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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – KuSi
    Commented Nov 5, 2023 at 21:47
  • $\begingroup$ @KuSi: Yes, that's right. $\endgroup$ Commented Nov 6, 2023 at 5:13
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 6, 2023 at 8:37

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