# 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 $\overline{\alpha \wedge \beta} = \overline{\alpha} \wedge \overline{\beta}$.

## 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 arenâ€™t holomorphic maps necessarily conformal?