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Please could someone check my work on this exercise (from book I am reading). Thanks!

Exercise:

Prove that $GL_n (\mathbb K)$ is non-compact when $n \ge 1$. Prove that $SL_n (\mathbb K)$ is non-compact when $n \ge 2$. What about $SL_1 (\mathbb K)$?

My solution:

Note that $nI \in GL_n$ and $\|nI\| = n \cdot \sqrt{n}$ where $\|\cdot\|$ denotes the Euclidean metric hence $GL_n$ is unbounded.

Let $D_n$ be the diagonal matrix with entries $d_{11} = n, d_{22}={1\over n}$ and $d_{ii}=1$ otherwise. Then $\det D_n = 1$ and $\|D_n\| \ge n$ hence $SL_n$ is unbounded.

$SL_1 (\mathbb R) = \{-1,1\}$ hence is bounded. $SL_1 (\mathbb C) = S^1$ hence is bounded. For the quaternions $\mathbb H$, $SL_1 (\mathbb H) = S^3$ hence is bounded.

learner
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  • That all looks good to me, at least for the cases you treat. I'm assuming you mean $\Bbb K\in {\Bbb R,\Bbb C,\Bbb H}$? – Adam Hughes Dec 16 '14 at 02:31
  • @AdamHughes Yes, that's right, $\Bbb K\in {\Bbb R,\Bbb C,\Bbb H}$. Thank you for your comment. – learner Dec 16 '14 at 02:37
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    Fyi, since $\mathrm{SL}_n(\mathbb K)$ is a subgroup of $\mathrm{GL}_n(\mathbb K)$ the non-boundedness of $\mathrm{SL}_n(\mathbb K)$ implies that $\mathrm{GL}_n(\mathbb K)$ is also non-bounded. So you can ditch the $nI$ example if you want, the $D_n$ example after it suffices. – Jim Dec 16 '14 at 02:38
  • @Jim That's right, thank you for pointing it out. – learner Dec 16 '14 at 02:42
  • Why proving that the groups aren't bounded is enough to prove they aren't compact? Exactly: because you're, apparently, taking the euclidean topology from $;\Bbb R^{n^2};$ (what is $;\Bbb K;$ , btw?), so perhaps it'd be worthwhile to mention this. – Timbuc Dec 16 '14 at 03:56
  • @Timbuc Good point, I should mention this. Thanks for pointing it out! – learner Dec 16 '14 at 04:02
  • $SL_1(\mathbb R)$ is not ${\pm 1}$; only ${1}$. – user5826 Sep 13 '19 at 04:52

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