How can I compute the full set of nodes of a surface? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:44:12Z http://mathoverflow.net/feeds/question/46185 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46185/how-can-i-compute-the-full-set-of-nodes-of-a-surface How can I compute the full set of nodes of a surface? node-question 2010-11-16T04:43:08Z 2010-11-16T15:00:20Z <p>The nodes of a surface are special cases of more general singularities. For example, the Cayley cubic has four nodes.</p> <p>The full set of singularities of a surface can be characterized by finding all points where the partial derivatives are all zero. However, not all singularities are nodes. Some are cusps or other kinds of singularities. For example, the Cayley cubic has five singularities, four of which occur at the nodes of the internal elliptope and the fifth which occurs at zero and is apparently not a node. </p> <p>Is there a simple way to check which singularities are surface nodes? Or, more interestingly, is there a way to compute the full set of nodes of a surface directly?</p> http://mathoverflow.net/questions/46185/how-can-i-compute-the-full-set-of-nodes-of-a-surface/46195#46195 Answer by Sándor Kovács for How can I compute the full set of nodes of a surface? Sándor Kovács 2010-11-16T05:36:24Z 2010-11-16T05:36:24Z <p>You should make this question more precise. </p> <p>There is a notion of a <em>node</em> of a surface which is, I believe, just an \$A_1\$ singularity, that is, analytically isomorphic to the vertex of the cone \$(xy-z^2)\subset \mathbb A^3_{x,y,z}\$. </p> <p>It seems that you want self intersections. Those are actually not normal singularities, so you can look at the non-normal locus, but that would still include other singularities. </p> <p>Then even if you restrict to double self-intersections, that is still not unique. Do you want (simple) normal crossing such as \$(xy=0)\subset \mathbb A^3_{x,y,z}\$, the non-Cohen-Macaulay \$(x=y=0)\cap (z=t=0)\subset \mathbb A^4_{x,y,z,t}\$, or something else?</p> http://mathoverflow.net/questions/46185/how-can-i-compute-the-full-set-of-nodes-of-a-surface/46216#46216 Answer by quim for How can I compute the full set of nodes of a surface? quim 2010-11-16T09:20:26Z 2010-11-16T09:20:26Z <p>A node (as in Cayley's surface) is a double point with nondegenerate tangent cone. To check whether a given point on a surface in A^3 is a node in this sense, change coordinates so that it is the origin and write the equation as 0=F_2+F_3+... with F_i homogeneous of degree i. The point is a node iff F_2 is irreducible. I am assuming the base field is algebraically closed.</p>