User marty - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:19:58Z http://mathoverflow.net/feeds/user/3545 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130777/could-the-jacobian-conjecture-be-undecidable Could the Jacobian conjecture be undecidable? Marty 2013-05-16T00:12:58Z 2013-05-17T17:34:54Z <p>Most of us know the Jacobian conjecture. Here's a version below for fixed positive integers $d$ and $n$:</p> <p>$J(d,n)$: If $f: C^n \rightarrow C^n$ is a polynomial map of degree $d$, and if the Jacobian determinant $\vert Jf \vert$ is nowhere vanishing (hence constant), then $f$ is injective (hence bijective).</p> <p>We know that the sentence "For all $n$, $J(3,n)$" implies the sentence "For all $d,n$, $J(d,n)$." In other words, the Jacobian conjecture has been reduced to degree $3$.</p> <p>We also know that, for any fixed $n$, $J(3,n)$ is provably true or provably false. This boils down to the completeness of the theory of algebraically closed fields of characteristic zero. </p> <p>But, do we know whether the sentence "For all $n$, $J(3,n)$" is provably true or false? In other words, might the Jacobian conjecture be... (oh no).. undecidable?!</p> <p>In other words, I could theoretically program my computer to set out to prove the Jacobian conjecture $J(3,1)$ (easy) and $J(3,2)$ then $J(3,3)$, etc.., and my theoretical computer would keep on going for epochs and epochs. But would it ever halt? Might this be undecidable?</p> http://mathoverflow.net/questions/126071/the-octonions-on-a-bad-day/126073#126073 Answer by Marty for The octonions on a bad day Marty 2013-03-31T05:06:18Z 2013-03-31T05:37:15Z <p>Suppose that $k$ is a field with $char(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.</p> <p>When $x \in A - k$, the trace $Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure -- $x$ is the root of a quadratic polynomial with coefficients $-Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not -- it doesn't really matter.</p> <p>It is an old theorem (Jacobson? Albert? I can't recall... check the "Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley-Dickson doubling process, the norm form $N$ is a Pfister form, i.e. $N$ is isomorphic to $&lt;1,-a> \otimes &lt;1,-b> \otimes &lt;1,-c>$ for some $a,b,c \in k$.</p> <p>It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $&lt;1,-1> \otimes &lt;1,-1> \otimes &lt;1,-1>$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.</p> <p>In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $&lt;1,-a> \otimes &lt;1,-b> \otimes &lt;1,-c>$ depends on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$, whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.</p> http://mathoverflow.net/questions/104435/where-do-nonstandard-elliptic-curve-angles-come-from Where do nonstandard elliptic curve angles come from? Marty 2012-08-10T20:11:32Z 2013-02-14T23:45:54Z <p>This is a question which has bounced around my head over the past few years. At the same time, I am answering <a href="http://mathoverflow.net/questions/104421/riemann-hypothesis-for-zeta-function-of-algebraic-curves-over-fields-of-infinite" rel="nofollow">http://mathoverflow.net/questions/104421/riemann-hypothesis-for-zeta-function-of-algebraic-curves-over-fields-of-infinite</a> with another question.</p> <p>Let $E$ be an elliptic curve over $Q$. Let $u$ be an nonprincipal ultrafilter on the set of prime numbers. </p> <p>For each prime $p$ (at which $E$ has good reduction, let's say), let $\pm \theta_p$ be the elliptic curve angle at $p$. In other words, $a_p = 2 \sqrt{p} \cdot \cos(\theta_p)$. Then, by the compactness of the interval $[-1,1]$, there is a nonstandard elliptic curve angle $\theta_u$ naturally associated to the set $(\theta_p)$ and $u$.</p> <p>I've been wondering if there's any other way to produce these nonstandard angles. For example, let $\sigma$ be a "generic" field automorphism of the complex numbers $C$, in the sense that $(C,\sigma)$ is a model of $ACFA$. Can one produce an elliptic curve angle $\theta$ directly from the data $(E, C, \sigma)$?</p> <p>What's so difficult here is that, in the transfer from characteristic $p$ to characteristic $0$, it is so difficult to figure out how to handle things like $\sqrt{p}$. The only hope, that I can see, would be to think of $a_p$ as a $p$-adic number (use $p$-adic cohomology), and then transfer the result to a Laurent series field (so $a_u$ might belong to $C((\varpi))$ and $a_u / \sqrt{\varpi}$ would be well-behaved). But this is all "pie in the sky" for now.</p> <p>Any ideas? Anyone thought about RH in models of ACFA?</p> http://mathoverflow.net/questions/118156/asymptotics-of-arithmetic-fuchsian-groups-and-shimura-curves Asymptotics of arithmetic Fuchsian groups and Shimura curves. Marty 2013-01-05T21:19:05Z 2013-01-05T21:19:05Z <p>I'm interested in what is known/expected about some families of arithmetic Fuchsian groups. Here is the simplest family that I'm interested in: Let $E = Z[\omega]$, where $\omega = e^{2 \pi i / 3}$. Consider the family of $Z$-valued binary Hermitian forms on $E$: $$H_n(x,y) = x \bar x - n y \bar y, \text{ for all } x,y \in E.$$ Here, we let $n$ range over all positive integers which are not norms from $E$, i.e., all positive integers for which $H_n$ does not nontrivially represent zero.</p> <p>Let $\Gamma_n$ be the special unitary group of $H_n$, i.e., $$\Gamma_n = { g \in SL_2(E) : H_n( g(x,y)) = H_n(x,y) \text{ for all } x,y \in E }.$$ Here $g(x,y)$ denotes the effect of matrix multiplication.</p> <p>As $H_n$ is an indefinite Hermitian form, these groups $\Gamma_n$ are discrete subgroups of the Lie group $SU(1,1)$. There's almost certainly another way of seeing these groups as coming from orders in indefinite quaternion algebras over $Q$. So I guess that these groups $\Gamma_n$ yield a family of Shimura curves of increasing "complexity" measured in any natural way. Associated to the groups $\Gamma_n$ are compact orbifold Riemann surfaces $X_n$, each with invariants including genus $g$, and a series of $t$ marked orbifold points, with indices $m_1, \ldots, m_t$.</p> <p>What do we know about the following: How do we expect the genus $g = g(X_n)$ to behave as $n \rightarrow \infty$? This I anticipate is known or well-studied.</p> <p>But also, how does the family of indices $m_1, \ldots, m_t$ behave asymptotically? For example, how many orbifold points of index $3$ do we expect on $X_n$, as $n \rightarrow \infty$?</p> <p>By some hyperbolic geometry, we can relate these indices to the volume of $X_n$. Can we use this to get some heuristics?</p> <p>Am I asking something silly? I know that one can eliminate torsion by passing to a finite-index subgroup, but I would anticipate that torsion doesn't disappear in a family of groups such as the $\Gamma_n$.</p> <p>References and speculations are welcome!</p> http://mathoverflow.net/questions/116666/triality-of-spin8/116679#116679 Answer by Marty for Triality of Spin(8) Marty 2012-12-18T04:02:26Z 2013-01-04T17:00:37Z <p>Here's my favorite way to answer your question. Hopefully the answer to Robert Bryant's question is "yes".</p> <p>Let $A$ be the ring of octonions (the "nonsplit" octonions over ${\mathbb R}$); it comes with an involution $\alpha \mapsto \bar \alpha$, from which there is a trace $Tr(\alpha) = \alpha + \bar \alpha$ and a norm $N(\alpha) = \alpha \cdot \bar \alpha$. From this, we get a trilinear form $T: A \otimes A \otimes A \rightarrow {\mathbb R}$ given by $$T(\alpha, \beta, \gamma) = Tr( (\alpha \beta) \gamma) = Tr(\alpha (\beta \gamma)).$$ (Multiplication is nonassociative, but the traces work out to the same result.)</p> <p>The group $Spin(8)$ can be constructed as the group of triples $(g_1, g_2, g_3) \in SO(A,N)^3$ of "rotation matrices" with respect to the norm quadratic form, such that for all $(\alpha, \beta, \gamma) \in A^3$, $$T(g_1 \alpha, g_2 \beta, g_3 \gamma) = T(\alpha, \beta, \gamma).