $k = 2$ the answer to the question is given by Fontaine--Laffaille theory; the quotient $(G/H)^{ab}$ corresponds to the projection $\widehat{\Q^{\times}_2} \rightarrow \Z^{\times}_2 \times \Z/2\Z$ sending $4$ to $1$ (for either twist).

EDIT: To respond to Kevin's comment, I think that $\overline{V}$ may well sometimes be reducible for some $k$. In general, we know that if $a_p$ is sufficiently small then $V$ is "close" to the corresponding representation with $a_p = 0$. On the other hand, if $a_p$ is small, then $-a_p$ is also small, and thus $V$ is "close" to its quadratic twist $V \otimes \eta$ where $\eta$ is unramified. Now an actual equality $V = V \otimes \eta$ would imply that $V$ is induced from the unramified quadratic extension of $\Q_p$, and this is indeed the case when $a_p = 0$. Yet it doesn't imply that $\overline{V}$ is irreducible, since the character one is inducing might be equal to its Galois conjugate modulo $p$. If $a_p = 0$, then $V$ is given by the induction of $\phi^{k-1}$ where $\phi \equiv \omega_2 \mod p$. Since $\omega^{p+1}_2$ is invariant under conjugation, it follows that $\overline{V}$ is reducible if and only if $p+1$ divides $k-1$. (When $p$ is odd and $k$ is even this doesn't happen very frequently.) Returning to small $a_p$, the fact that $\overline{V} = \overline{V} \otimes \eta$ is enough to say that there is always a surjection $$(G/H)^{ab} \rightarrow \Gal(\F_{p^2}/\F_p)$$ if $a_p$ is small enough. So one aspect of my question would be "is $a_2 = 2^{(k-1)/2}$ close enough to $a_2 = 0$ to deduce that $\overline{V}= \overline{V} \otimes \eta$? The result of Berger-Li-Zhu, even imagining that it applied with $p = 2$, would not be sufficient, because the bound there is something like $(k-2)/(p-1)$.

Answer by Lavender Honey for Is this a (well known) open problem?(infinitness and more on $anm \pm n\pm m$ )

2011-02-03T06:09:58Z

$\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$

(Caveat: normally I wouldn't answer a question with such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.)

Suppose that $A$ is a finite abelian group. Then I claim that given any set of at least $|A| + 1$ (not necessarily distinct) elements of $A$ one can find a proper subset whose sum is the identity. Proof: Denote the elements $a_i$ for $i = 1$ to $|A|+1$. By the pigeonhole principle, either one of the $|A|$ sums $\sum_{i=1}^{r} a_i$ for $r = 1$ to $|A|$ is the identity, or two of the sums are the same element of $|A|$, in which case consider the difference. We deduce from this the following: Let $n$ be any integer coprime to $a$ with more than $r:=|(\Z/a \Z)^{\times}|$ prime factors. Then $n$ has a proper divisor of the form $1 \mod a$.

Suppose that $k$ cannot be represented by the form $amn \pm m \pm n$, and suppose that $a > 2$. It is simple to deduce that this is equivalent to asking that $ak+1$ and $ak-1$ have no proper divisors of the form $\pm 1 \mod a$. It follows that $ak+1$ and $ak-1$ each have at most $r=|(\Z/a \Z)^{\times}|$ prime factors. The integers with at most $r$ prime factors are sometimes called $r$-almost primes. If $\pi_r(x)$ counts the number of $r$-almost primes $\le x$ then $$\pi_r(x) \sim \frac{x (\log \log x)^{r-1}}{\log(x)}.$$ (Compare this to the prime number theorem when $r = 1$.) In particular, we see that the $r$-almost primes have zero density (in any sense), and thus:

2) The density of integers that can not be represented in the form $amn + m + n$ is zero. Similarly, the density of integers that cannot be represented in the form $amn + m -n$ is zero. In particular, the density of the $a$-asterios numbers, the integers that can neither be represented in the form $amn+m+n$ nor $amn+m-n$, is zero.

