Let $X$ be smooth (real) manifold and let $T^{*}(X)_{\mathbb{C}}$ denote the complexification of the cotangent bundle. We define the complex valued differential r-forms on $X$ to be the smooth sections of the bundle $\wedge^r_{\mathbb{C}} T^{*} (X)_{\mathbb{C}} \to X$ (for instance see page 31 of Differntial Analysis on Complex Manifolds by Raymond Wells). Denote the vector space of these complex-valued r-forms by $\Omega^r(X)_{\mathbb{C}}$.

Apparently, we can define a version of the exterior derivative on complex-valued r-forms by taking the real exterior derivative and "extending by complex linearity" but I cannot understand why this is true.

One approach would be to prove that $\Omega^r(X)_{\mathbb{C}}$ is isomorphic as a vector space to $\Omega^r(X) \otimes_{\mathbb{R}} \mathbb{C}$ (where $\Omega^r(X)$ denotes the usual real differential r-forms) but I cannot see how to prove this.

  • $\begingroup$ Why not just do it locally? That's a standard way of defining the real exterior derivative, after all. $\endgroup$ – user98602 Aug 25 '15 at 21:09

A complex differential form can be thought of as two real differential forms - the real and imaginary parts, just like complex numbers. As you say, the exterior derivative is then the linear extension of the real exterior derivative. Namely,$$d(\alpha+i\beta)=d\alpha+id\beta.$$If you want a rigorous proof for the isomorphism mentioned in the last paragraph, you can do it locally, as Mike comments. This is the standard way to handle vector bundles.


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