# Why the codifferential depends only on the connection (and not the metric)?

Let $E$ be a vector bundle over a smooth Riemannian manifold $(M,g)$. Suppose $E$ is equipped with a metric $\eta$ and a metric connection $\nabla$.

Denote by $\Omega^k(M,E)$ the space of $E$-valued forms of degree $k$. $\nabla$ induces a covariant exterior derivative $d:\Omega^k(M,E) \to \Omega^{k+1}(M,E)$, which (together with the metrics $g,\eta$) induce the codifferential operator:

$\delta:\Omega^k(M,E) \to \Omega^{k-1}(M,E)$.

Even though the metric $\eta$ on $E$ plays a part in the definition of $\delta$, it turns out that $\delta$ is actually independent of it, and depends only on the connection $\nabla$ (there is an implicit dependence since $\nabla$ is required to be metric).

Indeed, one formula for $\delta$ is $\delta =\star d \star$ (up to sign), where $\star$ is the Hodge-dual $\Omega^k(M,E) \to \Omega^{d-k}(M,E)$ which is defined without any reference to the metric on $E$.

Is there any way to see why the codifferential depends only on the connection $\nabla$ and the metric on $M$, and not on the metric on $E$?

Of course, the derivation of the formula above for $\delta$ shows this, but I would like to find an argument without relying on this computation.

(I am looking for a more "conceptual explanation").

Note that it's easy to see from the definition of $\delta$ that it's invariant under scaling of the metric on $E$, however I do not see immediately why it's completely independent of it.

• Look at the Koszul formula for the de Rham differential of real-valued forms. It involves the action of a vector field on a real-valued function (i.e. differentiation). If you're talking about forms with values in a vector bundle, you need a differentiation action of a vector field on a section, i.e. a connection, to make sense of the formula. – Pedro Feb 12 '17 at 19:55
• @Pedro Yes, I know that. However, in the definition of the codifferential (which is the adjoint of $d$ w.r.t the induced metrics on the forms) the metric on $E$ plays a part. My question is whether there is a nice argument for the independence of $\delta$ in the metric on $E$? (It is trivial that it depends on the connection on $E$, since as you mentioned, $d$ is defined by it) – Asaf Shachar Feb 12 '17 at 19:58

Assume we have two metrics $\eta_1 = \left< \cdot, \cdot \right>_1$ and $\eta_2 = \left< \cdot, \cdot \right>_2$ on $E$ that are compatible with $\nabla$. Let $\Phi \colon E \rightarrow E$ be a vector bundle isomorphism such that

$$\left< s, t \right>_1 = \left< \Phi(s), t \right>_2.$$

In other words, for each $p \in M$, the map $\Phi_p \colon E_p \rightarrow E_p$ is the unique linear map representing the bilinear form $\eta_1$ with respect to $\eta_2$. One way to see that $\Phi$ exists is to note that $\Phi$ is obtained from the tensor $\eta_1 \in \Gamma(E \otimes E)$ by raising an index with respect to the metric $\eta_2$. The basic observation is that $\Phi$ is $\nabla$-parallel because it is obtained from a parallel tensor ($\eta_1$) by raising an index with respect to a parallel metric $\eta_2$. Alternatively, we can compute

$$\left< \nabla_X(\Phi)(s), t \right>_2 = \left< \nabla_X(\Phi(s)) - \Phi(\nabla_X s), t \right>_2 = \left< \nabla_X(\Phi(s)), t \right>_2 - \left< \nabla_X s, t \right>_1 \\ = \left( X \left< \Phi(s), t \right>_2 - \left< \Phi(s), \nabla_X t\right>_2 \right) - \left< \nabla_X s, t \right>_1 \\ = X \left< s, t \right>_1 - \left< s, \nabla_X t \right>_1 - \left< \nabla_X s, t \right>_1 = 0$$

where we used both $\nabla \eta_1 = 0$ and $\nabla \eta_2 = 0$.

Using $\nabla \Phi = 0$ and the definition of the covariant exterior derivative, one can then show that $d_{\nabla}$ commutes with $\Phi$ in the sense that

$$\Phi(d_{\nabla}(\alpha)(X_0, \dots, X_k)) = d_{\nabla}(\Phi(\alpha))(X_0, \dots, X_k)$$

where $\alpha, \Phi(\alpha) \in \Omega^{*}(M;E)$.

