# Obtaining the Dolbeault operator on the pullbak of the holomorphic tangent bundle.

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 $$\Phi^{*}(T^{1,0}X)$$.

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

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.