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!!!


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 $K \otimes E^\ast$.

  • $\begingroup$ thank you very much for your comments, I already learned a lot from them! there is just one question I was wondering about - you describe the canonical line bundle locally in terms of the vector $dz$, and on the other hand you span the tangent bundle $TM$ by two vectors, $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \overline{z}}$. In the last line (where you evaluate the map) this results in a "loss" of the coefficient $b(z)$. It looks a bit strange, can you maybe explain this a little? Thanks again! $\endgroup$ – harlekin Jul 13 '12 at 10:21
  • $\begingroup$ @harlekin I added a new second paragraph (as well as various other tweaks) which hopefully clears this up for you. $\endgroup$ – Henry T. Horton Jul 13 '12 at 22:19
  • $\begingroup$ this cleared up all problems! :) - thanks !! $\endgroup$ – harlekin Jul 14 '12 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.