# Adjoint of the covariant derivative on a Riemannian manifold

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?

• Well, it is the adjoint of the covariant derivative... with such a name that sounds like a nobiliary title, what more can it want?! :) Mar 5, 2012 at 4:41
• @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?
– Hans
Mar 5, 2012 at 12:32
• 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). Oct 1, 2012 at 3:22

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