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?


2 Answers 2


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".

  • 1
    $\begingroup$ This looks like a great example, which I'd really like to understand, but could you perhaps elaborate? I'm not yet comfortable with the notion of holonomy, and therefore can't translate the phrase "...whose holonomy is generated by the dilation..." into equations I understand. (I'm guessing the OP might be in a similar position, but this is just a guess.) In a similar vein, the "clearly" is also not clear to me. $\endgroup$ Oct 6, 2013 at 23:41
  • $\begingroup$ @JesseMadnick: I am adding details. $\endgroup$ Oct 7, 2013 at 4:36

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.


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