I have ran into a question while reading this paper by Witten. The question is mostly mathematical, so I thought it would be better posed here instead of on Physics SE.

Let $X$ be a Kahler manifold, $\Sigma$ a Riemann surface and $\Phi:\Sigma\to X$ be a smooth map. Let $TX=T^{1,0}X\oplus T^{0,1}X$ be the complexified tangent bundle of $X$.

I am interested in understanding what the Dolbeault operator on $\Phi^{*}(T^{1,0}X)$ looks like.

My first attempt was to assume that the Dolbeault operator on $\Phi^{*}(T^{1,0}X)$ should be the pullback of the Dolbeault operator on the holomorphic tangent bundle $T^{1,0}X$, which itself should be obtained from the splitting of the Hermitian connection $\nabla$ on $X$ into its $(0,1)$-form part.

I think I mostly understand how $\nabla$ works (it is similar to the Levi-Civita connection, with which I am somewhat familiar), and from there I seem to be able to understand the splitting, $\nabla s =\partial s + \bar{\partial}s$.

My issue is that $\bar{\partial}$ seems to be zero on $T^{1,0}X$. In particular, we can locally write a section of the holomorphic tangent bundle as $s=v^{i}\partial_{z^{i}}$ with holomorphic coefficient functions, then: $$ \nabla_{\partial_{z}^{i}} (s) = \left( \frac{\partial v^{j}}{\partial z^{i}} + v^{k}\Gamma^{j}_{ik} \right) \partial_{z^{j}} $$ Then as far as I can tell we have: $$ \nabla s = \left( \frac{\partial v^{j}}{\partial z^{i}} + v^{k}\Gamma^{j}_{ik} \right) dz^{j}\otimes \partial_{z^{i}} \in\Omega^{1,0}(X)\otimes T^{1,0}X $$ i.e. $\nabla s$ is a $(1,0)$ form-valued section, and thus $\bar{\partial}s=0$. This also fits with the intuitive idea that $\bar{\partial}$ of a holomorphic vector field should vanish.

I have good reason to believe that $\bar{\partial}$ on $\Phi^{*}(T^{1,0}X)$ is not trivial, so it seems like this is the wrong way to obtain this operator.

My next guess would be to pull back $\nabla$ via $\Phi$, and then separate the result into holomorphic and anti-holomorphic parts, but I am not so sure about this.

I would appreciate any help understanding how to correctly obtain/define $\bar{\partial}$ on $\Phi^{*}(T^{1,0}X)$, as well as any verification that what I have done above is correct.

EDIT: As far as I know the pullback of a holomorphic vector bundle by a non-holomorphic map needs not be holomorphic, so I don't even see why $\bar{\partial}$ makes sense on $\Phi^{*}(T^{1,0}X)$.

  • $\begingroup$ You’re wrong because $\bar\partial$ annihilates holomorphic $1$-forms (on a Riemann surface), but certainly not all $(1,0)$-forms. $\endgroup$ Nov 10, 2021 at 2:48
  • $\begingroup$ Thanks for your comment. I agree with your statement, but I'm not quite sure which part of my question you're referring to. $\endgroup$
    – CoffeeCrow
    Nov 10, 2021 at 3:15
  • $\begingroup$ Did you not say, "$\bar\partial$ seems to be $0$ on $T^{1,0}X$"? $\endgroup$ Nov 10, 2021 at 3:26
  • $\begingroup$ Yes, but $T^{1,0}X$ is the holomorphic tangent bundle, not the bundle of $(1,0)$-forms. $\endgroup$
    – CoffeeCrow
    Nov 10, 2021 at 4:25
  • $\begingroup$ Sorry. I misspoke. My comment still holds. Only holomorphic sections are annihilated, not general smooth sections. $\endgroup$ Nov 10, 2021 at 5:51

1 Answer 1


My misconception was thinking that sections of holomorphic bundles were automatically holomorphic. This is not the case. Additionally, it turns out that my second idea for how to construct $\bar{\partial}$ is the correct one. In particular, with $\bar{\partial}:=(\Phi^{*}\nabla)^{0,1}$, we have: $$ \bar{\partial}(s) = d\bar{z}\otimes(\partial_{\bar{z}}v_{j}+\partial_{\bar{z}}z^{i}\Gamma^{j}_{ik}v^{k})\partial_{z^{j}} $$ This is as expected.


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