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bio website plusepsilon.de
location Uni Hamburg
age 29
visits member for 4 years, 6 months
seen Jan 31 at 13:20

Marc Palm

Postdoc in Mathematics

http://www.plusepsilon.de


Mar
29
answered What is the logarithmic derivative of an (intertwining) operator?
Mar
24
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
@Marty I thought more about your claim about the intertwiner. You can see that JL are using the one that I describe as does Bump, etc. By Schur's lemma, there can be only one intertwiner up to a constant. It is the standard one for the adelic Eisenstein series as well. Yours does not match up with the standard one times a constant, or does it?
Mar
24
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
You get one quotient and one submodule, so yes that makes too. But the roles are interchanged in $I(\mu)$ and $I(\mu^w)$.
Mar
23
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
Are you sure? I can't see the invariance. The integral transfer needs to map $I(\mu)$ to $I(\mu^w)$.
Mar
22
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
Also there are other references for SL(2,C). Knapps book on semisimple groups or Wallachs book on reductive groups. Wallach seems more informative concerning things like temperedness and unitarity. SL(2,C) and GL(2,C) are very similar, not like SL(2,R) and GL(2,R).
Mar
22
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
I have chosen to delete my answer, because it seems not to address your question. As Marty points out $I(\chi_1,\chi_2)$ is isomorphic to $I(\chi_2,\chi_1)$ if both are irreducible. If they are reducible, they are not. What is an irreducible subquotient in one of them is an irreducible subrepresentation in the other. So that's why considering only unique subquotients hits all irr reps.
Mar
22
comment Local Langlands for $GL(2,\mathbf{C})$ and reducible principal series
The intertwiner seems wrong. You need $\int\limits_{U} f(wux) du$ for $w$ the Weyl element, or not?
Mar
21
awarded  Custodian
Mar
21
reviewed Approve phase portrait of system of differential equations
Mar
20
comment The representation of a group
What kind of maps? Group homomorphisms, I guess?
Mar
20
comment reference help about a result on representation theory
Ah okay, I see imaginary line modulo $x=2 \pi i / log(q)$, $q$ being the residue characteristic, that's isomorphic to $U(1)$, I guess:) I didn't see that before:\ But that's a confusing embedding of $U(1)$ into $\mathbb{C}^\times$ modulo $x$.
Mar
20
comment reference help about a result on representation theory
@WillSawin My issue is that unitary one-dimensional representation live on the imaginary line, not $U(1)$.
Mar
20
comment Simultaneously extending the functionals of a subspace of a Banach space to the whole space
Be careful, there exists Banach spaces which do not admit a Schauder Basis. The first examples are due to Enflo.
Mar
20
comment Simultaneously extending the functionals of a subspace of a Banach space to the whole space
There are results available, when the extension is unique, e.g. in a Hilbert space or more general results can be found here jstor.org/discover/10.2307/…. Of course, Hamel basis are not to be chosen in topological vector space, e.g. for Banach spaces one works with a Schauder basis.
Mar
20
revised reference help about a result on representation theory
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Mar
20
revised reference help about a result on representation theory
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Mar
20
revised reference help about a result on representation theory
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Mar
20
answered reference help about a result on representation theory
Mar
16
comment What can the theory of automorphic forms for $SL(n,\mathbb{Z})$ say about $SL(n,\mathbb{Z})$?
There is no Eichler-Selberg type trace formula for SL(3) because of the abscence of discrete series for SL(3,R).
Mar
16
comment What can the theory of automorphic forms for $SL(n,\mathbb{Z})$ say about $SL(n,\mathbb{Z})$?
there are certainly not easy ways to study this rep.theory... I meant to say easier ways in my last comment:\