Special connection of vector bundle over real manifold

Question

Let $E \rightarrow M$ be a vector bundle over a smooth manifold $M$ and let $g$ be a bundle metric. Does there exists a conection (maybe unique) $\nabla$ which is compatible with $g$. By this I mean: for two sections $s,s'$ of the bundle $E \rightarrow M$ and a vector field $X$ on $M$ we have $X(g(s,s'))=g(\nabla_{X}s,s')+g(s,\nabla_{X}s')$. I know that this is true for hermitian connection. Then the uniqueness is given by the holomorphic structure. But here I consider the smooth setting. So my question: Is there any special connection as described above? Is this connection unique? In what sense unique?

Greetings, Phillip

Answer 1

The existence follows from a very general fact about existence of a compatible connection in any principal bundle. That is to say, on one hand you have your vector bundle $E \to M$ endowed with $g$. In ther hand you also have the $GL(\mathbb{R},n)$, $n = rank(E)$ principal bundle $P(E)$ of frames of $E$. The metric $g$ gives you a so called reduction of $P(E)$ to a principal $O(n)$ subbundle. Then the existence of a connection compatible with $g$ on $E$ follows from the general existence theorem about existence of connections on any principal bundle. Of course, when you got the connection on the principal bundle you need to do an standard procedure to get the connection on the associated vector bundle $E$ i.e. to get the covariant derivative you wrote in your OP.

As the other user observed there are not hope to have uniqueness without more restrictive conditions e.g. the torsion. If you are interested in the Riemannian setting I suggest you to have a look to Simon Salamon's answer here: A geometric interpretation of the Levi-Civita connection?

Answer 2

Concerning the uniqueness question, it is definitely not unique in general. -On a Riemannian manifold there is a unique Levi Civita connection. I.e. a connection on the tangent bundle which is both metric and torsion free (this is the fundamental theorem of Riemannian geometry). But in fact one can prove another version of the fundamental theorem of Riemannian geometry which goes as follows:

Let $(M,g)$ be a Riemannian manifold and $A$ an antisymmetric 2-tensor on $M$. Then there exists a unique metric connection $\overline{\nabla}$ with torsion $A$.

I.e. the conclusion is that the torsion determines the connection uniquely (provided it is assumed to be metric).

Answer 3

The existence of a reduction $P_K$ of structure group of the associated principal bundle $P_G$ to its maximal compact subgroup $K \subset G$ follows from the fact that every Lie group with finitely many connected components admits a retraction to its maximal compact subgroup. A bump function argument makes a connection on any principal bundle, in particular on the reduction. Any two connections differ by a 1-form valued in the adjoint bundle, i.e. a section of $\Omega^1 \otimes (P_K \times^K \mathfrak{k})$, and any section of that bundle adds to a connection to give another connection. For pseudo-Riemannian metrics, the tensor lies in that same bundle, and getting rid of the torsion is therefore selecting the connection uniquely.