#### [SOLVED] Does the curvature determine the metric?

Hello,

I ask myself, whether the curvature determines the metric.

Concretely: Given a compact Riemannian manifold $M$, are there two metrics $g_1$ and $g_2$, which are not everywhere flat, such that they are not isometric to one another, but that there is a diffeomorphism which preserves the curvature?

Can we chose $M$ to be a compact 2-manifold?

Can we classify manifolds, where the curvature determines the metric?

What happens, if we also allow semi-riemannian manifolds for the above questions?

Thank you for help. #### @Robert Bryant 2012-06-22 16:07:28

It should not be surprising that, for a $2$-dimensional manifold, the Gauss curvature $K:M\to\mathbb{R}$ does not determine a unique metric $g$. After all, the former is locally one function of $2$ variables while the latter is locally three functions of $2$ variables. It would be remarkable if $K$ determined $g$, even up to isometry, and, of course, except for constant $K$, it does not, as the arguments of Weinstein and Kulkarni show.

Meanwhile, if one regards the curvature $\mathsf{R}(g)$ of of the Levi-Civita connection of the metric $g$ on a surface $M$ as a section of the rank $3$ bundle $\bigl(T\otimes T^\ast)_0\otimes \Lambda^2(T^\ast)={\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then the equation $\mathsf{R}(g) = {\mathsf{R}}_0$ for a given nonzero section ${\mathsf{R}}_0$ of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$ is a determined second-order equation for $g$. (Here, I am writing "$T$" for "$TM$", etc., to save space.) This then becomes a more reasonable question, one that has an easy answer: Namely, if $M$ is compact and connected and $g_1$ and $g_2$ are metrics on $M$ such that $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and such that these curvature tensors are nowhere vanishing, then $g_2 = c g_1$ for some constant $c>0$, and conversely. The reason for this is simple: If ${\mathsf{R}}_0$ is a nonvanishing section of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then requiring that $\mathsf{R}(g) = {\mathsf{R}}_0$ determines the conformal class of $g$ purely algebraically (because using $g$ to 'lower the first index' of ${\mathsf{R}}_0$ has to result in a tensor that is skewsymmetric in the first two indices). Thus, if $\mathsf{R}(g_1) = \mathsf{R}(g_2) = {\mathsf{R}}_0$, where the latter is nonvanishing, then $g_2 = e^u\ g_1$ for some function $u$ on $M$. Plugging this back into the equation $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and computing shows that $u$ must be harmonic with respect to the conformal structure determined by $g_1$ (which is also the conformal structure determined by $g_2$). If $M$ is compact and connected, then $u$ has to be constant.

As Misha points out, in higher dimensions, the situation is rather different and is well-described in the works of Kulkarni and Yau that he cites. However, just qualitatively, it's worth pointing out that specifying the sectional curvature of a metric is generally a very overdetermined problem in higher dimensions, which is why the rigidity results of Kulkarni and Yau should not be surprising.

To see the nature of this, recall that the Riemann curvature tensor $\mathsf{Rm}(g)$ of a metric $g$ on $M$ is got from $\mathsf{R}(g)$ by 'lowering an index'. By the first Bianchi identity, $\mathsf{Rm}(g)$ is a section of the subbundle ${\mathsf{K}}(T^\ast)\subset {\mathsf{S}}^2\bigl(\Lambda^2(T^\ast)\bigr)$ that is the kernel of the natural map induced by wedge product $$\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)\to \Lambda^4(T^\ast)$$ In fact, this map has a natural $\mathrm{GL}(T)$-equivariant right inverse, so that one has a canonical bundle splitting $$\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr) = {\mathsf{K}}(T^\ast)\oplus \Lambda^4(T^\ast)$$ and these two subbundles are $\mathrm{GL}(T)$-irreducible. (I'm using $\mathrm{GL}(T)$ to denote the 'gauge' group of bundle automorphisms of $T = TM$. I don't say that it's great notation, but it will be OK for this discussion.) An important interpretation of this splitting is the following one: Obviously, one can regard a section of $\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ as a quadratic form on the bundle $\Lambda^2(T)$. Then the elements of $\Lambda^4(T^\ast)\subset \mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ are exactly the quadratic forms that vanish on all of the decomposable elements in $\Lambda^2(T)$, i.e., the elements of the form $x\wedge y$ for $x,y\in T_xM$ for some $x$. In particular, a section $Q$ of ${\mathsf{K}}(T^\ast)$ is completely determined by its values on the 'cone bundle' $\mathsf{C}\subset \Lambda^2(T)$ that consists of the decomposable elements.

