Metric in terms of the connection

Question

The Levi-Civita connection can be written in terms of the metric as: $$ \Gamma^l_{jk}=\frac{1}{2}g^{lr}(\partial_kg_{rj}+\partial_jg_{rk}-\partial_rg_{jk}). $$ Can this relation be inverted for the metric?

Answer 1

In Riemann normal coordinates we have $x^i g_{ij}(x)=x^i\delta_{ij}$ (see here). In these coordinates it is easy to verify that $$ g_{jk}(x)=\delta_{jk}-\int_0^1 \alpha x^s\Gamma^{s}_{jk}(\alpha x)~d\alpha+\int_0^1 g_{js}(\alpha x)\alpha x^i\Gamma^{s}_{ik}(\alpha x)~d\alpha. $$ The metric appears linearly on the RHS in the final term. Therefore, this formula can be iterated repeatedly to obtain an expansion of $g$ in powers of $\Gamma$. As a consistency condition, one has to recompute the connection from $g$ and enforce that it be equal to the connection we started with. This places a condition on the allowed $\Gamma$ and ensures that such $\Gamma$ arise from a metric.

Answer 2

In polar Riemann normal coordinates $\xi^i$ (such that $\xi^1$ is the radial coordinate) the metric satisfies \begin{align} g_{1i}=g^{1i}=\delta_{i1} \end{align} This implies that $$ \Gamma^{1}_{ij}=-\frac{1}{2}\partial_1 g_{ij} $$ which can be integrated out from the origin to give $g_{ij}$.

Answer 3

Actually, there is a very simple idea. Fix the metric at just one point, $p$, on the manifold (this can be thought of as a 'gauge' (or choice of coordinates) if you like). To determine $<v,w>$ at an arbitrary point (for vectors $v,w$), parallel transport $v$ and $w$ to $p$ using the connection and take their inner product at $p$ using the fixed metric. This works because the metric is preserved by parallel transport.

Answer 4

The equations above are actually overdetermined. There are $\frac{n(n+1)}{2}$ unknowns (components of the metric tensor) but $\frac{n^2(n+1)}{2}$ equations.

You can rewrite it as $$\partial_ig_{jk} = g_{jp}\Gamma^p_{ik} + g_{kp}\Gamma^p_{ji}. $$ Since partials commute, a solution to the system has to satisfy $$ \partial_l(g_{jp}\Gamma^p_{ik} + g_{kp}\Gamma^p_{ji}) = \partial_i(g_{jp}\Gamma^p_{jk} + g_{kp}\Gamma^p_{jl}) $$ If you carry out the differentiation, replace each $\partial_ag_{ab}$ that appears in the resulting equation by the right side of the original equation, and replace each $\partial_a\Gamma^b_{cd}$ by the formula for its formula in terms of the curvature tensor and Christoffel symbols, you get new equations for the metric tensor, Christoffel symbols, and curvature tensor. These equations have to be appended to the original set of equations. Then one has to analyze the new system.

This is done more efficiently using differential forms. We start with a connection, which can be written in local coordinates in terms of the connection $1$-forms \begin{align*} \omega^a_b &= \Gamma^a_{bc}\,dx^c. \end{align*} This has a curvature tensor, which can be written in terms of the curvature $2$-forms $$ \Omega^a_b = d\omega^a_b + \omega^a_c\wedge\omega^c_b. $$ We want to solve for $g_{ab}\,dx^a\,dx^b$ such that $$ dg_{jk} = g_{pk}\omega^p_j + g_{pj}\omega^p_k. $$ Any such solution must also satisfy \begin{align*} 0 &= d(dg_{jk}) = \cdots = g_{pk}\Omega^p_j + g_{pj}\Omega^p_k. \end{align*} At this point, the situation depends on the curvature tensor. At one extreme, if the curvature tensors all vanish, then it follows by the simplest version of the Poincar'e lemma that if you specify $g$ at a single point, there is a unique solution. Note that $g$ need not be constant, because the components of the metric tensor, even the flat one, are not constant for most coordinates.

If the curvature form is nonzero, then you have to add these new zero-order equations to the original equations. At this point, things get very complicated and depend on the rank of the curvature tensor, appropriately defined. If the rank changes, then situation becomes even more complicated. I do not know what progress has been made on this.

Another question is to prescribe the Riemann curvature tensor as a $(0,4)$-tensor $R_{abcd}$. The curvature $2$-forms can be written as $$\Omega_{ab} = \frac{1}{2}R_{abcd}\,dx^c\wedge dx^d, $$ where $\Omega_{ab} = g_{ac}\Omega^c_b.$ The equation above for the curvature forms now becomes $$\Omega_{ab} + \Omega_{ba} = 0.$$ In dimensions $4$ and higher, this is still an overdetermined systems that is hard to analyze. On the other hand, in dimension $3$, it is not hard to see that both the metric and Riemann curvature tensors have $6$ independent components. Therefore, the number of equations equals the number of unknown functions. This is the system of PDEs that Bryant studied in the real analytic case and DeTurck and I studied in the smooth case.