Timeline for Examples of locally compact groups that do not admit enough finite dimensional representations
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2021 at 13:41 | comment | added | Rick Sternbach | I am now wondering what happens if we require the group to be simply connected. Do such groups always admit enough finite dimensional representations then? I am also curious about a "state of the art" survey of the topic of residually finite groups. Should I ask about these things in a separate question? | |
| S Dec 10, 2021 at 13:35 | vote | accept | Rick Sternbach | ||
| S Dec 10, 2021 at 13:35 | vote | accept | Rick Sternbach | ||
| S Dec 10, 2021 at 13:35 | |||||
| Dec 10, 2021 at 13:34 | vote | accept | Rick Sternbach | ||
| S Dec 10, 2021 at 13:35 | |||||
| Dec 7, 2021 at 19:52 | comment | added | YCor | OK, you've now addressed it. I think you mean "normed field". Also you don't have to exclude discrete normed fields, because it does not modify the question (since any discrete field $K$ embeds into the non-discrete fiel $K(\!(t)\!)$ as discrete subfield, and in any case it's usually "more difficult" to map continuously into $\mathrm{GL}_n$ of a discrete field. | |
| Dec 7, 2021 at 19:18 | history | edited | Rick Sternbach | CC BY-SA 4.0 |
make the meaning of "weaker form" precise.
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| Dec 7, 2021 at 17:26 | comment | added | YCor | You've ignored my previous comment while editing: I explained that the "weaker form" is not weaker. Unless you mean by "valued field" something other that what I imagine (which is in particular totally disconnected). | |
| Dec 7, 2021 at 16:27 | history | edited | Rick Sternbach | CC BY-SA 4.0 |
deleted the redundant "actually" in the last sentence
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| Dec 7, 2021 at 9:37 | answer | added | Wojowu | timeline score: 12 | |
| Dec 7, 2021 at 9:01 | answer | added | AlexIvanov | timeline score: 1 | |
| Dec 7, 2021 at 8:59 | history | became hot network question | |||
| Dec 7, 2021 at 8:32 | comment | added | YCor | As regards the difference between $K_G$ and $K'_G$: it seems to me that $K_G=G$ for every connected group $G$, since every valued field is totally disconnected (if I understand what you mean by valued field). While $K_G=\{1\}$ for many connected Lie groups. At the opposite, for $\mathrm{SL}_n(\mathbf{Q}_p)$ we have $K_G=G$ and $K'_G=\{1\}$. | |
| Dec 7, 2021 at 8:28 | comment | added | YCor | The object to understand, for a group $G$, is the intersection $K_G$ of all kernels of all finite-dimensional continuous representations over all valued fields (and similarly $K'_G$ defined considering finite-dimensional representations). You're asking when $K_G=\{1\}$ or $K'_G=\{1\}$. For instance for compact groups $G$, Peter-Weyl ensures $K'_G=\{1\}$. On the other hand, if $G$ is a finitely generated group then its "finite residual" (the intersection of all its finite index subgroups) is contained in $K_G\cap K'_G$. In particular if $G$ has no nontrivial finite quotient, $K_G=K'_G=G$. | |
| Dec 7, 2021 at 8:24 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Dec 7, 2021 at 1:57 | comment | added | Ben Wieland | $SL_2\mathbb C$ is simply connected, but $SL_2\mathbb R$ has infinite fundamental group. Its double cover is a counterexample. It is connected, so it doesn't have any $p$-adic representations. Its finite dimensional real reprentations yield representations of its Lie algebra, which complexify to complex representations of $SL_2\mathbb C$, which doesn't have a double cover so can't detect the kernel of the extension. | |
| Dec 7, 2021 at 1:55 | comment | added | Terry Tao | I didn't check carefully but I suspect Higman's group will serve as a (discrete) counterexample in the $k={\bf C}$ case. en.wikipedia.org/wiki/Higman_group | |
| Dec 7, 2021 at 1:48 | answer | added | Izaak Meckler | timeline score: 9 | |
| Dec 7, 2021 at 0:53 | history | asked | Rick Sternbach | CC BY-SA 4.0 |