Finally, let us show that if $\delta$ is the adjoint of $d_{\nabla}$ with respect to $\eta_2$ then it is also the adjoint of $d_{\nabla}$ with respect to $\eta_1$ and so it is independent of the metric $\eta$ as long as it is $\nabla$-compatible. Set $\Phi_k := \operatorname{id}|_{\Lambda^k(T^{*}(M))} \otimes \Phi$ and (abusing notation a little) denote the induced metric from $g$ and $\eta_i$ on $\Lambda^k(T^{*}M) \otimes E$ by $\left< \cdot, \cdot \right>_i$. Then $\Phi_k$ satisfies

$$\left < \alpha, \beta \right>_1 = \left< \Phi_k(\alpha), \beta \right>_2$$

for $\alpha,\beta \in \Omega^k(M;E)$ and the statement that $d_{\nabla}$ commutes with $\Phi = \Phi_0$ can be written as $\Phi_{k+1} \circ d_{\nabla} = d_{\nabla} \circ \Phi_k$.

Using all the properties above, we have

$$\int_{M} \left< d_\nabla(\alpha), \beta \right>_1 \, dV_g = \int_M \left< \Phi_{k+1}(d_{\nabla}(\alpha)), \beta \right>_2 \, dV_g = \int_M \left< d_{\nabla}(\Phi_{k}(\alpha)), \beta \right>_2 \, dV_g = \\ \int_M \left< \Phi_{k}(\alpha), \delta(\beta) \right>_2 \, dV_g = \int_M \left< \alpha, \delta(\beta) \right>_1 \, dV_g.$$

• Your observations are wonderful, as always. Thank you. BTW, I have tried to look at a slightly different approach, which revolves around an isometry between $(E,\eta_1)$ and $(E,\eta_2)$. This looked more reasonable to me, since any isometric preserving connection $(E,\nabla^E,\eta_E) \to (F,\nabla^F,\eta_F)$ commutes with $\delta$. However, it is not clear if one can choose an isometric morphism $(E,\eta_1)$ and $(E,\eta2)$ which preserves the connection. – Asaf Shachar Feb 15 '17 at 17:50
• @AsafShachar: Thanks! Actually I initially tried to do what you suggested but decided it was more complicated than necessary. You can get a isometric morphism between $(E,\eta_1)$ and $(E,\eta_2)$ - it will be the (unique) $\eta_2$- square positive root of $\Phi$ (which is $\eta_2$ symmetric and positive definite) and I suspect that the fact $\Phi$ is flat will imply that $\sqrt{\Phi}$ is flat but I haven't checked the details. – levap Feb 15 '17 at 19:45
• (Yeah, in a parallel $\eta_2$-orthonormal frame along an arbitrary curve, the components of both $\Phi$ and $\eta_2$ are constant and so the components of $\sqrt{\Phi}$ will be constant which will imply that $\sqrt{\Phi}$ is parallel). – levap Feb 15 '17 at 20:02
• I was thinknig exactly the same thing! Although my way to deduce $\Phi$ is parallel $\Rightarrow \sqrt \Phi$ is parallel was a little different (essentially using Leibnitz and some linear algebra facts. By the way, this is the same way one concludes the derivative of the matrix square root is defined by the natural equation...). Unfortunately, it seems that this more "natural" argument does not help for now: (1) We still need your argument to see $\Phi$ is parallel... – Asaf Shachar Feb 15 '17 at 20:23
• (2) We still need the last part of your argument, to deduce the result, since I do not see a way to use directly the fact $\sqrt \Phi$ is parallel. (You can see the last equation in my newly added answer, where we need $\delta^1, \sqrt \Phi_*$ to commute, However, I do not find any special property of $\sqrt \Phi$ when considered as a map $(E,\eta_1) \to (E,\eta_1)$ which implies this. – Asaf Shachar Feb 15 '17 at 20:24

The fiber metric $\eta$ plays a fundamentally different role to the Riemannian metric $g$ - the latter is the one of the two that interacts with the differentiation indices and thus will contribute terms to the $L^2$ adjoint.

Let's see explicitly how this works in the derivation for $\delta^{(1)}$. If we let $\nabla$ denote the natural extension of the connections on $TM$ and $E$ to all tensor bundles thereof, $\delta$ is the unique operator $\Omega^1(M,E) \to \Gamma(E)$ such that

$$\int \delta(s \otimes \theta)^\alpha w^\beta \eta_{\alpha \beta} = \int s^\alpha \theta_i \nabla_j w^\beta g^{ij} \eta_{\alpha \beta}$$

for all $\theta \in \Omega^1(M)$ and $s,w \in \Gamma(E)$. Rewriting the RHS with the product rule we get

$$\int \delta(s \otimes \theta)^\alpha w^\beta \eta_{\alpha \beta} = \int \nabla_j\left(s^\alpha \theta_i w^\beta g^{ij}\eta_{\alpha\beta}\right) - \int \nabla_j\left(s^\alpha \theta_i g^{ij} \right)w^\beta \eta_{\alpha\beta}$$

since $\nabla \eta = 0$. The first term vanishes by the divergence theorem, and this formula holds for all $w$; so we find the formula $\delta(s \otimes \theta)^\alpha=-\nabla_j(s^\alpha \theta_i g^{ij}).$

The reason $\eta$ does not appear in the formula is that it occurs in the same role on both sides of the adjoint equation, and commutes through all our derivatives (so long as it is compatible). This cannot be true of the Riemannian metric $g$ because in applying $d$ we tensor in a factor of $TM$, and thus $\delta$ depends upon $g$.

