# Locally Riemannian Connection

Let $\Gamma^a{}_{bc}=\Gamma^a{}_{cb}$ be a symmetric connection whose curvature is $$R^a{}_{bcd}=\partial_c\Gamma^a{}_{bd}-\partial_d\Gamma^a{}_{bc}+\Gamma^a{}_{ec}\Gamma^e{}_{bd}-\Gamma^a{}_{ed}\Gamma^e{}_{bc}.$$

What conditions can we observe locally on $\Gamma^a{}_{bc}$ or $R^a{}_{bcd}$ to conclude that it is locally Riemannian?

• Non local information as the holonomy group is not accessible so I wish to restrict to what can we know locally – Aureliano Skirzewski Jan 12 '18 at 16:43
• This question is related mathoverflow.net/questions/54434 – j.c. Jan 12 '18 at 16:59
• That question is not the same as yours, but unless I am misunderstanding something, I believe several of the answers and references will be helpful to you. See e.g. arxiv.org/abs/0804.2698 cited in John's answer mathoverflow.net/a/173399 – j.c. Jan 12 '18 at 17:16
• Note that $dg_{ab} = (g_{ap}\Gamma^p_{bc} + g_{bp}\Gamma^p_{ac})\,dx^c$, so you can use the Frobenius theorem. – Deane Yang Jan 12 '18 at 17:34
• You could employ the holonomy group in an arbitrarily small open set, which would already give a nontrivial condition, locally but not infinitesimally. – Ben McKay Jan 12 '18 at 17:39

Perhaps I can offer some information and comment on this problem. An essential part of the problem is how to interpret terms such as 'observe', 'accessible', 'identify', as the OP wants to know how to write down a computable criterion for a torsion-free connection to be the Levi-Civita connection of a Riemannian metric. Of course, this requires that one specify what one means by computable. For example, the OP does not consider the holonomy criterion (i.e., that the holonomy be a compact group) to be 'accessible', so I suppose that this means that tools to determine the holonomy group are not to be counted as computable, for the purposes of this question.

Various articles in the literature purport to solve this problem by providing an 'algorithm' for finding a compatible metric (if it exists) that only involves computing derivatives. However, all of these 'algorithms' depend on making constant rank assumptions for various systems of linear equations. As long as these constant rank assumptions hold, they work fine, but it is possible to 'fool' them by constructing examples in which the constant rank assumptions do not hold.

For example, take two closed disks with smooth boundary, $D_1$ and $D_2$, in $\mathbb{R}^2$ whose interiors are disjoint. Let $g_1$ be a metric on $\mathbb{R}^2$ that agrees with the flat metric $dx^2+dy^2$ outside of $D_1$ but has nonzero curvature somewhere inside $D_1$ and let $g_2$ be a metric on $\mathbb{R}^2$ that agrees with the flat metric $dx^2+2dy^2$ outside of $D_2$ but has nonzero curvature somewhere inside $D_2$. One can even arrange that the curvature of $g_i$ be nonzero in $D_i$ away from some closed subset $K_i$ that lies in the interior of $D_i$, so do this. Now let $\nabla$ be the connection that agrees with the Levi-Civita connection $\nabla_1$ of $g_1$ outside the disk $D_2$ and the Levi-Civita connection $\nabla_2$ of $g_2$ outside the disk $D_1$. (The connections $\nabla_i$ are equal outside the union of the interiors of $D_1$ and $D_2$.) Then $\nabla$ is not the Levi-Civita connection of any metric on $\mathbb{R}^2$. If the two closed disks do not intersect, then every point of $\mathbb{R}^2$ has an neighborhood that does not meet one of the disks, so there is a metric on that neighborhood (e.g., one of the $g_i$) compatible with $\nabla$. Hence $\nabla$ is locally Riemannian. However, if the closed disks are tangent at one point, then that point has no open neighborhood on which $\nabla$ is Riemannian.

Note that the above example shows that the condition of being locally Riemannian is not a closed condition on germs of torsion-free connections in the plane (since it can hold on the complement of a point) and hence it cannot be determined just as the satisfaction of some system of partial differential equations on the connection.

