User marc palm - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:05:37Z http://mathoverflow.net/feeds/user/10400 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Answer by Marc Palm for Plancherel formula for non-second-countable (non-unimodular) groups Marc Palm 2013-05-17T07:57:34Z 2013-05-17T07:57:34Z <p>Answer to the first question: <em>Jacques Dixmier, Les C-algèbres et leurs représentations.</em> Section 18.8.1</p> <p>Comment on the second question: I actually believe a decomposition of von-Neumann algebra into factors is only available for seperable vNas, which should be for the right regular representation equivalent to the group being second countable.</p> <p>Comment on the third question: I have no idea what could be meant. But the decomposition into factors will not be unique (this is probably what you mean with <em>not enough to consider irreducible representations</em>) and I don't even know what kind of traces should be involved. So for me, it seems unreasonable to expect something useful in this context, which has similar applications as the Plancherel formula.</p> http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc/130604#130604 Answer by Marc Palm for Finding spherical representations of $GL(n, \mathbb{C})$. Marc Palm 2013-05-14T16:10:25Z 2013-05-14T17:54:37Z <p>In the case of $GL(n, \mathbb{C})$, it is known that every unitary, irreducible, infinite-dimensional representation (the others are one-dimensional and factor through the determinant) is given as induced representation $\pi$ from a minimal parabolic associated to the Levi $M(\mathbb{C})$ (being the group of diagonal matrices). This one is spherical iff the restriction $\pi$ to $M(\mathbb{C}) \cap U(n)$ is trivial.</p> <p>The Mackey Induction Restriction formula plus the Iwasawa decomposition indicates that this is the same question for inducing the restriction of $\pi$ to $M(\mathbb{C}) \cap U(n)$ up to $U(n)$.</p> <p>This is the case if and only if the restriction of $\pi$ to $M(\mathbb{C}) \cap U(n)$ is trivial by Frobenius reciprocity.</p> <p>This strategy works more generally for all parabolically induced representation in real reductive groups. Then look possibly at the structure of their subquotients. It is a theorem of Casselman that for a real reductive Lie group all smooth, admissible reps are found as such subquotients of such parabolically induced representations.</p> http://mathoverflow.net/questions/130062/concrete-examples-of-noncongruence-arithmetic-subgroups-of-sl2-r Concrete examples of noncongruence, arithmetic subgroups of SL(2,R) Marc Palm 2013-05-08T07:34:11Z 2013-05-09T18:02:13Z <p>A subgroup of $SL_2(\mathbb{R})$ is called <strong>arithmetic</strong> if it is commensurable with $SL_2(\mathbb{Z})$.</p> <p>An arithmetic subgroup is called <strong>congruence</strong> if it contains a subgroup of type $\Gamma(N)$ for some $N\in \mathbb{N}$.</p> <blockquote> <p>Question: What are concrete examples of subgroups of $SL_2(\mathbb{R})$, which are arithmetic, but not congruence?</p> </blockquote> <p>I have heard the Belyi theorem produces some examples, but I have never seen a concrete one.</p> <blockquote> <p>Further Question: Can such things exist in higher rank Lie groups (real rank $\geq 2$)?</p> </blockquote> http://mathoverflow.net/questions/61263/modular-forms-of-weight-one/130101#130101 Answer by Marc Palm for Modular Forms of Weight One Marc Palm 2013-05-08T17:29:21Z 2013-05-08T17:29:21Z <p>Another natural reason is that there is no pseudo matrix coefficient for limit of discrete series representations, since they are not square integrable. If so, you could simply plugin this pseudo coefficient times the characteristic function associated to the congruence subgroup into the Arthur trace formula, and compute the geometric side.</p> http://mathoverflow.net/questions/53596/elementary-proof-of-algebraicity-of-hecke-eigenvalues-in-weight-1/130066#130066 Answer by Marc Palm for Elementary proof of algebraicity of Hecke eigenvalues in weight 1 Marc Palm 2013-05-08T09:15:20Z 2013-05-08T11:55:58Z <p>Here is another proof: Deligne and Serre have proven that the corresponding L-function equal the Artin L-function of a Galois representation. Deligne had proven similar facts known for weight $k \geq 2$ modular forms before that, and their proof essentially relies on the former results. This implies algebraicity and also is the only approach I know for the same result for Maass Hecke cusp forms of Laplace eigenvalue $1/4$, where the same algebraicity result is unknown.</p> <p><em>This is probably even harder than what you suggest, or equivalent(?). Quoting that result of Deligne and Serre would be a reasonable choice, and deducing algebraicity from it. The question is pretty old and didn't receive an answer so far, so I guess the class isn't running anymore anyways.</em></p> <p>There is no trace formula for Hecke eigenvalues of weight one forms available, since the limit of discrete series representations opposed to the discrete series representations are not square integrable, and have no pseudo-matrix coefficients.</p> http://mathoverflow.net/questions/129637/weyl-law-for-arithmetic-fuchsian-groups-known/129933#129933 Answer by Marc Palm for Weyl law for arithmetic Fuchsian groups known? Marc Palm 2013-05-07T08:41:09Z 2013-05-07T08:59:16Z <p>Consider the normalizer $\Gamma$ of $\Gamma_0(N)$ in $SL_2(\mathbb{Q})$ for $N$ squarefree, which is not a subgroup of $SL_2(\mathbb{Z})$, then you obtain the same Weyl law as classical. For my taste, these groups should also be called congruence subgroups, although they are not contained in $SL_2(\mathbb{Z})$. They are important if you want classically distinguish between nonisomorphic supercuspidal reps contained as factors of automorphic reps associated, which are associated to ramified quadratic extensions of $\mathbb{Q}_p$ at $p |N$.</p> <p><em>I don't know a single example what happens if $\Gamma$ does not contain a congruence subgroup, but is arithmetic.