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Let $L$ be a simple Lie algebra. Let $\beta (x ,y)$ and $\gamma (x ,y)$ be two symmetric associative bilinear forms on $L$. If $\beta,\gamma$ are nondegenerate, prove that $\beta$ and $\gamma$ are proportional.

(Here “associative” means that $\beta([x,y],z)=\beta(x,[y,z])$. )

[Hint:Use Schur's Lemma.]

Schur's Lemma: Let $\phi: L \rightarrow gl(V)$ be irreducible. Then the only endomorphisms of $V$ commuting with all $\phi(x) (x \in L)$ are the scalars.

This problem is an exercise on Page 31 of Introduction to Lie Algebras and Representation Theory by James E. Humphreys.

I have tried to construct a mapping $\beta(\cdot,y)$ from $L$ to $L^*$ and the similar "inverse" form of $\gamma$ so the lemma can be used. Also I considered the bilinear form from a nice-chosen basis of $L$ and discussed. But so far I have not found a good way to do with it.

Thank you for any help.

dailycrazy
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    What does “associative” mean here? You want to use Schur’s lemma in the following slightly different form: if $V$ and $W$ are two irreducibles (finite-dimensional, over an algebraically closed field), then either every map between them is zero or the space of maps is $1$-dimensional. – Qiaochu Yuan Sep 20 '20 at 20:45
  • Here “associative” means that $\beta([x,y],z)=\beta(x,[y,z])$. There are a lot of forms of Schur's lemma, and the one displayed in the discription is the only one referred in the book (and exactly in that section). So I think there exists one way to do the problem with this one. (And alsp, I still don't know how to do it with the one you referred.) Thank you. – dailycrazy Sep 21 '20 at 03:01
  • The more standard name for that condition is “invariant.” The different forms of Schur’s lemma are equivalent and they’re all easy to prove from themselves or each ofher. With the form I gave, you just observe that $\beta$ and $\gamma$ induce maps between the same pair of irreducibles, namely $L$ and $L^{\ast}$. – Qiaochu Yuan Sep 21 '20 at 04:46
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    Then using the lemma and the nondegeneracy of the bilinear form, the maps are proportional since the dimension of space of maps is one. Thank you! – dailycrazy Sep 21 '20 at 06:13
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    Yes. If you want to use your form of Schur's lemma you have to do a more annoying thing like compose the map $L \to L^{\ast}$ you get from $\beta$ with the inverse of the map you get from $\gamma$ and it's just annoying when the exact same proof gets you a more usable form of Schur's lemma. – Qiaochu Yuan Sep 21 '20 at 06:17
  • I see. So I think the Schur's lemma referred in the book is just for the proof of some theorems in the main body. The author assumes that the reader has already known the different forms of it :) – dailycrazy Sep 21 '20 at 06:22

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