Now, any metric $g$ on $M$ determines a unique section $\Lambda^2(g)$ of ${\mathsf{K}}(T^\ast)$ with the property that $$\Lambda^2(g)\bigl(x\wedge y\bigr) = |x\wedge y|^2_g = g(x,x)g(y,y)-g(x,y)^2,$$ and the sectional curvature of $g$ is simply the function $\sigma_g:\mathsf{Gr}(2,T)\to\mathbb{R}$ defined as the ratio $$\sigma_g\bigl([x\wedge y]\bigr) = \frac{\mathsf{Rm}(g)\bigl(x\wedge y\bigr)}{\Lambda^2(g)\bigl(x\wedge y\bigr)} \quad\text{for x,y\in T_xM with x\wedge y\not=0},$$ where we regard $\mathsf{Gr}(2,T)$ as the projectivization of the cone bundle $\mathsf{C}$.

The fact that the sectional curvature function is the ratio of two quadratic forms that are sections of $\mathsf{K}(T^\ast)$ shows that it is a very constrained function on $\mathsf{Gr}(2,T)$. In fact, unless the sectional curvature is constant on $\mathsf{Gr}(2,T_x)$, the numerator and denominator of this ratio at $x\in M$ are uniquely determined up to a common multiple. In particular, the set of such functions on $\mathsf{Gr}(2,T)$ can be regarded as the set of sections of a bundle over $M$ whose fibers are isomorphic to a space of dimension $d_n = \tfrac12 n(n{+}1) + \tfrac1{12}n^2(n^2-1) -1$ that is singular along the $1$-dimensional curve of constant functions.

In particular, specifying a candidate sectional curvature function $\sigma:\mathsf{Gr}(2,T)\to\mathbb{R}$ that is not constant on any fiber $\mathsf{Gr}(2,T_x)$ is equivalent to specifying $d_n$ functions of $n$ variables for the $\tfrac12 n(n{+}1)$ coefficients of $g$, which is highly overdetermined when $n>2$. This is why one has the level of rigidity for the sectional curvature that is indicated in Kulkarni's and Yau's results. In fact, as this analysis shows, if $\sigma_{g_1} = \sigma_{g_2} = \sigma$, where $\sigma$ is not constant on any fiber $\mathsf{Gr}(2,T_x)$, then one must have $g_2 = e^u g_1$ for some function $u$ on $M$ and that function $u$ will have to satisfy $\mathsf{Rm}(e^u g_1) = e^{2u} \mathsf{Rm}(g_1)$ which is, itself, a very highly overdetermined second order equation for the single function $u$. (It's kind of remarkable that there are any nonflat metrics $g$ that admit nonzero solutions $u$ to $\mathsf{Rm}(e^u g) = e^{2u} \mathsf{Rm}(g)$ at all, even without the compactness assumption. That is what makes Yau's counterexample in dimension $3$ so interesting. Indeed, though, it turns out that, up to constant scaling and diffeomorphism, there is exactly a $1$-parameter family of such exceptional metrics in dimension $3$, so they are extremely rare indeed. )

The upshot of all these observations is that specifying a nonconstant sectional curvature function $\sigma_g$ in dimensions above $2$ very nearly determines $g$ in all cases, so that, except for a very small set of degenerate cases, the sectional curvature does, indeed, determine the metric. However, it's such an overdetermined problem that this result is not all that surprising.

In particular, specifying $\sigma_g$ is specifying a greater quantity of information about $g$ than specifying, say, $\mathsf{Rm}(g)$. When $n>3$, even specifying $\mathsf{Rm}(g)$ is overdetermined, and, generally speaking, you'd expect $\mathsf{Rm}(g)$ to determine $g$ as well when $n>3$.