</em></p> <p>Btw, your question makes perfect sense, because the modified Rolecke-Selberg conjecture states (see the introduction <a href="http://link.springer.com/content/pdf/10.1007%2FBF02572621.pdf" rel="nofollow">http://link.springer.com/content/pdf/10.1007%2FBF02572621.pdf</a>) that apart from arithmetic lattices, there are at most finitely many discrete eigenvalues.</p> <p>There is some computational evidence by Hejhal for this conjecture: <em>Hejhal, Dennis A.(1-MN-SM) On eigenvalues of the Laplacian for Hecke triangle groups.</em></p> <blockquote> <p>Copy&amp;Paste from MathSciNet: This paper is part of a series of articles in which computer experiments are performed to numerically compute the eigenvalues of the Laplacian. In this paper the Hecke triangle groups generated by z→−1/z, z→z+2cos(π/N) are considered. The basic results are a computation of eigenvalues for the groups with N=3,4,6 (the congruence groups) with eigenvalue 14+R2 with R&lt;25, and the conclusion that no even cusp forms exist when N=5, R&lt;60 and N=7, R&lt;40. The last result, which is similar to results of Winkler, gives evidence in support of the Phillips-Sarnak conjecture that one should have few if any cusp forms for nonarithmetic groups (except for the obvious ones caused by symmetries). The procedure used is essentially the collocation method. One must also be careful in evaluating the K-Bessel functions that appear in the Fourier expansion of the forms.</p> </blockquote> <p>What is sure is that if $\Gamma$ does not contain a congruence subgroup, then the Maass cusps forms on $\Gamma \backslash \mathbb{H}$ can not be lifted to a vector of an automorphic adelic space $SL_2(\mathbb{A})$ in any obviuous way, and one cannot expect them to have a nice $L$-functions. Also the contribution of the continuous spectrum is most likely not expressable in terms of classical Dirichlet L-functions. Recall that this is the main point to be understood for Weyl laws of non-uniform lattices.</p> <p>I know that the Belyi's theorem generates an isomorphisms between nonsingular algebraic curves over $\mathbb{C}$ and $\Gamma \backslash \mathbb{H}$ compactified at the cusps, where $\Gamma$ is an arithmetic subgroup of $SL_2(\mathbb{Z})$, but noncongruence in general. This might be a source of such lattices and at least modular functions on them should exists (by Riemann Roch?).</p> http://mathoverflow.net/questions/129594/cuspidal-automorphic-representations-as-the-space-of-k-finite-vectors-in-a-unit/129630#129630 Answer by Marc Palm for Cuspidal automorphic representations as the space of $K$-finite vectors in a unitary cuspidal automorphic representation. Marc Palm 2013-05-04T10:34:15Z 2013-05-06T08:07:49Z <p><strong>Yes.</strong> All these viewpoints are equivalent. We write "rep" for irreducible representation.</p> <p>Edit on request:</p> <p>An analytic cuspidal rep is subrepresentation $\pi$ of $L_0^2(G(F) \backslash G(A), \omega)$ for a central character $\omega$.</p> <p>An algebraic cuspidal rep is an irreducible Harish-Chandra module and an irreducble $C_c^\infty(G(A_f))$-module. By Schur's lemma they automatically have a central character $\omega$. Moreover, you need appropiate growth conditions guaranteeing square-integrability. This guarantees also that the vectors are in the $L_0^2(G(F) \backslash G(A), \omega)$ vector space. Hence every algebraic cuspidal rep is a dense subspace of an analytic cuspidal rep. Let us now discuss the opposite inclusion.</p> <p><strong>Fact 1:</strong> Every analytic cuspidal rep $\pi$ of $G(A)$ factors into a tensor product $\otimes_v \pi_v$ over the places $v$ of $F$ of unitary reps of $G(F_v)$. We write $\pi_\infty \otimes \pi_f$. See Flath's article in the Corvallis Proceedings Theorem 4.</p> <p><strong>Fact 2:</strong> Two distinct unitary reps have distinct $tr \; \pi_v : C_c^\infty(G(F_v)) \rightarrow \mathbb{C}$ functional, and according to Fact 1, this holds for $tr \; \pi : C_c^\infty(G(A)) \rightarrow \mathbb{C}$ and $\pi$ cuspidal analytic as well. See this related question: <a href="http://mathoverflow.net/questions/88542/character-determines-the-representation" rel="nofollow">http://mathoverflow.net/questions/88542/character-determines-the-representation</a>. Hence the $C_c^\infty(G(A))$-module structure determines uniquely the analytic cuspidal representation.</p> <p><strong>Fact 3:</strong> Every unitary representation of a real reductive Lie group has a unique Harish-Chandra Module associated to it. So the $C_c^\infty(G(A_f)) \otimes C_c^\infty(G(A_\infty))$-module structure is encoded uniquely in the Harish-Chandra module $\times C_c^\infty(G(A_f))$-module structure. This can be found on the related Wikipedia page, but you seem to believe this already.</p> <p>So the missing point was fact 2, which allows an algebraic classification of unitary representations. This can be expressed in various languages, but I prefer the statement in terms of character distributions. </p> http://mathoverflow.net/questions/128678/motivating-geometric-representation-theory/128722#128722 Answer by Marc Palm for motivating geometric representation theory Marc Palm 2013-04-25T14:28:45Z 2013-05-04T13:49:24Z <p>I interpret GRT as explained here: <a href="http://ncatlab.org/nlab/show/geometric+representation+theory" rel="nofollow">http://ncatlab.org/nlab/show/geometric+representation+theory</a></p> <p>I give you two examples from number theory, in particular from the Langlands program and explain, how geometry might be useful.</p> <p>The main concern of the Langlands program are automorphic representations of a given reductive group $G$ over a global field $F$, how they can be transferred to different groups (<strong>functoriality</strong>) and how they are related to Galois representations/motives (<strong>correspondence</strong>). One of the key tool in this study is the Arthur trace formula: It relates </p> <p><strong>automorphic representations of $G/F$</strong> $\qquad\leftrightarrow\qquad$ <strong>conjugacy classes in $G(F)$.