Let $g\geq 1$ be an integer and define three rational functions and their sum by \begin{align*} H(q,t) &= f_1(q,t) + f_2(q,t) + f_3(q,t) \\ &:= \frac{t^{8g-4}q^{2g-1}(1+tq)^{2g-1}(1+q^2t^3)^{2g}}{(1-tq)(1-t^2q)} + \frac{t^{8g-4}(1+tq)^{2g-1}(1+tq^2)^{2g}}{(1-q)(1-tq)} -\frac{t^{8g-4}q^{2g-1}(1+tq)^{4g}}{(1-q)(1-t^2q)}. \end{align*} I am interested in what happens when we make the substitution $u=qt^2$. In particular, I have three questions:
- If we substitute $u=qt^2$ in each $f_i$, does each $f_i$ become a polynomial in $u$?
- (If the answer to $1.$ is no) If we mix the terms first and then substitute $u=qt^2$, does the overall function $H$ become a polynomial in $u$?
- (For each yes answer) What are the resulting polynomials obtained by the substitution $u=qt^2$?
I have performed some simplifications by hand. I have the following: \begin{align*} H(q,t) &= f_1(q,t) + f_2(q,t) + f_3(q,t) \\ &= \frac{u^{2g-1}(u+t)^{2g-1}(u^2+t)^{2g}}{(u-1)(u-t)} + \frac{t^{8g-1}(u+t)^{2g-1}(u+1)^{2g}}{(u-t^2)(u-t)} -\frac{u^{2g-1}(u+t)^{4g}}{(u-1)(u-t^2)} \\ &= \frac{u^{2g-1}(u+t)^{2g-1}(u^2+t)(u-t^2) + t^{8g-1}(u-1)(u+t)^{2g-1}(u+1)^{2g} - u^{2g-1}(u-t)(u+t)^{4g}}{(u-1)(u-t)(u-t^2)} \end{align*} I do not see how to remove the $t$ so that the overall function (or its three terms) are polynomials in $u$. I have also attempted simplifications with Sage, getting output such as:
u = var('u')
t = var('t')
g = var('g')
expr = (u^(2*g-1)*(u+t)^(2*g-1)*(u^2+t)*(u-t^2) + t^(8*g-1)*(u-1)*(u+t)^(2*g-1)*(u+1)^(2*g) - u^(2*g-1)*(u-t)*(u+t)^(4*g))/((u-1)*(u-t)*(u-t^2))
expr.simplify()
((t + u)^(2*g - 1)*t^(8*g - 1)*(u + 1)^(2*g)*(u - 1) - (t^2 - u)*(u^2 + t)*(t + u)^(2*g - 1)*u^(2*g - 1) + (t + u)^(4*g)*(t - u)*u^(2*g - 1))/((t^2 - u)*(t - u)*(u - 1))
Context: On page 559 of this paper, the authors say that we can obtain the 'pure part' of the mixed Hodge polynomial $H$ "by taking the monomials which are powers of $qt^2$." This pure part is conjectured to be the $E$-polynomial of a quiver variety. I am interested to see if this procedure can be performed term-by-term.