$$</p> <p>The full group of outer automorphisms is now <strong>almost</strong> clear -- cyclic permutations of $(g_1, g_2, g_3)$ give automorphisms of $Spin(8)$ defined above since $T(\alpha, \beta, \gamma) = T(\beta, \gamma, \alpha)$. </p> <p><strong>Edit below to reflect comments and correspondence with Daemi, and the comment of Bryant</strong></p> <hr> <p>The full $S_3$ action on $Spin(8)$ is obtained from cyclic permutations and the following slightly subtle action of transpositions. Let $C$ denote the main involution of $A$ (the one for which $Tr(\alpha) = \alpha + C(\alpha)$). For any $g \in SO(A,N)$, define $\bar g = C \circ g \circ C$; then $\bar g \in SO(A,N)$ as well. </p> <p>The action of a transposition on $Spin(8)$ follows: for the transposition $(12)(3)$, the associated outer automorphism of $Spin(8)$ sends $(g_1, g_2, g_3)$ to $(\bar g_2, \bar g_1, g_3)$. The other transpositions act in the analogous ways.</p> <hr> <p>The Jordan algebra is the exceptional Jordan algebra of 3x3 Hermitian matrices with octonion entries: $$J = \left\lbrace \left( \begin{array}{ccc} a &amp; \alpha &amp; \bar \beta \cr \bar \alpha &amp; b &amp; \gamma \cr \beta &amp; \bar \gamma &amp; c \ \end{array} \right) : \alpha, \beta, \gamma \in A, a,b,c \in {\mathbb R} \right\rbrace.$$</p> <p>The group $Spin(8)$ acts on the triple of octonions $(\alpha, \beta, \gamma)$ via the natural representation from above. It acts trivially on the real numbers $a,b,c$, and this gives an action of $Spin(8)$ on the exceptional Jordan algebra. The outer automorphism group $S_3$ acts by permuting $(a,b,c)$ and $(\alpha, \beta, \gamma)$ simultaneously. Together, these give an action of $S_3 \ltimes Spin(8)$ on the exceptional Jordan algebra.</p> <p><strong>Reference update:</strong></p> <p>The material above can be found in my paper on $D_4$ modular forms, Amer. J. of Math. 128 (2006), 849-898.</p> <p>The construction of $Spin(8)$ (over ${\mathbb Z}$) follows from Proposition 4.8 of M.-A. Knus, R. Parimala, and R. Sridharan, "On Compositions and Triality," J. reine angew. Math., 457:45–70, 1994.</p> http://mathoverflow.net/questions/113950/why-only-half-integral-weight-automorphic-forms/114088#114088 Answer by Marty for Why only half-integral weight automorphic forms? Marty 2012-11-21T18:58:40Z 2012-11-21T18:58:40Z <p>I think that modular forms (for $SL_2$) of integer and half-integer weights are most important for arithmetic, while modular forms of other (real or complex) weights are primarily objects of analytic interest. I admit, modular forms of other weights might have connections to arithmetic -- I don't know anything about harmonic Maass forms, VOA theory, etc., so I can't rule out a connection. But the most important connection to arithmetic -- the Euler product -- seems absent outside of half-integer and integer weights.</p> <p>The reason for this is the "metaplectic kernel" and a good reference is the paper "Computation of the Metaplectic Kernel" by Gopal Prasad and Andrei Rapinchuk, in Inst. Hautes Études Sci. Publ. Math. No. 84 (1996), 91–187 (1997). </p> <p>To have a good theory of Euler products, the modular forms should correspond to automorphic representations of some metaplectic group. So, for $SL_2$, you need a central extension of locally compact groups: $$1 \rightarrow U(1) \rightarrow \tilde G_{\mathbb A} \rightarrow G_{\mathbb A} \rightarrow 1,$$ where $U(1)$ is the circle, and you also need a splitting of the extension over $G_{\mathbb Q}$.</p> <p>When $G = SL_2$, the only nontrivial such extensions with splittings come from two-fold extensions of $SL_2({\mathbb A})$ -- the traditional metaplectic group. So you only see modular forms of half-integer weight if you want to work adelically, decompose automorphic representations into local pieces, get an Euler product, etc..</p> <p>For other groups over other fields, you can get other sorts of central extensions. This is the subject of Prasad-Rapinchuk's paper cited above.</p> <p>Personally, I prefer an even more restrictive class of "metaplectic groups" which arise from the algebraic framework of Brylinski and Deligne (central extensions of reductive groups by $K_2$). </p> http://mathoverflow.net/questions/113811/do-l-functions-exist-for-half-integral-weight-modular-forms/113888#113888 Answer by Marty for Do L-functions exist for Half-integral weight modular forms? Marty 2012-11-20T01:23:01Z 2012-11-20T01:23:01Z <p>Upon David Loeffler's request, here is a more fleshed out version of my former comments:</p> <p>In his comment, Nick Ramsey mentioned that the natural L-function for a half-integral weight modular form is really the entire family of quadratic twists of an L-function coming from a Shimura-type lift. I agree with Nick's perspective and it motivated me to work towards a more general framework. I don't mean this post as shameless self-promotion, but here's my preprint on <a href="http://front.math.ucdavis.edu/1108.1413" rel="nofollow">Split Metaplectic Groups and their L-groups</a>. I think this will probably not be widely read due to the excessive use of Hopf algebras and reliance on Lusztig's canonical bases. Fortunately, I've recently worked out ways to avoid these completely, and I will hopefully have another preprint up soon.</p> <p>Back to the question at hand: the general non-metaplectic perspective is that an L-function can be produced from a pair $(\pi, \rho)$ where $\pi$ is an automorphic representation of $G_{\mathbb A}$ and $\rho$ is an algebraic representation of the L-group ${}^L G$. For classical modular forms, one might take $G = PGL_2$ and ${}^L G = SL_2(C) \times \Gamma$ where $\Gamma$ is the absolute Galois group of ${\mathbb Q}$.</p> <p>My perspective on the metaplectic groups is that again, an L-function should be associated to a pair $(\pi, \rho)$ where $\pi$ is a genuine automorphic representation of $\tilde G_{\mathbb A}$ (this makes sense in a framework of Brylinski-Deligne, for example) and $\rho$ is an algebraic representation of a putative L-group ${}^L \tilde G$. In the simplest case, $\tilde G = Mp_2$ is the metaplectic group. My preprint is devoted to the construction of such an L-group.</p> <p>The key subtlety, observed by Nick and others, is the ambiguity if one uses the Shimura correspondence as guidance. Indeed, work of Shimura and Waldspurger requires the choice of an additive character, and thus for the definition of an L-function. In my construction, this choice gets wrapped up in the choice of algebriac representation $\rho$ of the L-group.</p> <p>Roughly speaking, the L-group ${}^L Mp_2$ of $Mp_2$ is <em>noncanonically</em> isomorphic to $SL_2(C) \times \Gamma$. It arises in my preprint from a somewhat magical/contrived twisting of both multiplication and comultiplication in the Hopf algebra of the direct product. A less contrived non-Hopfy approach will appear in a new paper sometime soon (I hope). I hope to tackle global issues as well.</p> <p>It turns out that every nontrivial additive character $\psi$ of ${\mathbb A} / {\mathbb Q}$ can be used to generate an isomorphism from the L-group ${}^L Mp_2$ to the direct product $SL_2(C) \times \Gamma$, whence a natural 2-dimensional representation $\rho_\psi$. The L-functions produced by various choices of additive character (various isomorphisms of the L-group to the L-group of $PGL_2$) should comprise an orbit, under quadratic twisting, of a single L-function.</p> <p>To summarize, the L-functions could be written $L(\pi, \rho_\psi)$ for $\pi$ a genuine automorphic representation of $Mp_{2n}$ and $\rho_\psi$ a two-dimensional representation of ${}^L Mp_{2n}$ coming from an additive character $\psi$.</p> http://mathoverflow.net/questions/108560/what-is-a-good-book-on-topological-groups/108561#108561 Answer by Marty for What is a good book on topological groups? Marty 2012-10-01T17:19:04Z 2012-10-01T17:19:04Z <p>How about Weil's classic: "L'intégration dans les groupes topologiques et ses applications"? You won't find Kazhdan's Property T nor Tannaka reconstruction, but it treats the other topics deeply and beautifully. Plus, it's good French practice if the 1st-year PhD student needs the practice.