User murat güngör - MathOverflow

How do these two Haar measures on SL(2,R) compare?

2013-05-10T09:32:31Z

By using the Iwasawa decomposition, one obtains a (bi-invariant) Haar measure on $G:=\mathrm{SL}(2,\mathbb{R})$ which can be symbolically written as $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, the measures appearing on the right-hand side being the usual ones. This actually means $$\int_G f(x)\,\mathrm{d}x=\int_K \int_N \int_A f(ank)\,\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$$ for suitable $f$. (Here the order in $ank$, which is expressed by the notation $\mathrm{d}x=\mathrm{d}a\,\mathrm{d}n\,\mathrm{d}k$, is important, whereas the order of the triple integral is immaterial.) Analogously, one can define another Haar measure $\mathrm{d}_1 x=\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a$ so that $$\int_G f(x)\,\mathrm{d}_1 x=\int_A \int_N \int_K f(kna)\,\mathrm{d}k\,\mathrm{d}n\,\mathrm{d}a.$$ We must have $\mathrm{d}_1 x=c\cdot\mathrm{d}x$ for some $c>0$. Is $c$ equal to 1?

Are $K$-finite vectors dense in irreducible Banach representations?

2013-04-21T18:10:29Z

Let $\pi$ be a continuous irreducible representation of $G:=\mathrm{SL}(2,\mathbb{R})$ in a Banach space $H$, and $\pi^1$ the representation of $\mathcal{C}_c(G)$ induced by $\pi$. Suppose $\mathcal{S}$ is an $\mathrm{L}^1$-dense subspace of $\mathcal{C}_c(G)$. Why does $\mathcal{S}$-invariance imply $\mathcal{C}_c(G)$-invariance for every closed subspace of $H$? In his book on $\mathrm{SL}(2,\mathbb{R})$, Serge Lang makes use of this in the proof of the fact that $K$-finite vectors are dense, $K$ denoting $\mathrm{SO}(2)$ as usual (Theorem 3, p. 24). The answer is clear when $\pi$ is a bounded (in particular, when $H$ is a Hilbert space and $\pi$ is unitary), since in this case $\pi^1$ has a continuous extension to $\mathrm{L}^1(G)$.

Does every irreducible Banach representation admit a $K$-finite vector?

2013-04-21T08:18:19Z

Let $G=\mathrm{SL}(2,\mathbb{R})$ and $K=\mathrm{SO}(2)$. Suppose $\pi$ is a continuous irreducible representation of $G$ in a Banach space $H$. Can one always find a nonzero $v\in H$ such that $\langle \pi(k)v : k\in K\rangle$ is finite-dimensional? Serge Lang uses this implicitly in the proof of Theorem 3 on p. 24 of his book on $\mathrm{SL}(2,\mathbb{R})$. (I know that the answer is affirmative when $H$ is a Hilbert space and $\pi$ is unitary on $K$.)

Do approximately the same polynomials have approximately the same roots?

2012-08-03T15:28:03Z

"If $U$ is an open subset of the complex plane, then matrices $X\in\textrm{M}(n,\mathbb C)$ all of whose eigenvalues belong to $U$ make up an open subset of $\textrm{M}(n,\mathbb C)$." Trying to prove this by using the argument principle, I was led to the following question: do approximately the same polynomials have approximately the same roots? More precisely, fix a complex polynomial $p(z)=a_0+a_1 z+\ldots+a_n z^n$ with $a_n\neq 0$. Let us say that a polynomial $q(z)=b_0+b_1 z+\ldots+b_n z^n$ is in the $\varepsilon$-neighborhood of $p(z)$ if $|b_i-a_i|<\varepsilon$ for all $0\leq i\leq n$. Now suppose all roots of $p(z)$ lie in, say, the open unit disk. Does $p(z)$ have an $\varepsilon$-neighborhood consisting of polynomials with the same property?

"No Small Subgroups" Argument

2012-08-02T14:08:10Z

What is the "no small subgroups" argument for $GL(n,\mathbb R)$? That is, how do we show that in $GL(n,\mathbb R)$ there exists a neighborhood of the identity which contains no subgroup other than the trivial one? I had some scribbling (for the $n=2$ case) but could not arrive at a clean proof.

A technical problem on the contragredient representation in the context of locally compact totally disconnected groups

2011-11-09T15:27:23Z

Let $\pi$ be an admissible representation of a locally compact totally disconnected group. I have a technical problem about the proof of

$\pi$ is irreducible if and only if its contragredient is so

given in 2.15(c) of the '76 article of Bernstein and Zelevinsky. There $\pi$ is assumed to have a nontrivial proper subrepresentation $E_1$, and it is asserted that the orthogonal complement of $E_1$ be a nontrivial proper subrepresentation of the contragredient, whence the result. What I cannot figure out is the nontriviality of this orthogonal complement. We simply have to find a nonzero smooth functional which vanishes on $E_1$; this shall follow from $E_1\neq E$ (as properness of the orthogonal complement follows from $E_1\neq 0$), but how?

Comment by Murat Güngör

2013-04-21T18:23:01Z

Thank you for your answer, professor. The argument you give essentially reduces my question to the following one: mathoverflow.net/questions/128268/….

Comment by Murat Güngör

2012-08-04T12:14:34Z

A typo: you are to claim that $2v\in V$, not $2v\in \exp(V)$.

Comment by Murat Güngör

2012-08-04T11:18:04Z

It simply solves the more-precisely-formulated question above, which generalizes at once to give a proof of the proposition in quotation marks.

Comment by Murat Güngör

2012-08-03T14:57:51Z

Qiaochu, you say that 1 has a neighborhood consisting of matrices having eigenvalues close to 1; do you have a simple proof of this?