Let $\pi_{r,2}(x)$ denote the number of twin $r$-almost primes less than or equal to $x$, that is, the number of integers $n \le x$ such that $n$ and $n+2$ are both $r$-almost primes. (For example, $\pi_{1,2}(x)$ counts the number of twin primes less than $x$.) What do we know about this function? Brun was the first to give an upper bound for $\pi_{r,2}(x)$ using sieving techniques. Refinements by others (in particular Selberg) allowed one to obtain the estimate $$\pi_{1,2}(x) \ll \frac{x}{(\log x)^2},$$ which gives the correct (conjectural) order of magnitude. Without going into the Selberg sieve, let me say that what these arguments really give is decent upper and lower bounds of the following kind (for large $x$): $$\frac{A x}{(\log x)^2} < \left\{n < x, \ p \nmid n(n+2) \ \text{if} \ p < x^{\alpha}\right\} < \frac{B x}{(\log x)^2}$$ for non-zero constants $A$ and $B$, where $0 < \alpha < 1$ is some fixed small constant, which we might imagine for the sake of argument is $1/10$. Since every twin prime $>x^{\alpha}$ contributes to this sum, this gives the correct (up to a constant) upper bound for $\pi_{1,2}(x)$. It also gives a lower bound for $\pi_{10,2}(x)$, since if every factor of $n < x$ is at least $x^{1/10}$, then $n$ has at most $10$ prime factors.

The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape. However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the argument of the last paragraph that $\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$. The latter is comparable to the number of primes less than $x$, where the former has an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated by numbers with (a few) small prime factors. In fact, as Kowalski pointed out to me, it is not even obvious that one can easily obtain the correct upper bound for $\pi_r(x)$ simply by sieving over primes. From the asymptotic for $\pi_{r}(x)$, one expects that $$(*): \qquad \pi_{r,2}(x) =^{?} \ O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$ (EDIT: My resident expert reports that this is known. Here is a sketch of the idea in the simpler case where we want to count pairs $n$ and $n+2$ where $n$ is a $2$-almost prime and $n+2$ is prime. First, for a small prime $p$, we want to find an upper bound for the number of $n < x$ such that $n$ is divisible by $p$ and both $n/p$ and $n+2$ are prime. This is a similar problem to counting twin primes, and in a similar way one obtains a bound of the form $O(x/\log x)$ (key point: the implied constant does not depend on $p$). If we wish to bound the number of pairs $(n,n+2)$ such that $n+2$ is prime and $n$ is a $2$-almost prime, we may instead count the triples $(p,n,n+2)$ where $p < x$ is prime, $n < x$ is divisible by $p$, $n/p$ is prime, and $n+2$ is prime. If for each $p < x$ we have a upper bound of $Ax/\log x$ (for the same $A$), in total we obtain the upper bound: $$ \frac{Ax}{\log x} \cdot \sum_{p < x} \frac{1}{p} \sim \frac{Ax \log \log x}{\log x}.$$ Of course, the devil is in the details! END EDIT) All one needs to answer 3) is that the exponent of $\log(x)$ in the denominator is $> 1$.

3). Assuming the expected result (*), the inverse sum of the $a$-asterios primes converges.

Consider the set of integers $S_a$ which do not have any prime factors of the form $\pm 1 \mod a$. This is a reasonable thing to do whenever $|(\Z/a\Z)^{\times}| > 2$. This is a weaker condition, so there are more of these numbers and consequently obtaining upper bounds is harder. We may form the Dirichlet series $$L(s) = \sum_{S_a} \frac{1}{n^s}$$ which has an Euler product: $$L(s) = \prod_{p \not\equiv \pm 1} \left(1 - \frac{1}{p^s}\right)^{-1}$$ Now let $K = \Q(\zeta_a)^{+}$ be the totally real subfield of $\Q(\zeta_a)$. It has degree $r/2 = \phi(a)/2$, where $r > 2$ unless $a = 1,2,3,4$ or $6$. (What we say now only makes sense for $r \ge 2$, in which case $r/2 \in \Z$.) A prime splits completely in $K$ if and only if it is of the form $\pm 1 \mod a$. Looking at the Euler product of $\zeta_K(s)$, we see that, up to a constant which can be explicitly written as some product over primes, $$\zeta_K(s) L(s)^{r/2} \sim \zeta_{\Q}(s)^{r/2}$$ as $s \rightarrow 1^{+}$, and hence $L(s) \sim (s-1)^{(2-r)/r}$ (up to some constant) as $s \rightarrow 1^{+}$. We deduce (Perron's formula) that the number of integers $\le x$ all of whose prime factors are not of the form $\pm 1 \mod a$ is asymptotic to $$ \kappa \cdot \frac{x}{(\log x)^{2/r}},$$ for some non-zero constant $\kappa$. This is the same analysis that gives the asymptotic formula for the number of integers $\le x$ which can be written as a sum of two squares (a result of Landau). We immediately deduce:

4a) The number of integers $a$ such that $ak+1$ (or $ak-1$) has no prime factors of the form $\pm 1 \mod a$ has zero density.

