7
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\omega^i_i \equiv 0$.

$\endgroup$
  • $\begingroup$ @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 $\star 1$. $\endgroup$ – levap Nov 2 '16 at 22:27
  • $\begingroup$ 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. $\endgroup$ – Asaf Shachar Nov 3 '16 at 6:53
  • $\begingroup$ @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 $\endgroup$ – levap Nov 3 '16 at 11:00
  • $\begingroup$ $A$ 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. $\endgroup$ – levap Nov 3 '16 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.