So, I have a question concerning $T_{X,x}^{\ast0,1}$ (defined below) when $X$ is a Riemann Surface. Let me give a little bit of background to whet your appetite (for question answering!)


So, we are dealing with a Riemann Surface $X$ (a connected complex manifold of dimension one). Given a point $x\in X$ we can form the holomorphic cotangent bundle at $x$, denoted $T_{X,x}^{\ast1,0}$ for me, as follows: start by considering holomorphic function elements $(f,U)$ at $x$--namely, $U$ is some neighborhood of $x$, and $f\in\mathcal{O}(U)$ (holomorphic functions on $U$). We denote the set of all such function elements at $x$ by $\mathcal{H}_x$. We can then define an equivalence relation on $\mathcal{H}_x$ by declaring that $(f,U)\sim(g,V)$ if $f\mid_W=g\mid_W$ for some open neighborhood $W\subseteq U\cap V$ of $x$. The set of equivalence classes of $\mathcal{H}_x/\sim$ will be denoted $\mathcal{O}_{X,x}$--called the holomorphic germs at $x$.

We then define an equivalence relation on the germs, by saying that two germs $f$ and $g$ at $x$ (domains are unimportant) are equivalent if $(f\circ z^{-1})'(z(x))=(g\circ z^{-1})'(z(x))$ for some chart $(z,U)$ at $x$. We denote the quotient of $\mathcal{O}_{X,x}$ by this equivalence relation as $T_{X,x}^{\ast1,0}$ and call it, as I've already said, the holomorphic cotangent bundle at $x$. We denote the equivalence class of $f$ in $T_{X,x}^{\ast1,0}$ as $df(x)$. $T_{X,x}^{\ast1,0}$ carries a well-defined vector space structure which is defined (and it is well-defined!) by $df(x)+\lambda dg(x)=d(f+\lambda g)(x)$.

We can then glue all of the holomorphic cotangent spaces together to get the holomorphic cotangent bundle $T_X^{\ast1,0}$ which is a holomorphic line bundle with local trivilizations $\pi^{-1}(U)\to U\times\mathbb{C}:cdz(x)\mapsto (x,c)$ (where, as usual, $(z,U)$ is a chart on $X$--$\{dz(x)\}$ forms a basis of $T_{X,x}^{\ast1,0}$ so this makes sense) and transition functions $t_{\alpha\beta}=(z_\beta\circ z_\alpha^{-1})')\circ z_\alpha$.

Ok, so that's all fine. It makes one-hundred percent sense to me--both symbolically and intuitively. The issue then comes in with defining $T_{X,x}^{\ast0,1}$. I understand it symbolically, it's just the tangent bundle one gets by conjugating the holomorphic cotangent bundles transition functions. The issue is with the intuition. The holomorphic tangent and cotangent bundle were defined so concretely, it would be nice to have a concrete description of $T_{X,x}^{\ast0,1}$.

After a little bit of thought, it seems that there is a very natural/concrete way to define $T_{X,x}^{\ast0,1}$. Namely, perform almost the exact same procedure one did with the holomorphic cotangent bundle except one replaces holomorphic germs with antiholomorphic germs (i.e. replace normal holomorphic functions $f:X\to\mathbb{C}$ with functions whose conjugate is holomorphic) and replace all the derivatives with respect to $z$, to derivatives with respect to $\bar{z}$.

Now, I'm fairly convinced that this is all kosher, but it begs two questions: a) why isn't this the definition, why is the definition via just transition functions favorable and b) $T_{X,x}^{\ast0,1}$ is not a holomorphic line bundle, but it's damn close--it's some kind of "antiholomorphic line bundle"--does this idea hold any significance?



You can construct the antiholomorphic cotangent bundle in the way you suggest, and it is an antiholomorphic linebundle.

The fact that this description is not emphasized in what you're reading is (I guess) just for convenience; given a careful description of the holomorphic cotangent bundle, it's easy just to define the antiholomorphic tangent bundle by passing to the conjugate. Also, you don't need to particularly think of the conjugate as being taken via transition functions (although it's fine to do so); you can also think of $T^{* 1,0}$ as a complex subbundle of the complexification of the (2-dimensional) tangent bundle of $X$ (thought of as a real manifold), and then $T^{* 0,1}$ is just the conjugate subbundle, where the conjugation occurs on the coefficients.


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