# Complex vector bundle valued forms

Let $$X$$ be a smooth manifold and $$E \to X$$ a complex vector bundle. A connection on $$E$$ is a $$\mathbb{C}$$-linear map $$\nabla \colon \Gamma(X,E) \to \Omega^1(X,E)$$ where $$\Gamma(X,E)$$ denotes the space of sections of $$E$$ and $$\Omega^1(X,E)$$ is the space of the space of sections of $$T^{\vee} X \otimes_{\mathbb{R}} E$$, where $$T^{\vee} X = \mathrm{Hom}_\mathbb{R}(TX,\mathbb{R})$$.

Now if $$X$$ comes with an almost complex structure $$J$$, then one usually considers the complexified spaces. In particular $$T^{\vee}_{\mathbb{C}} X = T^{\vee} X \otimes_{\mathbb{R}} \mathbb{C} = \mathrm{Hom}_\mathbb{R}(TX,\mathbb{C})$$. This space comes with a splitting induced by the $$\pm \mathrm{i}$$-subspaces of $$J$$, i.e. $$T^{\vee}_{\mathbb{C}} X = T^{1,0}X^{\vee} \oplus T^{1,0}X^{\vee}$$. This induces a splitting $$\Omega_{\mathbb{C}}^1(X,E) = \Omega^{1,0}(X,E)\oplus \Omega^{0,1}(X,E)$$. In many textbooks it is then written that $$\nabla = \nabla^{1,0} + \nabla^{0,1}$$. But $$\nabla$$ was defined as a map into $$\Omega^1(X,E)$$. Is this the same space as $$\Omega^1_{\mathbb{C}}(X,E)$$? Because I believe that this is the space of sections of $$T^{\vee}_{\mathbb{C}}X \otimes_{\mathbb{C}} E$$, where the tensor product over $$\mathbb{C}$$ is used. So $$T^{\vee}_{\mathbb{C}}X \otimes_{\mathbb{C}} E = T^{\vee}X \otimes_{\mathbb{R}} \mathbb{C} \otimes_{\mathbb{C}} E = T^{\vee}X \otimes_{\mathbb{R}} E$$ and hence, $$\Omega^1_{\mathbb{C}}(X,E) =\Omega^1(X,E)$$. Or am I missing something? In many textbooks this construction is unfortunately not written clearly.

Yes, the situation is confusing and often not explained clearly. A connection on a real vector bundle is not quite the same as a connection on a complex vector bundle, because latter is required to be $$\mathbb{C}$$-linear.
As you say, a connection on $$E$$ is a $$\mathbb{C}$$-linear map $$\nabla \colon \Gamma(X, E) \to \Omega^1(X, E)$$. In order for $$\mathbb{C}$$-linearity to make sense, the codomain $$\Omega^1(X, E)$$ has to be a $$\mathbb{C}$$-vector space. Indeed, in the context of complex vector bundles $$E \to X$$, one usually defines $$\Omega^1(X; E) := \Omega^1(X; \mathbb{C}) \otimes_{\mathbb{C}} \Gamma(E) = \Gamma( T^\vee X \otimes_{\mathbb{R}} \mathbb{C}) \otimes_{\mathbb{C}} \Gamma(E) \cong \Gamma( \mathrm{Hom}_{\mathbb{C}}( TX^{\mathbb{C}}, E)).$$ Note that $$X$$ can be a real manifold: we don't need $$J$$ in order to write down $$T^\vee X \otimes_{\mathbb{R}} \mathbb{C}$$. We only need $$J$$ to decompose $$T^\vee X \otimes_{\mathbb{R}} \mathbb{C}$$ into $$J$$-eigenspaces.
• Thanks for your answer! I guess you mean the tensor product of $C^{\infty}(X,\mathbb{C})$-modules in your equation. Otherwise the sections would be globally in a product form.
• Alternatively, you can simply require that for any vector field $\xi$ on $X$ the operator $\nabla_\xi:\Gamma(X,E)\to \Gamma (X,E)$ is complex linear, then you don't need to complexify the tangent bundle. Commented Nov 6, 2023 at 8:37