Let $\nabla_X$ be the covariant derivative on a Riemannian manifold w.r.t. the vector field $X$. It is not clear to me what the (formal) adjoint of this operator is: I mean the operator $\tilde\nabla_X$ satisfying (for let's say $\alpha,\beta$ 1-forms with compact support)

$$\langle\nabla_X \alpha,\beta\rangle = \langle\alpha, \tilde \nabla_X \beta\rangle.$$

Does this operator have a special name or geometric meaning?

Many thanks for your help.

  • 3
    $\begingroup$ Well, it is the adjoint of the covariant derivative... with such a name that sounds like a nobiliary title, what more can it want?! :) $\endgroup$ Mar 5, 2012 at 4:41
  • $\begingroup$ @Mariano Suárez-Alvarez. Hmmm, ok, I'm happy with the name :-). I was wondering if there is some (well-knwon, standard, useful, or just bit more concrete,....) representation of it. For example, such representations are found in many books for the adjoint of the exterior differential $d$ on forms (involving explicitly the metric tensor, or the hodge star). Why the adjoint of the $d$ appears in many books and the adjoint of $\nabla_X$ does not? Is it just because I don't know well the literature? $\endgroup$
    – Hans
    Mar 5, 2012 at 12:32
  • 2
    $\begingroup$ Similarly to the role of the adjoint of $d$ in defining the Hodge Laplacian on $k$-forms, given a connection $\nabla$ on a vector bundle $E \rightarrow M$, the operator $\nabla^* \nabla$ is a second order elliptic operator called sometimes the Bochner Laplacian. It has many uses in Riemannian Geometry (for example, in the application of the Bochner technique. See Petersen). $\endgroup$
    – levap
    Oct 1, 2012 at 3:22

1 Answer 1


You can explicitly compute the adjoint by integrating by parts: the metric-compatibility of $\nabla$ gives $$ \begin{align} g(\nabla_X \alpha, \beta) &= X g(\alpha,\beta) - g(\alpha, \nabla_X \beta) \\ &=\text{div}(g(\alpha,\beta)X)-g(\alpha,\beta)\text{div}(X)-g(\alpha,\nabla_X \beta) \end{align}$$

and thus integrating over a region containing the supports of $\alpha$ and $\beta$ you get

$$\langle \nabla_X \alpha, \beta \rangle = \langle\alpha,-\text{div}(X) \beta-\nabla_X\beta\rangle$$

so $\nabla_X^* = -\text{div}(X) - \nabla_X$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .