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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?

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  • $\begingroup$ In the meantime I (re)found a relevant paper saying the sextic one of G2 (by "magic" extension, of E7 etc) is symmetric: arxiv.org/abs/math-ph/0103021 $\endgroup$
    – Hauke Reddmann
    Commented Dec 18, 2021 at 16:36

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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.

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    $\begingroup$ I myself do not know... but what is the relative priority of Gelfand and Harish-Chandra on such issues, if you know? It wouldn't surprise me if it was independent, etc... Bad communication in those days, due to Cold War (not to mention no internet...) $\endgroup$
    – paul garrett
    Commented May 21, 2019 at 21:56
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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"...

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    $\begingroup$ That's effectively the structure constant tensor, which is always fully antisymmetric, right? $\endgroup$
    – Hauke Reddmann
    Commented May 27, 2019 at 16:35
  • $\begingroup$ 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. $\endgroup$
    – Marco Farinati
    Commented May 28, 2019 at 12:51

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