User junkie - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T04:46:11Z http://mathoverflow.net/feeds/user/5267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110767/lotteries-turans-problem-and-minimization-of-risk/110947#110947 Answer by Junkie for Lotteries, Turan's problem, and minimization of risk Junkie 2012-10-28T23:11:21Z 2012-10-30T05:03:44Z <p>This is not a real "answer" but an observation. Each 6-tuple has ${6 \choose 4}$ 4-tuples, so it stands to reason that once $N$ is some smallish multiple $m$ of ${50\choose 4}/{6\choose 4}$ then Bob's your uncle, and you can come reasonably close to equidistribution. This is to be contrasted with (say) the binomial expectation. The number of fours (or more) you expect is ${6\choose 4}m$, and the reasonable size $N=138180$ gives $m=9$ and 135 fours. The binomial distribution gives a variance about of slightly under $135$. I expect that one can essentially ensure about 135, plus or minus a small amount, via some covering selection.</p> <p>ADDITION: In verities, the binomial model does not well model the random choices, as they have considerably higher variance, 180-185 compared to 134.8. I do not understand the theoretical concepts <em>in toto</em>, but an aspect is that the coverage of fours from the random sixes is already askew.</p> <p>UPDATE: OK, here's the skinny on covering. I did a rather simple process. Do the following 138180 times. Pick a 4-tuple that so far has not appeared 9 times. Append to it the 2 numbers for which the resulting six minimizes the sum of the current counts of its 15 sub-fours. Accumulate the counts of 4-tuples from this six.</p> <p>Then apply a bit of post-processing if you want (throw out populous sixes). This gives a set of 138180 6-tuples in which every 4-tuple appears between 7 and 11 times (the average is 135/15 or 9, with random choices of sixes the four-counts will range from 0 to 20 or more). Then simulate the ${50\choose 6}$ lotteries. These give an expectation of 135 fours, the minimum was 120 and the maximum was 149. The variance was a mere 5.8, versus 135 (binomial) or 180 (random). The binomial distro gives less than 120 a 8.9% chance (and more than 149 a 10.7% chance). As added above, the actual random distro is even worse than binomial.</p> <p>I think this shows that with a (small) bit of work, some quite good variance reduction is possible. You can try to further trim the ends if desired. In the actual example, their edge was about 20-25% when free bets were included (later comments suggest 15-20% over the history). The accounting on page 7 says "12.8%" for just the cash component, but I get 425840/400000 is 6.4%. This analysis also lacks the jackpot, which I guess is equally likely to help/hurt among the big players (it is slightly chancey that only 1 of the 45 jackpots was hit, given there are 2-4 groups each buying up to 1/30 of the pool every time).</p> <p><img src="http://www.freeimagehosting.net/newuploads/3tygt.jpg" alt="alt text"></p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72329#72329 Answer by Junkie for Conjugacy for $p$-adic matrices of finite order Junkie 2011-08-08T04:43:10Z 2011-08-08T13:24:50Z <p>Here is a trial proof for the question over $Q_p$.</p> <p>Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$).</p> <p>Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers).</p> <p>Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$ for $(u,p)=1$. Further note, if $\bar\Phi_u=\prod \bar g$ then $\bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$, and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split, in other words this is the minimal polynomial. This follows since the reduction (mod $p$) of the companion matrix of $f$ is itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$.</p> <p>So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$.</p> <p>From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. The general Jordan form classifies the conjugacy type over a field, as is $Q_p$.</p> <p>Note that, $\Phi_3\Phi_6$ and $\Phi_6^2$ give 4x4 matrices with order 6, failing for $p=2$.</p> http://mathoverflow.net/questions/69253/are-there-any-solutions-to-2n-3m1/69256#69256 Answer by Junkie for Are there any solutions to $2^n-3^m=1$ Junkie 2011-07-01T12:34:26Z 2011-07-01T12:34:26Z <p>Here is the proof of Gersonides [Levi ben Gershon] (1343) for $2^n-3^m=1$. It uses nothing more that arithmetic modulo $8$.</p> <p>Case I: $m$ is even. Then $3^m$ is 1 mod 4, so $2^n$ is 2 mod 4, implying $n=1$ and $m=0$.</p> <p>Case II: $m$ is odd. Then $3^m$ is 3 mod 8, so $2^n$ is 4 mod 8, implying $n=2$ and $m=1$.</p> <p>The alternative equation $3^m-2^n=1$ follows similarly when $m$ is odd, but is a bit more tricky when $m$ is even (hint, factor $2^n=3^m-1=(3^{m/2}+1)(3^{m/2}-1)$ and argue from there).</p> http://mathoverflow.net/questions/67308/reference-requested-for-lim-n-rightarrow-infty-frac-sum-i1n-bars/67316#67316 Answer by Junkie for Reference requested for $\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^{n} \bar{s}(i)}{n^2} = \frac{\pi^2}{30}$ Junkie 2011-06-08T21:51:16Z 2011-06-08T23:27:37Z <p>If my understanding is correct, for "squarefree part" can be "squarefree kernel" in other cases, the generating Dirichlet series is $${\zeta(2s)\zeta(s-1)\over\zeta(2s-2)}=\prod_p\biggl(1+{p\over p^s}+{1\over p^{2s}}+{p\over p^{3s}}+\cdots\biggr)=\sum_n{\bar s(n)\over n^s}$$ alternating $1$ and $p$ as the coefficients, which is a $\zeta$ quotient as indicated. The residue at $s=2$ is $\zeta(4)/\zeta(2)={\pi^4/90\over\pi^2/6}={\pi^2\over 15}$, so that by Perron's formula $$\sum_{n\le X} \bar s(n)={1\over 2\pi i}\int_{(\sigma)}{\zeta(2s)\zeta(s-1)\over\zeta(2s-2)}{X^s ds\over s} \sim {\zeta(4)\over 2\zeta(2)}X^2={\pi^4/90\over2\pi^2/6}X^2={\pi^2\over 30}X^2,$$ with usual conditions about convergence in vertical strips, which are OK here. Dividing by $X^2$ gives the desired limit.</p> http://mathoverflow.net/questions/65841/homomorphism-from-hat-mathbbz-to-mathbbz/65848#65848 Answer by Junkie for Homomorphism from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$ Junkie 2011-05-24T13:35:50Z 2011-05-24T13:35:50Z <p>Let $\phi$ be such a homomorphism, on additive groups $\hat Z\rightarrow Z$. Write $(\vec x,\vec y)\in\hat Z$ for the element that is $x$ on primes that are 1 mod 3, and $y$ on primes that are 2 mod 3. </p> <p>Then $\phi(\vec x,\vec 0)=0$ for all $x\in Z$, for $(\vec x,\vec 0)$ is $l$-divisible for any prime $l$ that is 2 mod 3. The symmetrical argument claims $\phi(\vec 0,\vec y)=0$ too.</p> <p>Without loss of generality, we can assume that a preimage of $1$ is given by $(\vec 1,\vec 1)$.</p> <p>Next, applying the group law and setting $\alpha=\phi(\vec 1,\vec{-1})$, we derive the system $$1+\alpha=\phi(\vec 1,\vec 1)+\phi(\vec 1,\vec{-1})=\phi(\vec 2,\vec 0)=0$$ $$1-\alpha=\phi(\vec 1,\vec 1)-\phi(\vec 1,\vec{-1})=\phi(\vec 0,\vec 2)=0$$ This is impossible, so $\phi$ does not exist.</p> http://mathoverflow.net/questions/63544/computing-on-a-computer-the-first-few-non-trivial-zeros-of-the-zeta-function/63552#63552 Answer by Junkie for Computing (on a computer) the first few (non-trivial) zeros of the zeta function of a number field. Junkie 2011-04-30T19:04:49Z 2011-05-02T02:54:01Z <p>Step I: Put the degree 24 polynomial into Magma, make it a number field, and call LSeries on it. This divides the $L$-function into a product of 7 distinct ones (Dokchitsers code, under an attribute called "prod" on the L-series object), given by Artin representations. So my plan was to compute zeros for each of these $L$-functions, with them being a subset of those of $\zeta_M$ naturally, and recombine. As representations for $SL(2,3)$ this is as $$1\oplus \omega\oplus\bar\omega\oplus 2\tau_2\oplus 2\tau_2\omega\oplus 2\tau_2\bar\omega\oplus 3\kappa_3.