bio | website | plusepsilon.de |
---|---|---|
location | Uni Hamburg | |
age | 29 | |
visits | member for | 4 years |
seen | Sep 25 at 13:24 | |
stats | profile views | 8,328 |
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
added 27 characters in body |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
generalisation of GL(3,R) polar decomposition
psoitive definite symmetric matrix and $SO_K$ on both sides, than I'd say yes. |
Jul 13 |
comment |
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)
added 18 characters in body |
Jul 13 |
revised |
To what extent do we know the representations of GL(2,Zp)
added 8 characters in body |
Jul 13 |
asked | To what extent do we know the representations of GL(2,Zp) |
Jul 12 |
revised |
Sigma Algebra that is not a topology
deleted 62 characters in body |