User br - MathOverflow

A slightly odd (integral of Whittaker functions / sum of characters of $GL_n(\mathbb C)$ / sum of Schur functions)

2013-01-10T23:21:21Z

Let $W$ be the normalized spherical Whittaker function attached to a spherical representation $\pi$ on $GL_n(k)$, where $k$ is a $p$-adic field and $n\ge 3$.

I'm faced with the slightly-odd integral $$\int_{k^\times} \left|W\left(\matrix{y&&&\cr &\ddots&&\cr &&y&\cr &&&1}\right)\right|^2 |y|^s\ dy$$ where ${\rm Re}(s)$ is sufficiently large to get convergence.

Does anyone have any ideas about calculating this?

Recall the Casselman-Shalika-Shintani formula, $$W(\varpi^J)=\delta_B^{1/2}(\varpi^J){\rm Tr}\left(\rho_J\left(A_\pi\right)\right)$$ where $J=(j_1,\ldots,j_n)\in\mathbb Z^n$ satisfies $j_1\ge j_2\ge\ldots\ge j_n$, $\varpi^J$ is the diagonal matrix with $j$-th entry $\varpi^{j_i}$, $\delta_B$ is the modular character of the Borel subgroup, $\rho_J$ is the representation of $GL_n(\mathbb C)$ with highest weight $(j_1,\ldots,j_n)$, and $A_\pi$ is the matrix of Satake parameters of $\pi$. Using this, the integral can be written as $$\sum_{i\ge 0} q^{-i(s+n-1)}{\rm Tr}\left(\rho_{(i,\ldots,i,0)}(A_\pi)\otimes\rho_{(i,\ldots,i,0)}(\bar A_\pi)\right)$$ where the shift by $n-1$ comes from $\delta_B(\varpi^{(i,\ldots,i,0)})=\prod_{j=0}^{n-2}|\varpi^i|^{n-1-2j}$.

This is basically a sub-sum of the formula used to get the Rankin-Selberg $L$-function for $\pi\otimes\tilde\pi$, so it should have a reasonable answer.

What is the support of the Whittaker function of a new vector on GL(2)?

2012-12-09T23:40:22Z

Let $W$ be the normalized Whittaker function associated to a new vector in an irreducible generic representation $\pi$ of $G=GL_2(k)$, where $k$ is a $p$-adic field. Let $c$ be the conductor of $\pi$, meaning that $c$ is the smallest integer with $W(gk)=W(g)$ for all $k\in K_1(c)$ where $K_1(c)$ is the subgroup of $K=GL_2({\mathfrak o})$ with bottom row congruent to $(0,1)$ modulo $\mathfrak p^c$ (with the convention that $K_1(0)=K$).

What is the support of $W$ restricted to $K$?

Clearly, the support contains $K_0(c)$, the subgroup of upper triangular matrices modulo $\mathfrak p^c$, since $W$ won't vanish on the center of $G$. I imagine it will depend on further information about $\pi$ beyond the conductor, but I'm having trouble finding (or proving) much of anything definitive.

Answer by BR for Reference on Casselman-Shalika formula for GL(n) and PGL(n)?

2012-11-26T22:17:36Z

In this paper, Shintani proves the Casselman-Shalika(-Shintani) formula for GL(n). This preceded Casselman-Shalika's paper by a few years. Several of Cogdell's expository articles on L-functions have easy-to-read descriptions of the formula.

Answer by BR for Where do the real analytic Eisenstein series live?

2011-10-24T23:07:01Z

This is kind of a complicated question, since there isn't really a single good answer.

We begin with a simple Lie group $G$ (for simplicity!). On the one hand, we hopefully have a description of the unitary representations of $G$. On the other hand, we may want to understand how spaces such as $L^2(H\backslash G)$, where $H$ may be trivial or discrete or maximal compact or etc), decompose into unitary representations of $G$ (that it will decompose is known on general (highly nontrivial) principles). At least two issues arise.

First, what does it mean for a representation to "appear" in the decomposition of $L^2(H\backslash G)$? We'd like it to mean that there exists an $f\in L^2(H\backslash G)$ such that $f$ generates the representation. This can't possibly work in general, and it already fails for $L^2(\mathbb R)$. Basically, whenever $H\backslash G$ is not compact, there will be a "continuous" part to the decomposition made up of unitary representations that can't be found as subrepresentations of $L^2(H\backslash G)$. Personally, a priori, it is surprising to me that you can integrate a bunch of stuff not in $L^2$ and wind up with something in $L^2$. But then, I think about Fourier inversion and Paley-Wiener theorems, and it's not so surprising. (In fact, if you believe your future will contain a nontrivial amount of harmonic analysis, you should try to become well-acquainted with Fourier theory.)
Now, there are functions on $H\backslash G$ that generate these representations and they usually aren't very far from being in $L^2(H\backslash G)$ (like $e^{ix}$ on $\mathbb R$ and Eisenstein series on $\Gamma\backslash\mathfrak H$), but there really is no way to force them in there. A person might wonder why a benevolent God would allow this to happen, but that is outside of my expertise.

Second, which representations will appear in $L^2(H\backslash G)$? For example, the trivial representation appears in $L^2(H\backslash G)$ if and only if $H\backslash G$ has finite volume. And complementary series representations don't seem to appear at all (usually)! (This is Selberg's Conjecture.)

On a hopefully more helpful note, with certain definitions of a Schwartz space on $H\backslash G$, you can realize these functions as tempered distributions (meaning continuous linear functionals on the Schwartz space). In fact, the space of functions with uniform moderate growth on $\Gamma\backslash \mathfrak H$ contains Eisenstein series and is contained in the dual of the Schwartz space for $\Gamma\backslash\mathfrak H$. See some of Casselman's work, here and here. In a different direction, there is Schmid and Miller's work on automorphic distributions, e.g. here.

Strata of K-types appearing in irreducible representations of p-adic GL(2)

2012-10-22T05:44:27Z

I'm trying to use the language of strata to organize $K$-types of irreducible smooth representations of $GL(2)$ (and then hopefully prove things). Unfortunately, I'm still new to it, so I might be making some mistakes.