</p> http://mathoverflow.net/questions/108505/mathematics-talk-for-five-year-olds/108512#108512 Answer by Marty for "Mathematics talk" for five year olds Marty 2012-10-01T02:53:22Z 2012-10-01T02:53:22Z <p>Here's a short list of activities that could be fun to try:</p> <p>Begin by giving the numbers 1 through 9 to 9 students. Here it's good to have a physical number to give them -- a piece of paper with the number written large will work. Ask them to line up in order. Ask 7 whether he or she is even or odd (you don't have to remember their names if they are holding up numbers). Ask 7 about the numbers next to her -- are they even or odd? (5-year olds won't automatically know that even numbers are surrounded by odd numbers. They might not know the meaning of even and odd until you run the activity.) </p> <p>Further activity: Have only 1-5 stand up in order. Then rearrange them using only transpositions (say "Number 2, switch with Number 5"). Have the students count each transposition. Then have a (well-chosen) student try to put them back in order using only "switches". How many switches did it take? Can 5-year olds discover the sign of a permutation? How about if you record the number of switches and point out even/oddness?</p> <p>Further activity: Have only 1-5 stand up in order. Have them shake hands in pairs and try to have the others count the handshakes. How many handshakes were there? An even number or an odd number of handshakes? Can 5-year olds discover how the parity of "n choose 2" depends on n?</p> <p>Further activity: Have the numbers 1-9 stand up in order again. Have them find a partner to add to 10. Then back in line again. Have the even numbers step forward, and the odd numbers step back. Then the even numbers back and the odd numbers forward. Then back in partners to add to 10. Are evens partnered with evens? Odds partnered with odds? You can ask lots of questions and keep the kids moving.</p> <p>Watch out -- you might have to bring extra numbers and modify groups so that all kids can participate.</p> <p>Good luck! When in doubt you can ask the 5-year olds why 6 is scared of 7. </p> http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106619#106619 Answer by Marty for Philosophy behind Mochizuki's work on the ABC conjecture Marty 2012-09-07T18:11:19Z 2012-09-08T10:07:01Z <p>I'll take a stab at answering this controversial question in a way that might satisfy the OP and benefit the mathematical community. I also want to give some opinions that contrast with or at least complement grp. Like others, I must give the caveats: I do not understand Mochizuki's claimed proof, his other work, and I make no claims about the veracity of his recent work.</p> <p>First, some background which might satisfy the OP. For years, Mochizuki has been working on things related to Grothendieck's anabelian program. Here is why one might hope this is useful in attacking problems like ABC:</p> <p>Begin with the Neukirch-Uchida theorem. See "Über die absoluten Galoisgruppen algebraischer Zahlkörper," by J. Neukirch, Journées Arithmétiques de Caen (Univ. Caen, Caen, 1976), pp. 67–79. Asterisque, No. 41-42, Soc. Math. France, Paris, 1977. Also "Isomorphisms of Galois groups," by K. Uchida, J. Math. Soc. Japan 28 (1976), no. 4, 617–620. </p> <p>The main result of these papers is that a number field is determined by its absolute Galois group in the following sense: fix an algebraic closure $\bar Q / Q$, and two number fields $K$ and $L$ in $\bar Q$. Then if $\sigma: Gal(\bar Q / K) \rightarrow Gal(\bar Q / L)$ is a topological isomorphism of groups, then $\sigma$ extends to an <em>inner</em> automorphism $Int(\tau): g \mapsto \tau g \tau^{-1}$ of $Gal(\bar Q / Q)$. Thus $\tau$ conjugates the number field $K$ to the number field $L$, and they are isomorphic.</p> <p>So while class field theory guarantees that the absolute Galois group $Gal(\bar Q / K)$ determines (the profinite completion of) the multiplicative group $K^\times$, the Neukirch-Uchida theorem guarantees that the entire field structure is determined by the profinite group structure of the Galois group. Figuring out how to recover aspects of the field structure of $K$ from the profinite group structure of $Gal(\bar Q / K)$ is a difficult corner of number theory.</p> <p>Next, consider a (smooth) curve $X$ over $Q$; suppose that the fundamental group $\pi_1(X({\mathbb C}))$ is nonabelian. Let $\pi_1^{geo}(X)$ be the profinite completion of this nonabelian group. Basic properties of the etale fundamental group give a short exact sequence: $$1 \rightarrow \pi_1^{geo}(X) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar Q / Q) \rightarrow 1.$$</p> <p>Now, just as one can ask about recovering a number field from its absolute Galois group ($Gal(\bar Q / K)$ is isomorphic to $\pi_1^{et}(K)$), one can ask how much one can recover about the curve $X$ from its etale fundamental group. Any $Q$-point $x$ of $X$, i.e. map of schemes from $Spec(Q)$ to $Spec(X)$ gives a section $s_x: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$. </p> <p>One case of the famous "section conjecture" of Grothendieck states that this gives a bijection from $X(Q)$ to the set of homomorphisms $Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$ splitting the above exact sequence. One hopes, more generally, to recover the structure of $X$ as a curve over $Q$ from the induced <em>outer</em> action of $Gal(\bar Q / Q)$ on $\pi_1^{geo}(X)$. (take an element $\gamma \in Gal(\bar Q / Q)$, lift it to $\tilde \gamma \in \pi_1^{et}(X)$, and look at conjugation of the normal subgroup $\pi_1^{geo}(X)$ by $\tilde \gamma$, well-defined up to inner automorphism independently of the lift.)</p> <p>As in the case of the Neukirch-Uchida theorem, there is an active and difficult corner of number theory devoted to recovering properties of rational points of (hyperbolic) curves from etale fundamental groups. Here are two dramatically difficult problems in the same spirit:</p> <ol> <li><p>How can you describe the regulator of a number field $K$ from the structure of the profinite group $Gal(\bar Q / K)$?</p></li> <li><p>Given a section $s: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$, how can one describe the height of the corresponding point in $X(Q)$?</p></li> </ol> <p>I would place Mochizuki's work in this anabelian corner of number theory; I have always kept a safe and respectful distance from this corner.</p> <p>Now, to say something not quite as ancient that I gleaned from flipping through Mochizuki's recent work:</p> <p>Many people here on MO and elsewhere have been following research on the field with one element. It is a tempting object to seek, because analogies between number fields and function fields break down quickly when you realize there is no "base scheme" beneath $Spec(Z)$. But I see Mochizuki's work as an anabelian approach to this problem, and I'll try to describe my understanding of this below.</p> <p>Consider a smooth curve $X$ over a function field $F_p(T)$. The anabelian approach suggests looking at the short exact sequence $$1 \rightarrow \pi_1^{et}(X_{\overline{F_p(T)}}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\overline{F_p(T)} / F_p(T)) \rightarrow 1.$$ But much more profitable is to look instead at $X$ as a surface over $F_p$ which corresponds in the anabelian perspective to studying $$1 \rightarrow \pi_1^{et}(X_{\bar F_p}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar F_p / F_p) \rightarrow 1.$$ But this is pretty close to looking at $\pi_1^{et}(X)$ by itself; there's just a little profinite $\hat Z$ quotient floating around, but this can be characterized (I think) group theoretically within the study of $\pi_1^{et}(X)$ itself.</p> <p>I would understand (after reading Mochizuki) that looking at curves $X$ over function fields $F_p(T)$ as surfaces over $F_p$ is like looking at only the etale fundamental group $\pi_1^{et}(X)$ without worrying about the map to $Gal(\overline{F_p(T)} / F_p(T))$.