$$ Or as an $L$-function product $$\zeta_M=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\tau_2)^2\cdot L(\tau_2\omega)^2\cdot L(\tau_2\bar\omega)^2\cdot L(\kappa_3)^3.$$</p> <p>EDIT: Oh I see now, you wanted it for $M$ w/o assuming analyticity of Artin constituent parts, but Magma automatically dissembles it, and assumes. But now at this point, I see you were trying to avoid this decomposition perhaps, but then I really don't see how you could work with $\zeta_M$ directly, as the conductor is too big, being $163^{16}$. From the standpoint of scientific evidence, it is probably enough that CheckFunctionalEquation in Magma for each of the above constituents gives an answer indistinguishable from zero, though the feel of zeros is also nice.</p> <p>Another way to test individual analyticity is to use subfields. There is a unique (up to isomorphism) quartic subfield $K_4\subset M$, with $$\zeta_K=\zeta\cdot L(\kappa_3).$$ The conductor of $\zeta_K$ is small enough to compute with it directly (no decomposition), so the analyticity of $L(\kappa_3)=\zeta_K/\zeta$ can be checked (numerically) by seeing if Riemann zeros are a subset of those of $\zeta_K$. EDIT: Of course, $\kappa_3$ is induced from the a nontrivial linear character of the $Q_8$ subgroup of $SL(2,3)$, so analyticity already follows by induction.</p> <p>There is also a unique (up to isomorphism) sextic subfield $L_6\subset M$ with $$\zeta_L=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3),$$ so the linear parts can be isolated here by $\zeta_L/\zeta_K$, though maybe superfluous ("easy" theorem)? In that matter, already the cubic subfield $C_3\subset M$ has $\zeta_C=\zeta\cdot L(\omega)\cdot L(\bar\omega)$ if desired.</p> <p>There is a unique (up to isomorphism) octic subfield $N_8\subset M$ with $$\zeta_N=\zeta\cdot L(\tau_2\omega)\cdot L(\tau_2\bar\omega)\cdot L(\kappa_3).$$ The integer ring has discriminant $163^4$ so $\zeta_N$ is likely still computable directly. By quotient $\zeta_N/\zeta_K$, this can tell about the product of conjugate 2-dimensional $L$-functions. EDIT: see below for another idea, using twists of $\zeta_N$.</p> <p>Finally the unique duodecic subfield $R_{12}\subset M$ has $$\zeta_R=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3)^3.$$ So this gives nothing new, and the discriminant is too large anyway.</p> <p>Note that none of the parts has $L(\tau_2)$ directly, only $\zeta_M$ itself that is too hard to compute. EDIT: However, by Rankin-Selberg I think(?) it follows that the analyticity of $L(\tau_2)$ is equivalent to that after twisting by $\omega$ to get $L(\tau_2\omega)$.</p> <p><strong>Answer?</strong>: For that matter, recurring to the octic subfield $N_8$, instead of using just the Dedekind $\zeta$-function of $N_8$, twisting it by $\omega$ could be profitable, achieving $$L(\sigma_8\omega)=L(\omega)\cdot L(\tau_2\bar\omega)\cdot L(\tau_2)\cdot L(\kappa_3),$$ $$L(\sigma_8\bar\omega)=L(\bar\omega)\cdot L(\tau_2)\cdot L(\tau_2\omega)\cdot L(\kappa_3),$$ where $\sigma_8=1\oplus\tau_2\omega\oplus\tau_2\bar\omega\oplus\kappa_3$ is for the Dedekind of $N_8$. Also $\kappa_3\omega=\kappa_3$ nicely. Noting here the twisting gives $$L(\sigma_8)L(\sigma_8\omega)L(\sigma_8\bar\omega)=\zeta_M,$$ this could provide a way to compute $\zeta_M$, as each part of the left is known as analytic I presume. The conductor of $L(\sigma_8)$ is $163^4$, and that for the twists is $163^6$. Or one can avoid $\zeta_M$ alternatively, as my computation is $$L(\tau_2)^2={L(\sigma_8\omega)L(\sigma_8\bar\omega)\zeta^2\over\zeta_{N_8}\zeta_{L_6}},$$ where each factor on the right should be known (easily?) to be holomorphic away from the $\zeta$-pole. Note this exploits the linear characters of $SL(2,3)$, and you have none for your next case of $SL(2,5)$.</p> <p>In all cases, zeros need to be computed, and the right tool is L-calc. But I don't know if it is really feasible to go too far for $L(\sigma_8\omega)$ of conductor $163^6$, without intensive effort.</p> <p><strong>Part II</strong>: Zeros of $L(\tau_2)$ computation (example): I compute its first few zeros for the 2-dimensional representation of real character. For this representation, with $10^5$ coeffs (taking 6sec in Magma), I exported these to Lcalc (a somewhat hackish tool of M. O. Rubinstein, included in Sage I suppose but I did it direct), which returns after 12 seconds, listing the imaginary parts of the first 10 zeros on the half-line (also proving RH up to that point by A Turing method):</p> <pre><code> 0 0.99014365233 1.38830360231 2.35103235859 3.45296629741 4.32708276131 4.73989005257 5.42392092883 5.99574967707 6.70167394143 </code></pre> <p>The first zero is at $s=1/2$ as the sign is $-1$. The second is about $s\approx 1/2+0.990i$</p> <p>For reference, here are the L-calc settings I used, hackish as I say:</p> <pre><code>1 0 100000 0 2 .5 0.5 0.0 .5 0.5 0.0 51.884511447957879460656106859439682023 -1.0 0.0 0 </code></pre> <p>As explained in their help, the first "1" says the coeffs are integers, the second "0" says they are not special, the 3rd "100000" says they are that many, the 4th "0" says not periodic, the 5th "2" says degree 2, the 6th ".5" and "0.5 0.0" say the form of the gamma factor, the "51.884" is $\sqrt{163^2/\pi^2}$ as the analytic conductor, the "-1.0 0.0" is the sign, and the final "0" says no poles. Then the 100000 coeffs are given as integers.</p> http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63649#63649 Answer by Junkie for Checkmate in $\omega$ moves? Junkie 2011-05-01T20:57:10Z 2011-05-01T20:57:10Z <p><img src="http://www.freeimagehosting.net/uploads/d2ac857a24.jpg" alt="alt text"></p> <p>Here is another one, hopefully it fits on one(!) board with no more modifiers. White (in check) plays Kh3xQg3, and Black threatened by Rxb7#, moves the Rf4 arbitrarily far to the east, uncovering check from the Bd6. White just takes the bishop (any way), and Black has no defense but to keep on checking White horizontally with the eastern rook, with the White king heading east until it (finally!) attacks the rook, when then White will win via Rxb7.</p> <p>Notes: White has no other way to avoid the annoying rook checks, for the self-guarding Black rooks on the e-file prevent king movement to the west, and no interposes are possible by geometry. White's king can simply move east on ranks 2 and 3, but it doesn't matter too much. Black's moving the rook north on move 1 (when uncovering the bishop check) is not effective, for then White can interpose a rook on future vertical checks. The double check Rg4+ on Black's move 1 is also easily defeated by capturing that rook with the king. The only loose end is then whether by White's 1. Ki2 (not taking the queen) a faster checkmate is possible. The answer is no, for White doesn't even win, for Black can check forever with the queen on the g-file, the White king restricted to rank 3 and below.</p> <p>Unless there is something missing, this seems to work also.</p> http://mathoverflow.net/questions/63529/universal-cover-of-sl2r-does-it-admit-central-extensions/63554#63554 Answer by Junkie for universal cover of SL2(R): does it admit central extensions? Junkie 2011-04-30T19:24:36Z 2011-05-01T20:27:06Z <p><strong>Non-answer I:</strong> just a reference for the difference between topological and abstract groups: I think the first part of Chapter I of the above Moore <a href="http://archive.numdam.org/article/PMIHES_1968__35__5_0.pdf" rel="nofollow">reference</a> addresses the question, but I'm not sure it exactly answers it w/o more work. He notes at the start of chapter 1 that he is talking about abstract groups there, and "For topological groups [the abstract fundamental group] need not coincide with the usual [topological] fundamental group although it does in the most important cases (e.g. semi-simple Lie groups)", but I see no specific proof of this latter fact.