bio | website | plusepsilon.de |
---|---|---|
location | Uni Hamburg | |
age | 29 | |
visits | member for | 3 years, 11 months |
seen | Sep 25 at 13:24 | |
stats | profile views | 8,285 |
Apr 8 |
answered | What is the intuition behind the definition of cuspidal representations? |
Apr 7 |
comment |
What is the logarithmic derivative of an (intertwining) operator?
Note that my computation apply only to the highest type. For the computations at the real places and the unramified complex cases, you can have a look at my PhD thesis. There is a good reason for working with highest types/smallest weights = irreducible $K$-reps as soon as you have pinned down the local conditions. |
Apr 6 |
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Homeomorphisms that admit a decomposition
Okay, my mistake;) I see now that it seems to more complicated... |
Apr 5 |
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Homeomorphisms that admit a decomposition
There are only two strictly monotone functions on $[0,1]$ up to conjugation by homeomorphisms. One doesn meet your criteria. |
Apr 5 |
comment |
Homeomorphisms that admit a decomposition
Do you know the answer for $[0,1]$? Modulo conjugating by homeomoprhism of $[0,1]$, there seems to be only one map. |
Apr 5 |
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Space with 720° / not 2$\pi$ rotational symmetry?
Perhaps the Möbiusband? |
Mar 29 |
revised |
What is the logarithmic derivative of an (intertwining) operator?
added 8 characters in body |
Mar 29 |
answered | Current Status on Langlands Program |
Mar 29 |
comment |
Current Status on Langlands Program
I would also distinguish between functoriality (maps) and correspondence (objects)... you seem to be interested in correspondence between automorphic reps and Galois reps. |
Mar 29 |
revised |
What is the logarithmic derivative of an (intertwining) operator?
added 217 characters in body |
Mar 29 |
comment |
Current Status on Langlands Program
I also think that stabilization is not an issue for GL(n). |
Mar 29 |
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Current Status on Langlands Program
mathoverflow.net/questions/10578/… mathoverflow.net/questions/127157/… |
Mar 29 |
comment |
Current Status on Langlands Program
There have been many similiar questions in the past. |
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 |
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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 |
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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 |
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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 |
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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 |
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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? |