I'm not sure how much this will help you conceptually. Perhaps one thing to do is think about things in terms of the dual section $\xi_\alpha = w^\beta \eta_{\alpha\beta} \in \Gamma(E^*)$: the compatibility $\nabla \eta = 0$ means the adjoint equation can be written

$$\int \delta(s \otimes \theta)^\alpha \xi_\alpha = \int s^\alpha \theta_i \nabla_j \xi_\alpha g^{ij},$$

which is manifestly independent of $\eta$.

• Thanks. I think indeed a "coordinate view" might help clarify the different roles played here. However, I am having a little trouble following your exact calculation: (1) $s$ is supposed to be a section of $T^*M$, not $E$? (2) If so, then in the expression $s^\alpha \theta_i \nabla_j w^\beta g^{ij} \eta_{\alpha \beta}$ something is clearly wrong since the $g^{ij}$ are not related to the $s^{\alpha}$. I have tried to do the computations a little more carefully in my "answer" above, if you would like to have a look. – Asaf Shachar Feb 13 '17 at 9:27
• Finally, I am not sure exactly what is the "product rule" you are using here (I know the Leibnitz rule for the exterior derivative, but I don't follow the coordinate version, prehaps because of indices mixing). Thanks for your patience. – Asaf Shachar Feb 13 '17 at 9:27
• @AsafShachar: $s$ is a section of $E$, $\theta$ is a (scalar-valued) $1$-form. I guess I should write $\theta \otimes s$ instead. My calculation can be interpreted as abstract index notation rather than a coordinate one - I just find it extremely unwieldy to use index-free notation when dealing with complicated contractions like this. The product rule here is just the Leibniz rule for connections - there's no need to think about exterior differentiation here since I chose the easy $k=1$ case. – Anthony Carapetis Feb 13 '17 at 10:16
• OK, I fixed my "answer" accordingly. Unfortunately, I am less used to the abstract index notation. Is there a way to interpret your computation in "usual" coordinates? (according to my analysis)? (I guess I should take the time to master some abstract index notation anyway, I heard Penrose's book might be a good source for that. If you have any other recommendation (or a way to clarify this case in coordinates) it would be great. – Asaf Shachar Feb 13 '17 at 15:46

I am expanding on some details in the computation of Anthony Carapetis:

Let $s,w \in \Gamma(E) , \theta \in \Gamma(T^*M)$. We work locally: Let $x_\alpha$ be coordinates on a subset $U \subseteq M$, and let $e_i$ be a local frame for $E$ over $U$. Then:

$$s=s^{\alpha}e_{\alpha}, w =w^{\beta}e_\beta, \theta =\theta_idx^i, \eta_{\alpha \beta}:=\langle e_\alpha, e_\beta \rangle_E,g_{ij}:=\langle \partial_i,\partial_j \rangle_{TM}$$

hence $\theta \otimes s =s^{\alpha}\theta_i (dx^i\otimes e_{\alpha})$,

so writing $\delta(\theta \otimes s)=\big(\delta(\theta \otimes s)\big)^\alpha e_\alpha$, we get

$$\langle \delta(\theta \otimes s),w \rangle _E=\big(\delta(\theta \otimes s)\big)^\alpha w^{\beta} \eta_{\alpha \beta} \tag{1}$$

On the other hand,

$$\langle \theta \otimes s,\nabla w \rangle _{TM,E}=\langle s^{\alpha}\theta_i (dx^i\otimes e_{\alpha}),(\nabla w)^\beta_j (dx^j\otimes e_{\beta}) \rangle _{TM,E}$$

$$=s^{\alpha}\theta_i (\nabla w)^\beta_j \langle dx^i, dx^j \rangle_{T^*M} \cdot \langle e_{\alpha} , e_{\beta}\rangle_E=s^{\alpha}\theta_i (\nabla w)^\beta_j g^{ij} \eta_{\alpha \beta} \tag{2}$$

So, equations $(1),(2)$ and the definition of $\delta$ imply

$$\int \delta(\theta \otimes s)^\alpha w^\beta \eta_{\alpha \beta} =\int \langle \delta(\theta \otimes s),w \rangle _E= \int \langle \theta \otimes s,\nabla w \rangle _{TM,E}=\int s^\alpha \theta_i (\nabla w)^\beta_j g^{ij} \eta_{\alpha \beta} \tag{3}$$

where $(\nabla w)^\beta_j \eta_{\alpha \beta}= \langle \nabla_{\partial j} w , e_{\alpha} \rangle_E$.

