# Applying $\bar \partial$ to a tensor product of holomorphic vector bundles bundles

Reading Griffiths and Harris, it is stated as an "obvious" fact that for a tensor product $E \otimes E'$ of holomorphic vector bundles, $(D_E \otimes 1 + 1 \otimes D_{E'})'' = \bar \partial$ where $D_E$ is the unique connection compatible with the complex structure of $E$ (so its complex component $D_E''$ is equal to $\bar \partial$).

However, I don't really understand how $\bar \partial$ is being applied to the tensor product of vector bundles.

A couple pages back it is explained how if we choose a local holomorphic frame we can essentially try to apply $\bar \partial$ "normally" (as if we were in $\mathbb{C}^n$), and in the case we are looking at (applying it to zero forms) I think it should locally look like $\sum_i d\bar z_i \frac{\partial}{\partial \bar z_i}$ but how do we get from here to the description in terms of tensor products, and from that to $(D_E \otimes 1) + 1 \otimes D_{E'})'' = \bar \partial$?

Clearly I am very confused so any better explanation of this would help.

$\newcommand{\dbar}{\bar{\partial}}\newcommand{\Cpx}{\mathbf{C}}$If I understand what you're asking, the equation in question is an expression of the Leibniz rule.

First off, if $E \to M$ is a holomorphic vector bundle of rank $k$, and if $s$ is a smooth section, then $\dbar s$ is a well-defined $E$-valued $(0, 1)$-form: With respect to a trivialization over a neighborhood $U_{\alpha}$, the section $s$ is represented by a function $s_{\alpha}:U_{\alpha} \to \Cpx^{k}$. If $s_{\beta}$ is a trivialization over an open set $U_{\beta}$, there exists a holomorphic transition function $g_{\alpha\beta}:U_{\alpha} \cap U_{\beta} \to \Cpx^{k \times k}$ such that $s_{\beta} = g_{\alpha\beta} s_{\alpha}$. By the Leibniz rule (and since $\dbar g_{\alpha\beta} = 0$), $$\dbar s_{\beta} = \dbar(g_{\alpha\beta} s_{\alpha}) = (\dbar g_{\alpha\beta}) s_{\alpha} + g_{\alpha\beta} (\dbar s_{\alpha}) = g_{\alpha\beta} (\dbar s_{\alpha}).$$ That is, the local representatitives of $\dbar s$ transform like sections of $E$.

Now suppose $F \to M$ is a holomorphic vector bundle. (Using $F$ instead of $E'$ since primes also denote the type decomposition.) If $s_{E}$ and $s_{F}$ are smooth sections of $E$ and $F$, then again by the Leibniz rule $$\dbar(s_{E} \otimes s_{F}) = (\dbar s_{E}) \otimes s_{F} + s_{E} \otimes (\dbar s_{F}) = (\dbar \otimes 1 + 1 \otimes \dbar)(s_{E} \otimes s_{F}).$$ More properly, using subscripts to denote the corresponding bundles, $$\dbar_{E \otimes F}(s_{E} \otimes s_{F}) = (\dbar_{E} s_{E}) \otimes s_{F} + s_{E} \otimes (\dbar_{F} s_{F}) = (\dbar_{E} \otimes 1 + 1 \otimes \dbar_{F})(s_{E} \otimes s_{F}),$$ or $D_{E \otimes F}'' = D_{E}'' \otimes 1 + 1 \otimes D_{F}''$.

• How are you applying the Leibniz rule to the tensor product like that? I only know the Leibniz rule (for connections) as applying to the product of a function and a section like $D(f s) = df *s + f * Ds$
– Carl
Jun 24, 2015 at 6:43
• In fancy terms, the Leibniz rule appears any time you differentiate a bilinear function. More concretely, if you write $s_{E} = \sum_{i} s_{E}^{i} e_{i}$ and $s_{F} = \sum_{j} s_{F}^{j} f_{j}$ as linear combinations of local holomorphic sections, then the components of $s_{E} \otimes s_{F} = \sum_{i,j} s_{E}^{i} s_{F}^{j} e_{i} \otimes f_{j}$ are the (ordinary) products of the components of $s_{E}$ and $s_{F}$. Jun 24, 2015 at 11:34
• Okay, so we are just using the fact that $\bar \partial (e_i \otimes f_j) = 0$ since they are holomorphic, and then using the calculus Leibniz rule on the $S_E^i s_F^j$?
– Carl
Jun 24, 2015 at 14:55
• Yes, exactly. :) Jun 24, 2015 at 15:31