As we all know, for a Riemannian manifold $(M,g)$, there exists a unique torsion free connection $\nabla_g$, the Levi-Civita connection, that is compatible witht metric.

I was wondering if one can reverse this situation: Given a manifold with $M$ with connection $\nabla$, when does there exist a Riemannian metric $g$ for which $\nabla$ is the Levi-Civita connection.

If this were true for complex projective manifolds it would make me be very happy.

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    $\begingroup$ Levi-Civita is one person --- not two; so do not write it "Levi--Civita" $\endgroup$ – Anton Petrunin Feb 5 '11 at 18:49
  • $\begingroup$ @Anton: Thanks for the correction, I wasn't aware of this convention. $\endgroup$ – Jean Delinez Feb 5 '11 at 19:37
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    $\begingroup$ @Anton, thanks! I had honestly thought all these years that it was named after two people (and had typeset it that way). $\endgroup$ – Joel Fine Feb 11 '11 at 2:12
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    $\begingroup$ Can someone with enough reputation change the $\Delta$'s in the question and in Bill Thurston's answer to $\nabla$'s? This is highly non-standard notation, since $\Delta$ is usually the Laplacian. $\endgroup$ – Spiro Karigiannis Apr 17 '11 at 18:50
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    $\begingroup$ @Spiro (honestly I am a bit surprised you don't already have the rep): since the notation between the question and the answer matches, perhaps it is okay? $\endgroup$ – Willie Wong Apr 17 '11 at 23:54

Bill and Willie have (of course) given correct answers in terms of the holonomy of the given torsion-free connection $\nabla$ on the $n$-manifold $M$. However, it should be pointed out that, practically, it is almost impossible to compute the holonomy of $\nabla$ directly, since this would require integrating the ODE that define parallel transport with respect to $\nabla$. Even though they are linear ODE, for most connections given explicitly by some functions $\Gamma^i_{jk}$ on a domain, one cannot perform their integration.

Although, as Bill pointed out, you cannot always tell from local considerations whether $\nabla$ is a metric connection, you can still get a lot of information locally, and this usually suffices to determine the only possibilities for $g$. The practical tests (carried out essentially by differentiation alone) were of great interest to the early differential geometers, but they don't get much mention in the modern literature.

For example, one should start by computing the curvature $R$ of $\nabla$, which is a section of the bundle $T\otimes T^\ast\otimes \Lambda^2(T^\ast)$. (To save typing, I won't write the $M$ for the manifold.)

Taking the trace (i.e., contraction) on the first two factors, one gets the $2$-form $tr(R)$. This must vanish identically, or else there cannot be any solutions of $\nabla g = 0$ for which $g$ is nondegenerate. (Geometrically, $\nabla$ induces a connection on $\Lambda^n(T^\ast)$ (i.e., the volume forms on $M$) and $tr(R)$ is the curvature of this connection. If this connection is not flat, then $\nabla$ doesn't have any parallel volume forms, even locally, and hence cannot have any parallel metrics.)

To get more stringent conditions, one should treat $g$ as an unknown section of the bundle $S^2(T^\ast)$, pair it with $R$ (i.e., 'lower an index') and symmetrize in the first two factors, giving a bilinear pairing $\langle g, R\rangle$ that is a section of $S^2(T^\ast)\otimes \Lambda^2(T^\ast)$. By the Bianchi identities, the equation $\langle g, R\rangle = 0$ must be satisfied by any solution of $\nabla g = 0$. Notice that these are linear equations on the coefficients of $g$. For most $\nabla$ when $n>2$, this is a highly overdetermined system that has no nonzero solutions and you are done. Even when $n=2$, this is usually $3$ independent equations for $3$ unknowns, and there is no non-zero solution.

Often, though, the equations $\langle g, R\rangle = 0$ define a subbundle (at least on a dense open set) of $S^2(T^\ast)$ of which all the solutions of $\nabla g= 0$ must be sections. (As long as $R$ is nonzero, this is a proper subbundle. Of course, when $R=0$, the connection is flat, and the sheaf of solutions of $\nabla g = 0$ has stalks of dimension $n(n{+}1)/2$.) The equations $\nabla g = 0$ for $g$ a section of this subbundle are then overdetermined, and one can proceed to differentiate them and derive further conditions. In practice, when there is a $\nabla$-compatible metric at all, this process spins down rather rapidly to a line bundle of which $g$ must be a section, and one can then compute the only possible $g$ explicitly if one can take a primitive of a closed $1$-form.

