2015-04-07 06:10:18 8 Comments

It is well known that the Riemann curvature tensor of a metric satisfies

\begin{eqnarray} R_{jikl}=-R_{ijkl}=R_{ijlk},(1)\\ R_{klij}=R_{ijkl},(2)\\ R_{i[jkl]}=0 \mbox{(1st Bianchi identity)}.(3) \end{eqnarray}

**The question is whether there are more (may be non-linear and/or differential) equations satisfied by $R_{ijkl}?$**
(I do not mention here the 2nd Bianchi identity $R_{ij[kl;m]}=0$ since it involves covariant derivatives and hence is equation not only on the tensor $R$ but also on the original metric $g$).

One can ask this question in the following more precise form.

**Question 1.** Let $R_{ijkl}$ be a collection of numbers satisfying (1)-(3) with the range of indices from 1 to $n$. Does there exist a metric $g$ in a neighborhood of 0 in $\mathbb{R}^n$ such that its curvature tensor at 0 is equal to the given collection of numbers?

**Question 2.** Let $R_{ijkl}(x)$ be a collection of smooth (real analytic?) functions in a neighborgood of 0 in $\mathbb{R}^n$ satisfying (1)-(3). Does there exist a metric in a neighborhood of 0 such that its curvature tensor is equal to the given collection of functions $R_{ijkl}(x)$?

**UPDATE:** As mentioned by Peter Michor below, the answer to Q1 is positive. As mentioned by him and Vladimir Matveev, in dimensions $n>3$, $g\mapsto R(g)$ is an overdetermined system of PDE. Hence the answer to Q2 is 'No' for $n>3$, though it is not clear to me how to write explicitly any constrain on the image of the map. In $n=2$ the answer is 'Yes' according to Vladimir Matveev.

**ANOTHER UPDATE:** According to Robert Bryant's answer below, in $n=3$ the answer to Q2 is 'No' in general. However under some non-degeneracy assumptions it is 'Yes'.

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## 4 comments

## @Deane Yang 2015-04-12 04:01:51

EDIT: Attempted to insert correct constants and fix indices.

Might as well write down the explicit answer to Question 1. Everything below is written with respect to local co-ordinates.

Suppose you are given a curvature tensor $R_{ijkl}$ at one point. This satisfies the following symmetries: $$ R_{ijkl} = R_{klij} = -R_{jikl} = -R_{ijlk} $$ and the first Bianchi identity $$ R_{ijkl} + R_{iklj} + R_{iljk} = 0. $$ It is straightforward using the definition of the curvature tensor to show that if the Christoffel symbols vanish at the point $p$, then the curvature tensor is given (up to a not necessarily positive constant factor) at $p$ by $$ R_{ijkl} = \frac{1}{2}(\partial^2_{ik}g_{jl} + \partial^2_{jl}g_{ik} - \partial^2_{il}g_{jk} - \partial^2_{jk}g_{il}). $$ It is also straightforward to check that this is satisfied (up to a constant factor), if you set $$ \partial^2_{ik}g_{jl}(p) = \frac{1}{3}(R_{ijkl} + R_{ilkj}). $$ Note that this is not the unique solution. I leave the determination of the correct constant factors to the reader. Note that you do not even need to assume that $g_{ij}(p) = \delta_{ij}$. Only that $\partial_kg_{ij}(0) = 0$, which is equivalent to having the Christoffel symbols vanish at $0$.

To summarize, given any tensor $R_{ijkl}$ satisfying the symmetries stated above, any smooth metric of the form $$ g_{jl}(x) = g_{jl}(0) + \frac{1}{3}(R_{ijkl} + R_{ilkj})x^ix^k + O(|x|^3) $$ has curvature tensor $R_{ijkl}$ at $x = 0$.

