Local vs. infinitesimal rigidity - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:42:33Z http://mathoverflow.net/feeds/question/34640 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34640/local-vs-infinitesimal-rigidity Local vs. infinitesimal rigidity Dave Futer 2010-08-05T15:13:25Z 2010-08-05T20:43:21Z <p>Can someone please explain the difference between local rigidity and infinitesimal rigidity? Does either version of rigidity imply the other?</p> <p>In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension \$\geq 3\$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction.</p> http://mathoverflow.net/questions/34640/local-vs-infinitesimal-rigidity/34684#34684 Answer by Nathan Dunfield for Local vs. infinitesimal rigidity Nathan Dunfield 2010-08-05T20:37:50Z 2010-08-05T20:37:50Z <p>Infinitesimal rigidity implies local rigidity, but not conversely. Local rigidity means a representation has no deformations, whereas infinitesimal rigidity means the natural tangent space to the character variety is 0-dimensional (this tangent space is a certain cohomology group with twisted coefficients). Weil proved both in the context you mention. See e.d. David Fisher's <a href="http://www.msri.org/communications/books/Book54/files/02fisher.pdf" rel="nofollow">survey paper</a>, where Weil's theorem is Theorem 3.2. </p> <p>See also Section 5 of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.em/1175789760" rel="nofollow">this paper</a>. </p> http://mathoverflow.net/questions/34640/local-vs-infinitesimal-rigidity/34685#34685 Answer by David Dumas for Local vs. infinitesimal rigidity David Dumas 2010-08-05T20:43:21Z 2010-08-05T20:43:21Z <p>Local rigidity means that the structure in question is an isolated point in its deformation space (which is typically an algebraic set). Infinitesimal rigidity means that there are no first-order deformations of the structure in question. A first-order deformation is a nonzero element of a certain cohomology group.</p> <p>Because you can take the derivative of a path of structures and get a first-order deformation, infinitesimal rigidity implies local rigidity.</p> <p>Because a first-order deformation may or may not correspond to an actual path (due to higher-order obstructions), local rigidity does NOT necessarily imply infinitesimal rigidity.</p>