For forms on a Riemannian $n$-manifold $(M,g)$ there is a notion of a codifferential $\delta$, which is adjoint to the exterior derivative:

$$\int \langle d \alpha, \beta \rangle \operatorname{vol} = \int \langle \alpha, \delta \beta \rangle \operatorname{vol} $$

$\langle \alpha, \beta \rangle$ is a scalar product of two differential forms induced by $g$, $\operatorname{vol}$ is a volume form, corresponding to $g$. Moreover, $\delta$ is defined through a Hodge star $*$ and exterior derivative $d$:

$$ \delta : \Omega^k(M) \to \Omega^{k-1}(M) \\ \delta = (-1)^{n(k+1)+1}*d* $$

But since $M$ is Riemannian, $d$ can be extended with a Levi-Civita connection $\nabla$ to an exterior covariant derivative $d^\nabla$ to act on tensor-valued differential forms.

Is there such a codifferential for $d^\nabla$ in such case, when scalar product of tensor-valued forms is considered? Is there an analogue of Hodge decomposition? And what happens in the special case of $M$ being Einstein manifold?

Weaker question What is the adjoint $\nabla^*$ of Levi-Civita connection $\nabla : \Gamma(TM) \to \Gamma(TM \otimes T^*M)$ when acting on vector fields?

$$\int \langle \nabla u, A \rangle \operatorname{vol} = \int \langle u, \nabla^* A \rangle \operatorname{vol} $$

$u \in \Gamma(TM)$ --- smooth vector fields, $A \in \Gamma(TM \otimes T^*M)$ --- a linear operator on vector fields. All the required conditions assumed on the boundary to get rid of boundary terms in integration by parts.


1 Answer 1


I think the answers to your first two questions are yes, there is always a codifferential; and no, there is not always a Hodge decomposition. I don't know anything about the Einstein manifold case.

Let's say $E$ is a vector bundle over $M$ with metric and compatible connection $\nabla$. ($E$ could be some tensor bundle with $\nabla$ induced from the Levi-Civita connection, for example.) As you say, $\nabla$ gives an exterior derivative $d^\nabla$ on $E$-valued differential forms that obeys the Leibniz rule $$ d^\nabla (\omega \otimes e) = d\omega \otimes e + (-1)^{\text{deg} \omega} \omega \wedge \nabla e ,$$ where $\omega$ is a homogeneous form and $e$ is a section of $E$, and $\omega \wedge \nabla e$ means $\sum_i (\omega \wedge dx^i) \otimes \nabla_{\partial_i} e$.

For each degree of forms $k$, you can view $d^\nabla$ as a map $d^\nabla: C^\infty( \Lambda^k \otimes E) \to C^\infty( \Lambda^{k+1} \otimes E)$. Using the metrics, each of those spaces of sections are endowed with $L^2$ inner products. Then $d^\nabla$ always has a formal $L^2$-adjoint $\delta^\nabla$ going the other way: $\delta^\nabla: C^\infty( \Lambda^{k+1} \otimes E) \to C^\infty( \Lambda^k \otimes E)$. (Note the space of smooth sections can be completed to the space of $L^2$ sections, but $d^\nabla$ and $\delta^\nabla$ are unbounded operators and can generally only be defined on some dense subspace of that $L^2$ space.)

So you can always define $d^\nabla$ and its formal adjoint $\delta^\nabla$. The issue is that $d^\nabla$ squares to zero if and only if the connection $\nabla$ on $E$ is flat, meaning its curvature is zero. (Flat vector bundles $E$, i.e., bundles on which there exists a flat connection, are relatively scarce.) We need $d^\nabla$ to square to zero to even attempt to define the de Rham cohomology of $E$-valued forms, and to have nice properties like the Hodge decomposition. (I want to say that the "Dirac operator" $D = d^\nabla + \delta^\nabla$ and the "Laplacian" $\Delta = D^2$ are not elliptic unless $d^\nabla$ squares to zero, but I need to think about that.)

I don't have a good reference for this at hand, although the Wikipedia article on bundle-valued forms might be useful.

  • 1
    $\begingroup$ Thank you. Right now even flat case would be nice. Do you have an explicit expression for $\delta^\nabla$? (it would probably require an extension of Hodge star operator). $\endgroup$
    – Yrogirg
    Oct 3, 2014 at 13:24
  • 1
    $\begingroup$ Yes, I think one can define the Hodge star appropriately in this setting, and the formula for $\delta^\nabla$ should be the usual one. I will try to elaborate on this later. $\endgroup$ Oct 3, 2014 at 13:52
  • $\begingroup$ mathoverflow.net/questions/142913/… seems to be related $\endgroup$
    – Yrogirg
    Oct 3, 2014 at 15:55
  • $\begingroup$ as well as math.stackexchange.com/questions/116554/… $\endgroup$
    – Yrogirg
    Oct 3, 2014 at 16:01
  • $\begingroup$ @PhillipAndreae: Is it possible to give a rigorous meaning to the statement that "flat vector bundles are relatively scarce"? $\endgroup$
    – Alex M.
    Sep 16, 2018 at 15:39

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