## @Renato G. Bettiol 2015-04-11 17:37:27

The answer to Q1 is YES, as explained by Michor, as metrics that realize a prescribed curvature tensor at a point can be constructed using Taylor series. However, for the sake of completeness, I think it is worth supplying a few more details regarding here (this construction is elementary, but has been quite useful to me before, and I guess others may also find it useful). It is also interesting to note that this manifold is realized as a submanifold of Euclidean space with an explicit upper bound on the codimension.

Proof.Since $R$ satisfies the first Bianchi identity, there exist symmetric linear operators $H_j\colon V\to V$, $1\leq j\leq k\leq \tfrac12 n(n-1)$, such that $R=-\sum_{j=1}^k (H_j\wedge H_j),$ where $(H_j\wedge H_j)(X\wedge Y):=H_j X\wedge H_j Y$, see e.g. Jacobowitz, p. 102 or Steiner, Teufel, Vilms, p. 422. Thus, \begin{equation} \langle R(X\wedge Y),Z\wedge W\rangle=%-\sum_{j=1}^k \langle (H_j\wedge H_j)(X\wedge Y),Z\wedge W\rangle= \sum_{j=1}^k \langle H_j X,W\rangle\langle H_j Y,Z\rangle -\langle H_j X,Z\rangle\langle H_j Y,W\rangle.\qquad (*) \end{equation} For each $1\leq j\leq k$, consider the quadratic functions $f_j\colon V\times\mathbb R\to\mathbb R$, given by $f_j(v,y_j):=y_j+\tfrac12\langle H_j v,v\rangle$, and set \begin{equation*} f\colon V\times\mathbb R^k\to\mathbb R^k, \quad f(v,y_1,\dots,y_k):=\big(f_1(v,y_1),\dots,f_k(v,y_k)\big). \end{equation*} Then $f(0,0)=0$ and $0\in\mathbb R^k$ is a regular value of $f$, hence $M:=f^{-1}(0)\subset V\times\mathbb R^k$ is a smooth submanifold of codimension $k$. We have $T_{(0,0)}M=V$, and the second fundamental form $II\colon V\times V\to \mathbb R^k$ of $M$ at this point is \begin{equation*} II(v,w)=-\sum_{j=1}^k \big\langle \nabla_v (\operatorname{grad} f_j),w\big\rangle\operatorname{grad} f_j=-\sum_{j=1}^k \big\langle H_j v,w\big\rangle\operatorname{grad} f_j. \end{equation*} Thus, $(*)$ is the Gauss equation of $M\subset V\times\mathbb R^k$ proving that its curvature operator at $(0,0)$, with the induced metric, is exactly $R$.## @Robert Bryant 2015-04-07 11:34:21

The answer to Q2 for $n=3$ is actually 'no, without some nondegeneracy hypotheses'. The reason is as follows:

The curvature tensor $\mathcal{R}= R_{ijkl}\,(\mathrm{d}x^i\wedge\mathrm{d}x^j)\circ(\mathrm{d}x^k\wedge\mathrm{d}x^l)$, with all its indices lowered, is a section of the subbundle $K(M)\subset S^2\bigl(\Lambda^2(T^*M)\bigr)$ that is the kernel of the natural linear mapping $$ S^2\bigl(\Lambda^2(T^*M)\bigr)\longrightarrow \Lambda^4(T^*M). $$ You are asking, for a given section $\mathcal{R}$, whether there exists a metric $g$ such that $\mathrm{Riem}(g) = \mathcal{R}$. There is no pointwise algebraic condition on the section $\mathcal{R}$ imposed by this equation (that was what Q1 was about), but, as you note, there is the