If $r > 2$ (so $a \ge 3$ and $a \ne 3,4,6$) then the power $2/r$ of $(\log x)$ is at most $1/2$. Thus, we actually are led to the following guess:

4b) If $r = \phi(a) > 2$, then one would heuristically expect the inverse sum of integers $k$ such that none of the prime factors of $ak-1$ and $ak+1$ are $\pm 1 \mod a$ diverges. If $r = 2$, so $a = 3$, $4$, or $6$, then (Brun) the series converges.

Here is a related problem of the very same kind: can one count the number of integers $n \le x$ such that both $n$ and $n+1$ can be expressed as the sum of two squares, and prove that there are $\sim x/\log(x)$ such integers (perhaps up to non-zero constant factors)?

Any integer $n$ can be written as the product of two numbers not of the form $\pm 1$ unless every prime factor of $n$ is of the form $\pm 1 \mod a$. The integers all of whose prime factors are $\pm 1 \mod a$ can be analyzed exactly as in the last paragraph. In this case, the number of integers all of whose prime factors are of the form $\pm 1 \mod a$ is asymptotic to $$ \kappa \cdot \frac{x}{(\log x)^{(1-2/r)}}.$$ Suppose that $r > 2$. Then the set of such integers has density zero, and thus the set of integers which have a factor not of the form $\pm 1 \mod a$ has density one. Any set of density one has infinitely many "twins" satisfying any fixed congruence condition. Hence:

5) If $r = \phi(a) > 2$, then there are infinitely many $k$ such that both $ak+1$ and $ak-1$ have a factor not of the form $\pm 1 \mod a$. Indeed, such numbers have density one.

Finally, I have nothing to say about problem 1) besides the remarks I made in my rephrasing of the original question here:

http://mathoverflow.net/questions/49751/chens-theorem-with-congruence-conditions

Answer by Lavender Honey for Counting isomorphism classes of elliptic curves with specific torsion

2011-01-29T02:02:20Z

$\newcommand\F{\mathbf{F}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\SL{\mathrm{SL}}$

Because of the existence of the Weil paring, elliptic curves with such a subgroup only exist when $p \equiv 1 \mod \ N$.

Let $S_N$ denote the set of elliptic curves over $\F_p$ such that $E[N]$ is defined over $\F_p$. It will be slightly easier to assume that $N \ge 3$. In this case, $Y(N)$ is a fine moduli space, and an $\F_p$-point on $Y(N)$ corresponds to a pair $(E,\alpha:E[N] \simeq \Z/N\Z \times \Z/N \Z)$ defined over $\F_p$. Given an elliptic curve $E \in S_N$, how many points does it contribute to $Y(N)$? For a curve $E$ whose automorphism group is $\Z/2\Z$, We see that out of the $|\SL_2(\Z/N\Z)|$ possible choices of $\alpha$ (technical remark, we have fixed a Weil pairing so that $Y(N)$ is connected), $(E,\alpha) \simeq (E,\alpha')$ only if $\alpha' = \alpha$ or $\alpha' = [-1] \alpha$. Thus $E$ contributes $|\SL_2(\Z/N\Z)|/2$ points to $Y(N)(\F_p)$. In general, $E$ may have slightly more automorphisms, and we deduce that (for $N \ge 3$): $$|Y(N)(\F_p)| = |\SL_2(\Z/N\Z)| \sum_{E \in S_N} \frac{1}{|\mathrm{Aut}(E)|}.$$ Note that the quantity on the right is very close to $|\SL_2(\Z/N\Z)| \cdot |S_N|/2$, one only has to worry about the elliptic curves with $j = 0$ or $j = 1728$, and this can be done by hand if one wants to cross all the i's and dot all the t's.

