Are Casimir elements $\mathfrak {g}$-invariant?

Question

Let $\mathfrak {g}$ be a Lie algebra equipped with a non-degenerate invariant bilinear form $\left \langle \cdot, \cdot \right \rangle.$ Let $r \in \mathfrak {g} \otimes \mathfrak {g}$ be a Casimir tensor with respect to $\left \langle \cdot, \cdot \right \rangle.$ Is $r$ $\mathfrak {g}$-invariant i.e. for all $x \in \mathfrak {g}$ do we have $\left (\text {ad}_{x} \otimes \text {id} + \text {id} \otimes \text {ad}_{x} \right ) (r) = 0$?

Answer 1

If a bilinear form ⟨⋅,⋅⟩ is non-degenerate and invariant, then a Casimir tensor is an element of the tensor product of the Lie algebra that satisfies the equation (adx⊗id + id⊗adx)(r) = 0 for all x∈g. In general,Casimir elements are g-invariant.

Answer 2

The Casimir's are invariant because the symmetric bilinear form you use to define them is invariant.

The condition that a bilinear form on $\mathfrak g$ is "invariant" is usually written as $b([x,y],z)+b(x,[y,z])=0$, but this is equivalent to the condition that $b$ is an invariant vector in $\text{Bil}(\mathfrak g)$ for the natural action of $\mathfrak g$ on $\text{Bil}(\mathfrak g)$ induced by the adjoint action. Indeed if $V$ is any $\mathfrak g$-representation then $\text{Bil}(V)$ is a $\mathfrak g$-representation where if $b \in \text{Bil}(V)$ then for all $x \in \mathfrak g$ and $v_1,v_2 \in V$ $$ \begin{split} x(b)(v_1,v_2) &= x(\tilde{b})(v_1\otimes v_2) = -\tilde{b}\big(x(v_1\otimes v_2)\big)\\ & = -\tilde{b}\big(x(v_1)\otimes v_2 + v_1\otimes x(v_2)\big) \\ &= -b(x(v_1),v_2) - b(v_1,x(v_2). \end{split} $$ Thus $b$ is invariant if, for all $x\in \mathfrak g$ and $v_1,v_2\in V$ we have $b(x(v_1),v_2)+b(v_1,x(v_2))=0$. (Note that if $b$ is nondegenerate, this is equivalent to the condition that $\mathfrak g$ acts by skew-adjoint operators on $(V,b)$.)

When $V=(\mathfrak g, \text{ad})$, we can rewrite this condition as follows:

$$ \begin{split} b \in \text{Bil}(\mathfrak g)^{\mathfrak g} &\iff &b(\text{ad}(y)(x),z) + b(x,\text{ad}(y)(z)) =0, \quad \forall x,y,z \in \mathfrak g \\ &\iff &b([y,x],z)+b(x,[y,z]) =0 \quad \forall x,y,z \in \mathfrak g \\\ &\iff &b([x,y],z) = b(x,[y,z])\quad \forall x,y,z \in \mathfrak g. \end{split} $$

Next note that there is a canonical identification

$$ \text{Bil}(V) \cong (V\otimes V)^* = \text{Hom}_{\mathsf k}(V\otimes V, \mathsf k) \cong \text{Hom}_{\mathsf k}(V,\text{Hom}_{\mathsf k}(V,\mathsf k)) = \text{Hom}_{\mathsf k}(V,V^*) $$

given by $b \mapsto \theta_b$, where if $v_1\in V$, then $\theta_b(v_1)$ is given by $\theta_b(v_1)(v_2) = b(v_1,v_2)$ for all $v_2 \in V$. The map $b \mapsto \theta_b$ is clearly an isomorphism with inverse $\phi \mapsto b_{\phi}$ given by $b_{\phi}(v_1,v_2) = \phi(v_1)(v_2)$, and if $V$ is a $\mathfrak g$-representation, then it is easy to see that this identification is an isomorphism of $\mathfrak g$-representations. Moreover, by definition $b$ is nondegenerate if and only if $\theta_b$ is injective, which, when $V$ is finite-dimensional, is equivalent to $\theta_b$ being an isomorphism. Thus if $b \in \text{Bil}(\mathfrak g)$ is a nondegenerate invariant bilinear form on $\mathfrak g$ the associated map $\theta_b \colon \mathfrak g \to \mathfrak g^*$ is an isomorphism of $\mathfrak g$-representations.

Now for vector space $V$ we have a canonical injective linear map

$$ m\colon V^*\otimes V^*\to (V\otimes V)^*, \quad m(f_1\otimes f_2) =f_1.f_2 \quad \forall f_1,f_2 \in V^* $$ where $(f_1.f_2)(v_1,v_2) = f_1(v_1).f_2(v_2)$ for all $v_1,v_2 \in V$. Comparing dimensions we see that if $V$ is finite-dimensional, then $m$ is an isomorphism.

Putting all of this together, if $\mathfrak g$ is finite-dimensional and $b$ is a nondegenerate symmetric bilinear form, we may define

$$ C_b = (\theta_b^{-1}\otimes \theta_b^{-1})\circ m^{-1}(b)\in \mathfrak g \otimes \mathfrak g, $$ and since $b$ is invariant, and $\theta_b$ is a $\mathfrak g$-homomorphism, it follows that $C_b$ is invariant.

To make this more explicit, suppose that $\{x_i:1\leq i \leq n\}$ and $\{y_j:1 \leq j \leq n\}$ are bases of $\mathfrak g$ dual with respect to $b$ (so that $b(x_i,y_j) = \delta_{ij}$). Then if $\{\delta_i:1\leq i\leq n\}$ and $\{\eta_j:1\leq j\leq n\}$ are the bases of $\mathfrak g^*$ dual to $\{x_i:1\leq i\leq n\}$ and $\{y_j:1\leq j\leq n\}$, we have $m(\sum_{i=1}^n \delta_i\otimes \eta_i) = b$. Moreover, $\theta_b(x_i)(y_j) = b(x_i,y_j) = \delta_{ij}$, and so $\theta_b(x_i) = \eta_i$, and similarly $\theta_b(y_j) = \delta_j$. It follows that $C_b = \sum_{i=1}^n y_i\otimes x_i$.

For example, for $\mathfrak{sl}_2(\mathsf k)$, the trace form $t(x,y) = \text{tr}(xy)$ is a nondegenerate symmetric bilinear form (assuming $\text{char}(\mathsf k)\neq 2)$, and it is easy to compute that $t(e,f)=1$, $t(h,h) =2$, and all other pairings of the standard basis $\{e=E_{12},f= E_{21},h= E_{11}-E_{22}\}$ are zero. It follows that the dual basis to $\{e,h,f\}$ is $\{f,\frac{1}{2}h,e\}$ and hence

$$ C_t = e\otimes f + f\otimes e + \frac{1}{2} h\otimes h. $$