</p> <p>So, the natural number field analogue would be the following. Consider a smooth curve $X$ over $Q$. In fact, let's make $X = E - { 0 }$ be a once-punctured elliptic curve over $Q$. Then the absolute anabelian geometry suggests that to study $X$, it should be profitable to study the etale fundamental group $\pi_1^{et}(X)$ <em>all by itself</em> as a profinite group. This is the anabelian analogue of what others might call "studying (a $Z$-model of) $X$ as a surface over the field with one element". </p> <p>Without understanding any of the proofs in Mochizuki, I think that his work arises from this absolute anabelian perspective of understanding the arithmetic of once-punctured elliptic curves over $Q$ from their etale fundamental groups. The ABC conjecture is equivalent to <a href="http://en.wikipedia.org/wiki/Szpiro%27s_conjecture" rel="nofollow">Szpiro's conjecture</a> which is a conjecture about the arithmetic of elliptic curves over $Q$.</p> <p>Now here is a suggestion for number theorists who, like myself, have unfortunately ignored this anabelian corner. Let's try to read the papers of Neukirch and/or Uchida to get a start, and let's try to understand Minhyong Kim's work on Siegel's Theorem ("The motivic fundamental group of $P^1 \backslash ( 0, 1, \infty )$ and the theorem of Siegel," Invent. Math. 161 (2005), no. 3, 629–656.)</p> <p>It would be wonderful if, while we're waiting for the experts to weight in on Mochizuki's work, we took some time to revisit some great results in the anabelian program. If anyone wants to start a reading group / discussion blog on these papers, I would enjoy attending and discussing.</p> http://mathoverflow.net/questions/106551/is-this-fragment-of-arithmetic-on-p-infty-mathbb-z-decidable Is this fragment of arithmetic on $p^{-\infty} {\mathbb Z}$ decidable? Marty 2012-09-06T21:51:59Z 2012-09-06T21:51:59Z <p>Let $p$ be a prime number. Consider the abelian group $p^{-\infty} {\mathbb Z} = \bigcup p^{-n} {\mathbb Z}$ consisting of rational numbers whose denominator is a power of $p$, under addition.</p> <p>View ${\mathbb Z}$ as a predicate on $p^{-\infty} {\mathbb Z}$, in the sense that ${\mathbb Z}(x)$ is true if $x$ is an integer.</p> <p>Let $\phi$ be the function from ${\mathbb Z}$ to $p^{-\infty} {\mathbb Z}$ given by $\phi(n) = p^n$.</p> <p>Is the first order theory of $( p^{-\infty} {\mathbb Z}, =, >, 0, 1, +, {\mathbb Z}, \phi )$ decidable?</p> <p>A bit of background: The theory of $({\mathbb Z}, =, >, 0, 1, +, \phi)$ is decidable (where here $\phi$ is restricted to natural numbers). This is a result of Semenov; see my answer to the question <a href="http://mathoverflow.net/questions/103896/beyond-presburger-arithmetic/103914" rel="nofollow">http://mathoverflow.net/questions/103896/beyond-presburger-arithmetic/103914</a> for more.</p> <p>On the other hand, what I've described is pretty close to undecidable theories. If I included a predicate $f: {\mathbb Z} \times p^{-\infty} {\mathbb Z} \rightarrow p^{-\infty} {\mathbb Z}$ sending $(n,x)$ to $2^n \cdot x$, it would be undecidable by a result of Delon ("${\mathbb Q}$ Muni de l'Arithmetique Faible de Penzin est Decidable," Proc. of AMS, vol. 125, no 9, 1997).</p> <p>Any references or results would be greatly appreciated. </p> http://mathoverflow.net/questions/105014/changing-basis-on-an-extension-of-a-free-z-module Changing basis on an extension of a free Z-module. Marty 2012-08-18T23:32:21Z 2012-08-18T23:32:21Z <p>Consider a finite-rank free $Z$-module $Y$. Let $c: Y \times Y \rightarrow Z$ be a $Z$-bilinear form. Assume that $c(y_1, y_2) + c(y_2, y_1)$ is even, for all $y_1, y_2 \in$. Then $c$ "incarnates" an extension of $Z$-modules: $$0 \rightarrow \mu_2 \rightarrow \hat Y \rightarrow Y \rightarrow 0,$$ with a distinguished section. Here $\mu_2 = { \pm 1 }$, and $\hat Y = Y \oplus \mu_2$ as a set; define addition in this set by $$(y_1, \epsilon_1) + (y_2, \epsilon_2) = \left( y_1 + y_2, \epsilon_1 \epsilon_2 \cdot (-1)^{c(y_1, y_2)} \right).$$</p> <p>First question: Does $\hat Y$ have a name in the literature? I know it's a special case of the construction of extensions by cocycles, etc., but maybe it has its own name? Do such abelian extensions arise naturally? For example, if $T$ is the topological torus $(Y \otimes R) / Y$ with fundamental group $Y$, is there a natural manifold with fundamental group $\hat Y$ that occurs in the literature?</p> <p>Now, many algebraists would dismiss these extensions, because they are "trivial". All extensions of $Y$ split, since $Y$ is a free Z-module. But the extension $\hat Y$ does not split <em>canonically</em>.</p> <p>What interests me most is the "change-of-basis" formula for splittings. Namely, consider a (ordered) basis $(y_1, \ldots, y_r)$ of $Y$. This gives a splitting $\phi$, using the section above: define $$\phi \left( a_1 y_1 + \cdots + a_r y_r \right) = a_1 \hat y_1 + \cdots + a_r \hat y_r.$$</p> <p>Now consider another $Z$-basis $(y_1', \ldots, y_r')$ with change of basis matrix $A = (\alpha_i^k)$: $$y_i = \sum_k \alpha_i^k y_k'.$$ This gives another splitting $\phi': Y \rightarrow \hat Y$: $$\phi' \left( a_1 y_1' + \cdots + a_r y_r' \right) = a_1 \widehat{y_1'} + \cdots + a_r \widehat{y_r'}.$$</p> <p>As any two splittings differ by an element of $Hom(Y, \mu_2)$, so $\phi'(y) - \phi(y) \in \mu_2$ for all $y \in Y$. This difference is given on basis elements by the formula $$\phi'(y_i) - \phi(y_i) = (-1)^{E_i},$$ $$E_i = \sum_k \left( {\alpha_i^k} \atop 2 \right) c(y_k', y_k') + \sum_{1 \leq m &lt; n \leq r} \alpha_i^m \alpha_i^n c(y_m', y_n').$$</p> <p>Second, most important, question: Has anyone seen a formula like this before in other contexts? It involves nothing more than a invertible matrix $A \in GL(Y)$ and symmetric bilinear form $C \in Hom(Y \otimes Y, Z / 2 Z)$. So what else does this linear algebraic quantity $E_i$ capture? Where else does $(-1)^{E_i}$ occur in the wild?</p> http://mathoverflow.net/questions/103896/beyond-presburger-arithmetic/103914#103914 Answer by Marty for Beyond Presburger Arithmetic Marty 2012-08-04T05:51:06Z 2012-08-05T06:31:45Z <p>From a <a href="http://www.math.osu.edu/~friedman.8/pdf/0AppB072710.pdf" rel="nofollow">paper by Françoise Point</a>, "On the expansion $(N, +; 2^x)$ of Presburger arithmetic," I learned of a much more general result of Semenov, “Logical theories of one-place functions on the set of natural numbers”, Izv. Akad. Nauk SSSR Ser. Mat., 47:3 (1983), 623–658. </p> <p>Semenov's result implies that Presburger arithmetic together with the function $f(x) = c^x$ is decidable for any fixed $c \geq 2$. Of course, $f$ is characterized by the fact that $f(0) = 1$ and $f(x) = c \cdot f(x-1)$ for $x > 0$. So $f$ is definable in $(N,+,\times,0,1)$.</p> <p>Something about the sparseness of powers of $2$, it seems... but I haven't read the details.</p> http://mathoverflow.net/questions/22908/does-anyone-want-a-pretty-maass-form Does anyone want a pretty Maass form? Marty 2010-04-28T23:33:06Z 2012-05-30T20:41:39Z <p>A few months ago, I was curious about some properties of Maass cusp forms, of nonabelian arithmetic origin. As a result, I went through a somewhat predictable process of finding a totally real $A_4$ extension of $Q$, lifting the resulting projective Galois representation to an honest Galois representation, and writing a short program to compute as many coefficients of the Artin L-function (thus coefficients of the Maass form) as needed.</p> <p>Well, as often happens, I didn't find anything particularly surprising in the end. </p> <p>But now I "have a Maass form". Its a pretty Maass form -- the simplest one of eigenvalue 1/4, of "nonabelian" origin (not arising from a dihedral Galois representation). Its conductor is 163 -- a very attractive prime number (though its appearance here seems coincidental). Some class number 1 coincidences make the computation of its coefficients extremely quick and simple. </p> <p>So, does anyone want the Maass form (i.e. code to output coefficients quickly)? It's fun to play with, and doesn't take up too much space. I guarantee its modularity. If not, any suggestions where to put it (a little journal that publishes such cute examples)? </p> http://mathoverflow.net/questions/98262/the-simplest-even-artin-representations-of-degree-2-and-the-corresponding-maass-f/98290#98290 Answer by Marty for The simplest even Artin representations of degree 2 and the corresponding Maaß forms Marty 2012-05-29T15:47:51Z 2012-05-29T15:47:51Z <p>I'll leave the dihedral and octahedral case to others, but for the tetrahedral ($A_4$) and icosahedral ($A_5$) case, I can give some answer.