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

## 1 Answer

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 $\star 1$. – levap Nov 2 '16 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. – Asaf Shachar Nov 3 '16 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 – levap Nov 3 '16 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. – levap Nov 3 '16 at 11:06