</p> <p>An abstract group $G$ is "simply connected" if for every central extension $$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$ there is a <em>unique</em> (splitting) homomorphism $\phi: G\rightarrow E$ that composes with the quotient map to yield ${\rm id}_G$. He shows this is equivalent to $H^1(G,T)=H^2(G,T)=0$ where $T$ is the circle group, and notes in passing that $H^1(G,T)=0$ is equivalent to $[G,G]=G$.</p> <p>A cover $E$ of $G$ (more properly defined on the morphism) is when $E=[E,E]$ and the kernel of the map from $E$ to $G$ is central in $E$. He then proves that any group with $G=[G,G]$ has a (unique) simply connected covering group $F$, and that this is characterized by the inflation property that $H^2(G,T)\rightarrow H^2(F,T)$ is the zero map. He then shows that that $F$ is universal in the sense expected. All of this is at the abstract group level from my reading.</p> <p>Notably the "universal central extension" at the group level is the "universal covering group" at this level. In section 2 of Chapter I, he then compares to the topological case, and with Theorem 2.3, noting the ("somewhat amazing") fact that: "if $$1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$$ is a topological extension of $G$ by $A$ which splits as extension of abstract groups, then it splits as extension of topological groups", assuming $G=[G,G]$ here (and not just that the derived subgroup is merely dense). In fact, he shows that a splitting map $\phi: G\rightarrow E$ must be continuous.</p> <p>However, as he assumes that the extension is topological, it doesn't seem to show (from what I can tell) that the abstract universal covering group is indeed the topological universal covering group as would be desired.</p> <p><strong>Answer II:</strong> Another reference, more specific to $SL_n$ is Chapter 1 of <a href="http://users.ictp.it/~pub_off/lectures/lns023/Rehmann/Rehmann.pdf" rel="nofollow">http://users.ictp.it/~pub_off/lectures/lns023/Rehmann/Rehmann.pdf</a> </p> <p>Therein (first page) the central extension $$1\rightarrow K_2(n,k)\rightarrow St_n(k)\rightarrow SL_n(k)\rightarrow 1$$ gives the Steinberg group as the universal central extension, and $K_2(n,k)$ as the (algebraic) fundamental group. There is a epimorphism surjecting $K_2(n,k)\rightarrow K_2(k)$, from the inductive limit $St(k)\rightarrow SL(k)$, and for $n>2$ this is isomorphic. Milnor notes that $K_2(R)$ is the direct sum of $Z/2$ and an uniquely divisible group, so in particular larger than the $Z$ from topological coverings. See Example 1.6 of "Algebraic K-theory and quadratic forms" <a href="http://www.springerlink.com/content/t025u1152j330325/" rel="nofollow">http://www.springerlink.com/content/t025u1152j330325/</a> Note: I think he speaks of the Milnor $K$-group in general, but that it is not a bother at $K_2$.</p> <p>In Theorem 1.2 of Rehmann's paper (and Theorem 4.2 for general Chevalley), the generators of the kernel are given, from Matsumoto's <a href="http://www.numdam.org/item?id=ASENS_1969_4_2_1_1_0" rel="nofollow">work</a>.</p> http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63517#63517 Answer by Junkie for Checkmate in $\omega$ moves? Junkie 2011-04-30T11:33:06Z 2011-04-30T11:33:06Z <p>Here is my first try at a solution. Your idea was a good one, but bishops are better than rooks, I surmise.</p> <p>The two pictures here are placed in some distinct parts of the infinite board. The first just ensures it is White to move (in check), and that White's king will never play a role, as capturing a black unit, which are nearly stalemated as is, will release heavy pieces.</p> <p><img src="http://www.freeimagehosting.net/uploads/3c8e277e7d.jpg" alt="alt text"> <img src="http://www.freeimagehosting.net/uploads/72ef1c9b7e.jpg" alt="alt text"></p> <p>So White is left to checkmate with the four bishops and pawns. White threatens checkmate via a check from below on the northwest diagonal, and Black can only avoid this by moving the bishop northeast some amount. Upon Black moving this bishop, White then makes the bishop check anyways, the Black king moves where the Black bishop was, the pawn moves with check, the Black king again retreats northeast along the diagonal, and then White alternately moves the dark-square bishops, giving checks until the Black bishop is reached when it is mate.</p> <p>The point of this second picture is that White cannot checkmate Black unless the Black bishop plays a role. Four bishops are not enough to checkmate a king on an infinite board, and hopefully I have set it up so that the White pawns play no part once Black starts the king running northeast. Pawns are not too valuable when they cannot become queens.</p> <p>In extended chess notation, White plays 1. Ke5 on board A, then Black plays 1...Bz26 on board B, followed by 2. Bg3+ Kf6 3. e5+ Kg7 3. Bi5+ Kh8 4. Bf10+ Ki9 5. Bk7+ Kj10 6. Bh12+ ..., as White successively cuts off NW-SE diagonals until the Black bishop is reached. By moving the bishop X squares northeast on move 1, Black can delay the checkmate for X moves, if I set this up proper.</p> <p>Other plans by White should be beatable by moving the Black king off the long diagonal or capturing the light White bishop with the pawn. Once Black's king exits the area with the pawns, the Black bishop must be a part of the mating pattern. I don't think the Black king can be forced back to that area.</p> <p>Well, this is a first try.</p> http://mathoverflow.net/questions/34290/what-are-some-interesting-corollaries-of-the-classification-of-finite-simple-gro/34458#34458 Answer by Junkie for What are some interesting corollaries of the classification of finite simple groups? Junkie 2010-08-04T00:37:26Z 2010-08-06T10:21:56Z <p>I have a few comments, but will make this an answer due to length, and I just got a brainstorm of a real answer anyhow. I've now rewritten this to more answer the question: how does the Classification attach itself to other mathematical areas.</p> <ul> <li><p>Group actions are quite common in mathematics. Showing that only finitely many types of group actions occur in a problem is a typical idea. The theorem of Fried involving indecomposable polynomials fits into this, as there is an action on branched covers.</p></li> <li><p>There are a number of corollaries of CFSG, which are essentially classifications in their own right. The above Fried result depends on a type of doubly transitive action being classified. Another example, rooted in Dunfield/Thurston (page 45), they note that for the orbit in question in their application, a result of Gilman suffices (they use CFSG to assert that a finite 6-transitive group action contains $A_n$). For some these, asking whether all of CFSG is needed could be apropos.</p></li> <li><p>Another type of corollary is that some bounds are lowered, due to the fact that we now know (for instance) that <strong>all</strong> groups have (say) a representation satisfying a certain bound. The existence of a <em>qualitatively</em> different bound under CFSG (say polynomial opposed to exponential) has more interest than just making numbers smaller. Looking at the Babai paper for graph isomorphism, they even note (Theorem 3.1) that a easier weaker result suffices. <a href="http://people.cs.uchicago.edu/~laci/papers/hypergraphiso.pdf" rel="nofollow">http://people.cs.uchicago.edu/~laci/papers/hypergraphiso.pdf</a></p></li> <li><p>The work of Aschbacher, followed by the book of Kleidman and Liebeck, on maximal subgroups of finite classical groups is another source. Here $SL, SO, Sp$ and $SU$ are all involved. Aschbacher's theorem (here is a <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.100.7645&amp;rep=rep1&amp;type=pdf" rel="nofollow">survey</a> by King) says that there are 8 types of subgroups (stablizers of: subspaces, direct sums, spreads, forms, extension fields, tensor products, subfields; extraspecial normalizers, plus the exotic ninth class). Another survey (precursor to their book) is by <a href="http://projecteuclid.org/euclid.bams/1183657067" rel="nofollow">Kleidman and Liebeck</a>. Once the degree is above 14, I think, the 8 classes become uniform in description (though the exotics persist). This relates directly to group theory of course, but many math branches use these constructs.