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In Buchdahl's paper Algebraic deformations of compact Kähler surfaces, the author made a remark that: the product of two Riemann surfaces of genus at least 5 satisfies the dimension of $H^1(X,T_X)$ < dimension of $H^2(X,\mathcal{O})$, but I can't see why, why the genus must larger than 5? How to compute the dimension of $H^1(X,T_X)$ of the product of two Riemann surfaces, by the way how can we know $H^2(X,T_X)$ should not be zero? Any comment is welcome, thanks!

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If $X = C \times D$ then $T_X = T_C \boxtimes \mathcal{O}_D \oplus \mathcal{O}_C \boxtimes T_D$ and by Kunneth formula $$ h^1(X,T_X) = h^1(T_C)h^0(\mathcal{O}_D) + h^0(T_C)h^1(\mathcal{O}_D) + h^0(\mathcal{O}_C)h^1(T_D) + h^1(\mathcal{O}_C)h^0(T_D) = (3g(C) - 3) + 0 + (3g(D) - 3) + 0. $$ Similarly, $$ h^2(X,\mathcal{O}_X) = h^1(C,\mathcal{O}_C)h^1(D,\mathcal{O}_D) = g(C)g(D). $$ It remains to note that $$ g(C)g(D) - 3g(C) - 3g(D) + 6 = (g(C) - 3)(g(D) - 3) - 3 $$ and if $g(C),g(D) \ge 5$ then this is positive.

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  • $\begingroup$ It's a good answer, but I still have 3 questions: 1. why $T_X = T_C \boxtimes \mathcal{O}_D \oplus \mathcal{O}_C \boxtimes T_D$ not $T_X = T_C \boxtimes \mathcal{O}_C \oplus \mathcal{O}_D \boxtimes T_D$ 2. why $h^1(X,T_X) = h^1(T_C)h^0(\mathcal{O}_D) + h^0(T_C)h^1(\mathcal{O}_D) + h^0(\mathcal{O}_C)h^1(T_D) + h^1(\mathcal{O}_C)h^0(T_D)$, can you elaborate it a bit? I have searched it but can't find the same formula. 3. why $h^0(T_C)=0$. thanks. $\endgroup$
    – Tom
    Jul 29, 2020 at 3:03
  • $\begingroup$ 1. Your formula doesn't make sense (you cannot pullback $\mathcal{O}_C$ from $D$). 2. This is Kunneth formula applied to the direct sum decomposition. 3. Because $g(C) > 1$. $\endgroup$
    – Sasha
    Jul 29, 2020 at 6:49

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