</p> <p>For the tetrahedral case, the smallest conductor is 163. See my question: <a href="http://mathoverflow.net/questions/22908/does-anyone-want-a-pretty-maass-form" rel="nofollow">http://mathoverflow.net/questions/22908/does-anyone-want-a-pretty-maass-form</a></p> <p>I have some (not very well documented) code to compute Artin L-function coefficients for this even tetrahedral Galois representation and thus the (provably, by Langlands) Maass form. I've posted this code on my webpage <a href="http://people.ucsc.edu/~weissman/" rel="nofollow">http://people.ucsc.edu/~weissman/</a></p> <p>For the icosahedral case, a totally real $A_5$ extension of the rationals is given by the splitting field of $x^5 + 5 x^4 - 7 x^3 - 11 x^2 + 10 x + 3$. In my 1999 undergraduate senior thesis, at <a href="http://people.ucsc.edu/~weissman/MWSenThesis.pdf" rel="nofollow">http://people.ucsc.edu/~weissman/MWSenThesis.pdf</a>, I found some mild numerical evidence that the associated <em>degree-3</em> L-function is entire. I can't find it written there, but I'm guessing it's the first, or among the first, $A_5$ extensions of $Q$. I would have chosen something of minimal conductor, to minimize the compute-time. I think I found this by looking in tables from J. Buhler's thesis.</p> <p>Note that I didn't lift the projective representation $Gal \rightarrow PGL_2(C)$ to an honest 2-dimensional representation. This is a bit subtle, and I wasn't capable of that work at the time. Using the 3-dimensional representation (since $A_5$ has a faithful 3-dim representation) avoids this issue, capturing the adjoint square lift of the putative Maass form. I'd guess that lifting the projective representation would be possible now, if someone wanted to do the work.</p> http://mathoverflow.net/questions/97962/has-anyone-used-this-theorem-of-p-cartier Has anyone used this theorem of P. Cartier? Marty 2012-05-25T17:30:21Z 2012-05-25T17:30:21Z <p>In "Groupes Algebriques et Groupes Formels", Conf. au coll. sur la theorie des groupes algebriques, Bruxelles 1962, P. Cartier proves the following in Section 9, Theoreme 1:</p> <p>(What follows is my English exposition/summary and translation)</p> <p>Fix a field $k$ of characteristic $p$. Let $\phi: k \rightarrow k$ be the Frobenius endomorphism $\phi(x) = x^p$.</p> <p>Let $U$ be the abelian category of unipotent commutative algebraic groups over $k$. For such a group $G$, let $G^{\phi} = G \times_{Spec(k)} Spec(k)$ be the fibre product, where $k$ is viewed as a $k$-algebra via $\phi$. There are Frobenius and Verschiebung homomorphisms for all $n > 0$: $$F^n: G \rightarrow G^{\phi^n}, V^n: G^{\phi^n} \rightarrow G.$$</p> <p>Let $W_n$ be the group scheme of Witt vectors under addition, truncated so that the underlying scheme is just affine $n$-space. For any object $G$ of $U$, let $$H_n(G) = Hom(G, W_n).$$ (Cartier calls this functor $V_n$, but this is easily confused with Verschiebung!)</p> <p>Let $U_n$ be the full subcategory of $U$ consisting of groups $G$ for which $V^n = 0$; the Witt group scheme $W_n$ is a prototypical example.</p> <p>Let $E_n = Hom(W_n, W_n)$. This is a familiar Dieudonne ring (truncated). Then $H_n(G)$ is a contravariant functor from the category $U$ to the category of left $E_n$-modules.</p> <p><strong>Theorem 1, part (c):</strong> The functor $H_n$ is a duality from the category $U_n$ to the category of left $E_n$-modules.</p> <p>Now, as Cartier notes afer the theorem, the category of left $E_n$-modules is well understood in the finite-length setting. This is the productive industry of Dieudonne modules. My question is...</p> <p>Has anyone been using this theorem <em>outside</em> the finite-length setting? Certainly there are many naturally occurring commutative unipotent groups which are not finite over $k$. Does anyone use the corresponding left $E_n$-modules? If so, references would be appreciated? If not, why not -- are the left (say, finitely-generated) $E_n$-modules so complicated to study outside of the finite-length setting?</p> http://mathoverflow.net/questions/95982/type-of-26-dimensional-representation-of-different-real-forms-of-the-complex-simp/96008#96008 Answer by Marty for Type of 26-dimensional representation of different real forms of the complex simple Lie algebra $F_4$ Marty 2012-05-04T18:50:31Z 2012-05-04T21:08:38Z <p>I think the best way to see the signature of these quadratic forms is by using the formula from "A Classification Theorem for Albert Algebras" by R. Parimala, R. Sridharan, and Maneesh L. Thakur, Trans. AMS 350 #3, March 1998.</p> <p>All forms of $F_4$ arise from Albert algebras. Over $R$, these are 27-dimensional algebras, whose unital automorphisms form groups of type $F_4$. They are classified, over fields of characteristic neither $2$ nor $3$, by cohomological invariants $f_3$ and $f_5$. These cohomological invariants determine 3-fold and 5-fold Pfister forms, $\phi_3$ and $\phi_5$ respectively.</p> <p>The formula of P-S-T (above), or maybe originally due to Serre, is that for an Albert algebra $A$ over $k$, $$Q_A \perp \phi_3 \cong &lt;2,2,2> \perp \phi_5.$$</p> <p>Now there are only two Pfister forms over $R$ for $\phi_3$ and $\phi_5$. The signature of $\phi_3$ is either $(8,0)$ or $(4,4)$. Similarly, the signature of $\phi_5$ is either $(32,0)$ or $(16,16)$. The signature of $&lt;2,2,2>$ is $(3,0)$. Hence the possibilities for the signature $(p,n)$ of $Q_A$ are: $$(p,n) + (8,0) = (3,0) + (32,0),$$ $$(p,n) + (8,0) = (3,0) + (16,16),$$ $$(p,n) + (4,4) = (3,0) + (32,0),$$ $$(p,n) + (4,4) = (3,0) + (16,16).$$</p> <p>Only three cases are possible: $(p,n) = (27,0)$ or $(p,n) = (11,16)$ or $(p,n) = (15,12)$.</p> <p>As $F_4$ acts on the orthogonal complement of the identity, and the identity has positive norm, the possible signatures for the 26-dimensional rep of $F_4$ are: $$(26,0), (10,16), (14,12).$$</p> http://mathoverflow.net/questions/89097/representation-theory-of-reductive-groups-in-characteristic-p-as-a-limit-of-the/89126#89126 Answer by Marty for Representation theory of reductive groups in characteristic $p$ as a limit of the theories in characteristic $0$ Marty 2012-02-21T18:41:47Z 2012-02-21T18:41:47Z <p>I think, although it's dated later than Deligne's paper that you mentioned, that the first written instance of Kazhdan's principle is in the paper "Representations of groups over close local fields", Journal d'Analyse Math\'ematique, vol. 47,1986, pp.175--179.</p> <p>This is in the same journal issue as "Cuspidal Geometry of p-adic Groups" (by Kazhdan) and "Trace Paley-Wiener Theorem for Reductive p-adic Groups" (by Bernstein, Deligne, Kazhdan). The book "Representations of reductive groups over a local field" appeared in 1984, and according to the MathSciNet review of the article "Le 'Centre' de Bernstein", the Trace Paley-Wiener Theorem paper was already a preprint in 1984. </p> <p>So it seems to me that Kazhdan's principle was probably "in the air" by 1984, but not written down by him until the "close local fields" article above. I second Jim Humphreys' suggestion to contact Kazhdan himself for less speculative history. </p> http://mathoverflow.net/questions/88145/uniform-setting-for-computing-orders-of-algebraic-groups-over-finite-quotients-of/88151#88151 Answer by Marty for Uniform setting for computing orders of algebraic groups over finite quotients of the integers? Marty 2012-02-10T22:10:00Z 2012-02-10T22:10:00Z <p>The places to look are: </p> <ol> <li>Steinberg, "Endomorphisms of linear algebraic groups." Memoir AMS 80, (1968), and </li> <li>Gross, "The motive of a reductive group" Invent. math. 130, 287 ± 313 (1997).</li> </ol> <p>(I learned about the former from the latter).