# Product of two Riemann surfaces $X$ with $H^1(X,T_X)<H^2(X,\mathcal{O})$

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!

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.
• 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.
• 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 $g(C) > 1$. Jul 29, 2020 at 6:49