how to understand the tensor product canonical line bundle $\otimes$ dual bundle

Question

Suppose we have a Riemann surface $M$ together with a holomorphic vector bundle $E \to M$ of rank n. let $K$ denote the canonical line bundle and let $E^*$ denote the dual bundle

I am trying to understand the tensor product $K \otimes E^*$.

I have lots of trouble, because I need to do this on my own and my background in differential geometry and multilinear algebra is not strong.

I shall try to explain how I would describe $K \otimes E^*$ locally, if somebody could comment on what goes wrong that would be immensely helpful!

The canonical bundle $K$ over a Riemann surface $M$ is the cotangent bundle, or the bundle of holomorphic $1$-forms. In local coordinates $(z,\overline{z})$ an element of $K$ can be written as $$ \omega = \frac{i}{2} f\,dVol_h $$ where $f$ is a holomorphic function on $M$ (or at least defined locally) and $dVol_h = \frac{i}{2}\, h dz\wedge d\overline{z}$, the Volume element induced by the metric $h\,dz\,d\overline{z}$ on $M$ (here again $h$ is a holomorphic function on $M$).

Question 1: Is this a correct way to describe the canonical bundle locally ?

The dual bundle $E^*$ can be described locally given a choice of basis $\{e_1,\dots, e_n\}$ for the bundle $E$. We take the dual basis $\{\phi^1,\dots\phi^n\}$ (so $\phi_k (e_l) = \delta^k_l$) and write $$ \sigma = \sum^n_{k = 1} g_k\,\phi_k $$ where the coeficients $g_k$ are holomorphic functions locally defined on $M$.

Question 2: is this description sufficient ? i fear the locality of what I want to show is not emphasized enough but i am not sure how the local coordinates $(z,\overline{z})$ on a patch $V \subset M$ (say) should be mentioned here. do i need to invoke local coordinates on $E$, or is this done by specifying the basis?

Therefore the bundle $K \otimes E^*$ consists of tensor fields which have local description $$ \omega \otimes \sigma = \sum^n_{k = 1} (\frac{i}{2}\,h\,f\,g_k)\, dz \wedge d\overline{z} \otimes \phi_k $$

Question 3: does this formula makes sens ? I am "very unsure" here - I have a wedge product and a tensor symbol and don't know how to write this in a correct way. Below I attempt to understand such an object, maybe my last lines give away more of my misunderstandings:

an element in $K \otimes E^*$ can be interpret as a map $TM \otimes E \to \mathbb{C}$. alternatively we can also think of it as something that can be integrated, that which would amount to a contraction of the tensor.

Question 4: do these interpretations make sense ? how would I write them out as rigorous definitions?

Many thanks for comments and help!!!

Answer 1

Your first and biggest misconception is that you seem to be mixing up real and complex dimensions. A Riemann surface is a $2$-dimensional real manifold, but as a complex manifold it is $1$-dimensional. Hence you have just one local complex coordinate on $M$, namely $z$. The local coordinates are not $(z, \bar{z})$.

There is more structure on the tangent bundle of $M$ and the bundle of $k$-forms on $M$. Here's the general picture. Let $X$ be a $2n$-dimensional complex manifold, and consider local coordinates $(x_1, \dots, x_n, y_1, \dots, y_n)$. Using the complex structure on the tangent bundle (which we denote $i$) we can locally define vector fields $$\frac{\partial}{\partial z_j} = \frac{1}{2} \left(\frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right),$$ $$\frac{\partial}{\partial \bar{z}_j} = \frac{1}{2} \left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right).$$ We also locally have the dual basis of $1$-forms $$dz_j = dx_j + i ~dy_j,$$ $$d\bar{z}_j = dx_j - i ~dy_j.$$ Then we get a splitting of the cotangent bundle $$T^\ast M = \Lambda^{1,0} T^\ast M \oplus \Lambda^{0,1} T^\ast M,$$ where pointwise $\Lambda^{1,0} T^\ast M$ is spanned by the $dz_j$ and $\Lambda^{0,1} T^\ast M$ is spanned by the $d\bar{z}_j$. More generally, we have the splitting $$\Lambda^k T^\ast M = \Lambda^{k,0} T^\ast M \oplus \Lambda^{k-1, 1} T^\ast M \oplus \cdots \oplus \Lambda^{1, k-1} T^\ast M \oplus \Lambda^{0,k} T^\ast M,$$ where for $p + q = k$, $\Lambda^{p,q} T^\ast M$ is spanned pointwise by the $dz_{j_1} \wedge \cdots \wedge dz_{j_p} \wedge d\bar{z}_{j_{p+1}} \wedge \cdots \wedge d\bar{z}_{j_k}$. Sections of $\Lambda^{p,q} T^\ast M$ are called $(p,q)$-forms, sections of $\Lambda^{k,0} T^\ast M$ are called holomorphic $k$-forms, and sections of $\Lambda^{0,k} T^\ast M$ are called antiholomorphic $k$-forms.