Suppose that $X(N)$ has $c_N$ cusps and genus $g_N$ (there are some explicit slightly unpleasant formulas for these numbers, which can be found (for example) in Shimura's book. All the cusps are defined over $\F_p$ (with $p \equiv 1 \mod N$) so by the Riemann hypothesis for finite fields, $$|Y(N)(\F_p) - (1+p) + c_N| = |X(N)(\F_p) - (1+p)| \le 2 g_N \sqrt{p}.$$ If $g_N = 0$ (which only happens if $N \le 5$), this leads to an exact formula for $|S_N|$. In general, at least for large $p$, we see that $$|S_N| \sim \frac{2p}{|\SL_2(\Z/N \Z)|}.$$

To make this all completely explicit for $N = 3$ (for example), one gets, presuming I have not made a horrible computational error which is quite possible: $$S_3 = \begin{cases} (p+11)/12, & p \equiv 1 \mod \ 12, \\ (p+5)/12, & p \equiv 7 \mod \ 12. \end{cases}$$ (note that $p \equiv 1 \mod 3$):

Of course, "exact formulas" will only exist for $N \le 5$. Some related and slightly more difficult counting problems are also nicely explained by Lenstra here (See 1.10):

https://openaccess.leidenuniv.nl/bitstream/1887/3826/1/346_086.pdf

Answer by Lavender Honey for Which polynomials arise as formulas for a conjugate

2011-01-28T23:45:50Z

$\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$

If we replace $\alpha$ by $\alpha + \lambda$ for $\lambda \in \Q$ (translation), we may replace $Q(X)$ by $Q(X - \lambda) + \lambda$. Similarly, if we replace $\alpha$ by $\mu \cdot \alpha$ (dilation), we may replace $Q(X)$ by $\mu \cdot Q(X/\mu)$.

Let's discuss the case $[\Q(\alpha):\Q] = 3$. By translation, we may assume that the coefficient of $X$ is trivial. By dilation, we may assume that the leading coefficient is $1$ (this specifically uses the fact that we are in degree $3$). Hence $Q(X) = X^2 + c$ for some $c \in \Q$.

If a degree $3$ field $K = \Q(\alpha)$ contains at least two conjugates of $\alpha$ then it contains all the conjugates, and is therefore Galois with cyclic Galois group. Thus $Q$ induces an isomorphism of $K$ of order three, and $\alpha$ must be a root of $Q(Q(Q(X))) - X$ but not a root of $Q(X) -X$. Hence it is a root of $F(X) = (Q(Q(Q(X))) - X)/(Q(X) - X)$, which is an explicit degree $6$ polynomial.

Suppose that $\beta = u + v \sqrt{d}$ is a root of $F(X)$, where $u$ and $v$ are rational and $d$ is not a square. We may write $F(u + v \sqrt{d}) = R(u,v) + S(u,v) \sqrt{d}$, where $R$ and $S$ are polynomials with coefficients in $\Q[a,b]$. Since $F(\beta) = 0$, then $R(u,v) = S(u,v) = 0$. Computing the resultant of these polynomials with respect to $c$, we obtain the equation: $$(1 + 2 u^2 + 4 u^2)(-1 - 18 u + 4u^2 + 8 u^3) v = 0.$$ Since $u$ is rational, it follows that $v = 0$. Thus $F(X)$ does not have any genuine quadratic solutions, and hence $F(X)$ has no factors that are quadratic. (If we assume that $F(X)$ has a cubic factor, we can prove this in a slightly cleaner way. Since $F(X)$ has a cubic factor, it has at most one quadratic factor. Yet $Q(\beta)$ and $Q(Q(\beta))$ are also quadratic roots of $F(X)$, and hence either $Q(\beta) = \beta$ or $Q(\beta) = \sigma \beta$. Both possibilities are impossible.)

We are assuming that $F(X)$ has a cubic factor corresponding to $\alpha$. Then either $F(X)$ factors as a product of two cubics, or as a cubic times three linear factors. In either case, the cubics correspond to fields which are Galois and hence cyclic of degree three. Hence the discriminant of $F(X)$ is a square (easy exercise). We compute explicitly that $$\Delta_F = - (7 + 4 c)^3 (7 + 4 c + 16 c^2)^2,$$ and hence we deduce the necessary relation: $$7 + 4 c = - t^2$$ for some $t \in \Q$. Making the substitution $c = (-7 - t^2)/4$, the polynomial $2^{6} \cdot F(X)$ factors as $A(X,t) A(X,-t)$, where $$A(X,t) = 1 + 7 t - t^2 + t^3 + 18 x - 4 t x + 2 t^2 x - 4 x^2 - 4 t x^2 - 8 x^3.$$ As long as either this polynomial or its cousin $A(X,-t)$ are irreducible, we obtain a cubic with a root $\alpha$ such that $Q(\alpha)$ is a conjugate of $\alpha$. The polynomial $A(X,t)$ is reducible in $t$ if and only if there exists a rational point $A(X,t) = 0$. This turns out to be a rational curve, and we deduce that it is reducible if and only if there exists a $u \in \Q$ such that $$t = \frac{1 + 2 u - u^2 - u^3}{u(u+1)}.$$ On the other hand, $A(X,-t)$ is reducible if and only if there exists a $v \in \Q$ such that $$-t = \frac{1 + 2 v - v^2 - v^3}{v(v+1)}.$$ Hence any forbidden $t$ corresponds to a solution to the equation $$\frac{1 + 2 u - u^2 - u^3}{u(u+1)} = - \frac{1 + 2 v - v^2 - v^3}{v(v+1)},$$ which correspond to points on the curve $$C:u + u^2 + v + 4 u v + u^2 v - u^3 v + v^2 + u v^2 - 2 u^2 v^2 - u^3 v^2 - u v^3 - u^2 v^3 = 0.$$ This curve is smooth in the affine locus. The corresponding projective curve has three points at $\infty$, and two of the corresponding points are singular.

