Does Hodge-star commute with metric connections?

Let $E $ be a smooth oriented vector bundle over a manifold $M$. Suppose $E$ is equipped with a metric $\eta$, and a compatible connection $\nabla$. Denote the dimension of $E$'s fibers by $d$.

Let $\Lambda_k(E)$ denote the exterior algebra bundle of $E$ of degree $k$. The orientation and metric on $E$ induce a Hodge-star operator: $ \star_k:\Lambda_k(E) \to \Lambda_{d-k}(E)$.

$\nabla$ induce a connection on $\Lambda_k(E)$ (which we also denote by $\nabla$). Note that this induced connection is compatible with the metric on $\Lambda_k(E)$ by $\eta$.

Question: Does $\star,\nabla$ commute?

i.e, is it true that $ \star_k (\nabla_X \beta)=\nabla_X (\star_k \beta)$ for every $\beta \in \Lambda_k(E),X \in \Gamma(TM)$?

It can be shown that the answer is positive if and only if $\nabla_X (\star_0 1)=0$ (for every $X \in \Gamma(TM)$):

Step I: We reduce the assertion to the case $k=0$ (see details below).

Step II: Further reduction to $\nabla_X (\star_0 1)=0$:

I think I can verify $\nabla_X (\star_0 1)=0$ in the case where $\nabla$ is flat (See step III below). However, I am interested in the general case.

Proof of Step II: Let $\beta \in \Lambda_0(E)=C^{\infty}(M)$. $$ (1) \, \, \star_0 (\nabla_X \beta)=\star_0 \big((\nabla_X \beta) \cdot 1\big) = (\nabla_X \beta) \cdot (\star_0 1) $$

Also,

$$ (2) \, \, \nabla_X (\star_0 \beta)= \nabla_X (\star_0 (\beta \cdot 1))=\nabla_X \big(\beta \cdot (\star_0 1)\big)=(\nabla_X \beta) \cdot (\star_0 1 )+ \beta \cdot \nabla _X (\star_0 1),$$

so by equations $(1),(2)$ above, $\star_0 (\nabla_X \beta)=\nabla_X (\star_0 \beta)$, if and only if $\nabla_X (\star_0 1)=0$.

Proof of the reduction to the case $k=0$ (Step I):

Recall the Hodge-star is defined via $$ \star_0 \langle v,w \rangle= w \wedge \star_k v$$ for every $v,w \in \Lambda_k(E)$.

Let $v,w \in \Lambda_k(E)$. Then,

$$ (1) \, \,\nabla_X (v \wedge \star_k w)=\nabla_X v \wedge \star_k w+ v \wedge \nabla_X(\star_k w)=\star_0 \langle \nabla_X v,w \rangle+v \wedge \nabla_X(\star_k w)$$

Moreover, $$ (2) \, \, \nabla_X (v \wedge \star_k w)=\nabla_X (\star_0 \langle v,w \rangle) \stackrel{(*)}{=} \star_0 \big(\nabla_X \langle v,w \rangle\big)=\star_0 \big( \langle \nabla_X v,w \rangle+ \langle v, \nabla_X w \rangle\big)=$$

$$ \star_0 \langle \nabla_X v,w \rangle + \star_0 \langle v, \nabla_X w \rangle$$

Where equality $(*)$ is exactly the statement for $k=0$.

Equalities $(1),(2)$ imply:

$$ v \wedge \nabla_X(\star_k w)=\star_0 \langle v, \nabla_X w \rangle=v \wedge \star_k (\nabla_X w).$$

The uniqueness of the Hodge_star imply $\nabla_X(\star_k w) = \star_k (\nabla_X w)$.

Proof of step III: $\nabla$ is flat $\Rightarrow$ $\nabla_X (\star_0 1)=0$.

We can work locally: Since $\nabla$ is flat, parallel transport is path-independent (see here), so we can build a positively-oriented parallel orthonormal frame for $E$ over a small enough neighbourhood around each point in $M$. Given such frame $E_i$, we have $\nabla_X E_i=0$. Thus, $$ \nabla_X (\star_0 1)=\nabla_X (E_1 \wedge \dots \wedge E_d )=\sum_i E_1 \wedge \dots \wedge \nabla_X E_i \wedge \dots \wedge E_d=0.$$

Solution 1:

I did not check all your arguments but indeed $\nabla_X(\star_0 1) = 0$ in your setting. To see it, choose $p \in M$ and a local orthonormal frame $(E_1,\dots,E_d)$ on a neighborhood $U$ of $p$. Then,

$$ (\nabla_X(\star_0 1))(p) = \nabla_X(E_1 \wedge \dots \wedge E_d) = \sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \nabla_X E_i \wedge E_{i+1} \wedge \dots \wedge E_d. $$

Write $\nabla = D + A$ on $U$ where $D$ is the trivial connection and $A = (\omega^i_j) $ is a $d \times d$ matrix of one-forms (the connection form) defined by $\nabla_X E_i = \omega_i^j(X) E_j$. Since $\nabla$ is metric, the matrix $D$ is anti-symmetric and thus we have

$$ \sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \nabla_X E_i \wedge E_{i+1} \wedge \dots \wedge E_d = \\ \sum_{i,j=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \omega^j_i(X)E_j \wedge E_{i+1} \wedge \dots \wedge E_d = \\ \sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \omega^i_i(X)E_i \wedge E_{i+1} \wedge \dots \wedge E_d = \left( \sum_{i=1}^n \omega^i_i(X) \right) E_1 \wedge \dots \wedge E_d = 0$$

as $\omega^i_i \equiv 0$.