# Complex-valued differential forms

In Assignment 4, you get to work with complex-valued differential forms. These work mostly the same as real-valued differential forms, but there are a couple additional features.

• Recall that the wedge product for real-valued two 1-forms $\alpha$, $\beta$ is defined as
$\alpha \wedge \beta (u, v) = \alpha(u)\cdot \beta(v) – \alpha(v)\cdot \beta(u),$
where “$\cdot$” in the usual product for real numbers. The wedge product for complex-valued 1-forms is identical, except that $\cdot$ is replaced with the complex product
$(a + bi) \cdot (c + di) = ac – bd + (ad + bc)i.$
• A special operator on the complex numbers is the conjugation map $z \mapsto \bar{z}$, where $\overline{a + bi} = a – bi$. This operator can be applied to complex-valued forms, too. For a $k$-form $\alpha$ the conjugate will be
$\overline{\alpha}(X_1, …, X_k) = \overline{\alpha(X_1, …, X_k)}$
Note that conjugations commutes with the exterior calculus operators $d$, $\star$ and $\wedge$. That is, $\overline{d\alpha} = d\overline{\alpha}$, $\overline{\star \alpha} = \star \overline{\alpha}$ and $\overline{\alpha \wedge \beta} = \overline{\alpha} \wedge \overline{\beta}$.

Please ask in the comments if there is anything else you need to be clarified about complex-valued differential forms.

## 5 thoughts on “Complex-valued differential forms”

1. plav says:

In exercise 29:
$\bar u v = u \cdot v + (u \times v) i$
I am confused with this expression. It says that $u$ and $v$ are complex numbers (so $\bar u$ is also a complex number). The left side expresses a regular multiplication, while the right side has dot and cross product. What is happening here? Should I interpret the the real and complex components as coordinates of a vector and treat the left side as the wedge product?

1. plav says:

Solved. It is possible to figure it out by doing the exercise.

2. Josh says:

Since some people had questions about the identity, $\star \overline{\alpha} = \overline{\star \alpha}$, here’s some justification for why it is correct.

If $z$ is a holomorphic map (possibly conformal), then the identity $\star d z = i d z.$ At first, this appears to be a counterexample, but observe that $\bar z$ is antiholomorphic, so it satisfies $\star d \overline{z} = -id \overline{z}.$ Thus, $\star \overline{dz} = -id\overline{z} = \overline{i d z} = \overline{\star d z}.$ Thus, the identity is consistent for conformal maps.

1. akwon says:

One small question-you start by assuming that z is a holomorphic map, and mention that it may be conformal. The phrasing seems to suggest that we do not assume it is conformal, but aren’t holomorphic maps necessarily conformal?

1. Josh says:

Good question. Conformal maps are always holomorphic maps, but a conformal map must always have a nonzero derivative. For example, $z(w) = w^2$ is holomorphic but not conformal at the origin. See https://en.wikipedia.org/wiki/Conformal_map for more details.