# What is the significance of a connection being metric?

Suppose a vector bundle is equipped with a pseudo-Riemannian metric tensor, i.e a smoothly varying family of non-degenerate symmetric bilinear forms on the fibers.

A connection on the vector bundle is metric if parallel transport along every path downstairs is an isometry.

The wiki entry on metric connections said the significance of being metric (at least in the Riemannian metric case) is the ability to define the Hodge star, the Hodge dual, and the Laplacian.

What else does preserving the metric give? Which basic facts of (pseudo) Riemannian geometry rely on the connection on the vector bundle being metric?

• You can always define the Hodge star, the Hodge dual, and the Laplacian of $E$-valued forms as long as $E$ (and $M$) are equipped with Riemannian metrics. That a connection is metric just says that it plays well with these structures w/r/t the metric it induces on $E$. For instance, it tells us that $\pm * d_A *$ is the formal adjoint of $d_A$ with respect to the natural inner product on $k$-forms (coming from the above metric), which then lets us define the $A$-Laplacian as $d_A d_A^* + d_A^*d_A$, and this is an elliptic operator. – user98602 Mar 5 '17 at 23:01
• So more or less it just says "anything you can do with the metric you can do to $d_A$ in a way that preserves the metric". – user98602 Mar 5 '17 at 23:01
• If the connection is metric then covariant differentiation commutes with the various musical isomorphisms (i.e with "raising/lowering indices using the metric"). – levap Mar 6 '17 at 1:05

As it was said in the comments, the Hodge-dual does not rely on the connection at all. It relies on the existence of a volume form and musical isomorphisms (raising and lowering of indices), which are generated by the metric.

Also, for this to exist, you need the fiber metric to be present on the tangent bundle, or alternatively you need to have the fibers $E_x$ to be $n$ dimensional (if $M$ is $n$ dimensional), and to have a soldering form $\theta\in\Omega^1(M,E)$, which for every $x\in M$ provides a linear isomorphism $\theta_x:T_xM\rightarrow E_X$, such that the vector bundle $(\mathcal{E},\pi,E,M)$ may be identified with the tangent bundle $(TM,\pi,\mathbb{R}^n,M)$.

If you are given a linear connection $\nabla$ on the tangent bundle (or as said, any vector bundle that is soldered to $TM$), and $\nabla$ is metric (wrt. the metric tensor $g$), then the significance is that

• the parallel transport preserves lengths and angles, meaning if $P_\gamma$ is the parallel transport map along the smooth curve $\gamma$ (I don't care where it transports now, it's all formal/smybolic here), we have for $X,Y\in TM$, $$g(X,Y)=g(P_\gamma X,P_\gamma Y),$$

• which then implies that if $\gamma$ is a closed loop starting and ending at $x\in M$, then $P_\gamma$ is a linear isometry, meaning that the $GL(n,\mathbb{R})$ structure group may be reduced to $O(k,n-k)$, where $k,n-k$ is the signature of $g$,

• and further implies that parallel transport preserves volumes, since the volume form is determined by $g$,

• and $\nabla$ commutes with the musical isomorphisms, in (abstract) index notation, $\nabla_a(A^b)=\nabla_a(A_cg^{cb})=g^{cb}\nabla_a(A_c)$. This is signficant, because in Riemannian geometry (and assorted areas, such as general relativity), we usually view the vector field $A$ and the associated 1-form $\flat A$ (or alternatively the 1-form $\omega$ and the associated vector field $\sharp\omega$ as representing the same geometric (physical) object. But if $\nabla$ did not commute with raising/lowering, this would mean that the rates of change associated with these objects would not agree, and would depend of the two representations you use, which would invalidate the concept of them representing the same object.