Let $(M, g)$ be a Riemannian manifold. Is it possible to construct two different affine (or metric) connections, say $\nabla$ and $\nabla'$, which induce the SAME curvature tensor, i.e. $R(X, Y)Z=R'(X, Y)Z$ for any $X, Y, Z\in\Gamma(TM)$? I am not interested about answers related to metrics conformal to $g$. Are there any references related to this question? Thank you very much.
$\begingroup$
$\endgroup$
9
-
1$\begingroup$ Yes: take two flat tori with different volumes. $\endgroup$– Paul SiegelCommented Aug 12, 2014 at 20:22
-
1$\begingroup$ Fix a topological surface of genus $g > 1$. According to Teichmuller theory, there is a $3g - 3$-dimensional space of conformally inequivalent metrics with constant curvature $-1$. $\endgroup$– Paul SiegelCommented Aug 12, 2014 at 21:18
-
2$\begingroup$ Probably, your question would be better posed as: Given a potential curvature tensor (i.e., 2-form with values in endomorphisms of the tangent bundle), when can it be written as the curvature of a connection and in how many ways? A relevant question (and answer) can be found at mathoverflow.net/questions/73439/… $\endgroup$– Robert BryantCommented Aug 12, 2014 at 21:25
-
4$\begingroup$ @PaulSiegel: While what you say is correct, it doesn't have much bearing on the OP's question. Just because two metrics on a surface have the same Gauss curvature (a function), that does not imply that they have the same (Riemann) curvature tensor (which has type $(1,3)$, as opposed to the Gauss curvature, which is a function). In fact, two metrics on a surface that have the same (nonvanishing) curvature tensor and the same Gauss curvature are equal. $\endgroup$– Robert BryantCommented Aug 13, 2014 at 1:14
-
6$\begingroup$ @StevenGubkin: Actually, just because the problem has many solutions in the degenerate case that the curvature vanishes, that doesn't mean that it has many solutions in the 'generic' case. In fact, most of the time, in dimension $3$ or more, the Riemann curvature tensor completely determines the metric and, for most affine connections in dimension $4$ or more, knowing the curvature tensor determines the connection completely. It's only in 'degenerate' cases or in the low dimensions that the curvature doesn't determine the connection. $\endgroup$– Robert BryantCommented Aug 13, 2014 at 1:20
|
Show 4 more comments