</strong></p> <p><strong>Functoriality</strong>: Automorphic forms per se are in general really hard to attack, specifically those related transcendental Maass cusps forms. Merely to show their existence, Selberg exploited the above comparison (Selberg trace formula). Also, if you want to proof that certain automorphic representation can be <em>mapped</em> to an other reductive group $G'$, you can try to <em>compare</em> the conjugacy classes of $G(F)$ and $G'(F)$. The Jacquet-Langlands correspondence is proven along these lines. Other famous conjectures, which are known to follow from such <em>maps</em> are the Selberg eigenvalue conjecture or more generally the Ramanujan-Petersson conjecture.</p> <p><strong>Correspondence</strong>: If you can realize Galois representations geometrically as operators on certain Homology classes, and compare the Lefschetz trace formula with the Arthur trace formula, you can prove an equality of the Artin L-function of the Galois representation with the L-function os some automorphic representation. A famous example is the Shimura-Taniyama conjecture and the proofs of the Langlands correspondence for global function fields by Drinfeld/Lafforgue. I don't know much abut Deligne's proof of the modularity of modular forms of weight $k\geq 2$ and generalizations by Harris-Taylor, etc., but I guess it is the same principle. These give you also special cases of the Ramanujan-Petersson conjectures, simply because the Artin L-function of a Galois representation necessarily satisfies it.</p> <p>At least morally, one should be interested in concrete models/geometric realizations of irreducible representations. Harish-Chandra has classified all discrete series of semi-simple connected real Lie groups via computing their traces way before Atiyah-Schmid found a geometric realization for all of them via K-theory. The known classification of admissible reps of reductive groups like $GL(n)$ over non-archimedean fields are geometric to my knowledge, i.e., given as induced reps. Also Harish-Chandra himself expressed the Plancherel formula of real reductive Lie groups in terms of geometric data like conjugacy classes, very similar to the ideas in the Arthur trace formula, yielding a tool for a similar local analysis as given in the Langlands program (local for local field). </p> http://mathoverflow.net/questions/129528/harish-chandra-modules-of-psl-2-mathbbr/129529#129529 Answer by Marc Palm for Harish-Chandra Modules of PSL_2($\mathbb{R})$) Marc Palm 2013-05-03T12:20:44Z 2013-05-03T12:38:37Z <p>Yes, the central character of the even $K$-types is trivial, and of the odd ones is the sign character.</p> <p>Also, this can easily be seen from the classification of irreducible representation on Hilbert spaces.</p> <p>More generally, the category of rep on Hilbert spaces of a locally compact group $G$ with trivial central character coincides with that of the rep theory on Hilbert spaces of the locally compact group $G/Z$, here $Z$ being the center of $G$. So although, the categories are not semisimple, one can always decompose with respect to the center. </p> <p>Quote from Wikipedia: <em>In 1973, Lepowsky showed that any irreducible (g,K)-module X is isomorphic to the Harish-Chandra module of an irreducible representation of G on a Hilbert space....</em>(http://en.wikipedia.org/wiki/Harish-Chandra_module)</p> http://mathoverflow.net/questions/129402/local-langlands-conjecture-for-gl2 Local Langlands conjecture for GL(2) Marc Palm 2013-05-02T10:35:38Z 2013-05-02T16:04:28Z <p>Let $F_v$ be a local field. Let $\sigma_v$ be a two-dimensional representation of $Gal(\overline{F}_v:F_v) \rtimes W_{F_v}$. Now, there exists an infinite-dimensional representation $\pi_v$ of $GL_2(F_v)$, whose local $L$-functions and root number coincides with that of $\sigma_v$ .</p> <blockquote> <p>How and to what extent can we read off the $\pi_v$ from $\sigma_v$? </p> </blockquote> <p>When is $\pi_v$ a supercuspidal rep, an unramified, or ramified principal series rep, a Steinberg representation, a discrete series representation? What properties of $\sigma_v$ are decisive?</p> http://mathoverflow.net/questions/129362/no-exceptional-eigenvalues-of-weight-1-2-maass-forms-on-gamma-04/129388#129388 Answer by Marc Palm for No Exceptional Eigenvalues of Weight 1/2 Maass Forms on $\Gamma_0(4)$? Marc Palm 2013-05-02T08:33:03Z 2013-05-02T12:44:58Z <p><em>Edit: Oh, I see only now that you are interested in half weight Maass forms (damn cellphone browsers;). My answer applies to weight zero Maass forms only. Probably the Shimura lift GH addresses generalizes, thus, translating half weight Maass forms to integer weight one of some different (lower?) level.</em></p> <p>For $\Gamma_1(n)$ and $n\leq 18$, the Selberg eigenvalue conjecture for <strong>weight zero/even</strong> Maass forms is due to Huxley (1985). All eigenvalues are $> 1/4$ here.</p> <p>Check for example page 12 in <em>Blomer, Brumley - The role of the Ramanujan conjecture in analytic number theory, Bulletin AMS 50 (2013), 267-320</em></p> <p>Booker and Strömbergsson verified the Selberg eigenvalue conjecture for weight zero Maass forms for $\Gamma_1(n)$ and $n \leq 857$ squarefree.</p> <p>For <strong>weight one/odd</strong> Maass forms, the generalization of the Selberg eigenvalue conjecture holds trivially, because the infinite component of the corresponding automorphic representation is a ramified principal series. These are all tempered. </p> <p>There exists an even Maass form of eigenvalue $1/4$ for $\Gamma(23)$, I was told, because the class group of $\mathbb{Q}( \sqrt{-23})$ is $\mathbb{Z}/3$.</p> <p>Using the Shimura lift (as GH) mentions, this yields similar results for <strong>half integer weight</strong> forms.</p> http://mathoverflow.net/questions/129376/gap-between-first-two-nonzero-laplacian-eigenvalues-on-closed-compact-surface/129411#129411 Answer by Marc Palm for Gap between first two nonzero Laplacian eigenvalues on closed compact surface? Marc Palm 2013-05-02T11:40:46Z 2013-05-02T11:49:31Z <p>I think that in general $vol(X)$ is the best possible upper bound for the multiplicity of the first non-trivial eigenvalue. This can be proved rigorously for compact Riemann surfaces with the Selberg trace formula, but I would guess this holds more general via an analysis related to Weyl laws.