Timeline for Topological type $x_0^2+x_1^2+x_2^2+ x_3^2+x_4^2 = 0$ in $\mathbb P^4$

5 events

when toggle format	what		by	license	comment
Mar 11, 2012 at 16:36	vote	accept	Nikita Kalinin
Mar 9, 2012 at 20:09	comment	added	Mariano Suárez-Álvarez		@Will, the cohomology is not that of $\mathbb P^3$, though, no?
Mar 9, 2012 at 19:28	comment	added	Will Sawin		You can chance coordinates to $a=x_0+ix_1$, $b=x_0-ix_1$, $c=x_2+ix_3$< $d=x_2-ix_3$ to get the equation $ab+cd+x_4^2=0$, which shows that the surface is birational to $\mathb P^3$. I don't know enough about 3-folds to tell if this all implies it is isomorphic to $\mathbb P^3$.
Mar 9, 2012 at 17:37	comment	added	Johannes Nordström		The hyperplane theorem follows from the fact that the complement of the variety in $\mathbb{C}P^n$ is homotopy equivalent to a cell complex whose (real) dimension is $n$. In this case, one can see explicitly that the complement deformation retracts to $\mathbb{R}P^3$, and deduce that $b_3(V) = 0$.
Mar 9, 2012 at 17:28	history	answered	David E Speyer	CC BY-SA 3.0