Explicit Kodaira-Spencer map of hyperelliptic curves

Question

Let $g\geq 2$, and $$\mathcal T=\{(t_1,\cdots,t_{2g+2})~|~t_i\neq t_j,\forall i\neq j\}.$$ For any $t=(t_1,\cdots,t_{2g+2})\in \mathcal T$, let $$Y_t=\left\{y^2=\prod_{i=1}^{2g+2}(x-t_i)\right\}.$$ Thus we get a family of hyperelliptic curves $\mathcal Y \to \mathcal T$. Let $t^0=(t_1^0,\cdots,t_{2g+2}^0) \in \mathcal T$ be a fixed point (we assume $t_i^0\neq 0,\forall i$). Then we have the Kodaira-Spencer map $$\Theta_{\mathcal T,t^0} \longrightarrow H^1(Y_{t^0}, \Theta_{Y_{t^0}}).$$ Take dual, we get a map (also called Kodiara-Spencer map): $$\kappa:H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}}) \longrightarrow \Omega^1_{\mathcal T,t^0}.$$ Now one can take $$\frac{x^i}{y^2}(dx)^2, 0\leq i \leq 2(g-1);~ \frac{x^j}{y}(dx)^2, 0\leq j \leq g-3,$$ as a basis of $H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}})$. And take $$dt_1, \cdots, dt_{2g+2},$$ as a basis of $\Omega^1_{\mathcal T,t^0}$. Then can we write down the Kodaira-Spencer map explicitly under the above two basis?

Answer 1

Caution: I was thinking over the calculation that led to my proposed answer below, and I realized that I had neglected a term that I haven't actually justified as being zero, so now I'm less sure that these formulae are correct. There may be another term in the formula for $\kappa(\phi_i)$ and I don't see how to rule it out. As a result, I am withdrawing my answer until I have had time to determine whether the extra term really does vanish. My apologies to those who trusted my claim and up-voted it!

Unfortunately, I am not allowed to delete an 'accepted answer', so I'm putting this caution at the top so that readers will know not to regard it as correct (yet).

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I haven't checked all of the details, but I think that the following is correct and gives you a basis for an answer: We can get rid of the action of the Möbius group of linear fractional transformations, by, instead, looking at the manifold $$ \mathcal{S} = \{(s_0,\ldots,s_{2g-2})\ |\ s_i\not=0,1\ \text{and}\ s_i\not=s_j\ \text{when}\ i\not=j \ \}\subset \mathbb{C}^{2g-1} $$ and letting $s\in\mathcal{S}$ correspond to the genus $g$ hyperelliptic curve $C_s$ defined by $$ y^2 = x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2}). $$ The hyperelliptic involution is $\iota(x,y) = (x,-y)$. A basis for the $\iota$-even holomorphic quadratic differentials on $C_s$ (a space of dimension $2g-1$ when $g>1$) is given by $$ \frac{x^i\ dx^2}{x(x-1)(x-s_0)(x-s_1)\cdots(x-s_{2g-2})}\ \text{when}\ 0\le i\le 2g-2, $$ and a basis for the $\iota$-odd holomorphic quadratic differentials on $C_s$ (a space of dimension $g-2$ when $g>1$) is given by $$ \psi_j = \frac{x^i\ dx^2}{y}\qquad\text{when}\ 0\le j\le g-3. $$ It turns out that a more convenient basis for the $\iota$-even holomorphic quadratic differentials is $$ \phi_i = \frac{dx^2}{x(x-1)(x-s_i)}\qquad \text{when}\ 0\le i\le 2g-2. $$

Then, it seems, in this case, that the (dual) Kodaira-Spencer map is given by $$ \kappa(\psi_j) = 0\qquad \text{when}\ 0\le j\le g-3 $$ while there is a (nonzero) constant $c$ (which might depend on $g$, but I don't think so) such that $$ \kappa(\phi_i) = \frac{c\ ds_i}{s_i(s_i-1)}\qquad \text{when}\ 0\le i\le 2g-2. $$ (This formula even works when $g=1$, by the way.)

I think you can get your desired formula by making this equivariant with respect to the action of the Möbius group $\mathrm{PSL}(2,\mathbb{C})$, but I suspect that the general formula in this case won't be particularly nice.