</p></li> </ul> <p>My specific example was a paper of Bachoc and Nebe that showed that an 80-dimensional lattice with a large minimal norm (of 8) had a known automorphism group (related to $M_{22}$) that was maximal finite in $GL_{80}(Z)$. They then used to show that their lattice was not isometric to a different one they constructed. More generally, if there is no possible common finite supergroup of the known automorphisms of two lattices, they are not isometric. To prove this can require CFSG in one form or another.</p> <p>I agree with what was stated in a comment above: "But probably the more interesting question is where the classification has impact on mathematics outside group theory." </p> http://mathoverflow.net/questions/31238/a-signature-inequality/31785#31785 Answer by Junkie for A signature inequality? Junkie 2010-07-14T01:54:35Z 2010-07-14T02:03:45Z <p>Here is Magma code that will construct a 10x10 example, with one eigenvalue about +10 and nine of -1. Note that Magma has little numerical linear algebra, so I do it myself. As can be seen in the final $P$ and $Q$, the numbers become of size about $10^{40}$ here. You can round to the nearest integer after multiplying by $10^{100}$ and get an integral answer of course. It seems much harder to require that $P$ and $Q$ have positive coefficients.</p> <pre><code>RF:=RealField(100); function ThetaRoot(M) h11,h12,h21,h22:=Explode(Eltseq(M)); t2:=h11^2; t1:=h22^2; t12:=(h12+h21)^2-2*h11*h22; a:=t1+t2+t12; b:=-2*t1-t12; c:=t1; R:=Roots(Polynomial([c,b,a]),RF); if #R eq 0 then R:=[&lt;-b/2/a,2&gt;]; end if; th:=Arccos(Sqrt(R[1][1])); return th; end function; function UnitarilySimilar(M) t:=Trace(M); d:=Degree(Parent(M)); S:=SymmetricGroup(d); M:=M-t/d*Parent(M)!1; if d eq 1 or M[1][1] eq 0 then return Parent(M)!1; end if; D:=Diagonal(M); POS:=[i : i in [1..d] | D[i] gt 0]; NEG:=[i : i in [1..d] | D[i] lt 0]; p:=POS[1]; n:=NEG[1]; u:=S!1; if p ne 1 then u*:=S!(p,1); end if; if n ne 2 then u*:=S!(n,2); end if; if p eq 2 and n eq 1 then u:=S!(1,2); end if; T1:=PermutationMatrix(BaseRing(M),u); M1:=T1*M*Transpose(T1); if Abs(M1[1][1]) gt Abs(M1[2][2]) then Tt:=Parent(M)!1; Tt[1][1]:=0; Tt[1][2]:=-1; Tt[2][1]:=1; Tt[2][2]:=0; T1:=Tt*T1; M1:=T1*M*Transpose(T1); end if; th:=ThetaRoot(Submatrix(M1,[1,2],[1,2])); T2:=Parent(M)!1; T2[p][p]:=Cos(th); T2[p][n]:=-Sin(th); T2[n][p]:=-T2[p][n]; T2[n][n]:=T2[p][p]; M2:=T2*M1*Transpose(T2); if Abs(M2[1][1]) gt 10^(-25) then T2[p][n]:=Sin(th); T2[n][p]:=-T2[p][n]; M2:=T2*M1*Transpose(T2); end if; U:=UnitarilySimilar(Submatrix(M2,2,2,d-1,d-1)); U:=DirectSum(&lt;DiagonalMatrix([BaseRing(U)!1]),U&gt;); return U*T2*T1; end function; M:=DiagonalMatrix([RF!10,-1,-1,-1,-1,-1,-1,-1,-1,-1]); d:=Degree(Parent(M)); U:=UnitarilySimilar(M); MT:=U*M*Transpose(U); I:=Parent(MT)!1; for a in [1..d] do for b in [a+1..d] do // Ballantine gives "too diagonal" I[a][b]:=Random([-2^25..2^25])/2^32; end for; end for; // perturb it PERTURB:=I*MT*Transpose(I); B:=PERTURB; TRANS:=I*U; Diagonal(PERTURB); for a in [1..d] do for b in [a+1..d] do B[a][b]:=0; end for; end for; for a in [1..d] do B[a][a]:=B[a][a]/2; end for; function EpsilonKernel(M) d:=Degree(Parent(M)); S:=Vector([BaseRing(M)!0 : i in [1..d]]); for e in [1..d] do T:=M; for f in Reverse([1..e-1]) do S[f]:=T[e][f]/T[f][f]; T[e]-:=T[f]*S[f]; end for; if Abs(T[e][e]) lt 10^(-50) then S[e]:=-1; return S/Sqrt(Norm(S)); end if; end for; end function; function MagmaBrainDeadEigenvectors(M) R:=[r[1] : r in Roots(CharacteristicPolynomial(M))]; return DiagonalMatrix(R),Matrix([EpsilonKernel(M-r*Parent(M)!1) : r in R]); end function; D,V:=MagmaBrainDeadEigenvectors(B); H:=Transpose(V)*D*V; P:=Transpose(B)*H; Q:=H^(-1); P:=(P+Transpose(P))/2; Q:=(Q+Transpose(Q))/2; // cheat for symmetry Roots(CharacteristicPolynomial(P*Q+Q*P)); </code></pre> <p>EDIT: In larger dimensions, the $Q=H^{-1}$ becomes time-consuming, perhaps for stability.</p> http://mathoverflow.net/questions/31238/a-signature-inequality/31744#31744 Answer by Junkie for A signature inequality? Junkie 2010-07-13T18:30:24Z 2010-07-13T18:30:24Z <p>Overnight search (about 25 million random 5x5) found:</p> <pre><code>[ 835 791 -119 -1 981] [ 791 755 -113 0 931] [-119 -113 17 0 -140] [ -1 0 0 1 2] [ 981 931 -140 2 1166] [ 5 76 -4 2 -14] [ 76 2849 0 75 -531] [ -4 0 17 0 0] [ 2 75 0 2 -14] [ -14 -531 0 -14 99] </code></pre> http://mathoverflow.net/questions/31238/a-signature-inequality/31620#31620 Answer by Junkie for A signature inequality? Junkie 2010-07-12T21:31:34Z 2010-07-12T21:37:15Z <p>I get about 0.1% random in 3 for 4x4. Here is a "patterned" one:</p> <pre><code>[146 37 12 0] [ 37 10 3 0] [ 12 3 1 0] [ 0 0 0 1] [ 1 0 0 0] [ 0 4221 202 -857] [ 0 202 10 -41] [ 0 -857 -41 174] </code></pre> <p>Here is a nonnegative one:</p> <pre><code>[139 3 47 325] [ 3 1 0 6] [ 47 0 18 111] [325 6 111 761] [ 1 0 2 2] [ 0 5 12 5] [ 2 12 33 16] [ 2 5 16 10] </code></pre> <p>A completely positive one:</p> <pre><code>[ 757 1288 87 3416] [ 1288 2193 148 5809] [ 87 148 10 393] [ 3416 5809 393 15567] [ 1 4760 192 1776] [ 4760 32021426 1291596 11946513] [ 192 1291596 52097 481867] [ 1776 11946513 481867 4456990] </code></pre> http://mathoverflow.net/questions/31238/a-signature-inequality/31240#31240 Answer by Junkie for A signature inequality? Junkie 2010-07-09T19:32:28Z 2010-07-11T23:35:00Z <p>Example found randomly:</p> <p>EDIT: Make one with positive coefficients: $A=\pmatrix{1&amp;2&amp;3\cr2&amp;5&amp;6\cr3&amp;6&amp;10}$ and $B=\pmatrix{1&amp;1&amp;2\cr1&amp;2&amp;6\cr2&amp;6&amp;21}$.</p> <p>EDIT: Here's how, with Magma I get about a 25% probability with this Magma code:</p> <pre><code>R := RealField(30); function FindCounterExample() S := RandomSLnZ(3,5,5); A := S*Transpose(S); S := RandomSLnZ(3,5,5); B := S*Transpose(S); ROOTS := Roots(CharacteristicPolynomial(A*B+B*A),R); ROOTS := [r[1] : r in ROOTS | r[1] ge 0]; if #ROOTS eq 1 then A; B; end if; return #ROOTS; end function; </code></pre> <p>I get about a 25% probability of 1 positive eigenvalue, 75% of 2, and 0.15% of 3.</p> <pre><code>OUTPUT := [FindCounterExample() : i in [1..100000]]; SequenceToMultiset(OUTPUT); // {* 1^^25563, 2^^74296, 3^^141 *} </code></pre> <p>EDIT: I think this can be described as saying that there is about a 75% chance of the determinant of $AB+BA$ being negative, and when it in the 25% positive case, the chance is not too great that all the eigenvalues are positive. It can also depends on what RandomSLnZ is doing. The split might only be close to 75-25 and not exact.</p> <p>EDIT: Yes when I did it with RandomSLnZ(3,3,3) I get a split of about $114+844+42$, so the 75-25 is meaningless.</p> http://mathoverflow.net/questions/11503/does-war-have-infinite-expected-length/31205#31205 Answer by Junkie for Does War have infinite expected length? Junkie 2010-07-09T15:37:17Z 2010-07-09T15:37:17Z <p>Sorry to make this an answer, but the formatting needs specification.</p> <p>Here is a stupid 8-card version of your rules, top player going first.</p> <pre><code>AD KH KS AC KD AH AS KC </code></pre> <p>If I read your rules correctly, after one round it looks like:</p> <pre><code>AD KD KC AC KH AH AS KS </code></pre> <p>and so on. These seems to generalise to 8n cards.