For example, take the case $n=2$, and assume that $tr(R)\equiv0$ but that $R$ is nonvanishing on some simply-connected open set $U\subset M$. In this case, the equations $\langle g, R\rangle = 0$ have constant rank $2$ over $U$ and hence define a line bundle $L\subset S^*(T^\ast U)$. If $L$ doesn't lie in the cone of definite quadratic forms, then there is no $\nabla$-compatible metric on $U$. Suppose, though, that $L$ has a positive definite section $g_0$ on $U$. Then there will be a positive function $f$ on $U$, unique up to constant multiples, so that the volume form of $g = f\ g_0$ is $\nabla$-parallel. (And $f$ can be found by solving an equation of the form $d(\log f) = \phi$, where $\phi$ is a closed $1$-form on $U$ computable explicitly from $\nabla$ and $g_0$. This is the only integration required, and even this integration can be avoided if all you want to do is test whether $g$ exists, rather than finding it explicitly.) If this $g$ doesn't satisfy $\nabla g = 0$, then there is no $\nabla$-compatible metric. If it does, you are done (at least on $U$).

The complications that Bill alludes to come from the cases in which the equations $\langle g, R\rangle = 0$ and/or their higher order consequences (such as $\langle g, \nabla R\rangle = 0$, etc.) don't have constant rank or you have some nontrivial $\pi_1$, so that the sheaf of solutions to $\nabla g = 0$ is either badly behaved locally or doesn't have global sections. Of course, those are important, but, as a practical matter, when you are faced with determining whether a given $\nabla$ is a metric connection, they don't usually arise.

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    $\begingroup$ @Robert: I'm glad to see your answer, thanks for giving an answer for the more typical generic circumstances. As you suggest, I ended up focusing on unlikely pathological cases --- I was curious. $\endgroup$ – Bill Thurston Apr 18 '11 at 8:52
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    $\begingroup$ @Bill: Actually, the kind of cautionary examples that you highlighted are very important, and I, too, always bring them up when I'm lecturing on the subject. The old sources (such as Cartan), though wonderful, tend to be cavalier about constant rank assumptions. (A HW exercise I like to give is to construct a metric connection on $R^4$ that is locally Kahler, but not Kahler.) I do like to counterbalance those kinds of examples with `practical advice' about computing holonomy, partly because it's interesting and partly because it tends not to be treated in most modern texts. $\endgroup$ – Robert Bryant Apr 18 '11 at 12:07


First, there's a very simple criterion for whether $\nabla$ is an orthogonal connection: look at the holonomy of $\nabla$ around closed loops in the manifold, and ask whether they preserve a quadratic form. The set of quadratic forms preserved by a linear transformation is a linear subspace of all quadratic forms, so there's some linear subspace of quadratic forms preserved by the holonomy.

The condition that $\nabla$ is torsion free doesn't depend on a metric, so it's straightforward to check. The necessary and sufficient condition for $\nabla$ to be a Levi-Civita condition is that its holonomy preserve at least one positive definite quadratic form, and that it be torsion-free.

Note that the condition on holonomy is global: it can't be reduced to some set of pointwise identities involving $\nabla$, or even the local behavior of $\nabla$. For instance, take $\nabla$ to be the standard flat connection in $\mathbb R^n \setminus 0$ modulo the linear transformation $x \rightarrow 2x$. Since $\nabla$ is preserved by $x \rightarrow 2x$, it descends to the quotient $S^{n-1} \times \mathbb R$. It can locally be expressed as a Levi-Civita connection, but there is no globally-defined metric for which it is the Levi-Civita connection.

It's also possible to concoct simply-connected examples with a connection that is locally Levi-Civita, but not globally Levi-Civita. For instance: inside $S^3$ embed a copy of $T^2 \times I$, and make a Riemannian metric that for which $T^2 \times I$ is isometric to $[0,1] \times \mathbb E^2$ modulo a discrete group of translations, and for which each component of the complement has holonomy (as usual) equal to the full $SO(3)$. Make a second, similar metric, but where the $T^2$ has a different shape. Make a hybrid of the two, combining half from one $S^3$ and the other half from the other $S^3$, glued together by an affine map of the torus. The flat connection is identified by the gluing map, but the holonomy does not globally preserve a Riemannian metric.

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    $\begingroup$ Exercise: make a connection on $S^2$ that is locally Levi-Civita but not globally Levi-Civita. $\endgroup$ – Bill Thurston Feb 5 '11 at 19:13
  • $\begingroup$ Thanks a lot for your answer. Just one question: what is an "orthogonal" connection? $\endgroup$ – Jean Delinez Feb 5 '11 at 19:40
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    $\begingroup$ An orthogonal connection with respect to a quadratic form is one that preserves that quadratic form. If you're given the quadratic form $g$, this is the identity $X(\left < Y, Z \right >_g = \left < \Delta_X Y , Z\right >_g + \left < Y, \Delta_X Z \right >$. Note that if $Y$ and $Z$ range over a local orthonormal basis, this reduces to skewsymmetry of the matrix for $\Delta_X$ expressed in terms of the basis. $\endgroup$ – Bill Thurston Feb 5 '11 at 20:23
  • $\begingroup$ @Bill Great. Thanks a lot for your help. $\endgroup$ – Jean Delinez Feb 11 '11 at 17:19
  • $\begingroup$ Changed $\Delta$ to $\nabla$. $\endgroup$ – Deane Yang Apr 29 '11 at 13:55

To start with, you need the connection to be torsion free. After that, there is a characterisation of metric connections given by Schmidt, CMP 29 (1973) 55-59, which states that the linear torsion-free connection is metric if and only if the holonomy group is a sub-group of the orthogonal group of the desired signature.

