Let $E \to M$ be a vector bundle over some Riemannian metric $(M, g)$ and endow it with some fibre metric. Assume that covariant derivative $\nabla$ is compatible with the metric.

It is essentially an application of Stokes theorem to derive the following identity $$\nabla^*_X = - \nabla_X - div(X)$$ for the formal adjoint of $\nabla_X$ with respect to the usual $L^2$-scalarproduct.

Question 1: Is there a closed expression for the formal adjoint of $\nabla$ regarded as a map $\Gamma^\infty(E) \to \Gamma^\infty(T^*M \otimes E)$, i.e. $\nabla^*: \Gamma^\infty(T^*M \otimes E) \to \Gamma^\infty(E)$.

I'm especially interested in the case where one drops the compatibility condition on $\nabla$ and the metric. The background for this question lies in the equation $\nabla^* \nabla \phi = - \text{Tr}_g (\nabla^2_{\cdot, \cdot} \phi)$ for the Bochner Laplacian. I can see where it steams from for metric connections, but not for arbitary $\nabla$. Is this forumla just taken as a generalization of the compatible case?

Question 2: The word 'formal' refers to the fact that the adjoint is only a 'true' adjoint operator after extension to appropriate Hilbert spaces. Exists a definition of adjointness on Fréchet spaces such that one can view $\nabla^*$ as a true adjoint operator? (Assume that $M$ is compact so that the section spaces are Fréchet spaces). I had the idea to represent $\Gamma^\infty(E)$ as the projective limit of Hilbert spaces and then piece the the adjoints on the building blocks together, but failed working out the details.


3 Answers 3


Ad 1: Yes, there is. The formula is $$\nabla^*(X^\flat \otimes u) = - \nabla_X u -\mathrm{div}(X) \cdot u,$$ as can easily seen by local computation. Here, $X$ is a vector field and $X^\flat$ is the dual one form w.r.t. the metric.

Note that unless you have a scalar product on the bundle $E$, too, the dual operator will be an operator $\Gamma^\infty(T^*M \otimes E^*) \longrightarrow \Gamma^\infty(E^*)$. The above formula holds in both cases (depending on which case you are in, $u$ will be a section of $E$ or $E^*$, and the connection on the right will be either your given connection on $E$ or the corresponding dual connection on $E^*$ (this connection is always well-defined by the formula $(\nabla u)(s) = d(u(s)) - u(\nabla(s))$).

A similar formula holds even in the case that you don't have a metric on $M$, in which case the dual operator will send sections of $TM \otimes E^* \otimes |\Lambda|$ to sections of $E^* \otimes |\Lambda|$, where $|\Lambda|$ is the density bundle.

Ad 2: Let me first say that in my opinion, one uses the word "formal" only to indicate that one doesn't bother about any functional analytic meaning of the word adjoint whatsoever, and one does not automatically want to say that there has to be any setting where there is an "actual" adjoint.

However, to give a more constructive answer: Yes, this operator is always an adjoint operator in the functional analytic sense.

First, this is true for elliptic differential operators on manifolds. On compact manifolds, they have a unique closed extension, whose adjoint operator is given on smooth functions by the formal adjoint.

In general, consider for example the space $\mathscr{D}=\mathscr{D}(M, E)$ of test sections (which is not a Fréchet space unless $M$ is compact!). It can be embedded into its dual $\mathscr{D^\prime}(M, E^*)$. Now any differential operator $A$ is a continuous operator on $\mathscr{D}$, and its adjoint $A^\prime$ is an operator on $\mathscr{D}^\prime$. However, on the elements $\mathscr{D}\subset \mathscr{D}^\prime$ $A^\prime$ acts just as the formal adjoint of $A$. In this sense, the formal adjoint is also an "actual" adjoint in some sense.

However, this is quite formal and not at all deep. It follows at once from the definitions of distributions etc.