</p> http://mathoverflow.net/questions/28066/are-periods-of-rigid-calabi-yau-threefolds-over-q-algebraic/28078#28078 Answer by Junkie for Are periods of rigid Calabi-Yau threefolds over $Q$ algebraic? Junkie 2010-06-14T01:29:04Z 2010-06-14T01:29:04Z <p>For rigid, at least in the modular case (known in many events), you can compute the periods of the form, though this supposes you can explicitly write down the weight 4 newform. For instance, Schutt ( <a href="http://arxiv.org/pdf/math/0311106" rel="nofollow">http://arxiv.org/pdf/math/0311106</a> ) gives examples of level 73, and using Magma you can compute the periods as</p> <pre><code>&gt; M:=NewformDecomposition(NewSubspace(CuspidalSubspace(ModularSymbols(73,4))))[1]; &gt; Periods(M,100); [ (0.902834199842382836695960181248 + 0.0526923557275574794028757363126*i), (0.285105536792331422114513708795 + 0.0175641185758524931342918404798*i) ] </code></pre> <p>Here $L$-functions are not applicable, as the $L$-functions vanishes at the central point. I extended the above to a few hundred digits and found nothing with PowerRelation.</p> http://mathoverflow.net/questions/23593/open-project-lets-compute-the-fourier-expansion-of-a-non-solvable-algebraic-maa/23913#23913 Answer by Junkie for Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. Junkie 2010-05-08T03:28:13Z 2010-05-08T03:28:13Z <p>Here is Magma code that gets you the answer in a few seconds. I made a special case for the bad primes, and did them by hand.</p> <pre><code>_&lt;x&gt; := PolynomialRing(Rationals()); f5 := 344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5; g24 := 14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 - 1381768039105642956*x^6 + 4291028045077743465*x^8 - 2050038614413776542*x^10 + 287094814384960835*x^12 - 9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 + 152929135*x^20 - 50726*x^22 + x^24; K := NumberField(f5); _,D := IsSquare(Integers()!Discriminant(f5)); prec := 30; CHAR_TABLE := CharacterTable(GaloisGroup(g24)); chi := CHAR_TABLE[2]; BAD_FACTORS := [ &lt;2,Polynomial([1,-1,1])&gt;, &lt;3,Polynomial([1,-ComplexField(prec)!chi[9],1])&gt;, &lt;5,Polynomial([1,0,1])&gt;, &lt;7,Polynomial([1,0,1])&gt;, &lt;71,Polynomial([1,0,1])&gt;, &lt;137,Polynomial([1,1,1])&gt;, &lt;163,Polynomial([1,-ComplexField(prec)!chi[5],1])&gt;, &lt;1951,Polynomial([1])&gt;, &lt;16061,Polynomial([1,-2,1])&gt;, &lt;889289,Polynomial([1,-ComplexField(prec)!chi[8],1])&gt; ]; BAD := [bf[1] : bf in BAD_FACTORS]; FACTORS := [bf[2] : bf in BAD_FACTORS]; function LOCAL(p,d : Precision:=prec) if p in BAD then return FACTORS[Position(BAD,p)]; end if; R := Roots(ChangeRing(f5,GF(p))); if #R eq 1 then return Polynomial([1,0,1]); end if; if #R eq 2 then ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); return Polynomial([1,ord eq 3 select 1 else -1,1]); end if; if #R eq 5 then ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); return Polynomial([1,ord eq 1 select -2 else 2,1]); end if; r := Roots(ChangeRing(f5,GF(p^5))); x := r[1][1]; prod := GF(p)!&amp;*[x^(p^i)-x^(p^j) : j in [(i+1)..4], i in [0..4]]; wh := prod eq GF(p)!D; ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); if ord eq 10 then class := wh select 8 else 9; // compatible with FACTORS else class := wh select 6 else 5; end if; return Polynomial([1,-ComplexField(prec)!chi[class],1]); end function; L := LSeries(1, [0,0], 1951^2, LOCAL : Precision:=prec); // s-&gt;1-s, Gamma(s/2)^2 psi := DirichletGroup(1951, CyclotomicField(10)).1; p1951 := Polynomial([1,-ComplexField(prec)!CyclotomicField(5).1]); TP := TensorProduct(L, LSeries(psi : Precision:=prec), [&lt;1951, 1, p1951&gt;]); CheckFunctionalEquation(TP); </code></pre> <p>Here is the special values:</p> <pre><code>ev := Evaluate(TP,0); // 2-1.453085056... rel := PowerRelation(ev,4 : Al:="LLL"); NF := NumberField(rel); Q5&lt;zeta5&gt; := CyclotomicField(5); assert IsIsomorphic(NF,Q5); Q5!NF.1; </code></pre> <p>So $L(\rho,0)=-4\zeta_5(1+\zeta_5)$ for Marty. I get $L(\rho_0,-1)=32(48723\sqrt{5} - 778741)$ as an algebraic. I get $L(\rho,-2)=8800\zeta_5^3 - 14444\zeta_5^2 + 35604\zeta_5 + 17412$ with more precision. I determined the TensorProduct factor at 1951 via trial and error, making the obvious guesses until one worked (the failure is at 100-110 digits). With this, I take it to 240 digits and I can even get $$L(\rho,-4)=-18475535360\zeta_5^3 - 11142861380\zeta_5^2 - 12091894020\zeta_5 - 7107607296$$ and $$L(\rho,-6)=25255057273186244\zeta_5^3 - 1015274469604000\zeta_5^2 - 15695788409197884\zeta_5 + 9459547822189412$$ The precision can go higher if you want more.</p> <p>Finally, the Maass form:</p> <pre><code>function MaassEval(L,z) x:=Real(z); y:=Imaginary(z); printf "Using %o coefficients\n", Ceiling(11/y); C := LGetCoefficients(L,Ceiling(11/y)); pi := Pi(RealField()); a := Sqrt(y)*&amp;+[C[n]*KBessel(0,2*pi*n*y)*Sin(2*pi*n*x) : n in [1..#C]]; return a; end function; zz:=0.0001+0.0001*ComplexField().1; MaassEval(TP,zz); // Using 110000 coefficients // -1.71477211817772949974178783985E-8 + 9.01673609747756708674470686948E-9*i MaassEval(TP,zz/(1951*zz+1)); // Using 161297 coefficients // -1.71477211817772949974179078240E-8 + 9.01673609747756708674496293450E-9*i </code></pre> http://mathoverflow.net/questions/23593/open-project-lets-compute-the-fourier-expansion-of-a-non-solvable-algebraic-maa/23852#23852 Answer by Junkie for Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. Junkie 2010-05-07T13:05:35Z 2010-05-07T13:05:35Z <p>I got it to work!</p> <p>First the preliminary data:</p> <pre><code>_&lt;x&gt; := PolynomialRing(Rationals()); f5 :=344 + 3106*x - 1795*x^2 - 780*x^3 - x^4 + x^5; g24 :=14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 - 1381768039105642956*x^6 + 4291028045077743465*x^8 - 2050038614413776542*x^10 + 287094814384960835*x^12 - 9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 + 152929135*x^20 - 50726*x^22 + x^24; K := NumberField(g24); _,D := IsSquare(Discriminant(f5)); D:=Integers()!D; </code></pre> <p>Then the long computation of ArtinRepresentations. All the work is in the 7 or so bad primes you listed.</p> <pre><code>SetVerbose("ArtRep",3); A:=ArtinRepresentations(K); // Four hours for computing BAD:= Kartinrepdatabadprimes cat [p[1] : p in Factorization(D)]; chi:=Character(A[2]); L:=LSeries(A[2]); </code></pre> <p>Then the changed local function.</p> <pre><code>FUNC:=Lcffun; function LOCAL(p,d : Precision:=30) if p in BAD then return FUNC(p,d : Precision:=30); end if; R:= Roots(ChangeRing(f5,GF(p))); if #R eq 1 then return Polynomial([1,0,1]); end if; if #R eq 2 then ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); return Polynomial([1,ord eq 3 select 1 else -1,1]); end if; if #R eq 5 then ord := Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); return Polynomial([1,ord eq 1 select -2 else 2,1]); end if; r:= Roots(ChangeRing(f5,GF(p^5))); x:=r[1][1]; prod:= GF(p)!&amp;*[x^(p^i)-x^(p^j) : j in [(i+1)..4], i in [0..4]]; wh:=prod eq GF(p)!D; ord:= Lcm([Degree(f[1]) : f in Factorization(Polynomial(GF(p),g24))]); if ord eq 10 then class:=wh select 8 else 9; // could be reversed, see BAD else class:=wh select 6 else 5; end if; return Polynomial([1,-ComplexField(30)!chi[class],1]); end function; </code></pre> <p>This is not optimal with the GF(p) I think, but who cares. A difficulty is that the BAD primes force an ordering onto the conjugacy classes on 5-cycles, so there are two such functions I suspect.</p> <p>Now change the coefficient function of L to be as desired.</p> <pre><code>Lcffun := LOCAL; psi := DirichletGroup(1951, CyclotomicField(10)).1; TP := TensorProduct(L,LSeries(psi), [&lt;1951, 1&gt;]); // tried psi^3 too CheckFunctionalEquation(TP); // -4.7331654313E-30 - works! LSetPrecision(L, 9); CheckFunctionalEquation(L); // L directly </code></pre> <p>I tried both $\psi$ and $\psi^3$ in the TensorProduct. The L-function is actually wrong at 1951 under this, but the computation only goes to $C\sqrt{1951}$ terms or about 700 so it does not matter. Also, you can check L directly, reducing the precision.