second Bianchi identity$$ \mu(\nabla^g\mathcal{R}) = 0, $$ where $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is the natural skewsymmetrization operation. Since $\nabla^g$ depends on one derivative of the metric $g$, the above equation with a given $\mathcal{R}$ can be regarded as a first-order system of equations on $g$. When $n=3$, this is at most $3$ equations, the rank of the bundle $\Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$.Now, for a point $p\in M$ satisfying $\mathcal{R}(p)=0$, the value $\nabla^g\mathcal{R}(p)$ does not depend on $g$ (just look at the formula for $\nabla^g\mathcal{R}$ in local coordinates). Thus, if $\mathcal{R}(p)=0$, but $\mu(\nabla^{g_0}\mathcal{R})(p) \not= 0$ for some metric $g_0$, then $\mu\bigl(\nabla^{g}\mathcal{R}(p)\bigr) \not= 0$ for all metrics $g$ and hence there is no open neighborhood of $p$ on which the equations $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ have a solution $g$. In particular, the original system $\mathrm{Riem}(g) = \mathcal{R}$ has no solution in a neighborhood of such a point $p$.

To convince yourself that such examples exist when $n=3$, just note that the rank of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$ in this case is $6$, so an arbitrary section that vanishes at $p$ will have $6\times 3 = 18$ independent first derivatives. Thus, the map $\mu:K(M)\otimes T^*M\to \Lambda^2(T^*M)\otimes\Lambda^3(T^*M)$ is surjective (and the rank of the target bundle is $3$), so that the generic section of $K(M)$ that vanishes at $p$ will not satisfy the second Bianchi identity at $p$ for any metric $g$.

However, suppose that $n=3$ and that $\mathcal{R}$ is a

nondegeneratesection of $K(M)=S^2\bigl(\Lambda^2(T^*M)\bigr)$. I proved (back in the early 1980s) that, when $\mathcal{R}$ is real-analytic, there always exist local solutions to the equation $\mathrm{Riem}(g) = \mathcal{R}$. Specifically, I showed that, in this case, in addition to the $6$ second-order equations that these equations represent on $g$ and the $3$ first-order equations on $g$ that $\mu\bigl(\nabla^{g}\mathcal{R}\bigr) = 0$ represents, there is one more first-order equation $Q_\mathcal{R}(g)=0$ on $g$ that is satisfied by any metric $g$ that satisfies $\mathrm{Riem}(g) = \mathcal{R}$. Then I proved that the combined overdetermined system of $6$ second-order equations and $4$ first-order equations for $g$ isinvolutive, so that an application of the Cartan-Kähler Theorem proves local solvability. Unfortunately, the involutive system is never either hyperbolic or elliptic, though it can be of real principal type.I never published my proof, but later, Dennis DeTurck and Deane Yang studied the overdetermined system that I wrote down and published a proof of its local solvability in the smooth category. See Deturck and Yang,

Local existence of smooth metrics with prescribed curvature, Nonlinear problems in geometry (Mobile, Ala., 1985), 37–43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986.## @Peter Michor 2015-04-07 06:50:54

Q1: Yes. In Riemannian normal coordinates (using the exponential map as chart), the Taylor expansion of the metric at 0 has: identity as constant part, vanishing linear part (Christoffels vanish at 0), and the quadratic part is the curvature. This is the way, Riemann found his tensor. So just write the second order polynomial as indicated, which gives a Riemannian metric near 0 with the prescribed curvature tensor at 0. It is a metric because of equations (1)-(3).

Q2: No, in general. $g\mapsto R(g)$ is an overdetermined system of PDE's; so there are integrability conditions (like the second Bianchi identity and its iterated covariant derivatives)

## @Vladimir S Matveev 2015-04-07 07:51:28

To Q2: in small dimensions (n=2,3) the mapping $g\to R(g)$ is not overdetermined and at least in dimension 2 there always exists a metric with a given $R_{ijkl}$ satisfying (1,2,3)

## @MKO 2015-04-07 08:04:20

@PeterMichor: I think I understand your answer to Q1. Regarding the 2nd Bianchi identity, it involves both tensors $R$ and $g$. Hence it is not an equation on $R$ only, but also on the unknown metric $g$. Hence, I think, it cannot be considered as a constrain on $R$ only.

## @MKO 2015-04-07 08:30:06

Also I agree that in higher dimensions $g\mapsto R(g)$ is overdetermined. But I do not know how to write explicitly any extra equation on the image without using the metric $g$.