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    $\begingroup$ In other words, the holonomy group must be pre-compact. Given that this condition is non-local, it is unlikely that there is a more manageable equivalent formulation. $\endgroup$ – Sergei Ivanov Feb 5 '11 at 19:05
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    $\begingroup$ The statement is essentially trivial, not clear what Schmidt could write on these 4 pages... $\endgroup$ – Anton Petrunin Feb 5 '11 at 19:11
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    $\begingroup$ @Sergei: Right. The connection is Levi-Civita for some Riemannian metric if and only if it is torsion-free with relatively compact holonomy group. But what about pseudo-Riemannian metrics? $\endgroup$ – George Lowther Feb 5 '11 at 19:45
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    $\begingroup$ @George, @Sergei: since the set of quadratic forms preserved by any single group element is a linear subspace, it's already an easy condition, to look at the intersection of these subspaces among all holonomy elements. The infinitesimal intesections are also linear, so it's a matter of parallel transporting these local invariant subspaces to compare globally. For the orthogonal case, I think this is easier than checking whether the holonomy is compact. Eigenvalues aren't enough: the holonomy might be the group that conserves some subspace, and is orthogonal on subspace and quotient. $\endgroup$ – Bill Thurston Feb 5 '11 at 22:17
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    $\begingroup$ Schmidt's article can be accessed freely on Project Euclid. $\endgroup$ – Alex M. Feb 1 '16 at 12:57

This question has been definitively answered in the following paper: http://nzjm.math.auckland.ac.nz/images/f/f4/When_is_a_Connection_Metric_Connection%3F.pdf

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    $\begingroup$ The paper (or a nearly identical version of it) seems to have a permanent home on arxiv: arxiv.org/abs/0804.2698 $\endgroup$ – Ben Crowell Jul 30 '17 at 23:01

For a connection with "full rank" curvature matrix I think I have a more complete answer in form of an algorithm that verifies whether the connection is metric. This is partly based on an observation made by Robert Bryant so at least that part is 100% correct.

Consider a connection $D$ in the vector bundle $E.$ We assume that the curvature matrix has " full rank"(I can elaborate on this condition on the curvature) on a open set $U.$ We'd like to verify if the connection is metric on $U.$

We have the following steps in deciding weather the connection $D$ is metric on $U$:

1.Take a local frame $\sigma=(\sigma_1, \sigma_2,...,\sigma_m )$ on $U$ and calculate the curvature matrix $\Omega$ with respect to this frame.

  1. Determine the matrices $S_{ij}$(with real smooth functions as entries) defined by the equation $$ \Omega=\sum_{i<j} (\sigma^i \wedge \sigma^j) S_{ij} .$$

  2. Form the linear system $$ XS_{ij}+S_{ij}^TX=0. $$

  3. If the linear system from the previous step doesn't have a symmetric, positive definite solution $X=A$, the $D$ is not metric.

  4. If the solution $X=A$ exists then take its square root $\sqrt{A}=B$.

  5. Form the frame $\sigma'$ defined by $$\sigma'=\sigma B$$ and test whether the connection forms with respect to $\sigma'$ are skew. If they are not then the connection is not metric on $U$.

  6. If the connection forms with respect to $\sigma$ are skew, then the metric that makes $\sigma$ orthonormal is compatible with the connection.

  • $\begingroup$ Your algorithm is not complete even in dimension $2$. For example: The solution $X=A$ that one must assume to exist in Step 4 is not ever unique; one can always multiply any solution $A$ by any positive function. The question then becomes how to tell whether there is some choice of $A$ in Step 4 so that the connection forms with respect to $\sigma'$ in Step 6 are skew. Your algorithm needs to deal with testing all possible choices of solutions $A$ in order to decisively answer the question. (For example, see my remark in the penultimate paragraph of my answer about finding $f$.) $\endgroup$ – Robert Bryant Jan 5 '18 at 14:26
  • $\begingroup$ Thank you Dr. Bryant. I am trying to understand why A has to be unique and hopefully I'll post a clarifying comment. $\endgroup$ – Mike Cocos Jan 12 '18 at 1:15

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