My main reference is the Bushnell-Henniart book. For anyone who doesn't have access to it and wants to "play along at home", this essay contains a distillation of the important ideas.

Let $k$ be a $p$-adic field with ring of integers $\mathfrak o$, $G=GL_2(k)$, $K=GL_2(\mathfrak o)$. The principal congruence subgroup of level $\varpi^N$ (or level $N$, by abuse of notation) is $K_N=1+\varpi^N M_2(\mathfrak o)$ (and $K_0=K$). Let $\psi$ be an additive character on $k$ with conductor $\mathfrak o$ and extend it to $M_2(k)$ by composing with the trace: $\psi_M(x):=\psi({\rm tr}\ x)$. To further abuse notation, we'll suppress the subscript $M$ in $\psi_M$.

We'll say that the level of an irreducible (smooth) representation $\sigma$ of $K$ is the largest $N$ such that $\sigma$ is nontrivial on $K_N$ and trivial on $K_{N+1}$. Assume $\sigma$ has level $N\ge 1$. Since $K_N/K_{N+1}$ is a finite abelian group, $\sigma|_{K_N}$ decomposes into a direct sum of characters. Using the isomorphism $K_N/K_{N+1}\simeq M_2(\mathfrak o/\varpi)$ (given by $x\rightarrow x-1$), these characters can be written in the form $\psi_a(x):=\psi\big(a(x-1)\big)$ for some $a\in M_2(\mathfrak o/\varpi)$. For our purposes, a stratum is the level $N$ and the character $\psi_a$ on $K_N$.

What strata appear in the $K$-types of irreducible representations of $G$?

My possibly-incorrect understanding is that for a representation that can be compactly-induced from $ZK$, the fundamental stratum of the representation will appear in a $K$-type of lowest level. What happens for higher levels? And what happens for representations associated to the other chain order (than $M_2(\mathfrak o)$)?

For facts about $K$-types of irreducible representations, see Casselman's Restriction paper (this version should be more easily available), or Henniart's Appendix to this paper. For the supercuspidal case, I am aware Hansen's paper (though I don't understand it). The subtext of this paragraph is that I know (in principle) the $K$-types I'm interested in, yet I am still unable to transform this into knowledge of the corresponding strata. Whether this is due to me overlooking something simple or to a more serious issue, I do not know.

As an example, let $\chi_1$ and $\chi_2$ be characters of $k^\times$, with $\chi_1$ unramified and $\chi_2$ ramified with conductor $N_0\ge 1$ (so $\chi_2$ is nontrivial on $1+\varpi^{N_0}\mathfrak o$ but trivial on $1+\varpi^{N_0+1}\mathfrak o$). Set $\chi=\chi_1\otimes\chi_2$, and let ${\rm Ind}_P^G\chi$ be the corresponding ramified principal series. Then ${\rm Ind}_P^G\chi$ contains $K$-types $\sigma_N(\chi)$ of level $N\ge N_0$, where $\sigma_N(\chi)$ is the subrepresentation of ${\rm Ind}_{P\cap K}^K(\chi)$ of level $N$ (we aren't distinguishing between $\chi$ and its restriction to $K\cap P$).

Take $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$, so that $c=u\varpi^N$, with $u\in\mathfrak o^\times$, then (for example) $$g=\bigg(\matrix{adu^{-1}-b\varpi^N&b\cr &d}\bigg)\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)$$ Thus, for $v\in \sigma_N(\chi)$ and $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$ $$\sigma_{N,\chi}(g)\cdot v=\chi_2(d)\sigma_{N,\chi}\Bigg(\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)\Bigg)\cdot v$$ Since $\chi_2$ has conductor $N_0$, there exists $a_2$ such that $\chi_2(d)=\psi\big(a_2(d-1)\big)$. When $d\in 1+\varpi^N$ for $N>N_0$, the character will be trivial. I'm happy with this, but I don't understand how to get the rest of the calculation to work out, though I feel it should be a straight-forward exercise.

Writing a basis of a representation for $GL_2(\mathbb Q_p)$ in terms of the new vector

2012-10-05T17:21:50Z

For an irreducible smooth (generic) representation $\pi$ of $G=GL_2(k)$ with central character $\omega$, where $k$ is a $p$-adic field, we define the conductor of a vector $v\in\pi$ as follows. Let $K_0({\mathfrak p}^N)$ be the subgroup of $K=GL_2({\mathfrak o})$ with lower left-hand entry congruent to $0$ modulo ${\mathfrak p}^N$. The conductor of $v$, $c(v)$, is the smallest $N$ such that $$\pi\bigg(\matrix{a&b\cr c&d}\bigg)v=\omega(a)v\ \ {\rm for\ all}\ \bigg(\matrix{a&b\cr c&d}\bigg)\in K_0({\mathfrak p}^N)$$ In other words, $v$ is fixed by $K_0({\mathfrak p}^N)$ up to the action of the center. The conductor of $\pi$, $c(\pi)$, is the smallest $N$ such that there is a $v$ with $c(v)=N$. The subspace of vectors with conductor $c(\pi)$ is one-dimensional and the unique vector $v_0$ that maps to a Whittaker function with $W_0(1)=1$ is called the new vector.

My question is: how can we write down a somewhat-explicit basis of $\pi$ in terms of the new vector $v_0$, indexed by the principal congruence subgroups $K({\mathfrak p}^M)=1_2+{\mathfrak p}^MM_2({\mathfrak o})$? In other words, how can we write a basis of $\pi^{K({\mathfrak p}^M)}$ in terms of $v_0$?

A simple nonexplicit version is this result, from Casselman's "The restriction of a representation of $GL_2(k)$ to $GL_2({\mathfrak o})$": the restriction of $\pi$ to $K$ decomposes as $$res_K^G\pi=\pi^{K({\mathfrak p}^{c(\pi)-1})}\oplus\sum_{n\ge c(\pi)}u_n(\omega)$$ where $u_n$ is the unique irreducible representation of $K$ which is trivial on $K({\mathfrak p}^n)$ but not on $K({\mathfrak p}^{n-1})$ and that contains a vector $v$ with conductor $n$ in the above sense.