The singularities at these points are nodes (I think), and thus using Plücker's formula, we deduce that $C$ has genus: $$ g = (d-1)(d-2)/2 - n = (4 \cdot 3)/2 - 2 =4.$$ Thus $C$ has finitely many rational points (Faltings). (Note: this calculation may have been wrong, but we won't actually use it.) The curve $C$ has some obvious points at $\infty$ and when $u(u+1)v(v+1) = 0$. Make the substitution $u = x+y$ and $v = x-y$. Then the equation becomes: $$-x - 3 x^2 - x^3 + 2 x^4 + x^5 + y^2 + x y^2 - 2 x^2 y^2 - 2 x^3 y^2 + x y^4 = 0.$$ This is a quadratic in $y^2$. Hence we obtain a degree two covering $C \rightarrow E$, where $E$ is the curve $$E:-x - 3 x^2 - x^3 + 2 x^4 + x^5 + z + x z - 2 x^2 z - 2 x^3 z + x z^2 = 0.$$ Given a rational point on this curve, the discriminant $\Delta$ is rational, and hence $E$ is isomorphic to: $$\Delta^2 = 4(4 x^4 + 3 x^3 + x^2 + 2 x + 1) = 4(1+x)(1+2x)(1-x+2x^2)$$ This is birational to an elliptic curve which turns out to have conductor $112$. Cremona's program mwrank tells me has Mordell-Weil group $\Z/2\Z$. Hence the only rational points correspond to $x = -1/2$ and $x = -1$, which pull back to the "obvious" rational points on $C$. Hence we have determined $C(\Q)$ completely, and we see that one of $A(X,t)$ or $A(X,-t)$ is always irreducible. Hence we deduce:

If $\alpha$ is a cubic irrationality with $Q(\alpha) = \sigma \alpha \ne \alpha$, then, replacing $\alpha$ by $\lambda \alpha + \mu$, $Q(X)$ is of the form: $$Q(X) = X^2 - \frac{7+t^2}{4}$$ with $t \in \Q$. Conversely, for any such $Q(X)$, there exists a cubic irrationality $\alpha$ such that $\sigma \alpha = Q(\alpha)$.

Example: $t = 1$, and $Q(X) = X^2 - 2$. Then $\alpha = 2 \cos(2 \pi/7)$.

For higher degree polynomials, things will be even more of a mess, because there will be various possibilities corresponding to what the Galois closure is, &. &. The answer will have the following flavor: It will correspond to the rational points on a bunch of varieties minus rational points on subvarieties corresponding to degeneracies (which in this case turn out to be empty).

Answer by Lavender Honey for Is there a connection between exceptional Galois groups and Ramanujan's partition congruences

2011-01-26T23:48:41Z

No.

$$ $$

Chen's Theorem with congruence conditions.

2010-12-17T20:17:15Z

I would like to revisit a closed question of asterios in a more MO kind of way, because it cuts quite close to a related question about sieving that might be of general interest.

The original question: http://mathoverflow.net/questions/49669/finite-or-infinite-again-closed amounts to the following. Given a positive integer $a \ge 3$, do there exist infinitely many integers $k$ such that (simultaneously):

$ak+1$ does not have any non-trivial factors of the form $\pm 1 \mod a$.
$ak-1$ does not have any non-trivial factors of the form $\pm 1 \mod a$.

Presumably the answer is yes, because one expects there to be infinitely many twin primes $p$, $p+2$ satisfying any reasonable congruence condition. If one wants to prove something unconditionally, however, one could look towards generalizing the result of Chen. Drifting away from the original problem slightly, and making things more explicit, one could make the following conjecture:

(*?) There exist infinitely many primes $p$ such that $p+2$ is either prime or a product $qr$ of two primes $q$ and $r$, where $q$ and $r$ are of the form $1 \mod 4$.

