Timeline for Can a reductive group act non-linearly on a vector group?
Current License: CC BY-SA 2.5
17 events
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| Jul 23, 2010 at 20:01 | comment | added | George McNinch | Well, those remarks sound negative, but I think you have made a good suggestion. My feeling before asking was that there was probably a counter-example that I somehow didn't know. Now you've made me optimistic that the answer to the "refined question" is yes. | |
| Jul 23, 2010 at 19:59 | comment | added | George McNinch | ...finally (3) I'm not sure I see why the image of $G$ should be in any complement to $R$ in GL$_n(k[F])$. It is possible to have a semidirect product $H = L \cdot R$ with a unique conjugacy class of complements to $R$ and a subgroup $G < H$ which intersects $R$ trivially for which $G$ is not a subgroup of any conjugate of $L$ -- see e.g. the example in 3.3 of front.math.ucdavis.edu/1007.2777 | |
| Jul 23, 2010 at 19:54 | comment | added | George McNinch | @Marty: Further comments... (1) My instincts say GL$_n(k[[F]])$ is pro-affine, but not GL$_n(k[F])$. Is this right? is it related to your exchange with Speyer yesterday? (2) The kernel of GL$_n(k[F]/F^2) \to $GL$_n(k)$ probably identifies with the Frobenius twist of $\mathfrak{gl}_n$ as a module, so there may be non-vanishing $H^1$ and hence no conjugacy of Levis. (But despite my comment yesterday, it isn't clear that full conjugacy is req'd (just need image of $G$ centralized by "scalars").) | |
| Jul 22, 2010 at 21:56 | history | edited | Marty | CC BY-SA 2.5 |
Clarifying fix.
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| Jul 22, 2010 at 21:49 | comment | added | Marty | @George: Uh oh. You're right. We might need a "conjugacy of Levi" theorem, in the pro-setting, to get this to work out. From your recent work, I think you're the expert on this part! I agree that it's problematic, and I'll edit my post accordingly. | |
| Jul 22, 2010 at 21:11 | comment | added | George McNinch | Let $R$ be the kernel of $\operatorname{GL}_n(k[F]) \to \operatorname{GL}_n(k)$. Your argument presumably shows that there are no maps $G \to R$ for reductive $G$. I don't see the next step, though. One needs to see that the image of $G$ is conjugate to a subgroup of $\operatorname{GL}_n(k) \subset \operatorname{GL}_n(k[F])$; why is that? | |
| Jul 22, 2010 at 20:55 | history | edited | Marty | CC BY-SA 2.5 |
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| Jul 22, 2010 at 19:50 | comment | added | Marty | That's what my "actually" comment speaks too. I think you're correct and it's simpler to just write $A \in Mat_n(k[F])^\times$. Glad this works out - I don't work with unipotent groups in $char = p$ very often. | |
| Jul 22, 2010 at 19:44 | comment | added | George McNinch | Hi Marty, this looks great. Somehow I believed that $\operatorname{Mat}_n(k[F])^\times$ was the automorphism group, but didn't know to expect anything about pro-unipotence of the kernel. How is the "power series" condition on $A$ different from just requiring $A \in \operatorname{Mat}_n(k)[F]^\times = \operatorname{Mat}_n(k[F])^\times$? Or is that what your subsequent comment ("Actually - ...") speaks to? | |
| Jul 22, 2010 at 19:14 | comment | added | David E Speyer | Oh, I see. I hadn't noticed that you had single brackets in some places and double brackets in others. In that case, I agree that it isn't obvious. Fortunately, as you point out, we don't need to care about this point. | |
| Jul 22, 2010 at 19:12 | comment | added | Marty | Actually - that sounds dumb. I guess these are just invertible polynomials in the noncommutative polynomial ring. I just copied from Tanaka-Kaneta without thinking. | |
| Jul 22, 2010 at 19:10 | comment | added | Marty | The point that I got from Tanaka-Kaneta is that one must work with power series which are polynomials in $F$, and whose inverse is also polynomial in $F$. That may imply some nilpotence condition on the coefficient matrices. | |
| Jul 22, 2010 at 19:08 | comment | added | David E Speyer | I have no idea whether it is written down, but isn't it clear? You just want to show that, for any $n \times n$ matrix $A$, the element $1-A F^j$ is a unit. It's inverse is $1+A F^j + A Fr^j(A) F^{2j} + A Fr^j(A) Fr^{2j}(A) F^{2j} + \cdots$. | |
| Jul 22, 2010 at 18:56 | comment | added | Marty | @David: How can you be sure that the successive quotients are $G_a^{n^2}$? Is this written down somewhere? To me, it seems clear that the quotients are closed subgroups of $G_a^{n^2}$ -- sufficient for the pro-unipotence result -- but not that the quotients are the "whole thing". | |
| Jul 22, 2010 at 18:42 | history | edited | Marty | CC BY-SA 2.5 |
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| Jul 22, 2010 at 18:06 | comment | added | David E Speyer | Hi Marty! You can filter $Aut(G_a^n)$ with $G^j$ being the group of such power series which are the identity modulo $F^{j}$. Then $G^0/G^1$ is $GL_n(k)$, and $G^j/G^{j+1}$ is $G_a^{n^2}$ for higher $j$. If we were dealing with finite dimensionsal group schemes, that would show $G$ is unipotent. I don't know, though, what subtleties may exist in the pro-unipotent case. | |
| Jul 22, 2010 at 17:47 | history | answered | Marty | CC BY-SA 2.5 |