Answer by BR for splitting field for a division algebra

2012-10-05T01:37:41Z

It depends on what you mean by "splitting field". What is true is that for a division algebra $D$ (over $F$) of dimension $n^2$, if $E$ is a field extension of degree $n$ over $F$ that splits $D$, then $E$ is isomorphic to a subfield of $D$. See the Theorem on page 16 of Paul Garrett's notes.

Obviously, an algebraic closure of $F$ will split $D$, but won't be isomorphic to a subfield for dimension-counting reasons.

Answer by BR for Weyl law for SL(2,C)

2012-07-25T14:46:43Z

For general compact manifolds (of dimension $n$), the error term (on the number of eigenvalues less than $T^2$, counted with multiplicity) is $O(T^{n-1})$. So for $\Gamma$ co-compact, the error term is $O(T^2)$.

For merely co-finite $\Gamma$, it should be possible to bound the error term, but I am unaware of any results. In Section 4 of this survey by Müller (by the way, equation 1.3 is Weyl's Law for compact manifolds with the error term, and Section 2 sketches a different approach for compact locally symmetric spaces), he sketches a proof of a strong form of the law (using the Selberg Zeta function) and implies that the argument would work for other rank-one groups.

The book Groups Acting on Hyperbolic Space seems to (using Google Preview) prove the strong form of the law in the co-compact case (Section 5.5), but only seems to prove the weak form in the co-finite case (Section 8.9).

Answer by BR for Restriction of discrete series representations

2012-07-23T16:23:34Z

The answer depends on both the nature of the subgroup and the nature of the representation. For example, with $G=SL_2(\mathbb R)$, a discrete series discretely decomposes when restricted to $H=SO(2)$, but decomposes continuously for $H=GL_1(\mathbb R)$ (though this group is possibly too trivial). And if $G=SL_2(\mathbb R)\times SL_2(\mathbb R)$ and $H=SL_2(\mathbb R)$, embedded diagonally in $G$, the restriction of a holomorphic discrete series (on $G$, so this is the tensor product of two holomorphic discrete series on $H$) decomposes discretely when restricted to $H$ (as $D_n\otimes D_m\simeq \bigoplus_{k\ge 0}D_{n+m+2k}$), but the mixture of a holomorphic and an anti-holomorphic discrete series also has a continuous component when restricted (see Repka's work).

Kobayashi has published many interesting papers on this topic. See this expository paper for some necessary results and sufficient results for representations to decompose discretely. Unitary highest-weight modules (including holomorphic discrete series) are the best behaved here, but some hypotheses are still needed on the subgroup (here and Section 7 here). Basically, the tensor product (or restriction under some hypotheses) of unitary highest-weight representations decompose into unitary highest-weight representations (and if the representation is a holomorphic discrete series, the decomposition will be, too).

As an advanced example (see Section 7 in this other paper by Kobayashi), set $G=U(2,2)$, $H=Sp(1,1)$. Then $G$ has six families of discrete series, but only two decompose discretely when restricted to $H$ (ten other families of unitary representations decompose discretely).

Answer by BR for Motivic proof of Weil-conjectures?

2012-07-09T22:27:05Z

See Theorem 5.6 in Kleiman's article "The Standard Conjectures", in the Motives volume (PSPM 55.1). I'm not quite sure what you mean by "written completely in the language of motives", so it might not be exactly what you are looking for.

I initially posted this as a comment, but it really should be an answer, so I'm re-posting it as one.

Answer by BR for Perron, Fourier

2012-03-28T17:42:39Z

Generally, Perron's formula is the calculation of the inverse Mellin (or Laplace or Fourier) transform of a particular function. When the function's representation as a Mellin transform is known, this is simple. Otherwise, some work in necessary. To a certain extent, I guess it also depends on what you mean by "Perron's formula".

One version of Perron's forumla calculates the inverse Mellin transform of $I_T(s)/s$, where $I_T$ is the indicator function of the strip $|{\rm Im}(s)|< T$, a statement being (from Patterson's book on the zeta function), for $c>0$, $${1\over 2\pi i}\int_{c-iT}^{c+iT}{x^s\over s}ds=\cases{O\big(x^c/T\log(x)\big)& $0< x< 1$\cr 1/2+O(T^{-1}) & $x=1$\cr 1+O\big(x^c/T\log(x)\big)& $x>1$}$$ Since it is a statement about a particular function, you can't really get a general proof. On the other hand, I bet you could extract a certain amount of information as you do in the Paley-Wiener theorem (and similar results).

Wikipedia's version of Perron's formula, which is also an application of the above formula, is amenable to "proof via inversion". Let $g(s)=\sum_{n\ge 1} a_n/ n^s$ be a Dirichlet series converging absolutely for ${\rm Re}(s)>\sigma$. Re-write this as $g(s)=s\int_0^\infty A(x)x^{-s-1}\ dx$, where $A(x)=\sum_{n\le x}a_n$ (with some complication when $x$ is an integer).

If we set $B(x)=A(1/x)$, after changing variables $x\rightarrow x^{-1}$, this becomes $g(s)=s\int_0^\infty B(x)x^{s-1}\ dx=s{\cal M}B(s)$, where $\cal M$ denotes "Mellin transform" (otherwise $g(s)={\cal M}A(-s)$). Divide by $s$ and apply Mellin inversion to both sides (this requires some bound on the decay of $g(s)$ in the region of convergence, which isn't difficult), with $c\gg\sigma$, $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{-s}\over s}\ ds=B(x)$$ Send $x\rightarrow x^{-1}$ to get $${1\over 2\pi i}\int_{c-i\infty}^{c+i\infty}g(s){x^{s}\over s}\ ds=A(x)$$

Answer by BR for Lie algebras with abelian Cartan subalgebras.

2012-02-18T04:59:44Z

The subalgebra of diagonal matrices in the (non-reductive) standard Borel subalgebra ${\mathfrak b}$ of ${\mathfrak gl}_2$ is an abelian Cartan subalgebra of ${\mathfrak b}$, since it is abelian (hence nilpotent) and its own normalizer (directly calculate the bracket of an element $X$ of ${\mathfrak b}$ with a diagonal matrix $A$ to see that the result is diagonal only when $X$ was in fact diagonal or $A$ was in the center).