The question is: Is (*?) amenable to known sieving techniques, or, in the other extreme, does the imposition of congruence conditions on $q$ and $r$ create a difficulty similar to the parity problem?

(Of course, one can easily modify the conjecture in various ways, imposing congruence conditions on $p$ and different congruence conditions on $q$ and $r$ and then ask the analogous conjecture, providing that the congruence conditions don't combine in unpleasant ways. This is slightly tricky: one would not want to insist that $p \equiv 1 \mod 4$ and that $p+2$ was either prime or had two prime factors, both of the form $-1 \mod 4$, not because the resulting conjecture is false, but because it would then be equivalent to the twin prime conjecture, and would fall prey to the parity problem.)

(*?) is almost an amalgam of two (non-trivial!) sieving problems. Drop the congruence conditions on $q$ and $r$ and one gets Chen's theorem. Simply requiring that every prime divisor of $p+2$ is of the form $1 \mod 4$ on the other hand is close (in fact, slightly stronger) to asking that $p$ be represented by the quadratic form $a^2+b^2-2$, and the question of counting primes represented by quadratic forms was answered by Iwaniec in '74.

Answer by Lavender Honey for Can finitely many values of a polynomial determine it?

2011-01-21T02:35:07Z

You are asking for a condition on $f$ and $g$ such that the curve: $$C: f(y) - g(x) = 0$$ has a uniformly bounded number of rational points. Note that if $f$ and $g$ are equivalent under an affine transformation, then $C$ is divisible by a linear factor and is not reducible. The converse is almost true. Namely, as long as the degree of $f$ and $g$ is sufficiently large, and $f$ and $g$ are not of the form $a \circ b$ for polynomials $a$ and $b$ of degree $> 1$, then $C$ will be irreducible. (This follows from CFSG. Of course, using composition of functions one can create many degenerate examples: $P(y)^2 - Q(x)^2$ is divisible by $P(y) - Q(x)$. The example in the comments giving a example in even degrees arises in this way, by taking a degree two example and using composition.) If the degree $d$ of $f$ (and $g$) is prime, then $f$ and $g$ are certainly indecomposible, so let's concentrate on that case, since there are no reductions to smaller degree. For convenience, let's also only consider the case when $C$ is irreducible (if $d$ is prime, this is automatic if $d$ is sufficiently large, by the remark above.)

If $C$ has genus at least two, then $C$ will have only finitely many rational points (Faltings). Work of Caporaso, Harris, and Mazur:

http://www.ams.org/journals/jams/1997-10-01/S0894-0347-97-00195-1/home.html

suggests that the number of solutions may be even be bounded in terms of the genus, and hence in terms of the degree. Whether you believe Lang's conjectures or not, you are unlikely to disprove Lang's conjectures easily, so any negative example to your claim should come from a pair of functions $f$ and $g$ so that $C$ has small genus.

In small genus, we may have many rational points, but as far as integral points we also have Siegel's theorem to content with. A projective model $\widetilde{C}$ of $C$ is given by $Z^d f(Y/Z) - Z^d g(X/Z) = 0$. Setting $Z = 0$, we obtain the equation "at infinity" $Y^d - X^d = 0$, which has $d$ points over the complex numbers. Hence, assuming $d \ge 3$, $$\# \widetilde{C} \setminus C \ge 3.$$ By Siegel's theorem we deduce $C$ has only finitely many integral solutions. Thus, when the degree $d$ is prime and sufficiently large (or more generally, providing one avoids degeneracies arising from the phenomena alluded to in the first paragraph), any $f$ and $g$ in different equivalence classes will only coincide on a fixed number of integers.