Here is an attempt to make levap's idea seem more natural:

We start with the following basic observation:

If $\Phi:(E,\nabla^E,\eta_E) \to (F,\nabla^F,\eta_F)$ is an isometric bundle morphism which respects the connections, then

$\delta^F (\Phi_*\alpha) =\Phi_* (\delta^E (\alpha))$ for every $\alpha \in \omega^k(M,E)$.

($\Phi_*: \omega^k(M,E) \to \omega^k(M,F)$ is the induced morphism which is the action of $\Phi$ on the values of the forms).

Define $\Phi:E \to E$ via $\langle \Phi(v),u \rangle_2=\langle v,u \rangle_1$.

Consider $\Phi$ as a map $(E,\eta_2) \to (E,\eta_2)$.

$\Phi$ is self-adjoint:

$$\langle \Phi(v),u \rangle_2=\langle v,u \rangle_1=\langle u,v \rangle_1=\langle \Phi(u),v \rangle_2=\langle v,\Phi(u) \rangle_2$$

$\Phi$ is positive:

$$\langle \Phi(v),v \rangle_2=\langle v,v \rangle_1$$

Thus, there is a unique self-adjoint positive square root $\sqrt \Phi: (E,\eta_2) \to (E,\eta_2)$ map, which is easily seen to be an isometry when considered as a map $(E,\eta_1) \to (E,\eta_2)$:

$$\langle \sqrt \Phi(v),\sqrt \Phi(u) \rangle_2=\langle v,\Phi(u) \rangle_2=\langle v,u \rangle_1$$

We claim $\sqrt \Phi$ is $\nabla$-parallel:

levap showed in his answer that $\nabla_x \Phi=0$. Thus, by a Leibnitz property it follows that:

$$0=\nabla_x \Phi=\nabla_x (\sqrt \Phi)^2= (\nabla_x \sqrt \Phi) \sqrt \Phi + \sqrt \Phi (\nabla_x \sqrt \Phi)$$

This is a Sylvester equation, hence there is a unique solution $\nabla_x \sqrt \Phi=0$.

This implies that $\sqrt \Phi:(E,\nabla,\eta_1) \to (E,\nabla,\eta_2)$ preserves all the structures, thus by the observation above:

$$\delta^2 (\sqrt \Phi_*\alpha) =(\sqrt \Phi)_* (\delta^1 (\alpha))$$

Replacing $\alpha$ with $\sqrt \Phi_*\alpha$, we obtain

$$\delta^2 (\Phi_*\alpha) =(\sqrt \Phi)_* (\delta^1 (\sqrt \Phi_* \alpha)),$$ so

$$\delta^2 (\Phi_*\alpha) = \delta^1 (\Phi_*\alpha) \iff \delta^1 (\Phi_*\alpha) =(\sqrt \Phi)_* (\delta^1 (\sqrt \Phi_* \alpha)),$$ that is if and only if $\delta^1$ commutes with $\sqrt \Phi_*$.

However, it is not clear how to show this...

Edit:

I had the following conjecture:

Let $(E,\nabla^E), (F,\nabla^F)$ be smooth vector bundles over $M$ (with connections).

Let $\Phi:E \to F$ be a bundle isomorphism which maps $\nabla^E$-compatible metrics to $\nabla^F$-compatible metrics. Then

$\delta^F (\Phi_*\alpha) =\Phi_* (\delta^E (\alpha))$ for every $\alpha \in \omega^k(M,E)$,

where $\delta^E$ is the corresponding adjoint to $d_{\nabla^E}$ taken w.r.t any $\nabla^E$-compatible metric, and $\delta^F$ is the corresponding adjoint to $d_{\nabla^F}$ taken w.r.t the any $\nabla^F$-compatible metric.

A special case of this conjecture, for $E=F,\Phi=\operatorname{Id}$ is exactly the claim that the adjoint does not depend on the $\nabla^E$-compatible metric chosen.

However, this conjecture is clearly false in general, since there are connections $\nabla^E$ with no $\nabla^E$-compatible metrics, so every bundle isomorphism $\Phi:E \to F$ satisfies the requirement vacuously, but there is no reason for it to commute with $\delta$.

What is true is the following statement:

Let $\Phi:E \to F$ be a parallel bundle isomorphism (i.e $\nabla \Phi=0$), then $\Phi$ commutes with $\delta$'s taken w.r.t $\nabla^E,\nabla^F$ compatible metrics. Indeed, given $\nabla^E$-compatible metric, $\Phi:(E,\nabla^E,\eta_E) \to (F,\nabla^F,\Phi_*\eta_E)$ preserve all the structures, hence commute with $\delta$.