Timeline for Is the metaplectic group not a matrix group - counterexample
Current License: CC BY-SA 3.0
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| Dec 7, 2015 at 10:17 | comment | added | Venkataramana | Theo Johnson Floyd has already $completely$ answered the question. The issue is the following: suppose $G$ is an algebraic group defined over $\mathbb R$ such that the complex algebraic group $G(\mathbb C)$ is simply connected. (Then $G(\mathbb R)$ is cnnected) and any continuous representation of the universal cover of $G(\mathbb R)$ factors through a representation of $G(\mathbb R)$. This is proved in Helgason's book. If we take $G=SO(n)$, then $G(\mathbb C)$ is not simply connected, so the above argument does not apply; if we take $G$ to be a real form of $Spin (n)$, then it does. | |
| Dec 7, 2015 at 9:40 | answer | added | Svat Krysl | timeline score: 0 | |
| Jan 16, 2013 at 17:15 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 8:40 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 8:26 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 8:14 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 7:45 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 7:26 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 7:26 | comment | added | Yemon Choi | -1 for persisting despite Theo's explanation (and any number of sources which state that the metaplectic group is not a matrix group) | |
| Jan 16, 2013 at 7:10 | comment | added | Jules Berracasa | Thanks kreck. I have altered the question. Now I need to show that indeed it is a subgroup of the full Pin group and not the Spin group. | |
| Jan 16, 2013 at 7:08 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 6:15 | vote | accept | Jules Berracasa | ||
| Jan 16, 2013 at 6:04 | comment | added | user29720 | @Alexander Schlering: You're just constructed the disconnected double cover of the symplectic group inside of the disconnected Pin group. This has nothing to do with the metaplectic group, and is not useful. Be more attentive to the importance of checking connectedness. | |
| Jan 16, 2013 at 5:46 | answer | added | Theo Johnson-Freyd | timeline score: 8 | |
| Jan 16, 2013 at 5:45 | comment | added | Jules Berracasa | Sorry about that. I hope this is better for everyone. | |
| Jan 16, 2013 at 5:43 | history | edited | Jules Berracasa | CC BY-SA 3.0 |
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| Jan 16, 2013 at 5:29 | comment | added | Theo Johnson-Freyd | This question is written in an argumentative tone of voice, which risks turning people off. I recommend that you revise it, taking into consideration advice from mathoverflow.net/howtoask. Among other things, you could post your putative faithful finite-dimensional representation. At the minimum, writing up your construction carefully will probably lead you to find an error. | |
| Jan 16, 2013 at 5:27 | comment | added | MTS | Perhaps you could include details of your proposed counterexample? | |
| Jan 16, 2013 at 5:02 | history | edited | Jules Berracasa |
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| Jan 16, 2013 at 4:53 | history | asked | Jules Berracasa | CC BY-SA 3.0 |