Timeline for Intertwining operator is not an isomorphism?
Current License: CC BY-SA 4.0
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| May 17, 2020 at 9:05 | comment | added | Monty | @Subhajit Jana, I mean if $M_w(s,σ)(f_s)$ has a pole at $s=s_0$ of order $r$, then $M_{1,w}(s,σ)(f_s)$ is defined by $\operatorname{lim}_{s \to s_0}(s-s_0)^rM_{1,w}(s,σ)(f_s)$. If $M_w(s,σ)(f_s)$ has a zero at $s=s_0$ of order $r$, then $\operatorname{lim}_{s \to s_0}(s-s_0)^{-r}M_{1,w}(s,σ)(f_s)$. That's what I am saying for the leading term. | |
| May 17, 2020 at 8:47 | comment | added | Subhajit Jana | Sorry I am not much familiar with this $M_{1,w}$ operator. Does it come from the Laurant expansion around $s=0$? How does the coefficient of the expansion depend on $s$ then? | |
| May 17, 2020 at 8:21 | comment | added | Monty | @Subhajit Jana, Right! $M_{w}(s,\sigma)$ may have a pole. That's why I introduced the operator $M_1$ in my question. Then $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$ does not hold when $M(s,\sigma)$ has a pole? I think this would have some sense in case $M_{w}(-s,w\cdot \sigma)$ has zeo. Have we only the equation $M_{w}(s,\sigma) \circ M_{w}(-s,w\cdot \sigma)=id$ where $M_{w}(s,\sigma) $ is holomorphic? | |
| May 17, 2020 at 8:01 | comment | added | Subhajit Jana | Intertwining operators are not holomorphic. They have (typically) simple poles at the ``points of reducibility''. For instance, for $\mathrm{SL}(2)$ they appear at integer $s$ and give rise to special representations (by sub or quotient representation). $M(s,\sigma) M(-s,w\sigma)=Id$ should be interpreted as if $M(s,\sigma)$ has a pole then $M(-s,w\sigma)$ has a zero, i.e. has a non-trivial kernel. | |
| May 17, 2020 at 7:26 | comment | added | LSpice | What book specifically refers to the non-trivial kernel? | |
| May 17, 2020 at 7:18 | comment | added | Monty | @GHfromMO, Oh sorry. I didn't write the definition of intertwining operator. I added its definition in my question. | |
| May 17, 2020 at 7:17 | history | edited | Monty | CC BY-SA 4.0 |
added 85 characters in body
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| May 17, 2020 at 6:28 | comment | added | GH from MO | For representations of a group $G$, the term "intertwining operator" simply means a linear map compatible with the $G$-action. Some intertwining operators are bijective, some are not. Homomorphism vs. isomorphism. | |
| May 17, 2020 at 5:18 | history | asked | Monty | CC BY-SA 4.0 |