# compute the Hilbert series of a free product of algebras

We work over a field K. Assume there are two connected graded algebras $$B, C$$. And $$A=B * C$$. The Hilbert series of A is

$$A(x) = \sum_{n \geq 0} dim(A_n) x^n$$

Why the Hilbert series satisfy the equation:

$$\frac{1}{A(x)} = \frac{1}{B(x)} + \frac{1}{C(x)} -1$$

This question comes from an article "Combinatorial Hopf algebras and generalized Dehn–Sommerville relations" (6.4). It says this is a well-known fact. But I don't know how to prove it.

• I guess this would work but haven't check details yet: $A$ is spaned by $1$, $b_1c_1...b_nc_n$ and $b_1c_1...c_{n-1}b_n$($b$s and $c$s not equal to 1). Let the Hilbert series of second kind be $h_B(x;A)$ and third kind $h_C(x,A)$. You'll get things like $h_B(x;A)=(h(x;B)-1)+(h(x;B)-1)*h_C(x;A)$ and $h_C(x;A)=(h(x;C)-1)+(h(x;C)-1)*h_B(x;A)$ (I'm not sure the constant term is correct) this would be suffice. Commented Sep 20, 2022 at 2:33

This is false without the hypothesis that $$B$$ and $$C$$ are connected, meaning that $$B(0) = \dim B_0 = C(0) = \dim C_0 = 1$$. With this hypothesis, the desired relation is equivalent to

\begin{align*} A(x) &= \frac{1}{\frac{1}{B(x)} + \frac{1}{C(x)} - 1} \\ &= \frac{B(x) C(x)}{B(x) + C(x) - B(x) C(x)} \\ &= \frac{B(x) C(x)}{1 - (B(x) - 1)(C(x) - 1)} \\ &= B(x) C(x) \sum_{k=0}^{\infty} (B(x) - 1)^k (C(x) - 1)^k \end{align*}

which we can prove as follows: the free product consists of the direct sum of all tensor products of the form

$$B_{i_0} \otimes C_{j_0} \otimes B_{i_1} \otimes C_{j_1} \otimes \dots \otimes B_{i_{k+1}} \otimes C_{j_{k+1}}$$

where $$i_0 \ge 0, j_{k+1} \ge 0$$ and all the other indices are strictly positive (this is where we use the connectivity hypothesis; without it $$\dim A_0$$ would be infinite because $$A_0$$ itself would already be the free product of $$B_0$$ and $$C_0$$). This summand lives in degree $$\sum_m (i_m + j_m)$$ and for fixed $$k$$ such summands correspond exactly to terms in the product

$$B(x) \underbrace{ (C(x) - 1)(B(x) - 1)(C(x) - 1) \dots (B(x) - 1)}_{k \text{ pairs}} C(x)$$

where $$B(x)$$ corresponds to $$i_0$$, $$C(x)$$ corresponds to $$j_k$$, and the other factors correspond to the other indices in order.

As a sanity check we can take $$B$$ to be the free algebra on $$m$$ generators and $$C$$ to be the free algebra on $$n$$ generators (in degree $$1$$ in both cases), with Hilbert series $$\frac{1}{1 - mx}, \frac{1}{1 - nx}$$. Then $$A$$ is the free algebra on $$m + n$$ generators, with Hilbert series $$\frac{1}{1 - (m+n)x}$$, and we can easily verify that

$$1 - (m+n) x = (1 - mx) + (1 - nx) - 1.$$

• Thank you very much. You are right. But I think you can write C(x) in the final position in the equation. That will help for your explanation. Commented Sep 20, 2022 at 4:47
• Yes, that's a good idea, I reordered the factors to make this clearer. Thanks! Commented Sep 20, 2022 at 4:50