When $n=3$, specifying $\mathsf{Rm}(g) = \mathsf{R}$ is locally $6$ second order equations for $6$ unknowns, so it's a determined system. Back in the 80s, by using the Cartan-Kähler theorem, I showed that, when $\mathsf{R}$ is appropriately nondegenerate and real-analytic, the equation $\mathsf{Rm}(g) = \mathsf{R}$ is locally solvable for $g$, and the general solution depends on $5$ functions of $2$ variables. Later, Dennis DeTurck and Deane Yang proved this result in the smooth category as well (see Local existence of smooth metrics with prescribed curvature, in Nonlinear problems in geometry (Mobile, Ala., 1985), 37--43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986).

In higher dimensions, the natural determined curvature problem is to specify the Ricci tensor $\mathsf{Rc}(g)$ as a section of $S^2(T^\ast)$ (or some trace-modified version of it), and, there, Dennis DeTurck has the best results. #### @Michael Bächtold 2012-06-22 19:25:17

Interesting, so in the last paragraph you are in 2d and the difference to Mishas answer is that one interprets curvature as a different type of tensor? Or did I miss the point completely? And everyone has mentioned compactness, could someone give me a hint of how this enters in the proofs of these statements? #### @Robert Bryant 2012-06-22 19:30:25

Yes, I'll edit the answer to clarify that my remarks up to that point concern only the $2$-dimensional case. If I have time, I'll put in some comments about higher dimensions and/or other versions of the problem. #### @Michael Bächtold 2012-06-22 20:48:54

Thank you for the response! Does changing the way we interpret curvature also affect the answer in higher dimension? Concerning the local version of the question: am I right assuming that any of these ways of interpreting curvature has generically more infinitesimal symmetries than a metric tensor? Sorry for the many questions, they are not meant to be all adressed to you directly. #### @Misha 2012-06-22 21:23:26

@Michael: Yau proved in his paper "Curvature preserving diffeomorphisms" (Annals, 1974) that in dimension 3 there is a counter-example to Kulkarni's theorem for open manifolds, while Kulkarni's theorem still holds for closed 3-manifolds. In dimensions $\ge 4$ compactness is irrelevant (but compactness is assumed in Zank's question). #### @Michael Bächtold 2012-06-23 09:03:19

@Misha: Thank you for the clarification and the references. #### @Ilia Smilga 2013-12-10 10:53:05

Just a detail: shouldn't the "left inverse of the map $S^2 (\Lambda^2 (T^*)) \to \Lambda^4(T^*)$ induced by the wedge product" be a right inverse instead? (If I remember correctly, a right inverse of $f$ is a function $g$ such that $f \circ g = \operatorname{Id}$; in other words, a "section".) #### @Robert Bryant 2013-12-10 14:23:45

@IliaSmilga: Yes, of course, I should have written 'right inverse' there. I will make the change. #### @Misha 2012-06-21 20:49:15

For concrete 2-dimensional counter-example see page 328 of Kulkarni's paper "Curvature and metric", Annals of Math, 1970. Weinstein's argument there shows that every Riemannian surface provides a counter-example (using flow orthogonal to the gradient of the curvature).

On the positive side, if $M$ is compact of dimension $\ge 3$ and has nowhere constant sectional curvature, then combination of results of Kulkarni and Yau show that a diffeomorphism preserving sectional curvature is necessarily an isometry.

Concerning 2-dimensional counter-examples: First of all, every surface which admits an open subset where curvature is (nonzero) constant would obviously yield a counter-example. Thus, I will assume now that curvature is nowhere constant. Kulkarni refers to Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164, for a counter-example attributed to Stackel and Wangerin. You probably can get the book through interlibrary loan if you are in the US. Here is the Weinstein's argument. Consider a Riemannian surface $(S,g)$ and let $K$ denote the curvature function. Let $X$ be a nonzero vector field on $S$ orthogonal to the gradient field of $K$. (Such $X$ always exists.) Now, consider the flow $F_t$ along $X$. Then $F_t$ will preserve the curvature but will not be (for most metrics $g$ and vector fields $X$) isometric. For instance, $F_t$ cannot be isometric if genus of $S$ is at least $2$ or if $X$ has more than 2 zeros. #### @Bernhard Boehmler 2012-06-24 18:47:05

Thank you for your answer. Maybe someone could explain the Weinstein argument in more detail here, since I am an undergraduate student, who is completely new to Riemannian geometry: math.stackexchange.com/questions/162472/…. That would be very nice.