Compatibility of a connection and metric

Question

Every Riemannian manifold admits a metric connection. Suppose $M $ is a manifold and $\nabla $ is an arbitrary connection on the tangent bundle. Does $M$ necessarily admit a metric such that $\nabla$ is compatible with this metric?

Answer 1

Take $M=S^1$ and let $\nabla$ be the flat connection on the trivial line bundle over $M$ whose holonomy is generated by the dilation by $\lambda >1$. Then, clearly, $\nabla$ cannot admit a compatible Riemannian metric.

Edit: Here are some details which are quite standard. I will use the associated bundle construction which I will describe for a general manifold $M$ and general vector space $V$ (in our case, $M=S^1$ and $V={\mathbb R}$).

Consider the universal cover $\tilde M$ of $M$ and regard the product $\tilde M \times V$ as a vector bundle $\tilde E$ over $\tilde M$ with the fiber $V={\mathbb R}^n$. Equip this bundle with the trivial connection where leaves of the horizontal foliation are the products $x\times V$, where $x\in \tilde M$. Suppose we also have a linear representation $h: \pi_1(M)\to GL(V)$. Let the fundamental group of $M$ act on the product by deck transformations on the first factor (the universal cover) and via the linear representation $h$ on the second (fiber) factor. The quotient of $\tilde M \times V$ by the action is naturally a vector bundle $E$ over $M$ with the fiber $V$. The trivial connection on $\tilde E$ is invariant under the action of $\pi_1(M)$ (since it sends horizontal leaves to horizontal leaves) and hence descends to a flat connection $\nabla$ on $E$. Now, if you simply follow the definition of holonomy of a connection (as defined via parallel transport along loops on the base), you see that the holonomy homomorphism of $\nabla$ is $h$. Suppose now that $h$ has the property that its image $G$ is an unbounded subgroup of $GL(V)$. Then $G$ is not contained in any compact subgroup of $GL(V)$, in particular, it is not contained in any conjugate of the subgroup $O(n)< GL(V)$. Now, if $\nabla$ were to admit a compatible Riemannian metric, the holonomy group of $\nabla$ would be contained in $O(n)$, identified with the group of linear isometries of $E_x$, where $E_x$ is the fiber of $E$ over the base-point $x\in M$ (equipped with the Riemannian metric you have). This is a contradiction.

Now, for a specific example, consider $M=S^1$, $V={\mathbb R}$ and $h$ given by sending the generator of $\pi_1(S^1)$ to dilation by $\lambda>0$ on ${\mathbb R}$. The bundle $E$ we defined above is clearly oriented (since $\lambda >0$). Hence, $E$ is trivial as a vector bundle and, thus isomorphic to the tangent bundle $TS^1$. The image of the homomorphism $h$ is, of course, unbounded. Thus, the above reasoning implies that $\nabla$ is not compatible with any metric on $TS^1$.

For in depth discussion of connections, holonomy and so on, see for instance, Kobayashi and Nomizu "Foundations of Differential Geometry" or Petersen's "Riemannian Geometry".

Answer 2

Take a look at https://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita/, especially Thurston's answer, although it doesn't answer your question, but a more special situation.