Your question, however, asks whether there is a uniform bound. There is certainly no uniform bound for Seigel's theorem, at least when the genus is $\le 1$. There is a standard ``renomalization'' trick which takes a curve with infinitely many rational points and produces a curve with many integral points. This trick works in this case. Specifically, suppose that $C: f(y) - g(x) = 0$ has infinitely many rational points. Then there certainly exists some integer $N$ such that $C$ has a bizillion points of the form $(u/N,v/N)$ (take $N$ to be a common denominator). We may then write down the different integral model: $$C': N^d f(y/N) - N^d g(x/N) = 0,$$ which now has a bizillion integral points $(u,v)$. This also allows one to answer your question in general degrees, simply by choosing $f$ and $g$ so that $C$ has infinitely many rational points, and then renomalizing appropriately. The easiest specific example would be to take $C$ of genus zero. For example, take $f = t^n$ and $g = t^{n-1}(t-1)$. Then $C: f(y) - g(x) = y^n - x^{n-1}(x-1)$ has genus zero, as can be seen from the paremetrization $$x = \frac{1}{1 - t^n}, \qquad y = \frac{t}{1 - t^n}.$$ From the above construction, there will exist positive integers $N$ such that the polynomials $t^n$ and $t^{n-1}(t - N)$ will take on the same bizillion values. This answers your question in the negative.

EDIT: I guess the last example can be made quite concrete. Let $$N = (1 - 2^d)(1 - 3^d)(1 - 4^d) \ldots (1 - M^d).$$ Then $t^d$ and $t^{d-1}(t-N)$ both take on the values $\displaystyle{\left(\frac{aN}{1 - a^d} \right)^d}$ for $a = 2, \ldots, M$.

Answer by Lavender Honey for Can the factorisation of (p) in a number field K be described by the minimal polynomial of a primitive element?

2011-01-15T01:34:22Z

The discriminant of the polynomial is equal to the discriminant of the field times the square of the index $[\mathcal{O}_K: \mathbf{Z}[\alpha]]$. Hence, if the $p$-adic valuation of the discriminant of $f(x)$ is $0$ or $1$, then one already has an integral basis $p$-adically. Yet already the examples $x^2 - p^2$ and $x^2 - a p^2$ where $a$ is a quadratic non-residue modulo $p$ (and $p$ is odd) show that the situation is hopeless in general, even if one knows that $p$ is unramified.

Since, I imagine, you are trying to determine the behavior of Hecke algebras from computing characteristic polynomials of small Hecke operators, you might be able to get mileage out of the following observation, which says that "small" discriminants must come from ramification rather than index. Suppose, for example, that $f(x)$ modulo $p$ is exactly divisible by $a(x)^2$ for some irreducible polynomial $a(x)$ over $\mathbf{F}_p$. By Hensel's lemma, $f(x)$ is divisible by a lift $A(x)$ of $a(x)^2$. There are the following possibilities:

$A(x)$ is irreducible, and corresponds to an unramified prime of residue degree $2d$.
$A(x)$ has two irreducible factors, corresponding to a pair of unramified primes of residue degree $d$.
$A(x)$ is irreducible, and corresponds to a ramified prime of residue degree $d$ and ramification index $2$.

In the first two cases, the index of $\mathbf{Z}[\alpha]$ in $\mathcal{O}_K$ is divisible by $p^d$, and so the discriminant is divisible by $p^{2d}$. Hence, if you know that the valuation of the discriminant is less than $2d$, you can deduce that $A(x)$ corresponds to a ramified prime.

Row reduction of sparse matrices

2011-01-10T19:38:21Z

Let $p$ be prime (of size roughly $100$, say). Suppose that $M$ is a matrix with coefficients in $\mathbf{F}_p$ with roughly $An$ rows and $n$ columns, where $A>1$ is some fixed small constant. Suppose, moreover, that

Every row of $M$ has at most $B$ non-zero entries, which are all $\pm 1$, for some small $B$.
The rank of this matrix is small, say, of order $O(\sqrt{n})$ (this may or may not be relevant.)

I am interested in computing the rank of $M$. Does the fact that $M$ is sparse imply that there are relatively efficient ways of computing this rank? How large should I expect to be able to take $n$ and still be able to compute the rank? The current program I am running (which was kindly written [very quickly without a view to optimization] for me by someone else) chokes up with $n \sim 5000$ or so, and for $n \sim 4000$ seems to take around $20$ minutes.

Answer by Lavender Honey for Old question of Serre on discriminants of a sequence of polynomials

2011-01-14T17:05:32Z

Some people seem a little confused by the wording of this question. The better way of phrasing the question is: Is there a lower bound on the root discriminants of polynomials of degree $d$ as $d \rightarrow \infty$. The survey paper by Odlyzko (noted by Gerry in the comments above):

http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1990__2_1/JTNB_1990__2_1_119_0/JTNB_1990__2_1_119_0.pdf

mentions this as a question of Serre and others (it is a pretty obvious question to ask), and also points out that essentially nothing is known about this question. I don't think the situation has changed considerably in the intervening $20$ years.

