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 nontrivial 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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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! 


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. 