</p> http://mathoverflow.net/questions/23593/open-project-lets-compute-the-fourier-expansion-of-a-non-solvable-algebraic-maa/23824#23824 Answer by Junkie for Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. Junkie 2010-05-07T05:38:39Z 2010-05-07T05:38:39Z <p>Actually, I should read the help with Magma, for the direction to ArtinRepresentations is there.</p> <pre><code> f:=14488688572801 - 2922378139308818*x^2 + 134981448876235615*x^4 - 1381768039105642956*x^6 + 4291028045077743465*x^8 - 2050038614413776542*x^10 + 287094814384960835*x^12 - 9040633522810414*x^14 + 63787035668165*x^16 - 158664037068*x^18 + 152929135*x^20 - 50726*x^22 + x^24; K:=NumberField(f); A:=ArtinRepresentations(K); // Four hours for computing KartinrepdataFrob(7); // fails </code></pre> <p>Then A[2] has degree 2, though the conductor is $1951^2$ as you say. Magma does tensor product L-functions, but maybe not automatically at bad primes. But the hassle is, it cannot compute the Frobenius for many primes, probably due to the lack of cycle type distinction. I don't think Magma is the right tool for such a specific problem. Maybe the internal Frob function can be overridden and it will work via the Serre-Poonen trick.</p> http://mathoverflow.net/questions/23593/open-project-lets-compute-the-fourier-expansion-of-a-non-solvable-algebraic-maa/23675#23675 Answer by Junkie for Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. Junkie 2010-05-06T07:47:05Z 2010-05-06T07:47:05Z <p>Hmm, a problem. In 4 hours Magma can compute this:</p> <pre><code>&gt; f:=x^24 - 50726*x^22 + 152929135*x^20 - 158664037068*x^18 + 63787035668165*x^16 - 9040633522810414*x^14 + 287094814384960835*x^12 - 2050038614413776542*x^10 + 4291028045077743465*x^8 - 1381768039105642956*x^6 + 134981448876235615*x^4 - 2922378139308818*x^2 + 14488688572801; &gt; L:=LSeries(NumberField(f) : Method:="Artin"); </code></pre> <p>But the problem is that the decomposition of Lprod contains no 2-dimensional factors...</p> <pre><code>&gt; [&lt;Character(x[1]parent),x[2]&gt; : x in Lprod | #x[1]lpoles eq 0]; [ &lt;( 3, 3, 0, -1, -zeta(5)_5^3 - zeta(5)_5^2, zeta(5)_5^3 + zeta(5)_5^2 + 1, 0, zeta(5)_5^3 + zeta(5)_5^2 + 1, -zeta(5)_5^3 - zeta(5)_5^2 ), 1&gt;, &lt;( 3, 3, 0, -1, zeta(5)_5^3 + zeta(5)_5^2 + 1, -zeta(5)_5^3 - zeta(5)_5^2, 0, -zeta(5)_5^3 - zeta(5)_5^2, zeta(5)_5^3 + zeta(5)_5^2 + 1 ), 1&gt;, &lt;( 5, 5, -1, 1, 0, 0, -1, 0, 0 ), 1&gt;, &lt;( 6, -6, 0, 0, 1, 1, 0, -1, -1 ), 2&gt; ] </code></pre> <p>So we achieved a decomposition $1+\rho_3+\bar\rho_3+\sigma_5+2\eta_6$ with this, but no degree 2 factor? Is this correct? Maybe the splitting field is necessary here (when all reps will appear). The guess is that the above computation already did this in essence, so I can re-run it after adding more intelligence.</p> http://mathoverflow.net/questions/23593/open-project-lets-compute-the-fourier-expansion-of-a-non-solvable-algebraic-maa/23657#23657 Answer by Junkie for Open project: Let's compute the Fourier expansion of a non-solvable algebraic Maass form. Junkie 2010-05-06T02:02:24Z 2010-05-06T02:02:24Z <p>I will copy a comment of mine about Jehanne over here.</p> <p>His paper is: <a href="http://dx.doi.org/10.1006/jnth.2001.2656" rel="nofollow">http://dx.doi.org/10.1006/jnth.2001.2656</a></p> <p>Jehanne has 4 totally real Examples 2-5 (page 353-6), including the one of Booker. He also gets the degree 24 field. The method to compute the $L$-series is listed in Section 6.</p> <p>Jehanne gives Poonen's trick as being due to Serre (from Buhler's book). Booker says the same (page 332).</p> <p>As Kevin Buzzard pointed out, the application to computational Stark's conjecture was only for complex $A_5$ though, so the explicit totally real computation might not have been done outside of Booker.</p> <p>I don't think that Dokchitser does anything in his Magma code that is not in essence the same as you do here, as the splitting types are just realized in a alternative manner, usually more complicated due to generality. I think his code could probably do $A_4$ as a black box w/o much modification, but $A_5$ requires a degree 240 number field.</p> <p>I had forgotten where's Booker's thesis had been published. Thanks for the link.</p> http://mathoverflow.net/questions/22908/does-anyone-want-a-pretty-maass-form/23555#23555 Answer by Junkie for Does anyone want a pretty Maass form? Junkie 2010-05-05T07:30:30Z 2010-05-05T07:30:30Z <p>Here it is from Dokchitser in Magma:</p> <pre><code>&gt; L:=LSeries(HilbertClassField(QuadraticField(145)) : Method:="Artin"); &gt; Lprod; [ &lt;L-series of Riemann zeta function, 1&gt;, &lt;L-series of Artin representation of Number Field with defining polynomial x^8 - 636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with character ( 1, 1, 1, -1, -1 ) and conductor 5, 1&gt;, &lt;L-series of Artin representation of Number Field with defining polynomial x^8 - 636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with character ( 1, 1, -1, -1, 1 ) and conductor 145, 1&gt;, &lt;L-series of Artin representation of Number Field with defining polynomial x^8 - 636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with character ( 1, 1, -1, 1, -1 ) and conductor 29, 1&gt;, &lt;L-series of Artin representation of Number Field with defining polynomial x^8 - 636*x^6 + 135214*x^4 - 11109740*x^2 + 290532025 over the Rational Field with character ( 2, -2, 0, 0, 0 ) and conductor 145, 2&gt; ] &gt; L5:=Lprod[5][1]; &gt; CheckFunctionalEquation(L5); // LCfRequired(L5) demands 161 terms 1.57772181044202361082345713057E-30 &gt; [&lt;p,-Integers()!Coefficient(EulerFactor(L5,p),1)&gt; : p in PrimesUpTo(100)]; [ &lt;2, 0&gt;, &lt;3, 0&gt;, &lt;5, -1&gt;, &lt;7, 0&gt;, &lt;11, 0&gt;, &lt;13, 0&gt;, &lt;17, 0&gt;, &lt;19, 0&gt;, &lt;23, 0&gt;, &lt;29, -1&gt;, &lt;31, 0&gt;, &lt;37, 0&gt;, &lt;41, 0&gt;, &lt;43, 0&gt;, &lt;47, 0&gt;, &lt;53, 0&gt;, &lt;59, -2&gt;, &lt;61, 0&gt;, &lt;67, 0&gt;, &lt;71, -2&gt;, &lt;73, 0&gt;, &lt;79, 0&gt;, &lt;83, 0&gt;, &lt;89, 0&gt;, &lt;97, 0&gt; ] </code></pre> <p>These are the same as your ap(p), essentially.</p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23390#23390 Answer by Junkie for Class Field Theory for Imaginary Quadratic Fields Junkie 2010-05-04T00:00:40Z 2010-05-04T00:00:40Z <p>Here is a case where it is non-Abelian. I use $K$ of class number 3. If I use the Gross curve, it is Abelian. If I twist in $Q(\sqrt{-15})$, it is Abelian for every one I tried, maybe because it is one class per genus. My comments are not from an expert.</p> <pre><code>&gt; K&lt;s&gt;:=QuadraticField(-23); &gt; jinv:=jInvariant((1+Sqrt(RealField(200)!-23))/2); &gt; jrel:=PowerRelation(jinv,3 : Al:="LLL"); &gt; Kj&lt;j&gt;:=ext&lt;K|jrel&gt;; &gt; E:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]); &gt; HasComplexMultiplication(E); true -23 &gt; c4, c6 := Explode(cInvariants(E)); // random twist with this j &gt; f:=Polynomial([-c6/864,-c4/48,0,1]); &gt; poly:=DivisionPolynomial(E,3); // Linear x Linear x Quadratic &gt; R:=Roots(poly); &gt; Kj2:=ext&lt;Kj|Polynomial([-Evaluate(f,R[1][1]),0,1])&gt;; &gt; KK:=ext&lt;Kj2|Polynomial([-Evaluate(f,R[2][1]),0,1])&gt;; &gt; assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here &gt; f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1]; &gt; GaloisGroup(f); /* not immediate to compute */ Permutation group acting on a set of cardinality 12 Order = 48 = 2^4 * 3 &gt; IsAbelian($1); false </code></pre> <p>This group has$A_4$and$Z_2^4$as normal subgroups, but I don't know it's name if any.</p> <p>PS. 5-torsion is too long to compute most often.</p> http://mathoverflow.net/questions/23092/class-field-theory-for-imaginary-quadratic-fields/23150#23150 Answer by Junkie for Class Field Theory for Imaginary Quadratic Fields Junkie 2010-05-01T00:53:54Z 2010-05-01T00:53:54Z <p>Magma is not facile here but works, but maybe SAGE can do the same. You get$K(j,E[3])/K$to be a degree 12 and cyclic Galois group, for the$E$I think you want.