Why is tb < 0 for boundary of a convex surface? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:13:46Z http://mathoverflow.net/feeds/question/98640 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98640/why-is-tb-0-for-boundary-of-a-convex-surface Why is tb < 0 for boundary of a convex surface? kln 2012-06-02T08:08:18Z 2012-06-02T10:13:52Z <p>Why is \$tb(K)\$ (Thurston-Bennequin invariant) of a Legendrian knot \$K\$ which is the boundary of a convex surface \$\Sigma\$ is negative in a contact 3 manifold?</p> http://mathoverflow.net/questions/98640/why-is-tb-0-for-boundary-of-a-convex-surface/98648#98648 Answer by Marco Golla for Why is tb < 0 for boundary of a convex surface? Marco Golla 2012-06-02T10:13:52Z 2012-06-02T10:13:52Z <p>If the boundary of a convex surface is Legendrian, then we can see the Thurston-Bennequin number directly from the dividing curves \$\Gamma\$, in the sense that: <code>\$tb(L) = \#(\Gamma \cap L)/2\$</code>.</p> <p>This is proved, for example, in Etnyre's <a href="http://people.math.gatech.edu/~etnyre/preprints/papers/surfaces.pdf" rel="nofollow">notes</a>, Theorem 2.30.</p>