Timeline for Schur lemma and Whittaker functions
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2021 at 17:45 | history | became hot network question | |||
| Nov 15, 2021 at 16:19 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 15, 2021 at 11:47 | comment | added | GH from MO | The translates $\pi\left(\begin{smallmatrix}a& \\ &1\end{smallmatrix}\right)v_0$ span an infinite dimensional space, so the representation of $GL_1(\mathbb{Q}_p)$ on it is not irreducible. Infinite dimensionality is clear if you look at the support of each translate within the Kirillov model, which is the Whittaker model restricted to your subgroup $GL_1(\mathbb{Q}_p)$. See (6.12) on page 500 in Bump's book to get an idea of the support of the spherical vector $v_0$ in the Kirillov model. Then see what happens for $\pi\left(\begin{smallmatrix}a& \\ &1\end{smallmatrix}\right)v_0$. | |
| Nov 15, 2021 at 11:41 | vote | accept | Aersk | ||
| Nov 15, 2021 at 11:33 | answer | added | Spencer Leslie | timeline score: 9 | |
| Nov 15, 2021 at 11:26 | comment | added | Ehud Meir | I assume that you meant that $V_{v_0}$ is the span of these vectors? Even in this case, as other people have written here, the fact that it is a cyclic representation does not mean that it is irreducible. | |
| Nov 15, 2021 at 11:19 | comment | added | Aurel | The set $V_{v_0}$ as you define it is still not a vector space (not stable under linear combinations!). In my comments I interpreted it as the vector space generated by the set your wrote. | |
| Nov 15, 2021 at 11:18 | comment | added | Spencer Leslie | Are you assuming $v_0$ is a spherical vector? That might be the issue: the spherical vector is scaled by a character when acted on by compact diagonal elements, but not the full torus. The Whittaker formula (a special case of the Casselman-Shalika-Shintani formula) tells you how the full torus acts. | |
| Nov 15, 2021 at 11:17 | comment | added | Aersk | There was a typo in the definition of $V_{v_0}$, now i have corrected it. What is the statement about units? | |
| Nov 15, 2021 at 11:16 | comment | added | Aurel | You seem to be claiming that a cyclic representation (i.e. generated by a single element) is always irreducible. This is clearly false: think about the regular representation of $C_2$ (the cyclic group of order $2$). | |
| Nov 15, 2021 at 11:15 | comment | added | Spencer Leslie | I don’t understand your first claim. What is $V_{v_0}$ supposed to be? As currently written, it isn’t a vector space. Even once that is clarified, I don’t get the irreducibility argument. What does the statement about units tell you? | |
| Nov 15, 2021 at 11:14 | comment | added | Aersk | Because the vector space $V_{v_0}$ is given by the translates of $v_0$. This representation does not admit a subrepresentation. For example given elements $v_1,\;v_2\in V_{v_0}$ by definition we can express $v_1 = \pi\begin{pmatrix}a_1& \\ & 1\end{pmatrix}v_0$ and $v_2 = \pi\begin{pmatrix}a_2& \\ & 1\end{pmatrix}v_0$. Therefore $\pi\begin{pmatrix}a_1^{-1}a_2& \\ & 1\end{pmatrix}v_1 = v_2$. | |
| Nov 15, 2021 at 11:12 | history | edited | Aersk | CC BY-SA 4.0 |
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| Nov 15, 2021 at 11:02 | comment | added | Aurel | Hi! Why should your representation of $GL_1(\mathbb{Q}_p)$ be irreducible? | |
| Nov 15, 2021 at 9:41 | history | asked | Aersk | CC BY-SA 4.0 |