On a Riemann surface, the canonical bundle is the bundle of holomorphic $1$-forms, i.e. $$K = \Lambda^{1,0} T^\ast M.$$ Therefore in terms of a local coordinate $z$, a section $\alpha$ of $K$ is described by $$\alpha(z) = f(z) ~dz$$ for some holomorphic function $f$. Your expression of the form $$\omega = F ~dz \wedge d\bar{z}$$ would locally describe a section of $\Lambda^{1,1} T^\ast M$, not a holomorphic $1$-form.

Your local description is closer to being ok. Here I would let $\{e_1, \dots, e_n\}$ be a local frame for $E$, i.e. a set of $n$ local sections that form a basis for each fiber $E_z$ of $E$ for $z$ in our coordinate neighborhood. Then $\{\phi_1, \dots, \phi_n\}$ would be the dual frame defined by $$\phi_i(z)(e_j(z)) = \delta_{ij}$$ for all $z$ in our coordinate neighborhood. Then locally a section $\sigma$ of $E^\ast$ would look like $$\sigma(z) = \sum_{k = 1}^n g_k(z) ~\phi_k(z) \tag{$\ast$}$$ for each $z$ in our coordinate neighborhood.

Putting the above together, a section of $K \otimes E^\ast$ is a linear combination of sections of the form $\alpha \otimes \sigma$, where $\alpha$ is a section of $K$ and $\sigma$ is a section of $E^\ast$, and locally such a $\alpha \otimes \sigma$ looks like $$(\alpha \otimes \sigma)(z) = \sum_{k = 1}^n f(z)g_k(z) ~dz \otimes \phi_k(z)$$ for each $z$ in our coordinate neighborhood.

If $TM$ is the full holomorphic tangent bundle of $M$, then a section $\alpha \otimes \sigma$ of $K \otimes E^\ast$ can be considered as a map from sections of $TM \otimes E$ to the space of holomorphic functions on $M$ as follows. Let $\alpha \otimes \sigma$ have the form $(\ast)$ determined above. A section of $TM \otimes E$ is a linear combination of sections of the form $v \otimes s$, where $v$ is a section of $TM$ and $s$ is a section of $E$. Locally we have $$(v \otimes s)(z) = \left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \otimes \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right).$$ Then we have \begin{align*} (\alpha \otimes \sigma)(v \otimes s)(z) & = \alpha(v)(z) \sigma(s)(z) \\ & = f(z) ~dz\left( a(z) \frac{\partial}{\partial z} + b(z) \frac{\partial}{\partial \bar{z}} \right) \sum_{k = 1}^n g_k(z) ~\phi_k(z) \left( \sum_{j = 1}^n h_j(z) ~e_j(z) \right) \\ & = f(z) \left( a(z) \cdot 1 + b(z) \cdot 0 \right) \sum_{k = 1}^n \sum_{j = 1}^n \delta_{kj} g_k(z) h_j(z) \\ & = f(z)a(z) \sum_{k = 1}^n g_k(z)h_k(z). \end{align*} When the above is considered over a single point, we see how to map an element of $TM \otimes E$ to $\Bbb C$ using an element of $K \otimes E^\ast$.