Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:30:32Z http://mathoverflow.net/feeds/question/24175 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24175/why-does-the-euler-characteristic-of-a-toric-variety-equal-the-number-of-vertices Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope? Steven 2010-05-10T22:36:56Z 2010-05-12T05:18:30Z <p>In <a href="http://www.math.leidenuniv.nl/scripties/Trevisan.pdf" rel="nofollow">this link</a>, Corollary 3.2.2, page 59 the author claims that: The Euler characteristic of the toric variety $X_K$ associated to a convex polytope $K$ is the number of vertices of $K$.</p> <p>I want to see how it works. Could someone please illustrate this for me by using this method to compute the Euler characteristic of $\mathbb{P}^{2}$ and $\mathbb{P}^{1}\times \mathbb{P}^{1}$? Thanks so much.</p> http://mathoverflow.net/questions/24175/why-does-the-euler-characteristic-of-a-toric-variety-equal-the-number-of-vertices/24179#24179 Answer by VA for Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope? VA 2010-05-10T23:04:12Z 2010-05-10T23:04:12Z <p>Merely observe that a toric variety is the union of torus orbits $(\mathbb C^*)^r$ for various dimensions $r$, and that the Euler characteristic of $(\mathbb C^*)^r$ is zero if $r>0$ and $1$ if $r=0$. </p> <p>Vertices of a polytope correspond to 0-dimensional orbits, $r$-dimensional faces -- to $r$-dimensional orbits.</p> <p>$\mathbb P^2$ corresponds to a triangle, $\mathbb P^1\times\mathbb P^1$ to a square. It is not very hard to count their vertices.</p>