# Are there mixed-symmetry, primitive invariant tensors for simple Lie algebras?

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!

• 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. – Qiaochu Yuan Nov 18 '18 at 20:47
• 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! – user119921 Nov 18 '18 at 22:24