</p> <p>To any (quasi-split) group $G$ with maximal torus $T$, there is an associated Artin-Tate motive $M$ (i.e., a Galois representation). </p> <p>The point-count over finite fields is related to the action of Frobenius on the twisted dual motive: equation (3.1) of Gross's article (citing Steinberg) gives us: $$\vert G(k) \vert / q^{dim(G)} = \prod_{d \geq 1} det(1 - Fr \cdot q^{-d} : V_d ),$$ where $Fr$ denotes the Frobenius, $q$ the order of the finite field $k$, and $V = \bigoplus V_d$ a graded ${\mathbb Q}$ vector space as follows: $V = R / R^+$ where $R$ is the graded algebra $Sym^\bullet(E)^W$, where $E$ is the character lattice of $T$ tensored with ${\mathbb Q}$ (section 1 of Gross).</p> http://mathoverflow.net/questions/80094/the-schwartz-space-on-a-manifold/80112#80112 Answer by Marty for The Schwartz Space on a Manifold Marty 2011-11-05T04:55:21Z 2011-11-06T01:41:57Z <p>For Lie groups, at least for those that embed into $GL_n(R)$ for some $n$, my favorite treatment of the Schwartz space is in Casselman's paper "Introduction to the Schwartz Space of $\Gamma \backslash G$", Can. J. Math. XL, No 2, 1989. There Casselman defines an appropriate Schwarz space on $\Gamma \backslash G$ whenever $G$ is the Lie group obtained by taking the $R$-points of an affine algebraic group over $R$, and $\Gamma$ is any discrete subgroup of $G$ (including the trivial subgroup). </p> <p>I think this is the right place to look, before studying things like the Fourier transform (i.e. Plancherel and Paley-Wiener theorems).</p> http://mathoverflow.net/questions/80060/embedding-commutative-associative-rings-in-non-associative-rings/80066#80066 Answer by Marty for Embedding commutative associative rings in non associative rings Marty 2011-11-04T16:41:00Z 2011-11-04T16:41:00Z <p>Yes. The "can" question is not so interesting: think of the image of the integers inside any nonassociative algebra. But specific cases, and counting embeddings, are interesting. See work of Gross and Gan, "<a href="http://www.springerlink.com/content/9c6mqcxn5j2cvcfd/" rel="nofollow">Commutative Subrings of Certain Non-associative Rings</a>", Math. Ann. v.314 n.2, 1998. </p> http://mathoverflow.net/questions/77728/can-local-duality-for-elliptic-curves-be-proven-with-big-rings Can local duality for elliptic curves be proven with "big rings"? Marty 2011-10-10T19:05:34Z 2011-10-11T00:22:31Z <p>From Exercise 5.14, Ch. V of Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", I learned that the local duality for elliptic curves over $p$-adic fields can be proven for <em>Tate curves</em> by a relatively easy argument in Galois cohomology. Essentially, when the elliptic curve is $E_q = G_m / q^Z$ over a $p$-adic field $K$, one can find various long exact sequences connecting the Galois cohomology $H^1(K, E_q(\bar K))$ to the cohomology of $G_m$, $Q/Z$, etc., which are well-known by class field theory.</p> <p>Without being an expert in $p$-adic cohomology, Hodge theory, etc., I know that by passing to a big ring ($B_{dR}$ will work), one can find a pair of periods for an elliptic curve over $K$ with good reduction. There might not be any interesting nontrivial uniformization of such elliptic curves, but the periods carry the information instead.</p> <p>So can one exhibit (some of) the duality between $H^1(K, E(\bar K))$ and $Hom(E(K), Q/Z)$ when $E$ has good reduction, by using a big ring like $B_{dR}$? </p> <p>When I see period rings, they are always used as linear algebraic gadgets. But since $B_{dR}$ is a $K$-algebra with Galois action, might someone consider $H^1(K, E(B_{dR}))$? In other words, rather than taking a linear algebraic gadget over $\bar K$, and tensoring up to $B_{dR}$, might one study a variety over $\bar K$ and base change to $B_{dR}$ or at least take $B_{dR}$-points? Might $H^1(K, E(B_{dR}))$ pick up the Weil-Chatelet group in the spirit of my first question?</p> <p>Any references including $B_{dR}$-points of varieties would be greatly appreciated, as well as answers to the questions.</p> http://mathoverflow.net/questions/71727/is-there-a-basic-number-theory-for-elliptic-curves/71753#71753 Answer by Marty for Is there a "Basic Number Theory" for elliptic curves? Marty 2011-07-31T19:14:38Z 2011-08-01T22:45:29Z <p>I don't think that such a survey paper or textbook exists, but the closest thing I know of is "A note on height pairings, Tamagawa numbers, and the Birch and Swinnerton-Dyer conjecture" by Spencer Bloch, Invent. Math. v.58, no.1, pp. 65-76, 1980.</p> <p>Here's an abbreviated history, picking up where you left off: Takashi Ono wrote a paper "On the Tamagawa number of algebraic tori", Annals of Math., v.78, no. 1, July 1963. In that paper, Ono computes the volume of $T^1(A) / T(F)$, where $T$ is an algebraic torus over a number field $F$, and $A$ is the adele ring, and $T^1(A)$ denotes the intersection of kernels of $\vert \chi \vert$ as $\chi$ ranges over $F$-rational characters of $T$. Ono's formula states that this volume (called a Tamagawa number, but <em>not</em> to be confused with the local Tamagawa numbers $c_v$) equals $\vert Pic_{tor}(T) \vert / \vert Sha(T) \vert$. </p> <p>The numerator is the order of the torsion subgroup of the Picard group of $T$. The denominator is the order of the Tate-Shafarevich group of $T$. Most of the arithmetic is contained in the normalization of the measure on the quotient space $T^1(A) / T(F)$ -- this normalization of measure uses the L-function (an Artin L-function) of $T$, and the special case $T = G_m$ corresponds to the Dirichlet class number formula for $F$. </p> <p>From looking at Ono's paper (an earlier Annals paper from 1961), it appears that Weil and Tate were influential in his work.</p> <p>Fast forwarding to 1980 (skipping lots of great things for reductive groups), here's a brief summary of what Bloch does (in the Inventiones paper mentioned above). He begins with an abelian variety $E$ over a global field $F$ (I already used $A$ for the adeles). Using the fact that the dual abelian variety $\hat E$ can also be viewed as $Ext(E, G_m)$, Bloch uses the Mordell-Weil lattice $L$ of $F$-rational points on $\hat E$ to construct an extension of algebraic groups over $F$: $$1 \rightarrow T \rightarrow X \rightarrow E \rightarrow 1$$ in which $T$ is an $F$-split torus with character lattice $L$.</p> <p>Remarkably, Bloch proves that $X(F)$ is discrete and cocompact in $X(A)$. Moreover, most suggestively, Bloch proves that the BSD conjecture for $E$ is equivalent to the conjecture that the volume of $X(A) / X(F)$, with respect to a suitably normalized measure, equals $\vert Pic_{tor}(X) \vert / \vert Sha(X) \vert$.</p> <p>Of course, the meat of Bloch's approach is in the normalization of measure, which uses the L-function of $E$. I once gave a truly disastrous talk as a graduate student about Bloch's paper, in which all this normalization of measure stuff completely escaped me. I still find Bloch's paper very difficult and mysterious. It seems that it is mostly cited for its novel construction of height pairings, but not much has been done (publicly) with its interpretation of BSD.</p> http://mathoverflow.net/questions/70762/hilbert-symbol-and-weil-index-beyond-the-quadratic-case Hilbert symbol and Weil index, beyond the quadratic case? Marty 2011-07-19T18:23:57Z 2011-07-19T21:44:24Z <p>Let $F$ be a local nonarchimedean field. Let $n$ be a positive integer for which the group $\mu_n(F)$ of $n^{th}$ roots of unity in $F$ has order $n$. Let $\epsilon: \mu_n(F) \rightarrow C^\times$ be a faithful character.</p> <p>The $n^{th}$ order Hilbert symbol, composed with $\epsilon$ will be denoted by $(x,y)_n$ for all $x,y \in F^\times$.</p> <p>Somewhat remarkably, when $n = 2$, there is a gadget called the Weil index (depending on a nontrivial additive character of $F$), which is a function $w: F^\times \rightarrow \mu_4(C)$ satisfying $$(x,y)_2 = w(xy) w(x)^{-1} w(y)^{-1}.$$ In other words, the quadratic Hilbert symbol, viewed as a cocycle in $Z^2(F^\times, \mu_2)$ becomes a coboundary after applying the unique injective map on coefficients $\mu_2 \rightarrow \mu_4$.</p> <p>The Weil index has many lovely properties, but I'm interested merely in the extent to which it generalizes to $n > 2$. Namely, does there exist a "higher-order Weil index" $w_n: F^\times \rightarrow \mu_{2n}(C)$ satisfying $$(x,y)_n = w_n(xy) w_n(x)^{-1} w_n(y)^{-1}?$$ Or perhaps does one need to pass to a larger group than $\mu_{2n}(C)$ for the "Hilbert cocycle" to become a coboundary?