</p> <p>For certain arithmetic compact Riemannian surfaces (associated to division algebras), the first non-trivial eigenvalue is assumed to be larger or equal $1/4$. Some better upper bounds in this particular case are due to Michel and Venkatesh (the Jacquet-Langlands correspondence has to be applied). <a href="http://math.stanford.edu/~akshay/research/MV.pdf" rel="nofollow">http://math.stanford.edu/~akshay/research/MV.pdf</a></p> <p>The Langlands correspondence also suggests that the multiplicity can be arbitrary large.</p> http://mathoverflow.net/questions/129403/conceptual-reason-behind-shimura-lifts Conceptual reason behind Shimura lifts Marc Palm 2013-05-02T10:59:33Z 2013-05-02T10:59:33Z <p>Shimura lifts are correspondence between integer weight and half-integral weight automorphic forms. Half integral weight things are associated to representation of a double cover of $G =SL_2(\mathbb{R})$, where as integer things live already in a rep of $G$.</p> <blockquote> <p>Q: Is there a conceptual (representation-theoretic) explanation for this phenomenon? Is there a local (in terms of places) analogue of this phenomenon? </p> </blockquote> http://mathoverflow.net/questions/129196/orbital-integrals-of-pseudo-coefficients-of-supercuspidal-reps Orbital integrals of pseudo coefficients of supercuspidal reps Marc Palm 2013-04-30T09:20:46Z 2013-04-30T11:20:26Z <p>Let $\pi$ be a supercuspidal representation of $G =GL_2(F)$ for a non-archimedean local field $F$, then there exists a maximal subgroup $K$ of $G$, which is compact modulo the center, and a representation $\rho$ of $K$ such that $\pi = Ind_K^G \rho$. </p> <p>It is possibly to show that $tr\; \sigma( \phi) \neq 0$ iff $\sigma \cong \pi$ for $\phi$ being equal to $tr(\rho)$ on $K$ and zero off $K$. This means $\phi$ is a constant multiple of a pseudo-matrix coeffient of $\pi$. </p> <p>Now, one can compute that given an elliptic element $\gamma \in GL_2(F)$, i.e., the characteristic polynomial is irreducible, the corresponding elliptic orbital integral vanishes iff the conjugacy class of $\gamma$ doesn't meet $K$ and equals a constant multiple of $tr \rho(\gamma)$ with $\gamma$ conjugated inside $K$ otherwise.</p> <p>There exists a classification/construction of those $\rho$'s respective $\pi$'s, see eg. <em>Bushnell-Henniart --- Local Langlands conjecture for GL(2).</em></p> <blockquote> <p>Question: Does there exists a reference for the explicit value of $tr \rho(\gamma)$ depending on the strata of $\rho$ and the characteristic polynomial of $\gamma$?</p> </blockquote> <p>Remark: The depth-zero case is well documented in the representation theory of $GL_2(o/p)$.</p> http://mathoverflow.net/questions/129081/steinberg-reps-of-reductive-groups-over-local-fields-vs-finite-fields Steinberg reps of reductive groups over local fields vs finite fields Marc Palm 2013-04-29T10:48:19Z 2013-04-30T11:09:48Z <p>Let $G$ be a reductive group over a non-archimedean field $F$ with reisdue field $f$. </p> <p><em>Edit: The statements only make sense modulo tensoring by one-dimensional representations.</em></p> <blockquote> <p>Are the unitary, square-integrable representation (modulo tensoring by one-dimensional reps) of $G(F)$, which are not supercuspidal, in one-to-one correspondence with a certain subclass of representations of $G(f)$ (modulo tensoring by one-dimensional reps)?</p> </blockquote> <p>I am mostly interested in the case of $G=GL(n)$. The question has an easy answer if $n=2$.</p> <p><strong>For the case GL(2):</strong> The sq.int, non-supercuspidal reps are isomorphic to the Steinberg tensored by a one-dimensional representation $G(F)$. There is precisely one irreducible rep of $G(f)= G(o/p)$ contained in the Steinberg rep of $G(F)$, i.e., it is the Steinberg of $G(f)$.</p> <blockquote> <p>More concretely, do they also in general admit a $\Gamma(p) = \{ \gamma \in G(o) : \gamma = 1 \bmod p \}$-invariant vector modulo tensoring by one-d. reps? Is their restriction to the Iwahori(= pullback of $B(f)$ to $G(o)$ for a fixed Borel subgroup) or its "Levi component" a one-dimensional representation?</p> </blockquote> http://mathoverflow.net/questions/1972/langlands-dual-groups/128500#128500 Answer by Marc Palm for Langlands Dual Groups Marc Palm 2013-04-23T15:21:41Z 2013-04-23T15:21:41Z <p>I was googling for the root datum. This site and a very nice, conscise treatment of Calder Daenzer turned up: <a href="http://www.math.psu.edu/daenzer/Root_Data_for_Reductive_Groups.pdf" rel="nofollow">http://www.math.psu.edu/daenzer/Root_Data_for_Reductive_Groups.pdf</a> His notes deal with complex, reductive groups only (as requested by the OP). In the end, he states the definition of the Langlands dual group.</p> <p>Perhaps, you will also like Casselman: <a href="http://www.math.ubc.ca/~gor/Notes.pdf" rel="nofollow">http://www.math.ubc.ca/~gor/Notes.pdf</a></p> <p><em>I am aware that this question is pretty old!</em> I add this as answer because I was not quite satisfied with the remaining suggestions in relation of brevity/precision. E.g. Springer's book requires a massive book keeping of notation. I am aware that Springer's/the SGA proof works over any basefield, and is thus more general.</p> http://mathoverflow.net/questions/128110/categorified-versions-of-mackeys-functor/128111#128111 Answer by Marc Palm for Categorified versions of Mackey's functor Marc Palm 2013-04-19T17:14:25Z 2013-04-19T17:14:25Z <p>Have a look at page 4 of <a href="http://www.math.umn.edu/~webb/Publications/GuideToMF.ps" rel="nofollow">http://www.math.umn.edu/~webb/Publications/GuideToMF.ps</a> for a few examples of Mackey functors in different categories.