# D-finiteness of Hilbert series of non-commutative invariant ring under reductive group

Let $G$ be reductive group over a field of characteristic $0$ ($GL_n$ fine for this question). Let $V$ be a linear representation of $G$. Then $G$ acts on the tensor algebra $T(V) = \bigoplus_{n \ge 0} V^{\otimes n}$ which is graded by $\deg(V) = 1$. Now form the Hilbert series of the ring of invariants:

$H(t) = \sum_{n \ge 0} \dim (V^{\otimes n})^G t^n$.

I believe I have a proof that this is a D-finite function, i.e., the derivatives of $H(t)$ with respect to $t$ form a finite-dimensional vector space over the field of rational functions. Is this result already stated in the literature?

• Steven, I am puzzled - I thought I have outlined the following to you in Herstmonceux last summer: I think you can consider the highest weight words as walks, which yields a rational series. The highest weight words of weight zero are then a diagonal of this series, and hence D-finite. Is something wrong with this argument? May 28 '17 at 22:10
• @MartinRubey : I completely forgot about this discussion actually. Anyway, it's a different proof that I have in mind but this result is just a consequence of something else being studied, so I wanted to know if I should cite something. May 28 '17 at 23:17
• And also, I am a little confused by this argument: I don't want the dimension of the weight 0 piece. This is in general larger than the space of invariants. Could you be more specific about how I think of invariants as lattice paths? May 28 '17 at 23:19
• By very basic representation theory, highest weight vectors of weight 0 coincide with the invariants. May 28 '17 at 23:41
• @VictorProtsak: Yes. I can see why the weight 0 piece is given by walks in a lattice but I don't see how to isolate the highest weights using such a method. May 29 '17 at 0:23

For the defining representation of $SL_n$ (if I'm not mistaken, for $GL_n$ there aren't any) the invariants are rectangular standard Young tableaux, so I think you could cite