1
$\begingroup$

Let $M$ be a Riemann surface, the Hodge star is defined as $ \alpha \wedge *\beta = \langle \alpha, \overline{\beta} \rangle vol$ where $\langle \cdot , \cdot \rangle$ is the Hermitian product on the complexified cotangent bundle extending the Euclidean one. This definition implies that $*$ is complex linear, and therefore in local coordinates: $*dz= *(dx + idy) = dy + i(*dy) = -i dz$. The problem that I have is that I have read that $*$ maps $(1,0)$-forms to $(0,1)$-forms (in general $*: \Omega^{p,q} \rightarrow \Omega^{n-q,n-p}$) and therefore $*dz$ should be proportional to $d\overline{z}$

$\endgroup$
  • 2
    $\begingroup$ But with your formula you get $dz\wedge \star dz = 0$, don't you? Oh, that's consistent, I guess. Many authors write $\bar\star$ to get the second mapping. $\endgroup$ – Ted Shifrin Aug 28 '18 at 20:38
  • $\begingroup$ @TedShifrin yes I obtain $dz\wedge *dz = 0$ the definition in the first line is that of page 33 in Huybrechts (actually the equivalent $\alpha \wedge \overline{\beta} = \langle \alpha, \beta \rangle$ vol). But later on the same page he states that $*: \Omega^{p,q} \rightarrow \Omega^{n-q,n-p}$ $\endgroup$ – inquisitor Aug 28 '18 at 20:49
  • 1
    $\begingroup$ That's actually consistent. Then $\bar\star$ maps $\Omega^{p,q}\to\Omega^{n-p,n-q}$ so that $\alpha\wedge\bar\star\alpha$ is something you integrate. But note that when $p=1$ and $q=0$, then $1-q=1$ and $1-p=0$, so Huybrechts is self-consistent. Your last sentence is actually wrong. I didn't pay attention earlier. Sorry. $\endgroup$ – Ted Shifrin Aug 28 '18 at 21:16
  • $\begingroup$ Thanks, I was actually stupidly doing wrong the arithmetic! $\endgroup$ – inquisitor Aug 28 '18 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.