Timeline for Property of lattices in Lie groups
Current License: CC BY-SA 3.0
12 events
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| Mar 4, 2013 at 14:35 | history | edited | Venkataramana | CC BY-SA 3.0 |
added 14 characters in body
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| Mar 4, 2013 at 12:27 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Mar 4, 2013 at 12:01 | history | edited | Venkataramana | CC BY-SA 3.0 |
ades a summary
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| Feb 25, 2013 at 8:16 | history | edited | Venkataramana | CC BY-SA 3.0 |
edited body
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| Feb 25, 2013 at 7:12 | comment | added | Venkataramana | @J.Martel, let us consider (for the sake of simplicity) the case $\Sigma \subset SL_3({\mathbb Z})$ of finite index. Strong approximation says essentially, that for all but a finite set of primes, the composite map $\Sigma \subset SL_3({\mathbb Z})\rightarrow SL_3({\mathbb F}_p)$ is a $surjection$. This is a non-trivial statement in general (but easily proved for finite index subgroups $\Sigma $ of $SL_3({\mathbb Z})$. If $\Sigma$ contains all $n$-th powers, this says then that the $n$-th power map on the finite group is surjective. By choosing suitable primes, we can see that this can't b | |
| Feb 24, 2013 at 22:38 | comment | added | JHM | I know nothing about strong approximation or profinite completion; what does the hypothesis on $\Sigma$ being finite index amount to in your example? | |
| Feb 24, 2013 at 12:37 | comment | added | Venkataramana | in the last comment, "divisible by $p$" is to be replaced by "divisible by $n$". | |
| Feb 24, 2013 at 12:18 | comment | added | Venkataramana | If the $n$-th power map is surjective on the finite group $G_p=SL_3({\mathbb F}_p)$, then it is also injective; by Dirichlet's theorem on infinitude of primes in arith progressions, there are infinitely many primes such that $G_p$ has order divisible by $p$; then the n-th power map cannot be injective. | |
| Feb 24, 2013 at 9:36 | comment | added | Venkataramana | In general, suppose $\Gamma = G({\mathbb Z})$ is a "higher rank" arithmetic group. You need only work with the congruence completion. If for some $n$, such a $\Sigma$ exists, then for the same reasons as above, the $n$-th power map on $G({\mathbb F}_p)$ would be surjective, for almost all $p$ by strong approximation for $\Sigma$. But you can get anisotropic tori in $G({\mathbb F}_p)$ to replace the cubic extension in the above. | |
| Feb 24, 2013 at 9:36 | vote | accept | CommunityBot | ||
| Feb 24, 2013 at 9:24 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Feb 24, 2013 at 9:05 | history | answered | Venkataramana | CC BY-SA 3.0 |