For a Riemannian manifold $(M,g)$ with exterior derivative d, the codifferential d$^\ast$ is defined to be the unique map for which $$ g(\omega,d\omega') = g(d^* \omega,\omega'), ~~~ \omega,\omega' \in \Omega^{\bullet}. $$ Now if $\ast$ is the Hodge map for $g$, then it is not too difficult to show that d$= (-1)^k\ast$ d $\ast$, when acting on $\Omega^k(M)$.

When $M$ is a complex manifold with holomorphic and anti-holomorphic partial derivatives $\partial$, and $\overline{\partial}$, we have a similarly defined $\partial^\ast$, and $\overline{\partial}^\ast$, and a similar relation between these objects and the original derivatives involving the Hodge map (well actually there's a reversal but no matter). For the Lefschetz map something similar also happens.

What I would like to know is whether the adjoint $\nabla^*$ of the Levi--Civita connection $\nabla$ has some similar re-expression? Or is this too naive?

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    $\begingroup$ Um, yr defining equation for $\mathrm{d}^*$ is not quite right: for example, the left hand side is tensorial in $\omega$ and the right hand side is not. The point is that the two sides differ by a divergence and so coincide after integrating against a volume form (so long as yr manifold is compact and without boundary). $\endgroup$ Jun 11, 2012 at 21:53

2 Answers 2


Yes, even in a more general situation: Let $E\to M$ be a vector bundle with metric and metric connection $\nabla.$ Then there exists $d^\nabla\colon\Omega^k(M,E)\to\Omega^{k+1}(M,E)$ satisfying $d^\nabla(s\otimes\omega)=\nabla s\wedge\omega+s\otimes d\omega,$ and also the Hodge star extends to $\Lambda T^*M\otimes E.$ With this it is a nice exercise to compute that the adjoint of $d^\nabla$ is $\delta^\nabla=(-1)^? * d^\nabla *,$ where the sign is the same as for the usual codifferential.


Unfortunately this does not work, because the Hodge $*$ operator commutes with the Levi-Civita connection. Indeed, we have $$ \nabla (\langle u,v \rangle dV) = \langle \nabla u, v \rangle dV + (-1)^m \langle u, \nabla v \rangle dV = \nabla u \wedge * v + (-1)^m u \wedge *\nabla v $$ because $\nabla dV = 0$ since the metric $g$ is parallel with respect to $\nabla$. The left hand side of this formula is again $$ \nabla (u \wedge * v) = \nabla u \wedge *v + (-1)^m u \wedge \nabla(*v), $$ from which we get $\nabla * = * \nabla$. It follows that $* \nabla * = ** \nabla = (-1)^l \nabla$ for some $l$.

There is an expression for $\nabla^*$ in terms of a local orthonormal frame in Werner Ballmann's Lectures on Kahler manifolds (Proposition 1.27, Chapter 1, p. 11) that says that if $(X_1, \ldots, X_n)$ is such a frame, and if $\hat\nabla$ is the dual connection, then $$ \nabla^*u = - \sum_j X_j \llcorner \hat\nabla_{X_j} u. $$ I don't know if this is what you're looking for, if not you might have more luck with Bocher-Weitzenböck type identities.

  • $\begingroup$ Hmm... I just saw Sebastian's answer and am confused now. I think he's right, since if the metric is flat my answer would imply that the Laplacian was zero, which is absurd. On the other hand I can't quite figure out where I went wrong. $\endgroup$ May 16, 2012 at 8:19
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    $\begingroup$ Dear Gunnar, of course, the Hodge star is parallel, but the wedge product does not commute with the Hodge star, but becomes the insertion operator. $\endgroup$
    – Sebastian
    May 16, 2012 at 8:22

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