Answer by BR for Polar decomposition for quaternionic matrices?

2012-02-16T04:52:01Z

A key ~~word~~phrase here is "Cartan decomposition". Since $G=SL_n(\mathbb H)$ is a semisimple group, there is a diffeomorphism $$K\times \mathfrak p\rightarrow G$$ taking $(k,X)$ to $k\cdot {\rm exp}(X)$, where $K$ is a maximal compact subgroup of $G$ (i.e. the compact symplectic group) and $\mathfrak p$ is the vector subspace of ${\mathfrak sl}_n(\mathbb H)$ fixed by the Cartan involution $X\rightarrow \bar X^t$ (the quaternionic version of Hermitian).

Answer by BR for Character determines the representation?

2012-02-15T20:43:23Z

For a reductive Lie group, the character characterizes an irreducible admissible representation up to infinitesimal equivalence. Referring to Knapp's "Representation Theory, etc", Proposition 10.5 says that two infinitesimally-equivalent irreducible admissible representations have the same character, and Theorem 10.6 says that infinitesimally-inequivalent irreducible admissible representations have linearly independent characters.

For reductive $p$-adic groups, the character characterizes irreducible admissible representations., in that inequivalent irreducible admissible representations have linearly independent characters. See, e.g., Section 17 of Murnaghan's notes.

Answer by BR for Dedekind Zeta function: behaviour at 1

2012-02-09T03:14:41Z

We can actually do a good bit in the function field case, because the zeta function is of the form $$\zeta_F(s)={P(q^{-s})\over (1-q^{-s})(1-q^{1-s})}$$ where $P$ is a polynomial of degree equal to twice the genus of the underlying curve. When the genus of the curve is zero (e.g., $F=\mathbb F_q(t)$), $P(x)=1$. In this case, we can calculate the Laurent expansion of $\zeta_F(s)$ to be (using WolframAlpha to avoid thinking) $${q\over (s-1)(q-1)\log(q)}+{(q-3)q\over 2(q-1)^2}+O(s-1)$$ For the general case, we can multiply the above by the Laurent expansion for $P(q^{-s})$. For a genus-$g$ curve, the corresponding polynomial is $P(q^{-s})=1+a_1q^{-s}+\ldots+a_{2g}q^{-2gs}$. The Laurent expansion of $P(q^{-s})$ is $$\big(1+a_1 q^{-1}+\ldots+a_{2g}q^{-2g}\big)-(s-1)\log(q)\big(a_1 q^{-1}+2a_2q^{-2}\ldots+2g\cdot a_{2g}q^{-2g}\big)+O\big((s-1)^2\big)$$ Multiplying through, we get that the zero-th term in the Laurent expansion of $\zeta_F(s)$, where $F$ is the function field of a genus-$g$ curve, is $${(q-3)q\over 2(q-1)^2}\cdot P(q^{-1})+{q\over (q-1)\log(q)}\cdot {d\over ds}P(q^{-s})\bigg|_{s=1}$$

Answer by BR for Three-dimensional simple Lie algebras over the rationals

2011-10-19T05:33:55Z

Let me flesh out my comments. First, why there are infinitely many three-dimensional simple Lie algebras over $\mathbb Q$. One of the major steps in proving class field theory is to prove that we have an exact sequence of Brauer groups $$0\rightarrow {\rm Br}(\mathbb Q)\rightarrow\oplus_{p\le\infty}{\rm Br}(\mathbb Q_p)\rightarrow \mathbb Q/\mathbb Z\rightarrow 0$$ where for $p<\infty$, ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$, and of course ${\rm Br}(\mathbb R)\simeq \mathbb Z/(2)$. This says that given a finite set $S$ of primes (regarding $\infty$ as a prime) and division algebras $D_p$ over $\mathbb Q_p$ for all $p\in S$, if the corresponding invariants (in ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$) sum to zero in $\mathbb Q/\mathbb Z$, then there exists a division algebra $D$ over $\mathbb Q$ giving rise to the $D_p$ in the sense that $D\otimes_\mathbb Q\mathbb Q_p\simeq M_{n_p}(D_p)$ (we need to use matrices over a division algebra because the dimensions of $D$ and $D_p$ don't have to match).

Now, for quaternion algebras over $\mathbb Q$, the situation is simpler because (1) a central simple algebra of dimension 4 over $\mathbb Q_p$ is either a quaternion division algebra or $M_2(\mathbb Q_p)$ and (2) the hard algebraic number theory can be done "by hand" (it basically follows from quadratic reciprocity). Note this implies that the invariants for a central simple algebra of dimension 4 is either $0$ or $1/2$, hence the need for an even set of primes to get a division algebra over $\mathbb Q$.

So we have infinitely many quaternion algebras over $\mathbb Q$, all of which split over $\mathbb C$ to be isomorphic to $M_2(\mathbb C)$. To get three-dimensional Lie algebras out of this, you restrict to elements with "reduced trace" equal to zero.

Second, we wonder if we have found all three-dimensional simple Lie algebras over $\mathbb Q$. This is somewhat outside of my comfort zone. We switch to talking about simple algebraic groups of dimension three. We are interested in classifying forms of $SL_2(\mathbb Q)$. These are classified by the (non-abelian) Galois cohomology group $H^1(G(\bar{\mathbb Q}/\mathbb Q),{\rm Aut}_\bar{\mathbb Q}(SL_2(\mathbb Q))$. These can further be split into two classes, inner forms and outer forms, depending on whether the corresponding automorphism is inner or outer. Inner forms correspond to quaternion algebras. Outer forms correspond to certain unitary groups. This is from Platonov and Rapinchuk's "Algebraic Groups and Number Theory", section 2.3.4 (propositions 2.17 and 2.18). So it seems like we might be missing a bit, but the simple nature of our situation may mean that the outer forms are isomorphic to the inner forms, like $SU(1,1)\simeq SL_2(\mathbb R)$ (I have no clue).

