Timeline for A convergence problem about integral operator in the space of representations
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2012 at 0:23 | comment | added | GH from MO | @unknown: I think if you want $\pi(f)$ act on the set of smooth vectors in a unitary representation (i.e. $\pi(f)v$ not only exists in $V$ but it is also a smooth vector), then you need that $L_X f\in L^1(G)$ for any $X\in\mathfrak{g}$. This is because for $f\in C_c^\infty(G)$ we have that $\pi(X)\pi(f)v=\pi(L_X f)v$. | |
| Jan 16, 2012 at 12:00 | comment | added | user1832 | Alain: Of course we can always add zero seminorm. And the question I asked is that among all the seminors there, trivial or nontrivial, can we just pick up one particular one , and check the condition for it, then conclude that $\pi(f)v$ exists in $V$? Or we need to check the condition for all seminorms? | |
| Jan 16, 2012 at 11:57 | comment | added | user1832 | Alain: Thanks for your comment. It should be $\pi(f)v$ converges to a vector in V for any $v\in V$. Sorry for the confusion. | |
| Jan 16, 2012 at 10:58 | comment | added | Alain Valette | @unknown: I don't understand what you mean by "$\pi(f)$ converges to certain vector in $V$", as this mixes objects of different nature. Now remember that we can always add the zero seminorm to the family of seminorms defining the topology of $V$. For this one, the condition $|\pi(f)v|_\mu <\infty$ is trivial, so you can get no satisfaction with just one seminorm. | |
| Jan 16, 2012 at 1:52 | comment | added | GH from MO | I extended my response to include more general $f$'s. I don't think the integral makes sense for every $f\in L^1(G)$, one needs some "decay at infinity", see below. | |
| Jan 16, 2012 at 0:12 | answer | added | GH from MO | timeline score: 1 | |
| Jan 15, 2012 at 20:10 | history | edited | Yemon Choi |
Added fa tag
|
|
| Jan 15, 2012 at 17:43 | comment | added | user1832 | Thanks for your comment. Here I'm more interested in if $\pi(f)$ converges to certain vector in $V$. Do we need the continuity prior to this question? | |
| Jan 15, 2012 at 16:43 | comment | added | Alain Valette | There are two things. First you need to show that the integral for $\pi(f)v$ converges, usually in the weak sense (i.e. $\int_G f(g)h(\pi(g)v)dg$ exists in $\mathbb{C}$ for every $h\in V'$, the topological dual of $V$). Then you need to show that $\pi(f)$ is continuous as an operator $V\rightarrow V$, i.e. for every seminorm $\mu$ on $V$, there is a constant $C>0$ and seminorms $\nu_1,...,\nu_k$ such that $|\pi(f)v|_\mu \leq C.\max_{1\leq i\leq k} |v|_{\nu_i}$. | |
| Jan 15, 2012 at 15:07 | history | asked | user1832 | CC BY-SA 3.0 |