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Let's consider a Riemann surface $M$. The $(0,1)$-tangent bundle is locally spanned by $\frac{\partial}{\partial z}$. Suppose we have a deformation of $M$, then the new $(0,1)$-tangent bundle is given by $$\frac{\partial}{\partial\bar{z}}+\varphi\frac{\partial}{\partial z}, \varphi\in\mathbb{C}.$$ Taking dual, we should obtain $$d\bar{z}-\bar{\varphi}dz.$$ But then $$1=(d\bar{z}-\bar{\varphi}d\bar{z})(\frac{\partial}{\partial\bar{z}}+\varphi\frac{\partial}{\partial z},\varphi\in\mathbb{C})=1-|\varphi|^2.$$ Hence we should have $\varphi\equiv 0$. Therefore, we have no deformation at all. But this cannot be true because, for example, every compact Riemann surface of genus $g>0$ has non-trivial deformation of its complex structure. Page 12 of the following take elliptic curves as example. http://www.math.sunysb.edu/~cschnell/pdf/notes/kodaira.pdf

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What your calculation shows is that your formula for the dual $(1,0)$-form is wrong. –  Robert Bryant Apr 8 '13 at 14:42
    
Let's be a little bit charitable here. Why do you want that the application of your "dual" form to your vector field give 1? It should be 0. Which is the case if you read carefully your linked document. –  Loïc Teyssier Apr 8 '13 at 18:22
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@Loic: I did not mean for my comment to be mean or uncharitable; I was pointing out that the OP had a formula wrong and needed to correct it before continuing. Indeed, that turned out to be the case. Once that was corrected and the OP reformulated the question, I was going to point out that one cannot take $\phi$ to be a constant, in general (though, of course, that works for when $g=1$), and this wouldn't make sense anyway because there is no global holomorphic coordinate nor a unique holomorphic differential up to constant multiples (again, except in the case $g=1$). –  Robert Bryant Apr 9 '13 at 4:34

2 Answers 2

I have found my mistake. $d\bar{z}-\varphi dz$ just span the $(0,1)$-cotangent space but not the dual of that vector field. My $\varphi$ is not a constant, I want to say it is a $\mathbb{C}$-valued function. Sorry for another mistake. But anyway, thank you both of you!

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@Suen: A comment on proper use of the tools of MO: When you are responding to comments and not providing an answer, you should either edit your question or else give your response in the form of a comment to the Question (or Answer) to which it is responding. You should not submit your comment in the form of an Answer, as you have done here. This makes it seem as though you have received an Answer to your question when you have not. –  Robert Bryant Apr 9 '13 at 14:58
    
Sorry for that, because this is my first post and I am not familiar with that. Thanks for reminding me. –  Suen Apr 9 '13 at 15:16
    
Suen, you cannot edit the question, because you have created two new user accounts separate from the one that created the question. You should register an account, and flag for moderator attention, so we can merge your accounts. –  S. Carnahan Apr 9 '13 at 23:54

Opps, a typing mistake in the computation above, it should be $d\bar{z}-\varphi dz$ actting on that vector. Because the "dual" should be the dual form of the above $(0,1)$-vector field. And that is also what the lecture document says I think.

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