As Vladimir notes in the comments to this answer, I am assuming that if two quaternion algebras are non-isomorphic, then their Lie algebras are non-isomorphic. This is a legitimate worry, as if $K$ is a quadratic extension of $\mathbb Q$, then ${\rm Lie}(K)\simeq {\rm Lie}(\mathbb Q^2)$. What happens for quaternion algebras? First, note that for a field $k$ and a quaternion division algebra $D$ over $k$, ${\rm Lie}\big(M_2(k)\big)$ is not isomorphic to ${\rm Lie}(D)$, since, for example, ${\rm Lie}\big(M_2(k)\big)$ has a solvable three-dimensional subalgebra and ${\rm Lie}(D)$ does not (seeing this by explicitly calculating the brackets of basis elements). Second, for two quaternion algebras $D_1$ and $D_2$, if a quadratic extension $K$ splits $D_1$ but not $D_2$, then $D_1\otimes_\mathbb Q K\simeq M_2(K)$, but $D_2\otimes_\mathbb Q K$ remains a division algebra. Since ${\rm Lie}(D_1\otimes_\mathbb Q K)\not\simeq {\rm Lie}(D_2\otimes_\mathbb Q K)$, we have ${\rm Lie}(D_1)\not\simeq {\rm Lie}(D_2)$. Finally, since quaternion algebras over $\mathbb Q$ are determined by the primes where they split, we can always find a quadratic extension $K$ so that exactly one of the $D_i\otimes_\mathbb Q K$ splits.

This argument only partly extends to general division algebras, since there are non-isomorphic division algebras with the same splitting field.

After I typed the above, I happened to see that Chapter X of Jacobson's "Lie Algebras" is devoted to classifying simple Lie algebras over arbitrary fields, which he also does in this paper. In particular, he proves that for central simple algebras $A$ and $B$ over a field $k$,

An isomorphism between ${\rm Lie}(A)$ and ${\rm Lie}(B)$ extends uniquely to either an isomorphism or the negative of an anti-isomorphism between $A$ and $B$. If they are quaternion algebras, it is always an isomorphism.

Answer by BR for On the Paley-Wiener theorem

2011-12-10T04:40:49Z

I think your question is true or nearly true. I interpet it as "Can we find a non-negative even Paley-Wiener function $f$ with $f'$ non-negative on $\mathbb R_{>0}$?" As in AH's answer, $$f(y)=\int_0^\infty g(x)\cos(xy)\ dx$$ where $g$ is a smooth function of compact support. I'll make a general statement that is not quite what you want, then give an example that is nearly (at least) what you want.

It has been proven (see here) that if $g(0)\ne 0$, then $f^{(n)}$ has only finitely many zeros. (Conversely, if $g^{(n)}(0)=0$ for all $n$, $f$ changes sign infinitely often, see here.)

This isn't too hard to prove. Take $n\ge 1$, then $$f^{(2n)}(y)=(-1)^{n}\int_0^\infty x^{2n}g(x)\cos(xy)\; dx$$ Integrating by parts $2n$ times makes this $$\Bigg[{\cos(xy)\over y^{2n+1}}\big(x^{2n}g(x)\big)^{(2n)}\Bigg]_0^\infty-y^{-2n-1}\int_0^\infty \big(x^{2n}g(x)\big)^{(2n+1)}\cos(xy)\; dx$$ The last integral is the Fourier transform of a smooth compactly-supported function, so decays in $|y|$. Thus we have, as $|y|\rightarrow\infty$, $$|f^{(2n)}(y)|=y^{-2n-1}(2n)!|g(0)|\big(1+h(y)\big)$$ where $h(y)\in o(1)$. Since $|h(y)|<1$ for $y$ sufficiently large, $f^{(2n)}$ can only have finitely many zeros. Noting that if $f^{(n)}$ has infinitely many zeros, Rolle's Theorem would imply that $f^{(n-1)}$ has infinitely many zeros, we see that $f^{(n)}$ only has finitely many zeros for all $n\ge 0$.

So you can easily find $f$ with every derivative having finitely many zeros. But, it turns out, $f^{(n)}$ always has at least $n$ zeros. This isn't as bad as it sounds, as $f$ being even implies that $f'$ has a zero. And just because a function has zero doesn't mean it is negative. Unfortunately, I can't find anything that gives a definitive answer here.

From here (p. 82), for $m\ge 4$ an even integer, $$f(x)=-\int_{-\infty}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt$$ is a positive PW function (at least with the more relaxed definition of $L^2$ on the line), with $f'$ non-negative on $\mathbb R_{>0}$. Since the integrand is odd, $f$ is even as $$f(x)=-\int_{-\infty}^{-x} {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt-\int_{-x}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt=f(-x)$$

Answer by BR for Degree conjecture and automorphic L-functions

2011-11-20T06:53:39Z

Going off wikipedia, it is true that automorphic $L$-functions for $GL_n$ over a number field have non-negative integral degree, where by degree I mean the number $2\sum_{i=1}^k \omega_i$, where the $\omega_i$ are the coefficients of $s$ appearing in the gamma factor, which is more-or-less $$L_\infty(s,F)=Q^s\prod_{i=1}^k\Gamma(\omega_i s+\mu_i)$$ We know that for $GL_n(\mathbb R)$ and $GL_n(\mathbb C)$ the $\omega_i$ are either $1/2$ or $1$ (see, e.g., Knapp's "Local Langlands Correspondence: The Archimedean case", in Motives, vol 2), so twice the sum will always be an integer (of course, only a few of these $L$-functions are known to be in Selberg's class).

For general $G$, it depends on whether someone has written done the $L$-factors for general real reductive groups. I don't know if this has been done, or if it is technically known but difficult to write out, or if it is not known.

Answer by BR for reference help on a result of Whittaker functions of supercuspidal representations

2011-11-13T08:08:50Z

I think this can be seen directly. Let's work over $G=PGL(n)$, so I don't have to keep repeating "modulo the center". Recall that supercuspidal representations can be realized as subrepresentations in $L^2(G)$ consisting of compactly supported functions.

