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

Question

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.

Answer 1

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

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

Answer 2

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 $\eta$.

Answer 3

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

Answer 4

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).

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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...

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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 $\delta$.