Consider genus $g$ Riemann surfaces, and thier moduli space $\mathcal M_g$.

To determine dimension of $T\mathcal M_g$, start with a complex structure, which in some coordinates can be written $$J=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$$ and consider small deformations $J \to J+\epsilon$. The condition $(J + \epsilon)^2=-1$ yields $$\epsilon=\begin{pmatrix} 0 & \epsilon_1 \\ \epsilon_2 & 0 \end{pmatrix}.$$

Now comes the point I don't understand: it is claimed that vanishing of Nijenhuis tensor for $J+\epsilon$ implies $\bar\partial \epsilon_1=0=\partial \epsilon_2$. Can anybody please explain why this follows from $N_{J+\epsilon}=0$ ?

(The final part when one applies Riemann-Roch to study dimension of $\epsilon_1$ space called Beltrami differentials is clear to me)

What I tried is the following: write $$ \delta N = [\epsilon X,JY]+[JX,\epsilon Y]-\epsilon[ X,JY]-J[X,\epsilon Y]-\epsilon [JX,Y]-J[\epsilon X,Y]$$ and notice that $J\partial=+i\partial$, $J\bar \partial=-i\bar\partial$ and $\epsilon \partial = \epsilon_2 \bar \partial$, $\epsilon \bar\partial = \epsilon_1 \partial$. Then it is enough to evaluate the tensor for $X=\partial=Y$ and $X=\bar\partial$, $Y=\partial$, and the other two, but all these give zero, not the desired above conditions.

Also the above derivation smells a bit, since we know that in $\dim_\mathbb C \Sigma=1$ there is no obstruction to the deformation of the complex structure, so I would expect that the conditions $\bar \partial \epsilon_1=0=\partial\epsilon_2$ follow from just $\epsilon$ being off diagonal.

Reference: Collinucci's lectures page 25.

Note this is crossposted on mo


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