So we can define an intertwining integral from $\pi$ into the Whittaker space by $$\Big(g\rightarrow f(g)\Big)\longrightarrow \Big(g\rightarrow\int_N \bar\psi(n)f(ng)\ dn\Big)$$ ("compactly supported" gives convergence).

Let $W$ denote the Whittaker function of an $f\in\pi$. If the support of $f$ is $C\subset G$, then the support of $W$ is $NC$. To see this, take $g\notin NC$. So, for each $n\in N$, $ng\notin C$. So $f(ng)=0$ for all $n$. Hence $W(g)=0$. Since $C$ is compact, $NC$ is compactly supported modulo $N$.

[Added: I realized I was tacitly assuming that for some $f$, the above integral is not identically zero. Since $f$ has compact support, the integral being identically zero implies that $f$ lies in the kernel of the twisted Jacquet functor. Since the representation has a Whittaker module, the twisted Jacquet module must be non-zero, hence there is an $f$ whose integral does not vanish.]

I don't know a reference off-hand for this basic fact (it also follows from the fact that the functions in the Kirillov model of a supercuspidal representation have compact support, which is in a lot of sources), though Bushnell and Henniart's "Supercuspidal Representations of $GL_n$: Explicit Whittaker Functions" gives a more, well, explicit version of it. [Added: I just realized that this is the final result of Casselman-Shalika's Whittaker function paper.]

Answer by BR for Decomposition of distributions

2011-11-02T14:41:10Z

The Dirac Comb, an infinite sum of delta functions, is an example of a tempered distribution that cannot be thusly decomposed (its Fourier transform is another Dirac Comb).

[Added:] There is a positive result in this direction that I (among others) only partly-remembered: Any distribution can be written as a locally finite sum of derivatives of continuous functions. If the distribution has finite order, then the sum is finite. See Rudin's Functional Analysis, Theorem 6.28.

Answer by BR for How do Brauer groups relate to zeta functions?

2011-10-31T06:16:23Z

Global class field theory typically refers to the existence of a particular isomorphism (along with various compatibilities) $G(K/k)^{ab}\simeq C_k/N_{K/k} C_K$, where $K$ is a finite Galois extension. This can be recast, by dualizing, to say something about a correspondence between characters on $G(\bar k/k)$ and characters on $C_k$. These statements will be equivalent, but different proofs may not be...

I think you can prove it via Taylor-Wiles method (Kowalski mentions this in a survey paper referring to unpublished notes from a class Tunnell taught). But the proofs of global class field theory that I know are essentially cohomological (though they may avoid saying it). The statement about Brauer groups that you made is essentially what is necessary to prove that $(G(\bar k/k),C_{\bar k})$ is a class formation, from which a bit more work is necessary to prove class field theory (though you can deduce reciprocity laws from it).

Answer by BR for Is the integrality of the zeta function easy?

2011-10-25T22:48:45Z

It is not necessarily obvious that this suffices, though it is true (fortunately). You can find a proof in Milne's Lectures on Etale Cohomology. (I happened to just be looking at this while thinking about this question.)

It is Lemma 27.9, phrased as: Let $k\subset K$ be fields and $f(t)\in k[[t]]$. If $f(t)\in K(t)$, then $f(t)\in k(t)$.

Answer by BR for A question about zeros of Tate type integral

2011-10-18T05:55:48Z

The Beta function $B(s,s_0)={\Gamma(s)\Gamma(s_0)\over\Gamma(s+s_0)}$ has the correct outcome (infinitely many zeros and poles of the proper type). But it is the Mellin transform of a discontinuous function (with compact support): $$B(s,s_0)=\int_0^\infty (1-x)^{s_0-1}\mathbf 1_{(0,1)}(x)x^s\ {dx\over x}$$ (here $\mathbf 1_{(0,1)}$ is the characteristic function of the open unit interval, and we assume $s$ and $s_0$ are greater than zero), instead of the Mellin transform of a Schwartz function.

Can we do better than that? Maybe. We would expect there to be a correspondence between the smoothness of $\phi$ and the decay of $F(s)$ in vertical strips (like in Paley-Wiener theorems or Sobolev embedding), so maybe something that decays faster than $B(s,s_0)$ (which decays like $|s|^{-{\rm Re}(s_0)}$) would correspond to a smoother function. You could also try using a mollifier to smooth out the kernel for the Beta function, though this would require either translating everything into the Fourier transform setting or using the Mellin-compatible version of convolution.

Answer by BR for a question about Kirillov model of unitary representations over GL_n(R)

2011-10-11T03:23:30Z

I happened to glance through Cogdell's "L-functions and Converse Theorems for GL(n)" (in Sarnak/Shahidi "Automorphic Forms and Applications), which cites Jacquet and Shalika's "On Euler Products and the Classification of Automorphic Forms I", available from Jacquet's website, as answering your question in the affirmative (see Section 1.3): the map $\xi\rightarrow W_\xi|_P$ in injective (where $P$ is the mirabolic subgroup), moreover, the Kirillov model (of a generic irreducible unitary representation), defined as the space of restrictions of Whittaker functions, is isomorphic to ${\rm Ind}_N^P(\psi)$ as representations of $P$.

Answer by BR for Different cuspidal automorphic representations with same representations at infinity

2011-10-10T13:51:48Z

This is precisely the content of Harish-Chandra's theorem ("Automorphic forms on Semisimple Lie Groups", LNM 68, 1968), proven for general reductive groups:

Fix a finite-dimensional representation $\delta$ of $K_\infty$, an ideal $J$ of finite co-dimension in ${\mathcal Z}({\mathfrak g})$, a compact open subgroup $L$ of $K_{\rm fin}$ (the maximal compact subgroup of $G_{\mathbb A_{\rm fin}})$, and a central character $\omega$, then the space of $K$-finite automorphic forms $f$ with central character $\omega$ and $K_\infty$-type $\delta$ that are right $L$-invariant and annihilated by $J$ is finite dimensional.

