Timeline for Does $SL_3(R)$ embed in $SL_2(R)$?
Current License: CC BY-SA 3.0
16 events
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| Apr 14, 2016 at 7:09 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 7, 2016 at 17:52 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 7, 2016 at 17:47 | comment | added | Uri Bader | I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you @Andrei Smolensky for repeating correcting my embarrassing mistakes here. Thank you also Max Horn for your correction. | |
| Apr 7, 2016 at 17:45 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 7, 2016 at 13:27 | comment | added | Andrei Smolensky | $SL(3, R)$ contains an epimorphic image of $SL(3, \mathbb{Z})$, which is still perfect. Concerning the edits: $SK_1(R)$ is not known to be always nilpotent (it most likely is not). | |
| Apr 7, 2016 at 13:16 | comment | added | Max Horn | Oops, I meant to write: "... assume that R has characteristic 0, ..." | |
| Apr 7, 2016 at 12:16 | comment | added | Max Horn | Your answer seems to implicitly assume that $R$ is infinite, else your starting claim that $SL_3(R)$ contains $SL_3(\mathbb{Z})$ is false (e.g. if $R$ is finite). | |
| Apr 7, 2016 at 6:48 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 7, 2016 at 5:43 | comment | added | Uri Bader | Thanks, @Andrei Smolensky, I was careless here and in some other places. I will edit my answer soon. | |
| Apr 6, 2016 at 22:43 | comment | added | Andrei Smolensky | Great answer! I would just note that $\mathrm{SK}_1(n, R)=\mathrm{SL}(n, R)/\mathrm{E}(n, R)$ is not always abelian. In fact, it can have arbitrary large nilpotency degree for a ring of finite Bass—Serre dimension. | |
| Apr 6, 2016 at 21:28 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 6, 2016 at 10:17 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 6, 2016 at 10:08 | history | edited | Uri Bader | CC BY-SA 3.0 |
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| Apr 6, 2016 at 9:19 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
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| Apr 6, 2016 at 9:14 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
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| Apr 6, 2016 at 9:10 | history | answered | Uri Bader | CC BY-SA 3.0 |