How to think about parabolic induction. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T03:14:03Z http://mathoverflow.net/feeds/question/9154 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction How to think about parabolic induction. Kevin McGerty 2009-12-17T05:04:00Z 2012-12-10T13:41:15Z <p>In studying representations of a reductive group G, a standard technique is to use parabolic induction. The idea is that one studies such groups as a family (or perhaps in smaller families, like say just taking (products of) GL_n's) and then trying to understand representations of larger groups via representations of their smaller subgroups in the same family. Induction is the natural functor for doing this, but inducing directly from a Levi subgroup L (a natural class of reductive subgroups) gives you very large representations -- the homogenous space G/L is not projective and so bundles on it will have lots of sections. The remedy is to use parabolic induction: take the bigger subgroup P which contains L as it's reductive part, and extend representations of L to representations of P by letting the unipotent radical U of P act trivially. Then you induce the result from P to G and (since G/P is projective) you get a much smaller representation, which you can feasibly study via Hecke algebras etc.</p> <p>Now in all of the above, it seems to me to make more sense to think of L not as a subgroup of G at all, but rather the reductive quotient of the parabolic group P -- that is, as a subquotient of G only. Now however, comes my question. One of the theorems you prove when studying parabolic induction (I'm begin sloppy about the context deliberately, but if you like, when studying finite group of Lie type say) is that it does not depend of the choice of a parabolic containing L. Since the same Levi can lie in two non-conjugate parabolic subgroups, this draws you back to the conclusion that it was better to think of parabolic induction as an operation on representations of subgroups of G after all. </p> <p>How do people think about this? "Why" is parabolic induction independent of the choice of parabolic? The proof of this in the context of finite groups of Lie type goes as follows: prove a formula for the composition of a parabolic induction followed by a parabolic restriction (using two arbitrary pairs of Levi inside parabolic) which expresses the composition as a sum of (parabolic) inductions and restrictions for smaller groups (as for the normal Mackey identity in finite group). Then consider the Hom space between two parabolic induction functors for the same Levi, and notice that Frobenius reciprocity lets you write this in terms of the Mackey identity you have, and then induction on rank finishes you off. </p> <p>This argument doesn't feel very enlightening me (though that might be because I don't really understand it), and I know another slick geometric proof in the context of character sheaves on a Lie algebra, which I'm not sure I understand the representation-theoretic content of. The result also is not just a curiosity -- the Harish-Chandra strategy of classifying representations of G via cuspidal data wants to associate to an irreducible representation of G a unique "cuspidal datum" $(L,\rho)$ consisting of a Levi subgroup L and a cuspidal representation of L (up to conjugation in G), and this doesn't make sense without the independence of parabolic. </p> <p>Also I don't remember the situation for p-adic groups: the "Mackey formula" argument I sketched should show that the parabolic induction functors you get don't depend on the parabolic at least in at the level of the K-group (which would be all you need to get a cuspidal theory running) but the functors themselves are perhaps not isomorphic (because the identity becomes some sort of filtration on the composition of the induction and restrictions, which is probably already is really in the finite groups type, so perhaps the same thing happens already for finite groups of Lie type when you study modular representations?</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/9203#9203 Answer by Ben Webster for How to think about parabolic induction. Ben Webster 2009-12-17T21:54:38Z 2009-12-17T21:54:38Z <p>Not an answer, but here's a rephrasing: this question is equivalent to asking why parabolic restriction doesn't depend on the parabolic (parabolic restriction is the adjoint, so you restrict to the parabolic, and then take invariants of the radical).</p> <p>Perhaps the projection map to invariants of one radical to invariants of the other is an isomorphism?</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/9241#9241 Answer by S Kitchen for How to think about parabolic induction. S Kitchen 2009-12-18T03:46:00Z 2009-12-18T03:46:00Z <p>Can't independence of parabolic be explained geometrically by the existence of isomorphisms between the partial flag varieties $G/P$? This takes care of the $GL(3)$ example in the comments, certainly, since the parabolics described give $G/P = Gr(1,3)$ and $Gr(2,3)$, respectively. </p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/9483#9483 Answer by David Bar Moshe for How to think about parabolic induction. David Bar Moshe 2009-12-21T16:10:10Z 2009-12-21T16:10:10Z <p>The following remarks refer only to private cases, although I think that they can be generalized:</p> <ol> <li><p>G = GL(n,C). In the GL(3) example given in one of the comments, The G/P's are Gr1(C^3) and Gr2(C^3) which are projectively dual as projective spaces. In the more general case of a maximal parabolic subgroup in GL(n,C), there is the Grassmann duality. In the general case of G/P is a partial flag manifold, I think that the duality can be understood as Grassmann dualities of the Grassmann fibers in the iterated Grassmann bundle picture of the partial flag manifold.