# Symmetry of Casimirs of Lie algebras

The dimensions of the invariant tensors (Casimirs) of the simple Lie algebras are known, but I nowhere could find whether they are completely symmetric or antisymmetric with respect to an variable swap. (The sextic one of the special ones interests me most.)
Also, can you have Casimirs transforming like a dim>1 irrep of the permutation group?

By theorem of Gelfand, the center of $$\mathfrak{U(g)}$$ is isomorphic to $$\mathfrak{g}$$-invariant elements in $$S(\mathfrak{g})$$, i.e. $$\mathfrak{g}$$-invariant symmetric tensors.
Please expand your definition of "Casimir". Is a $$g$$-invariant element in $$g^{\otimes n}$$?
If $$g$$ is a simple algebra and $$\kappa$$ its Killing form, then the map $$g\times g\times g\to k$$ $$(x,y,z)\mapsto \kappa([x,y],z)$$ is completely antisymetric and $$g$$-invariant, so, it gives you a nonzero element in $$((\Lambda^3g)^*)^g$$, and, using the killing-induced isomorphism $$g\cong g^*$$, you get a nonzero element in $$(\Lambda^3g)^g$$. So, you get a "full antisymmetric Casimir of tensor degree 3"...
• I don't know (I don't know what do you precisely mean by structure constant tensor, the map $[,]:\Lambda^2g\to g$ as element in $(\Lambda^2g)^*\otimes g$?). What I know, is that if $g$ is simple (over the complex), then the dimension of $(\Lambda^3g)^g$ is one. So, any procedure ending with an element in $(\Lambda^3g)^g$ must give the same element up to scalar. Maybe there is a more direct way to get it. – Marco Farinati May 28 '19 at 12:51