Any biinvariant metric proportional to Killing metric

In summary, the killing form on a Lie algebra is defined as the trace of the adjoint operator of two elements in the algebra. It can serve as a biinvariant metric on the corresponding Lie group, and it has been shown that if the group is simple, then all other biinvariant metrics are proportional to the killing form. This is proven using Schur's Lemma, which states that two symmetric bilinear forms on a simple Lie algebra must be scalar multiples of each other. While understanding Schur's Lemma may require some knowledge of representation theory, it is a relatively basic concept that can be learned in a short amount of time. Some suggested resources for learning it include the book "Liebeck and James" and the paper "The
  • #1
center o bass
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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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  • #2
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 [itex] B,B'[/itex] be two invariant symmetric bilinear forms. Use the fact that [itex] \mathfrak{g} [/itex] is simple to show [itex] B,B'[/itex] are both either zero or nondegenerate (use the properties of B to show [itex] \left\{ x\in \mathfrak{g} : B(x,y)=0~\forall y\in \mathfrak{g} \right\} [/itex] is an ideal.) We may as well assume each is nondegenerate since otherwise the statement is trivial. We have two different isomorphisms between [itex] \mathfrak{g} [/itex] and [itex] \mathfrak{g^*} [/itex], say [itex] \phi,\psi [/itex] corresponding to the two different forms. Apply Schur's lemma to conclude [itex] \psi^{-1}\circ\phi [/itex] is a scalar multiple of the identity and show this implies the forms [itex] B,B' [/itex] are themselves scalar multiples of each other.
 
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  • #3
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 [itex] B,B'[/itex] be two invariant symmetric bilinear forms. Use the fact that [itex] \mathfrak{g} [/itex] is simple to show [itex] B,B'[/itex] are both either zero or nondegenerate (use the properties of B to show [itex] \left\{ x\in \mathfrak{g} : B(x,y)=0~\forall y\in \mathfrak{g} \right\} [/itex] is an ideal.) We may as well assume each is nondegenerate since otherwise the statement is trivial. We have two different isomorphisms between [itex] \mathfrak{g} [/itex] and [itex] \mathfrak{g^*} [/itex], say [itex] \phi,\psi [/itex] corresponding to the two different forms. Apply Schur's lemma to conclude [itex] \psi^{-1}\circ\phi [/itex] is a scalar multiple of the identity and show this implies the forms [itex] B,B' [/itex] 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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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"?
 
  • #7
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?
 
  • #8
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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FAQ: Any biinvariant metric proportional to Killing metric

What is a biinvariant metric?

A biinvariant metric is a type of metric that is invariant under the action of a Lie group. This means that the metric remains unchanged when the group acts on it, and is preserved by both left and right translations.

What is a Killing metric?

A Killing metric is a type of metric that is invariant under the action of the Lie derivative. This means that the metric remains unchanged when the vector field generated by the Lie derivative acts on it.

What is the relationship between biinvariant and Killing metrics?

A biinvariant metric is proportional to a Killing metric, meaning that they are related by a constant factor. This constant factor is determined by the structure of the Lie group and the metric.

Why are biinvariant metrics important in physics?

In physics, biinvariant metrics are important because they are used to describe the geometry of spaces that are invariant under the action of a Lie group. These spaces are often used to model physical systems, such as in the theory of general relativity.

How are biinvariant metrics used in mathematics?

In mathematics, biinvariant metrics are used to study the geometry of Lie groups and their associated spaces. They also play a crucial role in the study of Riemannian geometry and differential geometry, as well as in the field of mathematical physics.

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