# Hodge decomposition on Riemann surface

On a compact Riemannian manifold $M$ the Hodge decomposition takes the form $$\Omega^k(M)=d\Omega^{k-1}(M)\oplus\mathcal{H}(M)\oplus d^*\Omega^{k+1}(M)$$ Where $d^*$ is the adjoint of $d$ w.r.t. the inner product induced by the metric.

Now on a Riemann surface we do not have a metric, but we do have a canonical conformal structure. This conformal structure is enough to give us a Hodge-$\star$ operator $$\star:\Omega^1(M)\to\Omega^1(M)$$ In this setting the Hodge decomposition takes on the form $$\Omega^1_2(M)=E\oplus\star E\oplus\mathfrak{h}(M)$$ Where $$E=\overline{\{df\mid f\in \Omega^0(M)\}}$$ $$\star E=\overline{\{\star df\mid f\in \Omega^0(M)\}}$$ $$\mathfrak{h}(M)=\{\omega\in\Omega^1(M)\mid d\star \omega=d\omega=0\}$$ Where closures are in the $L^2$ sense.

These two decomposition looks quite different to me. The term $d\Omega^{k-1}(M)$ looks like $E$. But I do not recognise an analogue of $d^*$ in the Riemann surface case. $$d^*:\Omega^k(M)\to \Omega^{k-1}(M)$$ is usually constructed as $\star d\star$, however this requires a $\star:\Omega^2(M)\to\Omega^0(M)$, but the conformal structure on $M$ is not enough to give us this $\star$.

Also, the $\mathcal{H}$ in the Riemannian manifold case is characterised by a condition involving $d^*$, which we do not have on a Riemann surface, so I don't see how $\mathcal{H}$ compares to $\mathfrak{h}$.

So my question is if these two forms of the Hodge decomposition are in some sense the same, or if these are fundamentally different decompositions? The absence of a metric on the Riemann surface suggests that we should get a slightly coarser decomposition, since a metric is more rigid then a conformal structure. However there are enough similarities between the decompositions to make me think there must be some connection.

The hodge star is an isomorphism $\star \colon \Omega^k(M) \rightarrow \Omega^{n-k}(M)$. If you apply this observation with $n = k = 2$, you will see that

$$d^{*}(\Omega^2(M)) = \{ \star d \star \omega \, | \, \omega \in \Omega^2(M) \} = \{ \star df \, | \, f \in \Omega^{0}(M) \} = \star \{ df \, | \, f \in \Omega^{0}(M) \}.$$

If you apply it to $n = 2, k = 1$, you will see that

$$d^{*}(\omega) = \star (d (\star \omega)) = 0 \iff d (\star \omega) = 0$$

and so $\mathfrak{h}(M) = \mathcal{H}(M)$. Thus, you get the same direct sum decomposition.

• Thank you for your answer. So I see how the Riemannian manifold case can be written in the same form as the Riemann surface case. But in a way that is remarkable right? Because we miss a lot of the structure that seems to be necessary, namely the metric. So it seems that the combination of the low dimension and the presence of the conformal structure is just enough to make this work? – user2520938 Jun 4 '16 at 13:50
• The hodge star is always conformally invariant in the middle dimension (if the dimension is $2n$, and you multiply the inner product by $\lambda^2$ then the inner product on $n$ forms scales by $\lambda^{2n}$ and the volume form scales by $\lambda^{-2n}$) so you can repeat this argument to obtain a decomposition of $\Omega^{n}(M^{2n})$ which depends only on a conformal class of metrics on $M$. We don't really miss the metric - the decomposition in the middle level sees only metrics up to conformal equivalence. – levap Jun 4 '16 at 14:16
• The nice feature of complex one-dimensional manifolds (Riemann surfaces) is that they come with a canonical conformal class - this is unique for (complex) dimension one. – levap Jun 4 '16 at 14:16
• But I think it's not just that we have a conformal structure on a Riemann surface. Given a conformal structure on a higher dimensional complex manifold we would still have a $\star$ operator on the mid dimensional forms, but this wouldn't be as useful; we would miss too many operators on the higher forms. The low dimension of the Riemann surface makes it so that the missing $\star$ on $0$ and $2$ forms doesn't really matter. That's the way I see it anyway. – user2520938 Jun 4 '16 at 14:39