</p> http://mathoverflow.net/questions/128005/what-are-the-invariant-pseudo-differential-operators-on-a-lie-group/128082#128082 Answer by Marc Palm for What are the invariant Pseudo-differential operators on a Lie group? Marc Palm 2013-04-19T13:05:52Z 2013-04-19T13:13:43Z <p>Rather then only considering invariant PSD operators, you might want to consider all $G$ invariant operators, i.e., $G$-intertwiner. I describe their functional calculus below. Their functional calculus can be realized/studied via convolution products and representation theory. What I describe is in the realm of the first answer by Pedro Lauridsen Ribeiro. I let you decide whether this classifies as "algebraic", but it is certainly of operator-algebraic/ representation-theoretic flavour. I claim everything else which is $G$-invariant operator will have a equivalent functional calculus.</p> <p>Here is an example. Assume $H$ is compact. We identify $L^2(G/H)$ with the induced representation $\pi = Ind_{H}^{G} 1$ or the $H$-invariant vectors in $L^2(G)$ and then uses the convolution operators for $\phi \in C_c^\infty(G//H)$: $$T_\phi f(g) = \int\limits_{G} \phi(x) f(xg) d x.$$ E.g. for $G =SL_2(\mathbb{R})$ and $H=SO(2)$, the algebra $C_c^\infty(G//H)$ is commutative by the Gelfand trick (this is not so essential) and the trace $T_\phi$ is an integral of the Harish-Chandra/Selberg transform of $\phi$ over the spectrum of the hyperbolic Laplacian or, alternatively, an integral $$\int\limits tr\; \pi(\phi) d_{Pl} \pi$$ over the irreducible unitary (tempered) reps $\pi$ of $G$ with $H$-invariant vectors (only principal series representations here). The measure $d_{Pl}$ is the Plancherel measure.</p> <p>If $H$ is not compact, you can still do something similar working with $C_c^\infty(G)$ and obtain a similar analysis. E.g. take the two important situations when $H$ is a lattice or a parabolic subgroup in a reductive Lie group. The advantage: this generalizes to locally compact groups. E.g. on reductive groups over non-archimedean fields, there are no differential operators in any obvious way, but this gives you the Hecke operators. These ideas are crucial also in the context of the Selberg trace formula.</p> http://mathoverflow.net/questions/127977/a-generalisation-of-the-birch-and-swinnerton-dyer-conjecture/127981#127981 Answer by Marc Palm for A generalisation of the Birch and Swinnerton-Dyer conjecture Marc Palm 2013-04-18T15:57:41Z 2013-04-18T15:57:41Z <p>The equivariant Tamagawa number conjecture generalizes the BSD conjecture. I am not sure if this is the most general conjecture available. </p> http://mathoverflow.net/questions/127906/gelfand-representation-and-functional-calculus-applications-beyond-functional-ana/127915#127915 Answer by Marc Palm for Gelfand representation and functional calculus applications beyond Functional Analysis Marc Palm 2013-04-18T04:30:23Z 2013-04-18T06:38:40Z <p>Fourier analysis, i.e., representation theory of abelian groups, has a nice interpretation in terms of the Gelfand transform. The functional calculus is given as convolution operators on a LCA group get sent to multiplication operators on the Pontryagin dual.</p> <p>Also the representation theory of other groups, such as Lie groups or p-adic groups has profited from Gelfand theory. At the very minimum, it has been a good guiding principle. People rather work with smooth functions in this context. The focus has been here on more explicit description, quantitative analysis and interpretation of C star methods (topology) and von Neumann algebras (measure theory), which are identical in some sense for the type 1 groups. So to say, the functional analysis provides the existence of the measure. Plancherel measure on a type 1 group comes e.g. von a decomposition of right regular rep into irreducibles (vNa decomposition into factors), similarly the spectral side of the Selberg's and Arthur's trace formula. These trace formulas can also be regarded as results in differential geometry/number theory.</p> <p>A direct application of the GNS-construction/ a concrete example is e.g. the Gelfand-Raikov theorem. Quote from Terry Tao's blog:</p> <blockquote> <p>"Nevertheless, in the important case of locally compact groups, it is still the case that there are “enough” irreducible unitary representations to recover a significant portion of the above theory. The fundamental theorem here is the Gelfand-Raikov theorem, which asserts that given any non-trivial group element $g$ in a locally compact group, there exists a irreducible unitary representation (possibly infinite-dimensional) on which $g$ acts non-trivially. Very roughly speaking, this theorem is first proven by observing that $g$ acts non-trivially on the regular representation, which (by the Gelfand-Naimark-Segal (GNS) construction) gives a state on the *-algebra of measures on $G$ that distinguishes the Dirac mass $\delta_g$ at from the Dirac mass $\delta_0$ from the origin. Applying the Krein-Milman theorem, one then finds an extreme state with this property; applying the GNS construction, one then obtains the desired irreducible representation."</p> </blockquote> <p><a href="http://terrytao.wordpress.com/2011/01/23/the-peter-weyl-theorem-and-non-abelian-fourier-analysis-on-compact-groups/" rel="nofollow">http://terrytao.wordpress.com/2011/01/23/the-peter-weyl-theorem-and-non-abelian-fourier-analysis-on-compact-groups/</a></p> http://mathoverflow.net/questions/127529/on-the-absolute-convergence-of-the-local-zeta-integral/127544#127544 Answer by Marc Palm for On the absolute convergence of the local-zeta integral. Marc Palm 2013-04-14T16:25:52Z 2013-04-15T08:22:51Z <p>No to the stronger statement. Here is the first simple example. If you take the zeta integral of a function f and a non trivial Dirichlet character at s, then take absolute values inside the integral you obtain the zeta integral of |f| with s`=Re s. The former has no pole at zero but the later. So the order of divergence might change. Tates thesis explains this pretty well.