Meanwhile, if one is willing to consider open conditions as well as closed conditions, then it is easy to write down sufficient conditions for a connection to be locally Riemannian that, beyond algebra, only require the ability to take derivatives and to test whether an expression is zero or not. These 'accessible' sufficient conditions are not necessary, though.

For example, consider the $2$-dimensional case: Let $\nabla$ be a torsion-free connection on a simply-connected domain $U$ in the $x^1x^2$-plane, with connection coefficients $\Gamma^{i}_{jk}=\Gamma^{i}_{kj}$, let $\gamma^i_j = \Gamma^{i}_{jk}\,\mathrm{d}x^k$ be the entries of the $2$-by-$2$ matrix $\gamma$, and write $\mathrm{d}\gamma+\gamma\wedge\gamma = R\,\mathrm{d}x^1\wedge\mathrm{d}x^2$, where $R= (R^i_j)$ is a matrix of functions on $U$.

The first necessary condition is that $\mathrm{tr}(R) = R^1_1+R^2_2 = 0$, otherwise there would be no parallel volume form (and hence no parallel metric). (This is one first-order differential equation on $\nabla$.)

Now, suppose the open condition that $\mathrm{det}(R) = r^2 > 0$ for some positive function $r$ on $U$ (which, if $\gamma$ is to be a Riemannian connection, would follow from $R\not=0$). Once this condition is imposed, define a symmetric matrix $H$ of determinant $1$ by the rule $$H = \pm\frac{1}r\begin{pmatrix}R^2_1& -R^1_1\\ -R^1_1 & -R^1_2\end{pmatrix},$$ with the sign chosen to make $H$ be positive definite. Then $H$ is the unique symmetric positive definite matrix of functions on $U$ that has determinant $1$ and satisfies the condition that $HR$ be skew-symmetric.

Finally then, $\nabla$ is a metric connection in $U$ if and only if the identity $$\mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0$$ holds. (This is only four second-order differential equations on $\nabla$, since, by construction, the trace of $H^{-1}\mathrm{d}H$ always vanishes. One can check that these four equations are independent.)

In fact, since $\mathrm{d}\bigl(\mathrm{tr}(\gamma)\bigr)=0$, and $U$ is supposed simply-connected, there exists a function $f$ on $U$ such that $\mathrm{tr}(\gamma) = \mathrm{d}f$, and then the symmetric matrix $G = \mathrm{e}^f H$ satisfies $$\mathrm{d}G = G\gamma + {}^t\gamma G,$$ and so the metric $g = G_{ij}\,\mathrm{d}x^i\mathrm{d}x^j$ is parallel with respect to $\nabla$, and, up to a constant factor, it is the unique such metric.

Thus, a sufficient condition in this case is comprised by a first-order differential equation on $\nabla$, a strict inequality on the first derivatives of $\nabla$ (which is an open condition), and then four second-order differential equations on $\nabla$.

• Thank you for your response, I suppose the higher dimensional set of conditions is equivalent to finding a non degenerated 0-form $H_{ij}$ such that the 2-form $H_{ki}R^i{}_j$ is antisymmetric in $k$ and $j$. That's enough to reduce the number of independent components of the curvature tensor to the components of a Riemann. It just rests to find when does $H$ exist and check which of the solutions for $H$ satisfies $$\mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0.$$ Am I missing something? – Aureliano Skirzewski Jan 18 '18 at 12:18
• @AurelianoSkirzewski: That's close. In dimension $n$, if you make the assumption that the curvature $2$-form $\rho=\mathrm{d}\gamma +\gamma\wedge\gamma$ satisfies the condition that there exists exactly one positive definite, determinant 1, symmetric matrix $H$ such that $H\rho$ is antisymmetric (which is a combination of open and closed first-order conditions on the curvature $\rho$), then you just need to check whether $$\mathrm{d}H +\tfrac2n\,\mathrm{tr}(\gamma)H-H\gamma-{}^t\gamma H=0.$$ – Robert Bryant Jan 18 '18 at 13:02