  • $\begingroup$ I would like to use the formula for $\nabla^*$ in a work of mine. In order to save journal space, I would prefer not to include its proof; on the other hand, not all readers might know it, so do you think I could find it in some book which I could use as a reference? $\endgroup$
    – Alex M.
    Commented Jun 20, 2022 at 14:55
  • $\begingroup$ In my opinion, it is completely fine to state this formula in a journal article without any proof, as it is an easy exercise at the level of an introductory PDE course. That being said, I don't know a reference. $\endgroup$ Commented Jun 20, 2022 at 19:36
  • $\begingroup$ According to your first equation, the formal adjoint does not depend on the scalar product on $E$, does it? $\endgroup$
    – Filippo
    Commented Mar 29, 2023 at 20:05

Being a bit late for the party, here is nevertheless a small answer. In fact, there is a rather explicit way to compute adjoints of every differential operator (any order) between (smooth, compactly supported) sections of vector bundles. Of course, this is (in general) not the Hilbert space adjoint, here you need a bit more analysis ;)

The main idea is to use a symbol calculus based on a covariant derivative. I explain here the scalar version (trivial vector bundles) but the whole thing can be done in full generality as well. First, you choose a covariant derivative, say torsion-free and you choose a density on your manifold in order to have an integration measure. As you probably know, the symmetrized covariant derivative allows you to establish a $C^\infty(M)$-linear bijection between symbols, i.e. smooth functions on the cotangent bundle being polynomial in the fiber directions, and differential operators. Note that this is a real bijection, not just taking into account the leading symbol. Of course, the symbol depends on the chosen covariant derivative.

In a second step you compute once and for all the adjoint of a differential operator $D$ with symbol $f$ by zillions of integrations by parts. The funny thing is that there is a fairly simple way how the symbol of the adjoint looks like. You need two ingredients for that:

First, the covariant derivative allows you to define a horizontal lift which in turn determines a maximally indefinite pseudo Riemannian metric on the cotangent bundle (horizontal spaces are in bijection to tangent spaces at the base point, vertical spaces are in bijection to the cotangent space, thus there is a natural pairing). This metric has a Laplace operator (better: d'Alembert operator) $\Delta$ with which you can act on the symbol $f$. In the flat situation this is just \begin{equation} \Delta_{\mathrm{flat}} = \frac{\partial^2}{\partial q^i \partial p_i} \end{equation} for a Darboux chart on $T^*M$ induced by a chart on $M$. In general, there are a couple of Christoffel symbols needed to make this globally defined ;)

Second, the density $\mu$ of your integration might not be covariantly constant. In any case, it defines a one-form by \begin{equation} \alpha(X) = \frac{\nabla_X \mu}{\mu}, \end{equation} which is now be used to cook up a new differential operator on $T^*M$. You can lift $\alpha$ vertically to a vector field $F(\alpha)$ on $T^*M$, completely canonical.

Having these two ingredients, the adjoint of $D^*$ has the following symbol \begin{equation} f^* = \exp\left(\frac{1}{2i}(\Delta + F(\alpha))\right) \overline{f}. \end{equation} The prefactor in the exponential depends a bit on your conventions concerning the assignment of symbol to operator. With this formula it is typically really just a computation to get adjoints of all kind of operators. In many cases, you can chose your density to be covariantly constant, so $\alpha = 0$.

You can find all this in much detail in my book on Poisson geometry, based on some old papers on quantization of cotangent bundles in the late 90s (together with Bordemann and Neumaier).

In the case of interesting bundles, the formula is essentially the same: you only have to choose covariant derivatives for the two vector bundles in question and modify the Laplace operator accordingly. Then you can proceed in the same way. This generalization is in a paper of mine with Bordemann, Neumaier and Pflaum.


To expand marginally, in regard to your remark about compatibility, on Matthias Ludewigs answer: if $s$ and $t$ are sections of $E$, we have $$ (\nabla_X g)(s,t) := Xg(s,t) - g(\nabla_X s,t) - g(s,\nabla_X t) $$ so, as $Xg(s,t) = \operatorname{div}(g(s,t) X) - g(s,t) \operatorname{div}(X)$, in general $\nabla^*(X^\flat \otimes s) = - \nabla_X s- \operatorname{div}(X) s + (\nabla_X g)(s)^\sharp$, where the compatibility condition is just $\nabla_X g = 0$.

  • $\begingroup$ Suppose that my covariant derivative is the gradient (that is, $E = \mathbb R$). How does $\nabla^\ast$ translate into the negative divergence? $\endgroup$
    – shuhalo
    Commented Dec 26, 2020 at 20:02
  • $\begingroup$ @shuhalo The first term is a total derivative and so should vanish after integrating out, depending on reasonable boundary conditions. The last term vanishes identically. $\endgroup$
    – Abel Stern
    Commented Dec 28, 2020 at 11:48

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