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. 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?

  2. 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. 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?

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