</p> <pre><code>&gt; jrel:=PowerRelation(jInvariant((1+Sqrt(-15))/2),2 : Al:="LLL"); &gt; K:=QuadraticField(-15); &gt; Kj&lt;j&gt;:=ext&lt;K|jrel&gt;; &gt; A:=AbsoluteField(Kj); &gt; C:=EllipticCurve([-3*j/(j-1728),-2*j/(j-1728)]); &gt; b, d := HasComplexMultiplication(C); assert b and d eq -15; &gt; E:=QuadraticTwist(C, 7*11); // conductor at 3, 5 &gt; E:=ChangeRing(WeierstrassModel(ChangeRing(E,A)),Kj); &gt; c4, c6 := Explode(cInvariants(E)); &gt; f:=Polynomial([-c6/864,-c4/48,0,1]); &gt; poly:=DivisionPolynomial(E,3); // Linear x Cubic &gt; r:=Roots(poly)[1][1]; &gt; Kj2:=ext&lt;Kj|Polynomial([-Evaluate(f,r),0,1])&gt;; // quadratic ext for linear &gt; KK:=ext&lt;Kj2|Factorization(poly)[2][1]&gt;; // cubic x-coordinate &gt; assert #DivisionPoints(ChangeRing(E,KK)!0,3) eq 3^2; // all E[3] here &gt; f:=Factorization(ChangeRing(DefiningPolynomial(AbsoluteField(KK)),K))[1][1]; &gt; // assert IsIsomorphic(ext&lt;K|f&gt;,KK); // taking too long ? &gt; // SetVerbose("GaloisGroup",2); &gt; GaloisGroup(f); Permutation group acting on a set of cardinality 12 Order = 12 = 2^2 * 3 &gt; IsAbelian($1); true </code></pre> <p>The Magma has as online calculator for this. <a href="http://magma.maths.usyd.edu.au/calc" rel="nofollow">http://magma.maths.usyd.edu.au/calc</a></p> http://mathoverflow.net/questions/22629/are-there-primes-of-every-hamming-weight/22773#22773 Answer by Junkie for Are There Primes of Every Hamming Weight? Junkie 2010-04-27T20:12:28Z 2010-04-27T20:12:28Z <p>For this context, though not so highbrow: Wagstaff, Prime Numbers with a Fixed Number of One Bits or Zero Bits in Their Binary Representation, Experiment Math 10 (2001), 267-273.</p> <p><a href="http://www.expmath.org/restricted/10/10.2/wagstaff.ps" rel="nofollow">http://www.expmath.org/restricted/10/10.2/wagstaff.ps</a></p> http://mathoverflow.net/questions/11349/are-most-cubic-plane-curves-over-the-rationals-elliptic/22344#22344 Answer by Junkie for Are most cubic plane curves over the rationals elliptic? Junkie 2010-04-23T12:56:37Z 2010-04-23T12:56:37Z <p>"One could try to estimate the size of the Tate-Shafarevich group of a "random" elliptic curve, to get an idea of how often local solvability implies global solvability, but even if one does this it is not clear that this is counting curves in the same way."</p> <p>Bhargava has reportedly proven the 3-Selmer group has average size 4. The assumption of a minimal rank (1/2 average) and Parity conjecture would account that 2 of the 4 come from rank, and 2 of the 4 come from Sha, so 50%. His counting is by $|c_4| &lt; X^2$ and $|c_6| &lt; X^3$ I think. The workers on 3-descent have some bounds that relate the invariants to the coefficient size.</p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/22055#22055 Answer by Junkie for modular eigenforms with integral coefficients [Maeda's Conjecture] Junkie 2010-04-21T13:18:37Z 2010-04-22T04:46:55Z <p>"He said (and I never understood this comment so feel free to fill me in) that S_k(1;Q) being irreducible as a Hecke module was related to (equivalent to?) a certain L-value not vanishing, and L-values tend to vanish occasionally when you look hard enough."</p> <p>I dispute the impression of Hida with vanishing L-values. To precise this, a density statement is needed. The standard L-function technology whizzes from random matrices should expect that it doesn't vanish ever. In the same vein, Conrey conjectures that quadratic twists of weight 6+ forms never vanish aside from sign, though he kindly phrases it as "finitely many" as pointed out above.</p> <p><a href="http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf" rel="nofollow">http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf</a></p> <p>For weight 6 we have rank 2 vanishing for a few forms, as Dummigan lists: 95k6, 122k6, 260k6.</p> <p><a href="http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi" rel="nofollow">http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi</a></p> <p>I expect no vanishing for weight 8+. To my knowledge, no rank 3 vanishing exists for weight 4+. My recollection (Stein 2000) is that, outside with Gamma1(N), there is one at level 122 (sic, as above) weight 2 form with quadratic sign that vanishes to order 1 with no self-dual functional equation sign (eps = -0.76822128 + 0.6401844i).</p> <p>I am editing this now to explain L-function methods. The right random matrix idea is that L-values have cumulative distribution with $\sqrt t$ for small $t$. It is probably unnecessary though.</p> <p>For rather look at the BSD analogue. There is $L(centre)/\Omega$ and the other side is up to few rational factors an integer. It is also a square. So it is "like" a random integral square up to size $\Omega$ as the Tamagawa and torsion and much smaller. The "probability" of an (even signed) L-function vanishing centrally can be thought as $\sqrt\Omega$ as a chance that a random integral square up to size $\Omega$ is 0 is just 1 in $\sqrt\Omega$.</p> http://mathoverflow.net/questions/21642/what-are-the-maximal-subgroups-of-gsp2g-l/22054#22054 Answer by Junkie for What are the maximal subgroups of GSp(2g,l)? Junkie 2010-04-21T12:46:48Z 2010-04-21T12:46:48Z <p>"I expect a classification for general g to be longer, but maybe its managable when g = 2?"</p> <p>The Experimental Mathematics paper of Dieulefait for g=2 uses this. He quotes Mitchell from 1914.</p> <p><a href="http://www.expmath.org/expmath/volumes/11/11.4/pp503_512.pdf" rel="nofollow">http://www.expmath.org/expmath/volumes/11/11.4/pp503_512.pdf</a></p> <p>Kleidman and Liebeck is on Google Books. There is also a brief survey of it.</p> <p><a href="http://www.springerlink.com/index/X631112H21U82040.pdf" rel="nofollow">http://www.springerlink.com/index/X631112H21U82040.pdf</a></p> http://mathoverflow.net/questions/21871/weight-4-eigenforms-with-rational-coefficients-is-it-reasonable-to-expect-they/22051#22051 Answer by Junkie for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus? Junkie 2010-04-21T12:34:47Z 2010-04-21T12:34:47Z <p>"At the level of L-functions, this should force the L-fn attached to the weight two form of g to vanish to order the order of vanishing of the L-fn of the weight 4 form f. How could two modular forms of different weights know about each other in such a way as for the orders of vanishing of their L-functions to be entwined?"</p> <p>A paper of Dummigan has congruences with vanishing, but not how you say.</p> <p><a href="http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi" rel="nofollow">http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi</a></p> <p>A paper of Schoen gives CM examples for the Abel-Jacobi map.</p> <p><a href="http://www.jstor.org/stable/2154210" rel="nofollow">http://www.jstor.org/stable/2154210</a></p> <p>A paper of Villegas has a [3,0] type of conductor $59^2$ that has vanishing order 2 (page 437).</p> <p><a href="http://www.math.utexas.edu/~villegas/publications/square-root-2.pdf" rel="nofollow">http://www.math.utexas.edu/~villegas/publications/square-root-2.pdf</a></p> <p>William Stein mentions the j-invariant calculation in his thesis (page 68).</p> <p><a href="http://wstein.org/papers/thesis/stein-thesis.pdf" rel="nofollow">http://wstein.org/papers/thesis/stein-thesis.pdf</a></p> http://mathoverflow.net/questions/19079/what-is-the-status-of-the-gauss-circle-problem/20914#20914 Answer by Junkie for What is the status of the Gauss Circle Problem? Junkie 2010-04-10T11:30:59Z 2010-04-10T11:30:59Z <p>Back in 2007 or so, at a tea I heard a noted expert in the field pooh-poohing it (for instance, sign errors in the Stokes analogue), and he seemed not to want to read any more re-hashes (he had seen more than one from these authors, who seemed to keep changing the argument). This expert is one of those they thank. It was unclear whether he thought their whole idea (to the extent the Intro explained this) was even capable of working. I do not know its submission status.