Prove that Casimir operators commute with the elements of Lie algebra

[JTFreitas]

Homework Statement

Consider a vector space of dimension $n$ of a semi-simple Lie algebra $\mathcal{L}$ with basis ${a_1, a_2, ..., a_n}$, and the composition law:

$$ [a_r, a_s] = \sum_{p = 1}^{n} c_{rs}^{p}a_{p} $$

Prove that the Casimir operator acting on the vector space commutes with all the elements of the Lie algebra, $a_r$:

Relevant Equations

The Casimir operator $C$ is defined as
$$C = \sum_{i,j} g_{ij}a_{i}a_{j} $$
where
$$g_{ij} = \sum_{l,k}c_{ik}^{l}c_{jl}^{k} $$

Where the numbers $c_{rs}^{p}$ are the structure constants of $\mathcal{L}$, defined according to

$$[a, a_{s}] = \sum_{p = 1}^{n} \text{ad}(a)_{ps}a_{p} $$
and
$$[a_{r}, a_{s}] = \sum_{p=1}^{n} c_{rs}^{p}a_{p}$$
which implies that
$${\text{ad}(a_r)}_{ps} = c_{rs}^{p}$$

I want to show that $[C, a_{r}] = 0$. This means that:
$$ Ca_{r} - a_{r}C = \sum_{i,j} g_{ij}a_{i}a_{j}a_{r} - a_{r}\sum_{i,j} g_{ij}a_{i}a_{j} = 0$$

I don't understand what manipulating I can do here. I have tried to rewrite $g_{ij}$ in terms of the structure constants:$$\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}a_{j}a_{r} - a_{r}\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}a_{j}$$

Now, I am not sure I can do this, since $a_r$ isn't necessarily part of the sums, but I know that

$$a_{j}a_{r} = [a_{j}, a_{r}] + a_{r}a_{j}$$

Hence the expression becomes

$$\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}([a_{j}, a_{r}] + a_{r}a_{j}) - a_{r}\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}a_{j}$$
And based on the composition law, the expression becomes

$$\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}\left(\sum_{p = 1}^{n} c_{jr}^{p}a_{p} + a_{r}a_{j}\right) - a_{r}\sum_{i,j}\sum_{l,k}c_{ik}^{l}c_{jl}^{k}a_{i}a_{j}$$

And this is where I am completely out of ideas about tackling the expression.
Is my work up to this point even valid? Any pointers on how to proceed from here would be very much appreciated.

 

[George Jones]

The Casimir operator $C$ is defined as
$C = \sum_{i,j} g_{ij}a_{i}a_{j}$
where
$g_{ij} = \sum_{l,k}c_{ik}^{l}c_{jl}^{k}$

Is this really how $C$ is defined? My references use the inverse of $g_{ij}$ in the definition of $C$.

 

[JTFreitas]

Is this really how $C$ is defined? My references use the inverse of $g_{ij}$ in the definition of $C$.

Thank you for taking a look at it.
I just double-checked, and according to my textbook, it is defined with $g_{ij}$ as is.

 

[fresh_42]

We have to show that $C(X)\in \operatorname{Z(U}(\mathfrak{g}))$. Wikipedia mentions that this is due to the invariance of the bilinear form the Casimir operator is defined by (which also might explain the two different definitions in your and @George Jones' book). In the case above it is the Killing-form $\operatorname{B}(X,Y)=\operatorname{trace}(\operatorname{ad}(X)\circ \operatorname{ad}(Y))$ of the semesimple Lie algebra $\mathfrak{g}.$ Thus invariance means
$$
B(\operatorname{ad}(Z)(X),Y)= B([Z,X],Y)=-B(X,[Z,Y])=-B(X,\operatorname{ad}(Z)(Y))
$$
where the Casimir operator is given by $C=\sum_{i=1}^n a_iB(a_i,-).$I assume if you take these equations and fight your way through the coordinates, then you will be able to prove that $C(X)$ is in the center of $\operatorname{U}(\mathfrak{g})$ for all $X\in \mathfrak{g}.$

 

[George Jones]

Homework Statement:: Consider a vector space of dimension $n$ of a semi-simple Lie algebra $\mathcal{L}$ with basis ${a_1, a_2, ..., a_n}$, and the composition law:

$$ [a_r, a_s] = \sum_{p = 1}^{n} c_{rs}^{p}a_{p} $$

Prove that the Casimir operator acting on the vector space commutes with all the elements of the Lie algebra, $a_r$:
Relevant Equations:: The Casimir operator $C$ is defined as
$$C = \sum_{i,j} g_{ij}a_{i}a_{j} $$
where
$$g_{ij} = \sum_{l,k}c_{ik}^{l}c_{jl}^{k} $$

Where the numbers $c_{rs}^{p}$ are the structure constants of $\mathcal{L}$, defined according to

$$[a, a_{s}] = \sum_{p = 1}^{n} \text{ad}(a)_{ps}a_{p} $$
and
$$[a_{r}, a_{s}] = \sum_{p=1}^{n} c_{rs}^{p}a_{p}$$
which implies that
$${\text{ad}(a_r)}_{ps} = c_{rs}^{p}$$

Since your Lie algebra is semi-simple, can you choose a basis such that $g_{ij}$ has a particularly nice form?

 

[fresh_42]

Since your Lie algebra is semi-simple, can you choose a basis such that $g_{ij}$ has a particularly nice form?

It already has a particularly nice form, namely the Killing form. I am pretty sure it would be a lot easier to prove without coordinates, just using the adjoint representation and the fact that the Killing form is non degenerate.