</p> <p>Also, no for the second statement. Take $G$ being a couple of copies of the multiplicative group and argue as above. In this manner, you obtain something with arbitrary high multiplicity at zero, although the limit of the original integrals is convergent. </p> http://mathoverflow.net/questions/127338/representations-over-mathbbq-p/127342#127342 Answer by Marc Palm for Representations over $\mathbb{Q}_p$ Marc Palm 2013-04-12T10:50:44Z 2013-04-12T13:52:18Z <p>For finite groups and a field $F$ of characteristic zero, you can identify the $G$-endomorphisms of two induced representations $$Hom_G( I_H^G \pi, I_K^G \sigma)$$ with the space of functions $f: H \backslash G / K \rightarrow Hom_F(V_\pi, V_\sigma*)$ with $f(hgk)= \pi(h) f(g) \sigma(k)$. If $H=G$ and $\pi = \sigma$, i.e., the case you are interested in, everything becomes a convolution algebra. If you find a basis, they give you projections to irreducible components, and you can decompose explicitly.</p> <p>Equivalently, you could use Frobenius reciprocity $$Hom_H( \pi, R_H I_K^G \sigma)$$ with Mackey induction restriction formula giving you a decomposition $$R_H I_K^G \sigma$$ in terms of $H\backslash G/K$.</p> <p>All this is very combinatorial. There is no general algorithm for general $G$ even in the case of complex representation, e.g., nobody knows how to decompose the parabolic induction in $GL_3(\mathbb{Z} / p^N)$. So, one might say decomposing induced representations into irreducibles is an art rather then a theory. </p> <p>The situation is slightly different in the special case, where $H$ is <strong>normal</strong>. Then induction exhaust all irreducible representations and it is easy to parametrize them. However, also here you encounter the same difficulties only later:( To understand this special case, I suggest you should learn Clifford theory.</p> <p><em>Edit in response to a comment:</em> Let $N$ be a normal subgroup. $G$ acts on the representation $\pi$ of $N$ via $g: \pi \mapsto \pi^g(x) = \pi(gxg^{-1})$. </p> <blockquote> <ol> <li><p>$Ind_N^G \pi$ decomposes with single multiplicity iff $End_G( I_N^G \pi)$ is abelian. </p></li> <li><p>$Ind_N^G \pi$ is irreducible, if the cardinality of the $G$ orbit $\pi$ has the same cardinality as $G/N$. </p></li> <li><p>More generally, the cardinality of representations contained in $I_N^G \pi$ is given by $C(\pi) /N$ for the centralizers $C(\pi) = \{ g \in G : \pi^g =\pi\}$. </p></li> </ol> </blockquote> <p>1 is obvious by Schur's lemma: Assume $I_N^G \pi = \bigoplus \sigma^{\oplus m_\sigma}$ then as algebra isomorphisms $$End_G( I_N^G \pi) = \bigoplus End_G( \sigma^{\oplus m_\sigma}) = \bigoplus M(m(\sigma), F).$$</p> <p>Proof of 2+3 via Mackey formula, Schur's lemma and Frobenius reciprocity: As vectorspace isomorphisms $$Endo_G(I_N^G \pi) = Hom_N(\pi , Res\, I_N^G \pi) = \bigoplus_{g \in G/N} Hom_N(\pi, \pi^g).$$</p> http://mathoverflow.net/questions/127157/status-of-global-langlands-conjecture-for-gl-2-over-mathbbq/127251#127251 Answer by Marc Palm for Status of (Global) Langlands Conjecture for $GL_2$ over $\mathbb{Q}$ Marc Palm 2013-04-11T14:26:49Z 2013-04-11T14:32:13Z <p>Additionally to what Joel was saying, Langlands conjectured the existence of a universal group $\widehat{G}$ (depending on the number field only) whose 2-dim'l representation correspond to automorphic representation of GL(2) in a suitable way. These would include automorphic forms with the archimedean factors not being <em>algebraic</em>, e.g., even Maass forms with Laplace eigenvalue $\neq 1/4$. From what I understand, Arthur suggests a definition of $\widehat{G}$ here: <a href="http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf" rel="nofollow">http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf</a>. The situation is similar to case (1c) described above.</p> http://mathoverflow.net/questions/126277/is-eisenstein-series-not-identically-zero/126279#126279 Answer by Marc Palm for Is Eisenstein series not identically zero Marc Palm 2013-04-02T15:25:56Z 2013-04-03T08:46:28Z <p>The space $P(k) \backslash G(k)$ has a discrete set of representatives in $U(A) M(k) \backslash G(A)$. See for example the discussion after Lemma 3.3 in Gelbart-Jacquet "Forms of GL(2) from the analytic view point" for $G=GL(2)$ . I am sure somewhere in Moeglin-Waldspurger a similar lemma is quoted/proved somewhere for the more general $G$. Arthur has something similar certainly for $G$ reductive, but I remember that M-W consider also more generally finite covers etc. The main concern of these lemmas is actually the absolute convergence, but also they also provide the non-triviality.</p> <p>A suitable set of representatives can be given via the Bruhat decomposition. Pick a compact set $K \subset U(A) M(k) \backslash G(A)$ containing only one representative and having non-empty interior. Let $\varphi :U(A) M(k) \backslash G(A) \rightarrow \mathbb{C}$ be a function which is positive, smooth, compactly supported on this set $K$, non-vanishing on the interior. Then your sum will not be zero only for $\gamma^{-1} g$ for $g \in K$. This works equally well if you work modulo the center.</p> http://mathoverflow.net/questions/126243/the-hasse-weil-l-function-and-some-equations/126247#126247 Answer by Marc Palm for The Hasse-Weil L-function and some equations Marc Palm 2013-04-02T09:12:12Z 2013-04-02T09:17:13Z <p>No, that is impossible. The $k$-th derivative of a L function has necessarily infinitely many zeros. So you can choose $s_j$ and $t_j$ inductively such that the products give distinct zeros of $f^j$. Moreover, if one of $s_j$ is zero you can't say anything clever either, but I assume that you simply have forgotten that condition in your question.</p> http://mathoverflow.net/questions/125925/a-question-on-twisted-l-function/125927#125927 Answer by Marc Palm for A question on twisted L-function Marc Palm 2013-03-29T16:55:22Z 2013-03-29T17:03:27Z <p>The answer is no. Consider G=Gl(n). We have $L(s,\pi)=L(s-s',\pi\otimes|det|^{s'})$. The absolute value is the adelic norm. In general,eg in Tate's thesis, the parameter s is sometimes avoided for this reason.</p> <p>So your conjecture violates GRH. For n=1 and $\pi$ trivial over the rational numbers gives you a concrete counter example, because there exist zeros on the critical line.