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I think this question should be pretty straightforward but through a combination of being self-taught in group theory and being awful at geometry, something is escaping me. I am trying to understand and prove the following:

Let $G=SL(n,\mathbb{R})$ be an algebraic group and $H$ a subgroup of $G$ isomorphic to the group of integers $\mathbb{Z}^k$ and I know that the Zariski closure of $\mathbb{Z}^k$ is the commutative group $\mathbb{R}^k$. I want to prove that the Zariski closure of $H$ has a commutative Lie algebra of dimension $k$ ( i-e., is the Lie algebra of $\bar{H}$ commutative? ).

Shaun
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Yushi MuGiwara
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    If you say that $\overline{H}=\Bbb R^k$, then of course its Lie algebra is commutative, i.e., has zero product, by definition. – Dietrich Burde Mar 15 '23 at 19:45
  • @DietrichBurde, We know that $H$ is isomorphic to $\mathbb{Z}^k$, is it ture that $\bar{H} = \bar{\mathbb{Z}^k}$? if yes how can I prove it? – Yushi MuGiwara Mar 15 '23 at 19:50
  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun Mar 15 '23 at 20:54
  • @Shaun 'provide context' in what sense? the question is precise and clear, you want me to formulate the question in a complicated way to be accepted? – Yushi MuGiwara Mar 15 '23 at 21:04
  • I mean that you need to give details of your background and of what attempts you have made on the question so far. This is standard here. Consider answering the following: What are you studying? What text is this drawn from, if any? If not, how did the question arise? What kind of approaches (to similar problems) are you familiar with? What kind of answer are you looking for? Basic approach, hint, explanation, something else? Is this question something you think you should be able to answer? Why or why not? – Shaun Mar 15 '23 at 21:16
  • It is really unclear what are you confused about. The first comment gives a trivial answer to your question. Your first comment is irrelevant for the hint (and also does not make sense since the closure is defined for subsets of algebraic sets and is undefined for abstract groups such as $Z^k$). – Moishe Kohan Mar 16 '23 at 00:35
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    @Shaun, I am sorry, but your comment helps no one. The question is a perfectly fine one, whose point is perfectly understandable by anyone who can answer it. – Mariano Suárez-Álvarez Mar 16 '23 at 04:08
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    That a comment gives (what to someone who knows the subject well is a trivial) answer applies to 78.12% of the questions in the site... – Mariano Suárez-Álvarez Mar 16 '23 at 04:12
  • Yes, my question may seem basic or dumb to someone who is well-educated. For me, I just wanna learn mathematics here and I am sorry if my questions don't fit the standards of this community. – Yushi MuGiwara Mar 16 '23 at 14:51
  • It's alright, @YushiMuGiwara; just be more careful next time. – Shaun Mar 16 '23 at 15:03

1 Answers1

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The Lie algebra of a commutative group is commutative. See for example answers to this question: link.

spin
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  • This question seems not to meet the standards for the site. Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments suggesting improvements. Please also read the meta announcement regarding quality standards. – Shaun Mar 16 '23 at 10:11
  • @spin I know this statement thanks, I found what I was looking for. – Yushi MuGiwara Mar 16 '23 at 14:39