# 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, $$\nabla_X (\star_0 1)=\nabla_X (E_1 \wedge \dots \wedge E_d )=\sum_i E_1 \wedge \dots \wedge \nabla_X E_i \wedge \dots \wedge E_d=0.$$

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 $\omega^i_i \equiv 0$.

• @Asaf Schahar: BTW, you seem to prefer the "characterization" of the Hodge star as an operator that satisfies $\star_0 (\left< v, w \right>) = v \wedge \star w$ instead of $v \wedge \star w = \left <v, w \right> \Omega$ where $\Omega$ is the orientation element (probably because it doesn't involve the volume element) but this doesn't characterize the Hodge star as any scalar multiple of $\star$ also satisfies the first property (in particular, the zero operator works). That is, the first property determines $\text{span} \{\star\}$ but there is no normalization and no knowledge about $\star 1$. Commented Nov 2, 2016 at 22:27
• You are right, I was implicitly assuming $\star_0 1 =\Omega$. BTW, I do not see why mentioning the decomposition $\nabla = D+A$ was helpful in any way. (The $w^i_j$ are defined in terms of $\nabla$ alone, as you wrote). I am also not sure what do you mean by a trivial connection on a vector bundle. Commented Nov 3, 2016 at 6:53
• @AsafShachar: Given a local frame $\sigma = (E_1, \dots, E_d)$ on $U$, you can always define a connection $D_{\sigma}$ on $U$ by representing sections with respect to $\sigma$ and differentiating their components. Namely, $D^{\sigma}_X(\xi) = d(X)(E^i(\xi)) E_i$ where $(E^i)$ is the dual basis. This connection turns $\sigma$ into a parallel frame and conversely, any flat connection is represented locally as $D^{\sigma}$ for some frame $\sigma$. Now, given any connection on $U$, we can look at $\nabla - D$ and since the difference of two connections is tensorial, we get $\nabla = D + A$ where Commented Nov 3, 2016 at 11:00
• $A$ is an $\operatorname{End}(E)$-valued one-form which is represented in local coordinates by a matrix. All this is a fancy way to split a covariant derivative in a local frame into a trivial part of order 1 (regular vector differentiation) and a zero-order part which is the generalization of the Christoffel symbols. Like you noted, I actually didn't have to mention all of this, just say that the "Christoffel matrix" $A(X)$ of a metric connection with respect to an orthonormal frame should be anti-symmetric. Commented Nov 3, 2016 at 11:06