</p> http://mathoverflow.net/questions/125647/riemann-z-function-bounds-on-number-of-non-trivial-zeros-along-horizontal-lines/125651#125651 Answer by Marc Palm for Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones. Marc Palm 2013-03-26T19:01:26Z 2013-03-26T19:18:51Z <p>Best bound is O(log T) also for the multiplicity of zeros. Under RH, slightly better O(Log T/log log T). Under Lindeloeff, o(log T). This is pretty bad, because the conjecture is one.</p> <p>Edit: There are slightly better bounds on the multiplicity of zeros see Ivic: arxiv.org/pdf/math/0501434</p> <p>The situation is similar to that for the Selberg Zeta function. Best bound here O( T/ log T). Here, the conjecture is O(1), one for the modular group.</p> http://mathoverflow.net/questions/125080/spectral-synthesis-for-central-functions-on-locally-compact-groups/125143#125143 Answer by Marc Palm for Spectral synthesis for central functions on locally compact groups Marc Palm 2013-03-21T10:11:48Z 2013-03-21T11:30:38Z <p><em>This a long comment, which indicates the difficulties and gives a decomposition of measures in terms of orbital integrals instead of irreducible reps.</em></p> <p>As you have noticed yourself, there do not exists many continuous functions, which are invariant under conjugation. This was my comment with the closure of conjugacy classes.</p> <p>What kind of object is $tr\; \pi$ (assuming it exists)?</p> <p>I give several suggestions for $G$ being the $F$-points of a reductive group ($F$ local field). These are type I, seperable, unimodular.</p> <ol> <li>The most common definition yields that it is a distribution on $C_c^\infty(G)$ satisfying $$tr\; \pi( \phi \ast \theta) = tr\; \pi( \theta \ast \phi)$$</li> <li>Equivalently, it is a distribution on $C_c^\infty(G)$ satisfying $$tr\; \pi( \phi^g) = tr\; \pi(\phi), \qquad \phi^g(x)= \phi(g^{-1}xg)$$ </li> <li>There eists a locally integrable function $\theta_\pi$ on $G$ with $$tr \pi(\phi) = \int_G \theta_\pi(g) \phi(g)\; dg.$$ Here, $\theta_\pi$ is necessarily conjugation invariant. Some people refer to the trace meaning the function $\theta_\pi$, which is like identifying a distribution and a generalized function.</li> </ol> <blockquote> <p>Suggested conjecture: Every locally integrable central function is a direct integral of $\theta_\pi$'s?</p> </blockquote> <p>Moreover, $tr\; \pi$ are extremal algebra states iff $\pi$ is irreducible. That means they can not be written in terms of linear combination of other things. On the other hand, there are orbital integrals, which have the somehow the same properties. Note that there are variants of Plancherel theorems in terms of orbital integrals by Harish-Chandra.</p> <p>A integral decomposition of measures (not distributions though, but functionals on $C_c(G)$) into extremal(=ergodic) measures is known as Chocquet theory, see e.g. <a href="http://mathoverflow.net/questions/73550/ergodic-decomposition-of-quasi-invariant-measure" rel="nofollow">http://mathoverflow.net/questions/73550/ergodic-decomposition-of-quasi-invariant-measure</a> respective Theo Buehler's suggestion: <a href="http://matwbn.icm.edu.pl/ksiazki/cm/cm84/cm84217.pdf" rel="nofollow">http://matwbn.icm.edu.pl/ksiazki/cm/cm84/cm84217.pdf</a></p> <p>Apply this to $G$ acting itsself by conjugation and you have a decomposition of the Haar measure (take the Haar measure multiplied by some non-vanishing function since the article assumes probability measure)</p> <p>This together with theorem 1 (5) in James Glimm's article <a href="http://www.ams.org/journals/tran/1961-101-01/S0002-9947-1961-0136681-X/S0002-9947-1961-0136681-X.pdf" rel="nofollow">http://www.ams.org/journals/tran/1961-101-01/S0002-9947-1961-0136681-X/S0002-9947-1961-0136681-X.pdf</a> applied to $G$ acting on itsself by conjugation yields that the measures are supported on a single orbit under additional hypothesis.</p> <blockquote> <p>Theorem: Let $G$ be a 2nd countable, locally compact group of with relatively open conjugacy classes, then for every conjugation invariant function $f$ on $G$ the measure $$f(g) \; dg$$ decomposes into a direct integral of measures, which are each supported only on one conjugacy class.</p> </blockquote> <p>This theorem applies to reductive groups over local field. I am not sure how to implement this to get something with irreducible rep instead of "orbital integrals" though. The unitary dual of a type I group is $T_0$ as requested by Glimm's theorem, but how to move on? Moving from conjugacy classes to irreducible reps can only be done via dualities, as you might know from the Arthur trace formula. </p> http://mathoverflow.net/questions/124754/what-is-the-difference-between-an-automorphic-form-and-a-modular-form/124784#124784 Answer by Marc Palm for What is the difference between an automorphic form and a modular form? Marc Palm 2013-03-17T13:40:32Z 2013-03-17T13:40:32Z <p>The most common definition of an automorphic form is that it a K-finite Z-finite vector in an automorphic representation. Often one requires the automorphic rep to be irreducible for being able to get a reasonable L function from it. For the purpose of the Langlands stuff, this is the only reasonable definition, also for Taniyama-Shimura conjecture etc. These are the only functions being analyzable by trace formulas etc. </p> <p>When we consider Hecke Maass or modular cusp forms , this is a specialization. For this, you can look up strong approximation, e.g. in Bumps book.</p> <p>Depending on their focus, some authors consider being automorphic not related to a congruence setting, but this is rare. (Note also that division algebras give rise to uniform lattices. At least, these forms must be called automorphic. ) Also one assumes reasonable growth conditions,i.e., no poles.</p> http://mathoverflow.net/questions/124529/volume-of-pgl2-f-pgl2-a Volume of PGL(2,F) \ PGL(2, A) Marc Palm 2013-03-14T15:17:28Z 2013-03-14T18:53:20Z <p>Let $F$ be a global field. What is the measure of $PGL_2(F) \backslash PGL_2(\mathbb{A})$?</p> <p>This depends of course on the normalizations of the Haar measures on $PGL_2(F)$ and $PGL_2(\mathbb{A})$. Probably its most likely available in the literature for $PGL_2(F)$ admits the discrete measure and $PGL_2(\mathbb{A})$ the Tamagawa measure, but I couldn't find!?