Any biinvariant metric proportional to Killing metric

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SUMMARY

The Killing form on a Lie algebra, defined as $$B(X,Y) = \text{Tr}ad_X \circ ad_Y$$, serves as a biinvariant metric on a simple Lie group G. It is established that all other biinvariant metrics are proportional to the Killing form, particularly when G is simple. The proof involves Schur's Lemma, demonstrating that two invariant symmetric bilinear forms B and B' are scalar multiples of each other due to the nondegeneracy of the forms and their corresponding isomorphisms between the Lie algebra and its dual. The discussion emphasizes the necessity of understanding Schur's Lemma for a complete grasp of this relationship.

PREREQUISITES
  • Understanding of Lie algebras and their properties
  • Familiarity with invariant symmetric bilinear forms
  • Knowledge of Schur's Lemma in representation theory
  • Basic concepts of simple and semisimple Lie algebras
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  • Study Schur's Lemma in detail to understand its implications in representation theory
  • Explore the properties of invariant symmetric bilinear forms in Lie algebras
  • Read "Linear Representations of Finite Groups" by Liebeck and James for practical examples
  • Investigate the relationship between Lie algebras and their duals, focusing on isomorphisms
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Mathematicians, physicists, and students of representation theory seeking to deepen their understanding of Lie algebras and their metrics, particularly those interested in the Killing form and its applications in theoretical physics.

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The killing form on a lie algebra is defined as
$$B(X,Y) = \text{Tr}ad_X \circ ad_Y$$
where ##ad_X: \mathfrak{g} \to \mathfrak{g}## is given by ##ad_X(Y) = [X,Y]##, where the latter is the lie bracket between X and Y in ##\mathfrak{g}##. Expressed in terms of components on a basis on ##\mathfrak{g}## we have
$$B_{ij} = c_{il}^{\ \ k} c_{jk}^{\ \ l}$$.

The killing form can serve as a biinvariant metric on the lie group G, and I've seen it stated several times that, if the group G is simple, then all other biinvariant metrics are proportional to the killing form. Especially the formula
$$B(X,Y) \sim \text{Tr}(\rho(X)\rho(Y))$$
where ##\rho## is a representation of ##\mathfrak{g}## is thrown a lot around.

So I wonder how this statement is proved?
 
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It's essentially an application of Schur's Lemma in the context of Lie algebras which I will just give a quick sketch and leave out all the minor details.

Let B,B' be two invariant symmetric bilinear forms. Use the fact that \mathfrak{g} is simple to show B,B' are both either zero or nondegenerate (use the properties of B to show \left\{ x\in \mathfrak{g} : B(x,y)=0~\forall y\in \mathfrak{g} \right\} is an ideal.) We may as well assume each is nondegenerate since otherwise the statement is trivial. We have two different isomorphisms between \mathfrak{g} and \mathfrak{g^*}, say \phi,\psi corresponding to the two different forms. Apply Schur's lemma to conclude \psi^{-1}\circ\phi is a scalar multiple of the identity and show this implies the forms B,B' are themselves scalar multiples of each other.
 
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Terandol said:
It's essentially an application of Schur's Lemma in the context of Lie algebras which I will just give a quick sketch and leave out all the minor details.

Let B,B' be two invariant symmetric bilinear forms. Use the fact that \mathfrak{g} is simple to show B,B' are both either zero or nondegenerate (use the properties of B to show \left\{ x\in \mathfrak{g} : B(x,y)=0~\forall y\in \mathfrak{g} \right\} is an ideal.) We may as well assume each is nondegenerate since otherwise the statement is trivial. We have two different isomorphisms between \mathfrak{g} and \mathfrak{g^*}, say \phi,\psi corresponding to the two different forms. Apply Schur's lemma to conclude \psi^{-1}\circ\phi is a scalar multiple of the identity and show this implies the forms B,B' are themselves scalar multiples of each other.

Can the statement be proved without referring to Schur's Lemma? I'm a phycisist, so I have not
had the time to go deep into representation theory. However, I have read up on ideals, simple and semisimple lie algebras.
 
center o bass said:
Can the statement be proved without referring to Schur's Lemma? I'm a phycisist, so I have not
had the time to go deep into representation theory. However, I have read up on ideals, simple and semisimple lie algebras.

Schur's lemma is really elementary. You can probably understand it if you spend 15 minutes at it.
 
center o bass said:
Can the statement be proved without referring to Schur's Lemma? I'm a phycisist, so I have not
had the time to go deep into representation theory. However, I have read up on ideals, simple and semisimple lie algebras.

Why don't you look into it and get back to us with any question you may have.
 
micromass said:
Schur's lemma is really elementary. You can probably understand it if you spend 15 minutes at it.

Any suggestions where to read up on it? Can it be done and still avoiding "modules"?
 
WWGD said:
Why don't you look into it and get back to us with any question you may have.
I have a deadline for something in which it would be nice have an explanation of the Killing form proportionality. Do you have any suggestions where I could read up on it without having to learn additional material?
 
I don't know many books, but Liebeck and James, it comes with solutions. BTW, representations can be seen as modules, and modules are somewhat like vector spaces, but use rings instead of fields. See the section for Schur's lemma http://www.mathematik.uni-kl.de/~taylor/PDF/jamesliebeck.pdf The book is broken into small, self-contained sections.
 
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