bio | website | plusepsilon.de |
---|---|---|
location | Uni Hamburg | |
age | 29 | |
visits | member for | 4 years, 6 months |
seen | Jan 31 at 13:20 | |
stats | profile views | 8,459 |
Jul 20 |
comment |
How does the right regular of GL(n, R) and GL(n,Qp) decompose?
Thanks for this answer. Lot of material here to be checked. But you don't know anything compromised as one can find e.g. in Gelbart-Jacquet for the global picture $L^2( GL(2, k) \backslash GL(2, \mathbb{A})=$ cuspidal rep. + characters + direct integrals involving parabolic induction? |
Jul 20 |
revised |
How does the right regular of GL(n, R) and GL(n,Qp) decompose?
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Jul 20 |
revised |
How does the right regular of GL(n, R) and GL(n,Qp) decompose?
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Jul 20 |
comment |
Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic?
Following your suggestion, I edited the title. |
Jul 20 |
revised |
Why is the double cover of $Sl(2,\mathbb{R})$ not algebraic?
added 10 characters in body; edited title; edited title |
Jul 20 |
asked | How does the right regular of GL(n, R) and GL(n,Qp) decompose? |
Jul 19 |
revised |
Discrete-compact duality for nonabelian groups
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Jul 19 |
revised |
Discrete-compact duality for nonabelian groups
added 272 characters in body; edited body |
Jul 19 |
answered | Discrete-compact duality for nonabelian groups |
Jul 17 |
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What is the relationship amongst all the different kinds of spectra?
The spectrum of a unital commutative $C^*$ algebra $A$ is defined such that $Hom_{top}(Spec(A), \mathbb{C}) = A$, the same holds for the spectra in homotopy theory with different hom. So I always understand this as a question about representable functors. Does this make sense? |
Jul 16 |
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Sigma Algebra that is not a topology
@Joel: Interesting, I edited the answer. |
Jul 14 |
accepted | To what extent do we know the representations of GL(2,Zp) |
Jul 14 |
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generalisation of GL(3,R) polar decomposition
@Carnahan: It is sometimes called polar decomposition, since $PSL_2(\mathbb{R}) / PSO(2)$ is the upper halfplane $H$, and the above decompositions give you the elements of $H$ in polar coordinates, but you are right in general. |
Jul 14 |
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generalisation of GL(3,R) polar decomposition
I just wanted to comment that on both sides $SO_K(3)$ works and I presumably thought you might had a typo in the question=) |
Jul 13 |
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generalisation of GL(3,R) polar decomposition
@em12: Btw, are sure it holds for $GL(2)$? |
Jul 13 |
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generalisation of GL(3,R) polar decomposition
I mean just use $\sqrt{K} m \sqrt{K}^{-1}$ and using the result for $SO(3)$. If $K$ has negative eigenvalues, this strategy does not apply, since there is a problem with $\sqrt{K}$. |
Jul 13 |
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generalisation of GL(3,R) polar decomposition
@Spice, I wanted to say, if $K$ induces a scalar product on $\mathbb{R}^3$, the decomposition result is the same, since all scalar products are the same, since there exists $\sqrt{K}$. Then every matrix $m$ in $End(\mathbb{R}^3)$ can be given as $m=o d o'$ for $d$ diagonal and $o,o'\in SO_K(3)$. |
Jul 13 |
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generalisation of GL(3,R) polar decomposition
psoitive definite symmetric matrix and $SO_K$ on both sides, than I'd say yes. |
Jul 13 |
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To what extent do we know the representations of GL(2,Zp)
Yes since all irreducible complex representation of $GL(2, o)$ factor through the finite group $GL(2,o/n)$ for some ideal $n$. Then using $Ind_B^G Ind_1^B 1 = Ind_1^G 1$ gives all irreducible representations. |
Jul 13 |
revised |
To what extent do we know the representations of GL(2,Zp)
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