As Franz notes, the existence of class field towers is no way implies the existence of polynomials with bounded discriminant - "most" fields are not monogenic.

Some have wondered - perhaps optimistically - whether asymptotically there is even a bound $$|\mathrm{disc}(P)| > (cd)^d$$ For irreducible $P$, where one can take $c$ to be some fixed constant $ > 1$ unless $P$ is essentially a cylotomic polynomial. If this was true, then it would imply that the Mahler measure of an algebraic integer that is not a root of unity is bounded away from $1$, answering a question of Lehmer.

Comment by Lavender Honey

2012-09-24T21:43:00Z

I don't see that JSE makes such a conjecture at all, which is just as well, because it is trivially false for any prime $p \equiv 1 \mod 4$ such that the class number of $\mathbf{Q}(\sqrt{p})$ is $> 1$. (say $p = 229$).

Comment by Lavender Honey

2012-08-18T20:38:47Z

For $\mathrm{dim}(V) = 2$, see: jstor.org/stable/40067979. For $\mathrm{dim}(V) \ge 2$, take symmetric powers.

Comment by Lavender Honey

2012-02-25T03:19:46Z

@GH: Rouché? Touché!

Comment by Lavender Honey

2012-02-02T22:17:30Z

What if $A \simeq E^d$?

Comment by Lavender Honey

2012-01-28T01:08:01Z

@FL: Of course you are right, which makes the result even more impressive.

Comment by Lavender Honey

2012-01-26T04:27:53Z

Possibly the most "natural" application of the Artin conjecture is the Cebotarev density theorem. Of course, Cebotarev famously proved his theorem only using the (known) case of Artin for cyclic extensions $L/K$.

Comment by Lavender Honey

2012-01-16T07:07:15Z

For the claim in the last sentence you should also assume that $p$ is unramified in $K$. (Otherwise, imagine $K = \mathbf{Q}[x]/(x^3-x-1)$ and $\alpha = -23$).

Comment by Lavender Honey

2012-01-16T00:14:10Z

"I assume that we have to look at the idelic formulation of global class field theory to find it.". Not really. There are many quadratic extensions $K(\sqrt{\alpha})$ which are clearly not cyclotomic (for example, because they are ramified at one prime above $p$ but not at a different prime above $p$.

Comment by Lavender Honey

2012-01-12T02:32:59Z

BTW, to add to Hansen's answer, it might be worth noting that, if $p$ divides $N$, your definition of $F$ and $G$ being connected by an edge coming from a prime above $p$ is equivalent to asking that the mod $p$ (or some prime above $p$) Galois representations $\overline{\rho}_G$ and $\overline{\rho}_F)$ have isomorphic semi-simplifications. Thus Hansen's remark that all forms are connected to a form of level prime to $N$ follows from a theorem of Ash-Stevens. You might want to ask Ash about it - or even better, ask his student(?) David Hansen.

Comment by Lavender Honey

2012-01-10T04:25:38Z

Talking to yourself is the first sign of madness...

Comment by Lavender Honey

2011-12-19T04:59:22Z

Dear Matt, Thanks! (For some reason, I always seem to get that wrong...)

Comment by Lavender Honey

2011-11-23T20:24:00Z

$A(q) = q \prod_{n=1}^{\infty} (1 - q^{8n})^3 = \sum_{0}^{\infty} (2n+1) (-1)^n q^{(2n+1)^2}$ is modular, and leads to quite a few "easy" examples.

Comment by Lavender Honey

2011-11-23T02:03:44Z

This doesn't quite fall under the scope of your question, but it's not so easy to decide, for example, whether the modular forms $A(q) = q \prod_{n=1}^{\infty}(1 - q^n)^{24}$ and $B(q) = 0$ have coefficients in common.

Comment by Lavender Honey

2011-11-10T23:44:25Z

The Tits alternative gives lots of free subgroups which are Zariski dense, and their abelianizations are long way from being finite (see Theorem 3 of Tits' paper). For the converse, Tom Church gave examples of finite groups, but there are infinite examples as well, for example, $\mathrm{Sp}_{2n}(\mathbf{Z})$ naturally contains a copy of $\mathrm{SL}_n(\mathbf{Z})$, and the latter has finite abelianization for all $n \ge 3$.

Comment by Lavender Honey

2011-11-07T01:18:05Z

@BSteinhurst: Really? You mean if someone said: "the real number $\pi^{\pi^{\pi}}$ is almost surely transcendental, but there are currently no possible approaches to proving this" you would take issue with that?