Variational characterization of curvature? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:00:02Z http://mathoverflow.net/feeds/question/10853 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10853/variational-characterization-of-curvature Variational characterization of curvature? fuzzytron 2010-01-05T22:19:09Z 2010-01-07T19:42:00Z <p>Consider a surface $S$ smoothly embedded in $\mathbb{R}^3$. Classically, the (Riemannian) curvature of $S$ is described by the second fundamental form, which is constructed from partial derivatives of a local parameterization.</p> <p>Alternatively, is there a "nice" variational characterization of surface curvature? (E.g., one that does not depend on local parameterizations but only on the metric $g$.) In other words, is there a scalar functional whose minimizer completely describes the Riemannian curvature?</p> <p>One idea that comes to mind is that Riemannian curvature is the curvature associated with the Levi-Civita connection -- hence, you might try to construct a functional over the set of metric connections on $(S,g)$ that penalizes torsion.</p> <p>(This question is motivated by discrete (e.g., piecewise linear or simplicial) differential geometry, where local differential quantities are ill-defined but metric quantities are available nonetheless.)</p> http://mathoverflow.net/questions/10853/variational-characterization-of-curvature/10874#10874 Answer by José Figueroa-O'Farrill for Variational characterization of curvature? José Figueroa-O'Farrill 2010-01-06T03:22:34Z 2010-01-06T03:22:34Z <p>The curvature <em>is</em> a local invariant. There is such a thing as the curvature at a point. The curvature is described as a tensor, after all. It is different in, say, symplectic geometry, where because of the Darboux theorem all symplectic manifolds of the same dimension are locally symplectomorphic; a fact usually paraphrased as "there is no symplectic curvature". This probably means that there is no "global invariant" formulation for the curvature.</p> <p>As for the variational formulation, one possible line of approach would be to set up an action functional on <em>algebraic curvature tensors</em>; that is, sections of <code>$S^2\Lambda^2T^*M$</code> which are in the kernel of the Bianchi map</p> <p><code>$$S^2\Lambda^2T^*M \to \Lambda^4T^*M$$</code></p> <p>cooked up in such a way that the Euler-Lagrange equations are the differential Bianchi identities, since then such a tensor would be the Riemann curvature tensor of the metric you use to define the action functional and whose Levi-Civita connection appears in the Euler-Lagrange equations.</p> <p>Your idea about the action functional on the space of connections is what usually goes by the name of the <em>Palatini</em> (or <em>first-order</em>) formalism in GR. It is convenient in action functionals to treat the conenction and the soldering forms as independent quantities and let the Euler-Lagrange equations impose the torsion-free condition on the connection.</p> <p>As a typical example, consider the Palatini action $$\int_M R(e,\omega) \mathrm{dvol}$$ where $R$ is formally the scalar curvature but written in terms of the soldering form $e$ and the connection $\omega$. If you vary the action with respect to $e$ and $\omega$ separately you find that $\omega$ has no torsion and that the $M$ is Ricci-flat. To see what you gain in this formalism you just have to contemplate the calculation of the Euler-Lagrange equations for the Einstein-Hilbert action for the same Ricci-flatness condition, namely, $$\int_M R(e) \mathrm{dvol}$$ where now the connection is written explicitly in terms of $e$.</p> http://mathoverflow.net/questions/10853/variational-characterization-of-curvature/11024#11024 Answer by Richard Montgomery for Variational characterization of curvature? Richard Montgomery 2010-01-07T07:46:38Z 2010-01-07T07:46:38Z <p>Take a circle of radius $r$ about a point $p$ (metric concept). Compute its circumference (metric concept). Compare $C(r)$ to $2 \pi r$ in the limit as $r \to 0$ to get the curvature $K(p)$ at $p$. Specifically, $C(r) = 2 \pi [ r - (1/6) K(p) r^3 + ...]$. There is a similar formulae involving the area $A(r)$ of the circle. </p> <p>Not variational, but quite metric, and quite parameterization independent. These formulae can be found in Spivak and many other d.g. texts.</p> http://mathoverflow.net/questions/10853/variational-characterization-of-curvature/11043#11043 Answer by Deane Yang for Variational characterization of curvature? Deane Yang 2010-01-07T15:41:22Z 2010-01-07T19:42:00Z <p>I don't particularly understand the point of trying to characterize curvature as the critical point or minimum of some functional, so let me answer the question differently.</p> <p>Curvature arises naturally as the second derivative of an energy functional evaluated at a critical point as follows:</p> <ul> <li>Fix two points on a Riemannian manifold and consider the following (standard) energy functional for curves joining the two points:</li> </ul> <p>$E[\gamma] = \int_0^1 |\gamma'(t)|^2\,dt$</p> <p>Note that the Riemannian structure is used to define the norm of the velocity vector.</p> <ul> <li><p>It is well known that the critical points of $E$ are constant speed geodesics</p></li> <li><p>It is also well known that if $\gamma$ is a critical point of $E$ <em>and $\gamma$ is deformed using a parallel vector field along $\gamma$</em>, then the second variation of $E$ is simply the integral along $\gamma$ of the sectional curvature evaluated on the $2$-plane spanned by $\gamma'$ and the variation of $\gamma$.</p></li> </ul> <p>So sectional curvature measures in a very precise way how geodesics behave when varied infinitesimally. This for me is the most concrete, direct, and useful way to understand what curvature is.</p> <p>EDIT: Corrected description of second variation</p>