To see how this corresponds to your case, fix a representation $\pi_\infty$. For simplicity, assume $\pi_\infty$ is spherical. Thus $\delta$ is the trivial representation. Any vector $\phi$ generating $\pi_\infty$ will be an eigenvector for the Casimir operator, $C\phi=\lambda\phi$, so we take $J$ to be the ideal generated by $(C-\lambda)$. And since $\chi(K)$ is a compact totally disconnected subgroup of ${\mathbb C}^\times$, the kernel of $\chi$ must be a compact-open subgroup $L$, so any $\phi$ will be right $L$-invariant. The only thing left is the central character. The archimedean part is fixed as well as the restriction to a compact open subgroup, so there should only be finitely many options left.

A proof for $SL_2({\mathbb R})$ can be found in Chapter 8 of Borel's "Automorphic forms on $SL_2({\mathbb R})$". The general argument is sketched in section 8 of Borel's contribution to Sarnak and Shahidi's "Automorphic Forms and Applications".

Answer by BR for classification of irreducible admissible (g,K)-module for GL(3,R)

2011-10-05T20:43:02Z

For a general real reductive group, all irreducible admissible $({\mathfrak g},K)$-modules are quotients of parabolically-induced discrete series (or limits thereof) representations (where we allow "trivial" parabolic induction ($P=G$) for discrete series on the group). See Theorem 14.92 in Knapp's Representation Theory of Semisimple Groups. This is a refinement of the Langlands Classification (which replaces "discrete series" with "tempered"). Knapp's paper "Local Langlands Correspondence: the archimedean case", in volume 2 of Motives, PSPM 55, gives an explicit classification for $GL_n$ (over $\mathbb R$ and $\mathbb C$). Also see Moeglin's article "Representations of GL(n) over the Real Field" in Representation Theory and Automorphic Forms, PSPM 61.

For $GL_n(\mathbb R)$, we can say that given an irreducible admissible $({\mathfrak g},K)$-module $V$, there exists a parabolic subgroup $P=MN$ of $GL_n$ with block sizes either $1$ or $2$ (since $GL_n$ only has discrete series for $n=1$ or $2$), and a discrete series representation $\sigma$ of $M$, such that $V$ is isomorphic to the unique quotient of the $({\mathfrak g},K)$-module underlying ${\rm Ind}_P^G(\sigma,s)$, where $s$ is a tuple of complex parameters, one for each block in $M$. Further analysis can tell you when two induced representations give you the same $({\mathfrak g},K)$-module, and when the induced representation is irreducible.

Which Shimura varieties are known to be automorphic?

2011-09-16T19:12:40Z

This seems like something that should be well-known, but as an outsider to the field, I'm having trouble locating precise statements.

Hasse-Weil zeta functions of Shimura varieties should be alternating products of automorphic $L$-functions. This seems to be known when the underlying group is $GL_2$ (or a quaternion division algebra) over a totally real field, $GSp_4$ over $\mathbb Q$ (maybe a totally real field), unitary groups in three variables over $\mathbb Q$ (maybe a totally real field), and maybe certain other unitary groups.

In "Where Stands Functoriality Today?", Langlands writes that the "principal factors" of the zeta function of (general) Siegel modular varieties are automorphic $L$-functions attached to spinor representations (which have not been analytically continued), so some more general calculations seem to be known...

My question: For which groups are the Hasse-Weil zeta functions of the associated Shimura varieties known to be alternating products of automorphic $L$-functions (maybe modulo "technical" restrictions and bad prime calculations)?

Answer by BR for "Space" of L-functions

2011-09-14T22:06:08Z

Just to summarize some of the comments:

Strong Multiplicity One says that if the local factors of two (cuspidal) automorphic L-functions agree at all but finitely many places, then they agree at all places ("all but finitely many places" can conjecturally be replaced with "all places in a set of Dirichlet density greater than $1-1/2n^2$"). So, in this sense, the space of automorphic L-functions is rigid (you can't make changes to a small number of places and get something automorphic). This can fail for residual representations but I don't have good examples to share.

In another sense, you could deform an L-function $L(s,\pi)$ by considering $L(s,\pi\otimes|\cdot|^z)=L(s+z,\pi)$, where $\pi\otimes|\cdot|^z$ is $\pi$ with its central character twisted by $|\cdot|^z$ (some care is necessary if $z$ is not imaginary). Also, Eisenstein series exist in meromorphically parametrized families, so their L-functions also exist in families. For example, the L-function for the classical non-holomorphic Eisenstein series $E_z$ is $\xi(s+z)\xi(s+1-z)$.

Regarding vector spaces, note that the sum of two automorphic L-functions will (generally) not have an Euler product, so won't be automorphic in the usual sense.

Finally, just to explicitly tie the above in with the original question, Artin L-functions are conjecturally cuspidal automorphic for $GL_n$, and motivic $L$-functions are conjecturally automorphic for some group $G$ which can them be conjecturally transfered to $GL_n$.

Answer by BR for Are the $L$-functions of $X_0(N)$ automorphic?

2011-09-03T19:13:33Z

The zeta function of the modular curve $X_0(N)$ is the product of the $L$-functions of a basis of cusp forms of weight 2 for $\Gamma_0(N)$ (the basis taken to be normalized eigenforms for the Hecke operators prime to $N$), up to a finite number of factors. See, e.g., Milne's notes on modular forms, Theorem 11.14 (p. 108).

Modular curves are the (or at least one of the) simplest examples of Shimura varieties (See Milne's notes on Shimura varieties). One of the main motivations for the study of Shimura varieties is showing that their Hasse-Weil zeta functions are products (allowing positive and negative powers) of automorphic $L$-functions (as part of a broader program to prove the same thing for general algebraic varieties, i.e. that motivic $L$-functions are automorphic). There are plenty of other reasons to study Shimura varieties, though (e.g. they are the most powerful tool for proving results about special values of automorphic $L$-functions, more advanced versions of $\zeta(2n)\in(2\pi)^{2n}{\mathbb Q}$)

The original version of Taniyama-Shimura-Weil is "for any elliptic curve $E$, there exists a non-constant map from some $X_0(N)$ to $E$ (defined over $\mathbb Q$). So, there are historical reasons for phrasing it that way.

