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Let $L$ be a Lie algebra, and let $a\in [L,L]$. How to prove that $trace(ad_a)=0$?

Ronald
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  • This is a recent duplicate. Give me a second. – Qiaochu Yuan Feb 02 '16 at 02:39
  • @QiaochuYuan: Were you perhaps thinking of your answer to this Question? – hardmath Feb 02 '16 at 03:57
  • @hardmath: no, someone asked this exact question recently. It might've been deleted though. – Qiaochu Yuan Feb 02 '16 at 04:31
  • $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, and the fact that $ad:L\to\mathfrak{gl}_n$ is a homomorphism. –  Feb 02 '16 at 05:34
  • @QiaochuYuan I also remember that we have answered the same question recently, and I think, also posted by the same user (who might have deleted the question along with both answers). – Dietrich Burde Feb 02 '16 at 08:01
  • I don't remember that I asked this question before.. And I didn't delete any question. It is not possible actually to delete a question with answers – Ronald Feb 02 '16 at 08:06
  • Ronald: sorry, I might be mistaken. But the question definitely was there a few days ago, with tag "Lie algebras". Can you find it ? – Dietrich Burde Feb 02 '16 at 08:09
  • No problem. But you may mean this question .. Which is asked by me and still active http://math.stackexchange.com/questions/1625573/3-dim-simple-complex-lie-algebra – Ronald Feb 02 '16 at 08:16
  • No, I mean the very same question. My answer was: $ad(L)=ad([L,L])=[ad(L),ad(L)]$. It might have been a user who was removed upon request. – Dietrich Burde Feb 02 '16 at 08:18
  • I think your answer is nice, why you don't add it here . interested people can see different answers – Ronald Feb 02 '16 at 08:27
  • Ronald, thanks for the compliment! I cannot add an answer, because the question is on hold. – Dietrich Burde Feb 02 '16 at 16:28

1 Answers1

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Hint: If $x=[a,b]$, Jacobi implies that $ad_x=ad_a(ad_b)-ad_b(ad_a)$. Since $trace(fg)=trace(gf)$ the result follows.

Tsemo Aristide
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