</p> http://mathoverflow.net/questions/120916/average-ranks-of-abelian-surfaces Comment by Junkie Junkie 2013-02-09T21:56:42Z 2013-02-09T21:56:42Z The paper &quot;Some heuristics on elliptic curves&quot; (sec 4), has a discussion of ordering by discriminant versus conductor, and the author concludes they should all be the same for the therein purposes (rank 2 curves). In 3.4 (second paragraph), the naive height and discriminant distinction is glossed barely. The paper does not describe Faltings height explicitly, though a real period (not volume) occurs in the analysis of rank estimates and $L$-function vanishings. There is an arXiv preprint, but it lacks some later revisions, I detect. <a href="http://magma.maths.usyd.edu.au/~watkins/papers/heur.pdf" rel="nofollow">magma.maths.usyd.edu.au/~watkins/papers/heur.pdf</a> http://mathoverflow.net/questions/121023/on-a-result-of-montgomery-and-vaughan-about-eulers-totient Comment by Junkie Junkie 2013-02-06T22:55:25Z 2013-02-06T22:55:25Z I don't think the integral you have written down has very proper convergence properties, on the vertical lines. If you smooth the counting function, with a Mellin transform that decays enough as $t\rightarrow\infty$, then I think I agree that the zeros are more closely linked. http://mathoverflow.net/questions/110767/lotteries-turans-problem-and-minimization-of-risk/110947#110947 Comment by Junkie Junkie 2012-10-30T08:55:54Z 2012-10-30T08:55:54Z Part of it maybe the 5s and 6s effect. In that random sample, the max is 222, in there are 9 fives and 1 six, giving an &quot;extra&quot; four count of 36+14 above independence. The import of duplicates in the 138180 is similar, as a twofer guarantees an extra 28 non-independent for some lottery occurrence. Triplicates are likely too (80% chance). http://mathoverflow.net/questions/105477/what-are-the-limits-of-the-erds-rankin-method-for-covering-intervals-by-arithmet Comment by Junkie Junkie 2012-10-25T02:37:43Z 2012-10-25T02:37:43Z Maier-Pomerance indicate that they expect $z(\log z)^2$ as the limit (see 1.5), if one knew prime $k$-tuples. They basically use an on-average version of that (in AP), in the paper improving the constant. Thus for large primes, they can't show that any of them individually sieves out more than 1 number, but on average they can show at least 1.31, and Pintz 2. When knowing prime $k$-tuples, at least with uniformity enough, the large primes would then be shown to be more optimal, in sieving out. http://mathoverflow.net/questions/109152/axiomatizing-gross-zagier-formulae Comment by Junkie Junkie 2012-10-08T19:37:51Z 2012-10-08T19:37:51Z My recollection is that if you take level 122, nontrivial real character, weight 2 newform over $Q(i)$ -- then $L(f,1)=0$ but the sign is merely some random number on the unit circle. It is an example to investigate. I don't know whether $L'(f,1)$ is meaningful. Usually I expect, as per David Loeffler that the 0th derivative is nonvanishing. What does this mean in the Heegner construction context? http://mathoverflow.net/questions/90784/cm-abelian-variety-from-an-algebraic-hecke-character Comment by Junkie Junkie 2012-03-10T03:19:02Z 2012-03-10T03:19:02Z See Schappacher's book, Periods of Hecke characters (chapter un on Motives). <a href="http://dx.doi.org/10.1007/BFb0082094" rel="nofollow">dx.doi.org/10.1007/BFb0082094</a> See also the thread <a href="http://mathoverflow.net/questions/33269/fontaine-mazur-for-gl-1" rel="nofollow" title="fontaine mazur for gl 1">mathoverflow.net/questions/33269/&hellip;</a> http://mathoverflow.net/questions/90532/any-software-to-compute-multivariable-resultant Comment by Junkie Junkie 2012-03-08T04:04:39Z 2012-03-08T04:04:39Z Fermat should do this <a href="http://home.bway.net/lewis/" rel="nofollow">home.bway.net/lewis</a> And I think Magma can too, via either eliminating variables sequentially, or possibly with EliminationIdeal. They have an online calculator, if the problem is not too bulky. http://mathoverflow.net/questions/89529/reference-request-program-to-work-with-cyclic-quotient-singularities Comment by Junkie Junkie 2012-02-28T04:31:24Z 2012-02-28T04:31:24Z Magma does too much more and different than you want. The $a_i$ shall come from the global desingularization. Then the HomAdjoints will give canonical sheaf sections. If you work just at one alone singularity, you might need to rip open the Magma code to see how they do it. The main file is package/Geometry/SrfHyp/surface_resolution.m http://mathoverflow.net/questions/87865/what-is-the-name-for-a-finite-group-representation-that-is-the-sum-of-all-the-irr Comment by Junkie Junkie 2012-02-08T03:31:14Z 2012-02-08T03:31:14Z You might mean a Gelfand model. See Garge and Oesterle, and references <a href="http://www.degruyter.com/view/j/jgth.2010.13.issue-3/jgt.2009.060/jgt.2009.060.xml" rel="nofollow">degruyter.com/view/j/jgth.2010.13.issue-3/&hellip;</a> http://mathoverflow.net/questions/87511/smart-elliptic-curve-rational-point-search-given-regsha/87549#87549 Comment by Junkie Junkie 2012-02-05T23:32:24Z 2012-02-05T23:32:24Z &quot;To make thing fast, they use overcongergent modular symbols to compute it. &quot; I guess it takes time $N^2$ or $N^3$ in the level. They work with twists, to soften this. The classical method can emulate the same idea for twists, as with Elkies (1994), and a paper of Delaunay and Roblot. <a href="http://jtnb.cedram.org/item?id=JTNB_2008__20_3_601_0" rel="nofollow">jtnb.cedram.org/item?id=JTNB_2008__20_3_601_0</a> http://mathoverflow.net/questions/87347/the-higman-group Comment by Junkie Junkie 2012-02-02T23:51:15Z 2012-02-02T23:51:15Z <a href="http://en.wikipedia.org/wiki/Higman_group" rel="nofollow">en.wikipedia.org/wiki/Higman_group</a> See the 1974 reference there if you want finitely presented infinite simple groups. http://mathoverflow.net/questions/87234/magma-coercion-syntax Comment by Junkie Junkie 2012-02-01T20:43:56Z 2012-02-01T20:43:56Z Usually you want to Evaluate to reduce the ring dimension. Here it seems you also need to move from a function field to a polynomial ring with Numerator.  R&lt;t&gt;:=PolynomialRing(Rationals());  Roots(Numerator(Evaluate(ysoln,[1,t,1])));  [ &lt;-2, 1&gt;, &lt;0, 2&gt;, &lt;2, 1&gt; ] http://mathoverflow.net/questions/11502/the-current-status-of-the-birch-swinnerton-dyer-conjecture/86994#86994 Comment by Junkie Junkie 2012-01-29T23:58:18Z 2012-01-29T23:58:18Z Bhargava, as a joke, wrote to the Clay institute, asking that since he had shown (in conjunction with Kolyvagin's work, eg) a positive proportion of ell curves over Q, have BSD true, whether he could have a positive proportion of $1 million. The hard work being though, in the higher rank case. http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana/83366#83366 Comment by Junkie Junkie 2011-12-14T05:10:33Z 2011-12-14T05:10:33Z &quot;Is the paper cited just saying, at the end of the day, that one can exchange the order of two finite sums?&quot; I don't think, the conclusion is just that. If you look at 6.2, they say the integral of the product measure is the double integral over each measure separately (for simple functions). So my take on the category theory, is that it is concluding the product measure acts functorially, as expected by the suspected diagram chase?! So I guess$\int_{A\times B}f(x,y)(\mu_A\times\mu_B)(x,y)=\int_A\int_B f(x,y)\mu_A(x)\mu_B(y)\$ is the Fubini theorem. I agree there is no analysis. http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul/83030#83030 Comment by Junkie Junkie 2011-12-09T06:46:40Z 2011-12-09T06:46:40Z How is this different from the tensor product of two L-functions (given as Euler products). There again, to form it you take all the products of roots, no?