</p> <p>I remember that there was a question about the measure of $SL_n(\mathbb{Z}) \backslash SL_n(\mathbb{R})$ here in the past, but couldn't find it.</p> <p>It should be related to special values of the Dedekind zeta function.</p> http://mathoverflow.net/questions/131083/continuous-class-functions-separating-conjugacy-classes-of-compact-groups Comment by Marc Palm Marc Palm 2013-05-19T11:07:43Z 2013-05-19T11:07:43Z But the conjugacy classes in compact groups are closed. http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Comment by Marc Palm Marc Palm 2013-05-17T14:49:04Z 2013-05-17T14:49:04Z Ah okay, first countability is necessary and sufficient for having a metric in a locally compact group. So correction: first countable implies second countable for lc groups if seperable:( http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Comment by Marc Palm Marc Palm 2013-05-17T13:14:30Z 2013-05-17T13:14:30Z ...for non-type 1 groups, then one could argue something. So far I am only saying you get a vNa decomposition into factors (not unique), and of course a state decomposition into extremal states by Chocquet's theorem(not unique though). http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Comment by Marc Palm Marc Palm 2013-05-17T13:12:19Z 2013-05-17T13:12:19Z Any second-countable space is separable: if is a countable base, choosing any from the non-empty gives a countable dense subset. Conversely, a metrizable space is separable if and only if it is second countable, which is the case if and only if it is Lindel&#246;f. Note that locally compact, second countable groups are always seperable. But that was not the point: I am pointing out that the Hilbert space $L^2(G)$ is seperable iff $G$ is second countable. I assume implicitly (wlog) that locally compact implies Hausdorff by definition. If you make a precise statement what you mean by a Plancherel http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Comment by Marc Palm Marc Palm 2013-05-17T10:16:59Z 2013-05-17T10:16:59Z Ignore no. I don't remember why my impulse was to say no. Seperable and second countable are often the same thing in special situation. http://mathoverflow.net/questions/130887/plancherel-formula-for-non-second-countable-non-unimodular-groups/130923#130923 Comment by Marc Palm Marc Palm 2013-05-17T09:47:54Z 2013-05-17T09:47:54Z No, I am saying the Hilbert space $L^2(G)$ is seperable in the sense that they have a countable orthonormal basis iff $G$ is second countable. Now, how do you define a state on say $C_c^\infty(G)$ or $C_c(G)$ from a representation $\pi$ if $\pi(\phi)$ is not Hilbert Schmidt, trace class or something analogous. What is the suggested analogon you have in mind? Sure, the integral decomposition exists, but is not unique and the unitary dual is not a nice space anymore. For type 1 e.g. it will be almost Hausdorff. http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc/130604#130604 Comment by Marc Palm Marc Palm 2013-05-14T16:18:46Z 2013-05-14T16:18:46Z I also would claim that the subquotients are never spherical, but I am not sure in the generality I have stated the results. http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc/130604#130604 Comment by Marc Palm Marc Palm 2013-05-14T16:17:26Z 2013-05-14T16:17:26Z Note, that I don't know which one of the parabolically induced ones have irreducible subquotients or are unitarizabile, though. I am only saying a classification of the former gives pretty easily a classification of the latter. http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc/130587#130587 Comment by Marc Palm Marc Palm 2013-05-14T15:01:33Z 2013-05-14T15:01:33Z No. What you state is simply the fact when $(G,K)$ is a Gelfand pair. The OP is search for a set of unitary representation, e.g., for $GL_2(\mathbb{C})$, it would be all unitary unramified continuous series representation and the $| \det |^s$ with $\Re s =1$. http://mathoverflow.net/questions/130522/finding-spherical-representations-of-gln-mathbbc Comment by Marc Palm Marc Palm 2013-05-14T06:50:43Z 2013-05-14T06:50:43Z I understand the OP is interested in a classification of the unitary (or smooth, admissible) representation, which are irreducible and have a invariant vector under the maximal compact subgroup, or equivalently the trivial representation is contained in the restriction to it. I think in his context, he wants to consider either the $\mathbb{R}$- or $\mathbb{C}$-points of these classical algebraic group, $U(n)$ making no sense over $\mathbb{R}$, though, and having a trivial answer over $\mathbb{C}$. Similarly, for $SO(n)$ over $\mathbb{R}$. http://mathoverflow.net/questions/21555/why-are-modular-forms-usually-defined-only-for-congruence-subgroups/21611#21611 Comment by Marc Palm Marc Palm 2013-05-13T08:35:20Z 2013-05-13T08:35:20Z Actually, it is a theorem that every unitary or even smoth, admissible repreentation of $GL_2(F_v)$ has a $\Gamma_0(p_v^N)$-invariant vector for $N$-sufficiently large. All modular forms turn up earlier or later for some $\Gamma_0(K)$ for $K$ large. Whether it makes always sense computational to go to large $K$, is a different one. http://mathoverflow.net/questions/129402/local-langlands-conjecture-for-gl2 Comment by Marc Palm Marc Palm 2013-05-13T08:14:00Z 2013-05-13T08:14:00Z Dear Matthew, I am totally satisfied with your explanation. Would you like to copy and paste it as an answer? Thank you=) http://mathoverflow.net/questions/130241/how-do-these-two-haar-measures-on-sl2-r-compare Comment by Marc Palm Marc Palm 2013-05-10T11:45:54Z 2013-05-10T11:45:54Z The modular function is $\Delta$ is constant one for $SL_2(\mathbb{R})$ and any other reductive group over a local field. http://mathoverflow.net/questions/130062/concrete-examples-of-noncongruence-arithmetic-subgroups-of-sl2-r Comment by Marc Palm Marc Palm 2013-05-10T09:08:11Z 2013-05-10T09:08:11Z Thanks Matthew for both your comments:-) http://mathoverflow.net/questions/130089/orders-in-number-fields Comment by Marc Palm Marc Palm 2013-05-08T17:05:15Z 2013-05-08T17:05:15Z Who voted to close? And why?