Answer by BR for Examples where it's useful to know that a mathematical object belongs to some family of objects

2011-09-01T06:52:09Z

A somewhat amusing example in the theory of automorphic forms is that the constant $1$ function is the residue of an Eisenstein series (considered as a meromorphic vector-valued function). So, any $1$, anywhere, can be replaced by $Res_{s=1}E_s$.

As a quick application, take a cusp form $f$ for $GL_2$ over a number field $k$ that generates an irreducible representation $\pi_f$. Identifying $1$ with the residue of an Eisenstein series, the $L^2$-norm of $f$ turns into the residue of a Rankin-Selberg L-function times the norm of the first Fourier-Whittaker coefficient of $f$, $\rho_f(1)$: $$||f||_{L^2}=\int_X |f|^2\ dx=Res_{s=1}\int_X |f|^2E_s\ dx=|\rho_f(1)|^2Res_{s=1}\Lambda(s,\pi_f\otimes\tilde \pi_f)$$ If we want, we can break up the Rankin-Selberg L-function a bit to get something more tangible. $$||f||_{L^2}=|\rho_f(1)|^2L_\infty(1,\pi_f\otimes\tilde \pi_f) L(1,Sym^2\pi_f)Res_{s=1}\zeta_k(s)$$ where $L_\infty$ is a certain product of Gamma functions (whose parameters depend on $\pi_f$), $\zeta_k$ is the Dedekind zeta function of $k$ (whose residue at $s=1$ we know from the class number formula), and $L(1,Sym^2\pi_f)$ is the symmetric-square L-function of $\pi_f$, which is more mysterious (though a certain amount is known about how it changes as $\pi_f$ varies). Typically, we assume either $||f||_{L^2}=1$ or $\rho_f(1)=1$, so the formula turns information about L-functions into information about Fourier-Whittaker coefficients or $L^2$-norms (or vice-versa).

Comment by BR

2013-02-11T00:40:13Z

@Joël: The table of contents and some of chapter 14 are available on google books -- books.google.com/…

Comment by BR

2013-02-10T23:18:50Z

Does chapter 14 of Wallach's "Real Reductive Groups II" have what you want?

Comment by BR

2013-02-08T04:19:34Z

Similar questions have been asked: mathoverflow.net/questions/95205/…, mathoverflow.net/questions/70318/…, mathoverflow.net/questions/28000/…, not that they answer your specific question, but maybe a good place to look...

Comment by BR

2013-01-11T20:23:46Z

Are you interested in infinite-dimensional representations or only finite-dimensional representations?

Comment by BR

2013-01-05T17:54:13Z

A) At ramified places the local representation could be a ramified principal series, which will still not be square integrable, and it is, in general, difficult to say when this does or does not occur. B) A tempered automorphic representation is indeed one that is locally tempered everywhere.

Comment by BR

2012-12-11T19:56:55Z

Hi Matthew, thanks for the series of comments. (By the way, the support of the new vector in the Kirillov model for ramified principal series (in both characters, not just one) or ramified special representations is ${\mathfrak o}^\times$, see the paper Paul links to in his answer, and, in fact, the general formula for the Kirillov model of the new vector can be "reverse engineered" from the fact that the Mellin transform is precisely the $L$-function.) I might be able to figure out the answer using the formal series description in JL.

Comment by BR

2012-12-11T19:35:36Z

Hi Paul, thanks for the answer. I do agree that those notes are very useful. In fact, I was looking at it when I thought of the question! :) Perhaps I'm just being dense, but I was unable to see how to use it to get the result I'm interested in.

Comment by BR

2012-12-10T14:37:45Z

Hi labirintas, thanks for the paper. A general (possibly nontrivial) fact, at least for GL(n), is that $W(1)\ne 0$ (this may require "newform-ness"). Since $W(z)=\omega(z)W(1)$, where $\omega$ is the central character of $\pi$, $W$ is nonzero on the center. More specifically, the "new vector" criterium can also be written as $W(gk)=\omega(d)W(g)$ for $k\in K_0(c)$, $k=\bigg(\matrix{a&b\cr c&d}\bigg)$.

Comment by BR

2012-11-01T19:54:13Z

Hi goci, you should be able to comment on answers to your own questions, even if you don't have enough reputation to comment more generally. The problem you seem to be having is that you are logging in under new accounts each time (with the same name). If you are able to find your way back to the account that asked the question, you should be able to leave comments here. To merge your different accounts, leave a message on the Meta discussion meta.mathoverflow.net/discussion/605/5/… (you will also need to sign up for a meta account).

Comment by BR

2012-11-01T19:42:03Z

I believe this conjecture does not have a specific name. Though if this paper (arxiv.org/pdf/1206.0149.pdf) is to be believed, it could be called Maillet's Conjecture (but it does not seem to be generally known by that name).

Comment by BR

2012-10-23T16:15:56Z

Mrc, Patterson derives the above product from the Hadamard product in his Zeta function book in Section 3.1. Mohammad, On the general principle that the zeroes of the zeta function are mysterious, I am skeptical that much of interest could be proven in this way. It seems very difficult to make sense of the inverse Mellin transform of $\pi^{s/2}{(1-s/\rho)\over 2s(s-1)\Gamma(1+s/2)}$, so I'm not sure anyone will be able to help answer your question.

Comment by BR

2012-10-22T17:23:02Z

Hi Marc, I sent you an email. I'll write some more comments here later (and I finally got around to commenting on your previous answer to a question of mine), but I have to go run some errands. Thanks for the answer!

Comment by BR

2012-10-22T16:42:05Z

Hi Marc, sorry for taking so long to respond! I am exactly trying to move through the types. In fact, Reeder's Oldforms paper does mostly what I want to do. This is related to the other question of mine you answered: I'm doing a calculation that runs through a basis of a representation and needs to be compatible with the $K$-type structure, and I'm trying to write it in a way that I can work with.

Comment by BR

2012-10-22T16:33:15Z

Hi Marc, I think you might have misread, though maybe I'm missing something: the one-dimensional space of vectors is fixed by $K_0(p^N)$ rather than $K(p^N)$ (i.e. they are newforms).

Comment by BR

2012-10-08T14:19:35Z

Thanks, David, I'll check it out.