Timeline for Baruch's proof of Kirillov's conjecture
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2012 at 15:13 | comment | added | GH from MO | This condition is included in the Harish-Chandra class of which $\mathrm{GL}_n(\mathbb{R})$ is a member. | |
| Jan 27, 2012 at 14:42 | comment | added | GH from MO | @Alain: I learned more about $Z(\mathcal{U}\mathfrak{g})$, where $\mathfrak{g}$ is the Lie algebra of $G$. It turns out that $Z(\mathcal{U}\mathfrak{g})$ equals the set of $G'$-invariants of $\mathcal{U}\mathfrak{g}$, where $G'$ denotes the subgroup of $\mathrm{Aut}(\mathfrak{g})$ generated by $\exp(\mathrm{ad}(x))$, $(\mathrm{ad}(x)$ nilpotent, $x\in\mathfrak{g}$. In particular, when the image of $\mathrm{Ad}:G\to\mathrm{Aut}(\mathfrak{g})$ lies in $G'$, we have that $Z(\mathcal{U}\mathfrak{g})$ is fixed by $\mathrm{Ad}(G)$. I continue in next comment. | |
| Jan 23, 2012 at 22:28 | comment | added | GH from MO | Actually, the center of $\mathcal{U}(\mathfrak{gl}(n,\mathbb{R}))$ lies in the center of $\mathcal{U}(\mathfrak{gl}(n,\mathbb{C}))$, because if an element of $\mathcal{U}(\mathfrak{gl}(n,\mathbb{C}))$ commutes with $\mathfrak{gl}(n,\mathbb{R})$, then it also commutes with $\mathfrak{gl}(n,\mathbb{C})$. So the argument in Goldfeld-Hundley, which is an elaboration of Alain's idea, justifies the original response above and takes care of Faisal's concerns as well. | |
| Jan 20, 2012 at 12:26 | comment | added | GH from MO | @Alain: It really isn't :-( I come from classical analytic number theory, and try to catch up with representation theory. | |
| Jan 20, 2012 at 6:59 | comment | added | Alain Valette | @GH And you claim that representation theory is not your expertise! (:-) | |
| Jan 20, 2012 at 0:33 | comment | added | GH from MO | @Alain: In the spirit of your second response, one can see that $d\pi(X)$ acts by scalars whenever $X$ lies in the center of $\mathcal{U}(\mathfrak{gl}(n,\mathbb{C}))$. This is because this (smaller) center is generated by $n$ symmetric Casimir-like operators, see Prop.12.3.15 in Goldfeld-Hundley: Automorphic representations and L-functions for the general linear group, Volume 2. I read somewhere that the same feature holds for any real reductive group (in which case $\mathfrak{gl}(n,\mathbb{C})$ is replaced by the complexified Lie-algebra), but I could not find a reference so far. | |
| Jan 14, 2012 at 1:25 | comment | added | GH from MO | @Alain, I tried again to complete your argument. Please read and check my response below. Thank you. | |
| Jan 13, 2012 at 10:14 | comment | added | Alain Valette | @GH: $\exp(G)$ is contained in the connected component of identity, and contains a neighborhood of identity, hence generates the connected component of identity. So if the group is connected (e.g. $GL_n(\mathbb{C})$, as mentioned in the OP), your argument + Schur's lemma do the job. Be aware that you take $X$ in the Lie algebra $\mathfrak{g}$, not in the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$. Let me also correct a sign: to be coherent with the left regular representation, you should define $(Xf)(g)$ as the derivative at $t=0$ of $f(e^{-tX}g)$. | |
| Jan 13, 2012 at 9:07 | comment | added | GH from MO | @Alain: In my last comment I attempted to complete your argument. Please tell me if it is OK. I am implicitly using that $\exp(G)$ contains the component of the identity. | |
| Jan 13, 2012 at 8:50 | comment | added | Alain Valette | @Faisal: You made a very good point: I overlooked the fact that $GL_n(\mathbb{R})$ is not connected (and connectedness is important, as GH's computation shows). If the restriction of $\pi$ to $GL_n^+(\mathbb{R})$ (=matrices with positive determinant) is still irreducible, then my argument goes through. If $n$ is odd, this will be the case (since $GL_n(\mathbb{R})$ is then a direct product). The case of even $n$ remains unclear to me. | |
| Jan 13, 2012 at 8:35 | comment | added | GH from MO | I completed the argument, please check. Assume that $f$ is supported on the coset $h\exp(G)$, so that the left translate $L(h^{-1})f$ has support in $\exp(G)$. By Schur's lemma we have $$ \pi(h)\pi(Xf) = \pi(L(h^{-1}) Xf) = \pi(X L(h^{-1}) f) = d\pi(-X) \pi(L(h^{-1}) f)=d\pi(-X) \pi(h)\pi(f). $$ It follows that $\pi(Xf)=d\pi(-X) \pi(f)$ for any $h$, whence for any $f\in C_c^\infty(G)$. Now the argument can be finished as before. | |
| Jan 13, 2012 at 5:18 | comment | added | GH from MO | @Alain: Thank you. At the moment I only see that $X$ commutes with $\pi(g)$ for $g\in\exp(G)$. Also, I think $\Lambda_R(Xf)=d\pi(-X)\Lambda_R(f)$. Indeed, the left hand side is the $t$-derivative at $t=0$ of the operator $\int_G f(e^{tX}g)\pi(g)\ dg$, using that $X$ commutes with $\pi(g)$, and the last integral equals $\int_G f(g)\pi(e^{-tX}g)\ dg$. Am I missing something? | |
| Jan 12, 2012 at 18:21 | comment | added | Faisal | I don't think "Schur's lemma" is completely obvious in this setting either, especially when $G = GL_n \mathbb R$ is disconnected. | |
| Jan 12, 2012 at 16:40 | comment | added | Alain Valette | @Paul: Right, this is really not trivial. On the other hand, the OP took it for granted. | |
| Jan 12, 2012 at 13:44 | comment | added | paul garrett | And for this to make sense, $\pi(f)$ is trace-class, which is not obvious. | |
| Jan 12, 2012 at 6:53 | history | answered | Alain Valette | CC BY-SA 3.0 |