</p> <p>The main difficulties occur when $n$ is a multiple of 4, I think. Any references or known results?</p> <p>------------EDIT BELOW----------------</p> <p>Oh no - I missed a dumb point. $(x,y)_n$ is skew-symmetric. So when $n$ is a multiple of $4$, we have $(x,y) \neq (y,x)$ quite often. Thus the existence of such a "higher-order Weil index" is not possible when $n$ is a multiple of $4$. Well... maybe this question will keep others from making similar mistakes.</p> http://mathoverflow.net/questions/68897/what-is-the-current-status-for-lusztigs-positivity-conjecture-for-symmetric-cart What is the current status for Lusztig's positivity conjecture for symmetric Cartan datum? Marty 2011-06-27T02:05:22Z 2011-06-30T02:43:37Z <p>This is related to the earlier <a href="http://mathoverflow.net/questions/39934/when-does-lusztigs-canonical-basis-have-non-positive-structure-coefficients" rel="nofollow">question here</a></p> <p>In Conjecture 25.4.2 in his "Introduction to Quantum Groups," Lusztig conjectures that "If the Cartan datum is symmetric, then the structure constants $m_{a,b}^c$, $\hat m_c^{a,b}$ are in $N[v,v^{-1}]$. </p> <p>In what cases is this proven? I only care about classical groups, specifically split reductive groups over $Z$ as constructed via Lusztig's canonical bases (specialize at $v=1$). I would love it if the structure constants were non-negative, so that there were no troublesome signs floating around in the antipode. When is this known to be the case? My googling leads me to suspect something is known in the simply-laced case. Anything more general yet? References? I'm not interested in the non-symmetric case. </p> http://mathoverflow.net/questions/65335/hopf-algebra-of-chevalley-group-from-the-root-system/65345#65345 Answer by Marty for Hopf algebra of Chevalley group from the root system Marty 2011-05-18T15:37:47Z 2011-05-18T15:37:47Z <p>A no-nonsense construction, over $Z$, following work of Kostant and Chevalley, is given in Lusztig's paper "Twelve bridges from a reductive group to its Langlands dual". The heart of the construction of the Hopf algebra is in Section 5. </p> <p>This is easy enough to find online, and according to Lusztig's webpage, it can also be found published in Contemp. Math. 478 (2009), 125-143.</p> <p>Good luck!</p> http://mathoverflow.net/questions/64617/mathematical-ideas-named-after-places/64807#64807 Answer by Marty for Mathematical ideas named after places Marty 2011-05-12T16:19:34Z 2011-05-12T16:19:34Z <p>Nowhere differentiable: named for Ainsworth, Nebraska, I believe.</p> http://mathoverflow.net/questions/63342/ring-of-continuous-functions-reference-request Ring of continuous functions, reference request. Marty 2011-04-28T21:04:57Z 2011-04-30T05:43:02Z <p>I am looking for a reference for the following facts in functional analysis and topology. (If these "facts" are not true, I suppose I'm looking for the closest approximation which is true.)</p> <p>Let $X$ be a locally compact Hausdorff topological space. Let $C(X)$ denote the ring of continuous complex-valued functions on $X$, endowed with the compact-open topology. Then $C(X)$ is a complete locally convex topological (complex) vector space (this can be found in Kothe, vol. 2, I think).</p> <p>Now let $Y$ be another locally compact Hausdorff topological space. From Kothe, vol. 2, I know that $C(X \times Y)$ is naturally isomorphic to $C(X) \hat \otimes C(Y)$, where $\hat \otimes$ denotes the completion with respect to the injective tensor product topology.</p> <p>I believe that pointwise multiplication $C(X) \times C(X) \rightarrow C(X)$ extends (uniquely) to a continuous linear map from $C(X) \hat \otimes C(X)$ to $C(X)$.</p> <p>If $f: X \rightarrow Y$ is a continuous function, then precomposition with $f$ yields a continuous $C$-algebra homomorphism $f^\ast: C(Y) \rightarrow C(X)$.</p> <p>I believe the following to be true:</p> <p>Theorem: For every continuous algebra homomorphism $\phi: C(Y) \rightarrow C(X)$, there exists a unique continuous map $f: X \rightarrow Y$ such that $\phi = f^\ast$.</p> <p>In other words, I wish that $C( \bullet )$ is a faithful functor from the category of locally compact Hausdorff spaces and continuous maps to the category of rings in the (symmetric monoidal under $\hat \otimes$) category of complete locally convex topological vector spaces and continuous linear maps.</p> <p>Any references and/or corrections would be very welcome!</p> <p>But an important note: I am <em>not</em> looking for well-known modifications, like "try the $C^\ast$-algebra instead" or the von Neumann algebra, etc.. I have good reasons for considering the ring $C(X)$ with the compact-open topology, and I don't wish to mess with it.</p> http://mathoverflow.net/questions/18538/semistable-filtered-vector-spaces-a-tannakian-category Semistable filtered vector spaces, a Tannakian category. Marty 2010-03-18T02:06:48Z 2011-04-14T00:17:26Z <p>Let $k$ be a field (char = 0, perhaps). Let $(V,F)$ be a pair, where $V$ is a finite-dimensional $k$-vector space, and $F$ is a filtration of $V$, indexed by rational numbers, satisfying: </p> <ol> <li>$F^i V \supset F^j V$ when $i &lt; j$.</li> <li>$F^i V = V$ for $i &lt;&lt; 0$. $F^i V = { 0 }$ for $i >> 0$.</li> <li>$F^i V = \bigcap_{j &lt; i} F^j V$.</li> </ol> <p>We define: $$F^{i+} V = \bigcup_{j > i} F^j V.$$</p> <p>The slope of $(V,F)$ (when $V \neq { 0 }$) is the rational number: $$M(V,F) = \frac{1}{dim(V)} \sum_{i \in Q} i \cdot dim(F^i V / F^{i+} V).$$</p> <p>The pair $(V,F)$ is called semistable if $M(W, F_W) \leq M(V, F)$ for every subspace $W \subset V$, with the subspace filtration $F_W$. </p> <p>A paper of Faltings and Wustholz constructs an additive category with tensor products, whose objects are semistable pairs $(V,F)$. A paper of Fujimori, "On Systems of Linear Inequalities", Bull. Soc. Math. France, seems to imply that the full subcategory of slope-zero objects (together with the zero object) is Tannakian (the abelian category axioms require semistability), with fibre functor to the category of $k$-vector spaces (though Fujimori considers quite a bit more).</p> <p>Does anyone know another good reference for the properties of this Tannakian category? Can you describe the associated affine group scheme over $k$? I'm particularly interested, when $k$ is a finite field or a local field.</p> <p>Update: I think the slope-zero requirement is too strong (though it is assumed in Fujimori). It seems to exclude almost all the semistable pairs $(V,F)$, if my linear algebra is correct. Anyone want to explain this to me too?</p> http://mathoverflow.net/questions/59626/semisimple-weil-deligne-representations Semisimple Weil-Deligne representations Marty 2011-03-26T01:13:30Z 2011-03-26T16:41:15Z <p>I've just realized that I don't understand something important and basic about the Weil-Deligne group and its representations. (I'm not very surprised by this).</p> <p>Following Deligne's article, Section 8 of "Les Constantes des Equations Fonctionnelles Des Functions L" from the 1972 Antwerp volume, one is led to consider representations of the Weil-Deligne group in the following algebraic sense: for any local nonarchimedean field, there is a group scheme $W'$, defined over $Q$, which is a semidirect product of the Weil group scheme $W$ and the additive group scheme $G_a$. This is a <em>non-affine</em> group scheme, and the Weil subgroup scheme $W$ is obtained as a countable disjoint union of affine subschemes (the cosets of inertia).</p> <p>So, as is now standard, we consider algebraic representations of the group scheme $W'$, over various fields $E$ of characteristic zero, as such representations provide a unified framework (thanks to results of Grothendieck, Deligne, Serre) for $\lambda$-adic representations that arise from arithmetic.</p> <p>A crucial piece of this is to restrict attention (or semisimplify) to the <em>semisimple</em> representations of the Weil-Deligne group. And presumably, such semisimple representations form a Tannakian category (over any base field of characteristic zero).</p> <p>And so to my question... what is the algebraic group associated to this Tannakian category? Or am I just confused? And how does the (non-affine) Weil-Deligne group scheme relate to this (affine) algebraic group obtained by restricting attention to these semisimple representations? Does this involve one of these awfully large group schemes like $Spec(E[E^\times])$, where $E$ is a characteristic zero field (something like the semisimple algebraic hull of the discrete group $Z$)? Any references?</p> http://mathoverflow.net/questions/130010/examples-of-exotic-moduli-problems-for-elliptic-curves/130020#130020 Comment by Marty Marty 2013-05-08T00:51:16Z 2013-05-08T00:51:16Z But such a cool-looking paper! http://mathoverflow.net/questions/130010/examples-of-exotic-moduli-problems-for-elliptic-curves/130020#130020 Comment by Marty Marty 2013-05-08T00:50:56Z 2013-05-08T00:50:56Z Either that's a bizarre TeX error, or else I have a new entry for the question <a href="http://mathoverflow.net/questions/18593/what-are-the-worst-notations-in-your-opinion-closed" rel="nofollow" title="what are the worst notations in your opinion closed">mathoverflow.net/questions/18593/&hellip;</a> http://mathoverflow.net/questions/129972/quasi-minuscule-representations Comment by Marty Marty 2013-05-07T15:06:35Z 2013-05-07T15:06:35Z <a href="http://en.wikipedia.org/wiki/Minuscule_representation" rel="nofollow">en.wikipedia.org/wiki/Minuscule_representation</a> http://mathoverflow.net/questions/129037/criterion-for-nilradical-of-a-maximal-parabolic-subalgebra-to-be-abelian Comment by Marty Marty 2013-04-29T05:36:30Z 2013-04-29T05:36:30Z The reference of (my) choice is Richardson, Rohrle, Steinberg, &quot;Parabolic subgroups with abelian unipotent radical,&quot; in Inventiones v.110, no. 3 (1992), p. 649-671. http://mathoverflow.net/questions/128479/can-i-use-both-of-setbuilder-notations-in-one-article Comment by Marty Marty 2013-04-23T18:57:37Z 2013-04-23T18:57:37Z I like using &quot;such that&quot; (i.e., \text{such that}) whenever there's room for it. And with semicolons, it might seem like the paper is sadly winking ); http://mathoverflow.net/questions/128038/great-mathematics-books-by-pre-modern-authors Comment by Marty Marty 2013-04-19T13:12:23Z 2013-04-19T13:12:23Z Way too broad a question for my taste. It's good the OP realized that older math books could be worthwhile, but asking for a list is kind of like asking for &quot;the greatest books of all time&quot;. There are just too many. But criticism aside, I found my horizons broadened by &quot;The Mathematics of Egypt, Mesopotamia, China, India, and Islam,&quot; edited by V. Katz. That's where I first realized that I could read and enjoy <i>much</i> older math texts, especially those from outside the Eurocentric canon. You could start with Katz's book as a source for excerpts, and look up full books when interested. http://mathoverflow.net/questions/126071/the-octonions-on-a-bad-day Comment by Marty Marty 2013-04-01T01:33:15Z 2013-04-01T01:33:15Z @Mariano: Not really. I guess my point is that matrix algebras are in general endomorphism rings of vector spaces -- the 2x2 case is just one example of the infinite family of examples of type $A_n$. But the octonions are really connected to $G_2$ -- no way around it, no easy shortcuts. I would still say that Zorn's split octonions are simpler than the non-split octonions. For example, identifying a maximal order in the non-split octonions is difficult (due to Coxeter after earlier mistakes), but the maximal order in the split octonions is the obvious choice. http://mathoverflow.net/questions/126071/the-octonions-on-a-bad-day Comment by Marty Marty 2013-03-31T19:43:50Z 2013-03-31T19:43:50Z @Mariano, I think the issue you're describing is that most mathematicians are happy to live without the octonions and exceptional groups. Matrix algebras and $GL_n$ (and classical groups) are sufficient for most people's work. The octonions and $G_2$ are not so universally studied, and maybe people think they are more difficult than they really are. I don't think Zorn's model of the split octonions (over $Z$) is too bad at all -- just 2x2 matrices with vectors in $Z^3$ off the diagonal. Hard to get much simpler than Zorn, I think. http://mathoverflow.net/questions/120984/rep-theory-consequences-of-bott-weil-borel Comment by Marty Marty 2013-02-06T18:19:49Z 2013-02-06T18:19:49Z First, I don't think that representations are coming from nowhere. When a group $G$ acts on a space $X$, and you have a $G$-equivariant bundle on the space $X$, then you get a representation of $G$ on the sections (and higher cohomology) of the bundle. Maybe the most impressive results beyond the theorem itself are how useful it is for generalizations. The generalization that comes first to my mind is Schmid's &quot;L&#178;-cohomology and the discrete series&quot; (Annals, 1976) which proved a conjecture of Langlands by using a geometric realization in the spirit of Borel-Weil-Bott. http://mathoverflow.net/questions/116947/center-of-mass-from-the-abstract-point-of-view-or-could-the-ancient-greeks-inven Comment by Marty Marty 2012-12-21T04:53:07Z 2012-12-21T04:53:07Z &quot;I assume that the ancient Greeks had an idea of a complete normed space (${\mathbb R}$ and ${\mathbb R}^2$ would be enough for our purposes for quite a while), a set, a linear transformation, and the center of mass.&quot; Really? Really??! http://mathoverflow.net/questions/116666/triality-of-spin8/116735#116735 Comment by Marty Marty 2012-12-18T21:42:12Z 2012-12-18T21:42:12Z I'm not sure if continuing questions are supposed to be given as answers... but here are a few comments. First, I don't know what you would mean by conjugating $g_1$, $g_2$, and $g_3$. They are real matrices. The fixed points of the triality automorphism on $Spin(8)$ form the subgroup $G_2$. Changing the lift should just conjugate the $G_2$ within the $Spin(8). http://mathoverflow.net/questions/116666/triality-of-spin8/116679#116679 Comment by Marty Marty 2012-12-18T16:18:43Z 2012-12-18T16:18:43Z Since kreck brought it up, it works over Z too, using Coxeter's maximal order in the octonions. Thanks also to Figueroa-O'Farrill for fixing the Tex! http://mathoverflow.net/questions/113811/do-l-functions-exist-for-half-integral-weight-modular-forms/113814#113814 Comment by Marty Marty 2012-11-19T17:02:54Z 2012-11-19T17:02:54Z So -- long story short -- I share Nick's intuition that the L-function is the entire family of quadratic twists, and I went so far as to cook up a theory of L-groups for (split, for now) metaplectic groups in order to understand this better. http://mathoverflow.net/questions/113811/do-l-functions-exist-for-half-integral-weight-modular-forms/113814#113814 Comment by Marty Marty 2012-11-19T17:01:26Z 2012-11-19T17:01:26Z When$G$is the simplest metaplectic group$\widetilde{SL}_2$, I argue that the L-group${}^L G$is a group scheme over$\mathbb{Z}$, which is isomorphic to$SL_2 \times \Gamma$($\Gamma$the Galois group), but <i>noncanonically</i> and not until base change to$\mathbb{ZZ}[i]$. However, each additive character of${\mathbb A} / {\mathbb Q}$gives such an isomorphism, and this realizes a bijection between quadratic twists and (L-)isomorphisms of the L-group to$SL_2 \times \Gamma$. Taking the standard representation of$SL_2$, you get the L-function of a quadratic twist. http://mathoverflow.net/questions/113811/do-l-functions-exist-for-half-integral-weight-modular-forms/113814#113814 Comment by Marty Marty 2012-11-19T16:56:26Z 2012-11-19T16:56:26Z Following up on Nick's question -- the Langlands correspondence for metaplectic groups is something I've thought about a lot in the past few years. I have a paper on L-groups for Metaplectic Groups on the ArXiv, for example. Since that paper is hard to read (I'm working hard to get rid of all the Hopf algebra machinery used there), I'll summarize here. An L-function should come from <i>two</i> pieces of data: an automorphic representation of$G$(for half-integral weight eigenforms,$G$is a metaplectic group) and an algebraic (finite-dimensional) representation of the L-group${}^L G\$.