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Just a disclaimer, I am not a mathematician so I am sorry for being less than formal.

Let us say we have a Hilbert series $\mathcal{H}(K[V]^G)$. Now I know that by eliminating some parameters in a given theory I get the freedom of a subgroup $H < G$. Now I may compute another Hilbert series $\mathcal{H}(K[V]^H)$. I have noticed that by computing $\mathcal{H}(K[V]^G) / \mathcal{H}(K[V]^H)$ I get what I am interpreting as the Hilbert series corresponding to the invariants of $G$ but not of $H$.

As an example, let us have complex invariants under $G$ and now I find that the $\mathbb{R}$ version of this Hilbert series corresponds to invariants under $H$. What is the significance of the quotient of these two Hilbert series?

M.Π.B
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    Hilbert series are multiplicative under tensor product, so if one Hilbert series divides another that could be indicative of a tensor product decomposition (although it's not a guarantee). – Qiaochu Yuan Feb 22 '23 at 18:41
  • So imagine I have a Lie algebra su(3) and a dim 8 irrep. I build this Hilbert series. Now I go to su(2) and a dim 5 irrep. I divide both Hilbert series. Is this valid? – M.Π.B Feb 23 '23 at 17:32
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    I don't know what you mean by "valid." If one is divisible by the other it's interesting but isn't by itself a proof that a meaningful tensor product decomposition exists. – Qiaochu Yuan Feb 23 '23 at 20:46

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