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I am interested in $\mathfrak{g}$-invariant tensors for a simple Lie algebra $\mathfrak{g}$. That is, in tensors $\kappa_{i_1\dots i_n}$ such that $$ \sum\limits_{s=1}^m f^\rho_{\nu i_s} \kappa_{i_1\dots\hat{i_s}\rho i_{s+1}\dots i_m} = 0~, $$ where $f^\rho_{\nu i_s}$ are the structure constants of $\mathfrak{g}$. These correspond to the invariant polynomials often denoted as $\mathcal{P}(\mathfrak{g})^{\mathfrak{g}}$.

I can only find sources (Humphreys, Tauvel-Yu etc.) that treat the completely symmetric or skew-symmetric cases. It is, e.g., known that there are $r$ symmetric, primitive invariant polynomials corresponding to the Casimirs of $\mathcal{G}$, where $\mathcal{G}$ is the Lie group integrating $\mathfrak{g}$ and $r$ is the Lie algebra's rank. There are also $r$ skew-symmetric primitive invariant polynomials that determine the non-trivial cocycles of the Lie algebra cohomology. See, e.g., this paper for nice and explicit expressions for these.

My question is this: Are there any mixed-symmetry invariant polynomials that cannot be written as a product of completely symmetric and anti-symmetric ones? If not, do you have any an idea on how to show this? If so, do you have an example, say for $\mathfrak{su}(n)$? Also, would there be restrictions on the order of those polynomials (the skew-symmetric ones, e.g., have order 2m-1)?

This feels like something that is just classically known, but I cannot find a reference, find an example nor show the converse, so I thought I'd ask here. Thank you very much for any help!

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  • $\begingroup$ I don't know what the indices are supposed to be ranging over here. Do you only want to consider tensor powers of $\mathfrak{g}$ itself or also other representations? There's a very nice answer for $\mathfrak{g} = \mathfrak{su}(2)$ and the standard representation. $\endgroup$ – Qiaochu Yuan Nov 18 '18 at 20:47
  • $\begingroup$ Apologies, the indices are supposed to range over the Lie algebra. I.e., I'm interested in tensor powers of $\mathfrak{g}$ itself. However, the nice answer you mention still sounds interesting! $\endgroup$ – user119921 Nov 18 '18 at 22:24

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