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 $$Y_t=\left\{y^2=\prod_{i=1}^{2g+2}(xt_i)\right\}.$$ 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 KodairaSpencer map $$\Theta_{\mathcal T,t^0} \longrightarrow H^1(Y_{t^0}, \Theta_{Y_{t^0}}).$$ Take dual, we get a map (also called KodiaraSpencer map): $$\kappa:H^0(Y_{t^0}, \omega^{\otimes 2}_{Y_{t^0}}) \longrightarrow \Omega^1_{\mathcal T,t^0}.$$ Now one can take $$\frac{x^i}{y^2}(dx)^2, 0\leq i \leq 2(g1);~ \frac{x^j}{y}(dx)^2, 0\leq j \leq g3,$$ 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 KodairaSpencer map explicitly under the above two basis?

$\begingroup$ I don't know the answer off the top of my head, but you might look in some of the classical books on algebraic curves; it's probably treated in some such treatise. One thing does seem clear: You won't need the quadratic differentials that are odd with respect to the hyperelliptic involution $(x,y)\to (x,y)$; indeed, $\kappa$ will only involve the differentials $x^i(dx/y)^2$, since these are a basis for the cotangent space to the $(2g{}1)$dimensional hyperelliptic locus $\mathcal{H}_g$ within the $(3g{}3)$dimensional moduli space $\mathcal{M}_g$ of all curves of genus $g$. $\endgroup$ – Robert Bryant Feb 7 '14 at 20:30

$\begingroup$ @Robert Bryant Thank you very much. I don't know any book treating this special problem. Do you have any good reference to this problem? $\endgroup$ – Pyramid Feb 10 '14 at 11:46

$\begingroup$ Already asked (more or less): mathoverflow.net/questions/22345/… $\endgroup$ – Qfwfq Feb 11 '14 at 18:06

$\begingroup$ @Qfwfq: But the answer there is not very explicit... $\endgroup$ – Robert Bryant Feb 11 '14 at 18:20

$\begingroup$ @RobertBryant: indeed I said 'already asked', not 'already answered' :) In fact, your answer below is very explicit. $\endgroup$ – Qfwfq Feb 12 '14 at 14:15
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 upvoted 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).
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 $$ \mathcal{S} = \{(s_0,\ldots,s_{2g2})\ \ s_i\not=0,1\ \text{and}\ s_i\not=s_j\ \text{when}\ i\not=j \ \}\subset \mathbb{C}^{2g1} $$ and letting $s\in\mathcal{S}$ correspond to the genus $g$ hyperelliptic curve $C_s$ defined by $$ y^2 = x(x1)(xs_0)(xs_1)\cdots(xs_{2g2}). $$ 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 $2g1$ when $g>1$) is given by $$ \frac{x^i\ dx^2}{x(x1)(xs_0)(xs_1)\cdots(xs_{2g2})}\ \text{when}\ 0\le i\le 2g2, $$ and a basis for the $\iota$odd holomorphic quadratic differentials on $C_s$ (a space of dimension $g2$ when $g>1$) is given by $$ \psi_j = \frac{x^i\ dx^2}{y}\qquad\text{when}\ 0\le j\le g3. $$ It turns out that a more convenient basis for the $\iota$even holomorphic quadratic differentials is $$ \phi_i = \frac{dx^2}{x(x1)(xs_i)}\qquad \text{when}\ 0\le i\le 2g2. $$
Then, it seems, in this case, that the (dual) KodairaSpencer map is given by $$ \kappa(\psi_j) = 0\qquad \text{when}\ 0\le j\le g3 $$ while there is a (nonzero) constant $c$ (which might depend on $g$, but I don't think so) such that $$ \kappa(\phi_i) = \frac{c\ ds_i}{s_i(s_i1)}\qquad \text{when}\ 0\le i\le 2g2. $$ (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.

$\begingroup$ Thanks. I wonder how could you get the formula? $$\kappa(\phi_i)=\frac{cds_i}{s_i(s_i1)}.$$ $\endgroup$ – Pyramid Feb 12 '14 at 11:10

$\begingroup$ My calculation is based on Example 1 (starting at the bottom of page 39 and continuing to the top of page 41) of Complex Manifolds by Morrow and Kodaira (first edition, 1971). I used their ideas as a guide, but I had to change the notation a bit and adapt their arguments. Assuming that I did the calculations correctly (and there was a normalizing constant that I wasn't sure about, hence $c$), the above is the answer that this yields. $\endgroup$ – Robert Bryant Feb 12 '14 at 12:21