Reference for equivalent definitions of the genus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:12:21Z http://mathoverflow.net/feeds/question/11768 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11768/reference-for-equivalent-definitions-of-the-genus Reference for equivalent definitions of the genus Qiaochu Yuan 2010-01-14T18:06:25Z 2010-01-14T20:31:36Z <p>Let $X$ be a (edit: nonsingular) projective complex algebraic curve. The genus of $X$ can be defined as the dimension of the space of holomorphic $1$-forms on $X$, which in turn can be defined either analytically or algebraically in terms of Kahler differentials. It can also be defined as the topological genus of $X$ considered as a surface, which in turn can be defined either topologically as the number of tori in a connected sum decomposition of $X$ or homologically in terms of the Betti numbers of $X$. Does anyone know of a reasonably self-contained reference where some or all of these equivalences are proven?</p> <p>(There is a <a href="http://mathoverflow.net/questions/152/how-do-you-see-the-genus-of-a-curve-just-looking-at-its-function-field" rel="nofollow">related question</a> about computing the genus of a curve from its function field as well as a nice <a href="http://lamington.wordpress.com/2009/09/23/how-to-see-the-genus/" rel="nofollow">post by Danny Calegari</a> explaining the relationship to the Newton polygon, but I am mostly interested in the algebraic-to-topological step of going from Kahler differentials to the number of tori in a connected sum decomposition.)</p> http://mathoverflow.net/questions/11768/reference-for-equivalent-definitions-of-the-genus/11769#11769 Answer by norondion for Reference for equivalent definitions of the genus norondion 2010-01-14T18:16:19Z 2010-01-14T18:23:26Z <p>For $\mathrm{dim} H^0(X, \Omega^1_X) = \dim H^1(X, \mathbb{Q})$ see <a href="http://en.wikipedia.org/wiki/Hodge_theory" rel="nofollow">http://en.wikipedia.org/wiki/Hodge_theory</a>. For $\dim H^1(X, \mathbb{Q}) =$ number of tori use induction and the Mayer-Vietoris sequence.</p> <p>(And for $\mathrm{dim} H^0(X, \Omega^1_X) = \mathrm{dim} H^1(X, \mathcal{O}_X)$ see <a href="http://en.wikipedia.org/wiki/Serre_duality" rel="nofollow">http://en.wikipedia.org/wiki/Serre_duality</a>.)</p>