</p></li> <li><p>There are cases where intertwining operators between geometrically realized representations on G/P1 and G/P2 (sharing the same Levi subgroup) are explicitely constructed, for example:</p></li> </ol> <p><a href="http://www.mathematik.uni-bielefeld.de/~hoffmann/preprint/snowbird.pdf" rel="nofollow">http://www.mathematik.uni-bielefeld.de/~hoffmann/preprint/snowbird.pdf</a></p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/56598#56598 Answer by Victor Ginzburg for How to think about parabolic induction. Victor Ginzburg 2011-02-25T05:19:07Z 2011-02-25T05:19:07Z <p>Two remarks:</p> <p>1) There is a notion of induction of nilpotent orbits (Lusztig-Spaltenstein) where a similar issue arises. In that case, there is a geometric argument.</p> <p>2) For p-adic groups look into two old papers by Bernstein-Zelevinsky.</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/115955#115955 Answer by Sam Gunningham for How to think about parabolic induction. Sam Gunningham 2012-12-10T07:05:21Z 2012-12-10T07:05:21Z <p>I have also been trying to understand the big picture here. As far as I can tell, the key point is that two parabolic subgroups $P$ and $Q$ of $G$ have conjugate Levis if and only if $P\backslash G/Q$ contains an orbit which is the preimage of a single Bruhat cell in $P\backslash G$ and $G/Q$.</p> <p><strong>Some more details</strong>: (My apologies if I have missed the point of the question and the following is already well understood to everyone (or just wrong!). I have tried to write down a general argument that would apply in various settings, but I am probably missing important features of the proof in any particular setting.)</p> <p>Suppose $P$ and $Q$ are parabolic subgroups of $G$ with Levi factors $L$ and $M$ respectively (I would also prefer to think about the Levis as being subquotients of $G$). I will not assume $L$ and $M$ are related to begin with.</p> <p>Subsets of the double coset space $P\backslash G/Q$ give rise to functors from representations of $L$ to representations of $M$. Taking the whole of $P\backslash G/Q$ corresponds to the functor $Res_{M,Q}^G Ind_{L,P}^G$. Taking smaller subsets picks out summands of this (note that the Mackey formula expresses restriction then induction as a sum over such double cosets).</p> <p>The parabolic $P$ gives rise to an embedding of the root systems of $L$ into the root system of $G$ (I am being slightly sloppy here about how/when I'm choosing Borels... this can be made more precise in a number of different ways, hopefully all equivalent) . The Weyl group $W_L$ gets idetified with a parabolic subgroup $W_P$ of the Weyl group $W$ (similarly for $(Q,M)$). The Bruhat decomposition idetifies $P\backslash G/Q$ with $W_P \backslash W/W_Q$ (as sets).</p> <p>By a minor abuse of terminology Levis $L$ and $M$ can be said to be conjugate if the root system of $L$ is conjugate to the root system of $M$ by an element of $W$. If we had chosen embeddings of $L$ and $M$ as subgroups of $G$, this is equivalent to them being conjugate as subgroups (independantly of the choice of parabolic). This is also equivalent to $W_P$ being conjugate to $W_Q$. </p> <p>It follows from the Bruhat decomposition that $L$ and $M$ are conjugate if and only if there is a double coset in $Q\backslash G/P$ which is the image of a single Bruhat cell in $P\backslash G$ and $G/Q$.</p> <p>Let me make things more concrete for a second: suppose we fix a Borel $B$, and assume $P$ and $Q$ contain $B$. Then $P$ and $Q$ are conjugate if and only if they are equal. In that case $P$ itself is a double coset in $P\backslash G/P$. which is already a Bruhat cell (i.e. a point). If $P$ and $Q$ are not conjugate, but $W_P$ and $W_Q$ are, then there is an element $a\in W$ such that $W_PaW_Q = W_Pa = aW_Q$. This means that $PaQ = BaQ = PaB$.</p> <p>Hence there is a canonical functor from $Rep(L)$ to $Rep(M)$. This functor is invertible (its inverse is the corresponding double coset in $Q\backslash G/P$). If we identify $L$ and $M$ compatibly with the identification of root systems then I claim this functor is the identity. Moreover, by its construction, this functor is a summand of $Res_{M,Q}^G Ind_{L,P}^G$. For a representation $V$ of $L=M$, the identification of $Ind_{L,P}^G$ with $Ind_{L,Q}^G$ is the adjoint of the inclusion $V\to Res_{L,P}^GInd_{L,Q}^GV$.</p> <p><strong>Remarks</strong>: The parabolic induction and restriction functors can be thought of as pull-push of sheaves along the following correspondence:</p> <p>$BL \leftarrow BP \rightarrow BG$.</p> <p>The composition of the correspondences associated to $(P,L)$ and $(Q,M)$ is</p> <p>$BL \leftarrow BP \times_{BG} BQ = P\backslash G/Q \rightarrow BM$.</p> <p>For the anologue of these ideas in the setting of character sheaves, one should replace $P\backslash G/Q$ with its loopspace, the corresponding relative Steinberg variety.</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/115970#115970 Answer by Alexander Braverman for How to think about parabolic induction. Alexander Braverman 2012-12-10T12:25:03Z 2012-12-10T13:41:15Z <p>Usually (for $p$-adic groups, real groups or automorphic forms) you can write an explicit "intertwining operator" between the two inductions. This is a very basic construction in representation theory - as basis as parabolic induction itself in some sense. Usually it is given by some sort of integral which converges only in some range and one needs to work in order to prove its meromorphic continuation (it does have both zeros and poles; because of this parabolic induction is only generically independent of the choice of representation of $L$).</p> http://mathoverflow.net/questions/9154/how-to-think-about-parabolic-induction/115977#115977 Answer by Wilberd van der Kallen for How to think about parabolic induction. Wilberd van der Kallen 2012-12-10T13:23:21Z 2012-12-10T13:23:21Z <p>Let me contribute some confusion. In the situation I am familiar with (algebraic groups), one may first restrict to a Borel subgroup $B$ with the property that $B\cap L$ is a Borel subgroup of $L$. Recall that if $P$ contains both $L$ and $B$, then restricting from $P$ to $B$ and next inducing back up to $P$ does nothing to $P$-modules. So instead of inducing up from $P$ one might as well first restrict to $B$ and then induce up from $B$ to $G$. And all Borel subgroups are conjugate. </p>