Blow-up of the affine cone

Question

I was trying to blow-up at the origin the cone $X$ given by the equation $x^2 + y^2 = z^2$. So by definition the blow-up is given by the solutions of the equation $x^2 + y^2 = z^2, xt = zu, xv = yu, yt = zv$ where $(x,y,z) \in X$ and $(u,v,t) \in \mathbb P^2$.

I take the usual affine charts $ u = 1, v = 1, t = 1$.

In the first chart with coordinates $(x,v,t)$, the equations become $z = xt, y = xv$ so $x^2(1+v^2) = x^2t^2$. So we obtain $x=0$ and $v,t$ are free or $1 + v^2 = t^2$. Similarly the second chart gives $\{(0,u,t) \mid u,t \in \mathbb A^1\} \cup \{ (y,u,t) \in \mathbb A^3 \mid 1 + u^2 = t^2 \}$. Finally if $t = 1$ we get $z=0$ or $u^2 + v^2 = 1$.

Is this correct ? If yes I don't understand how get a "global picture" of how the blow-up should like. It should be a cylinder but I can't see why. I also can't compute $E$ : from the equations I wrote it seems to me that any $(0,0,0,u,v,t)$ is solution of these equations.

Answer 1

I agree with your computations. Conceptually, what happened is that you found the total transform of $X$ after blowing up the origin inside $\mathbb{A}^3$. This consists of $X'\cup E$, where $X'$ is the strict transform and $E$ is the exceptional divisor, in this case, the $\mathbb{P}^2$ lying above the origin. (You don't say what $E$ is in the question, and you should, but I'm taking a guess here.)

As for the visualization question, the strict transform of your space might not exactly be a cylinder, but it's isomorphic to one in some sense which I don't want to make precise- consider that it might be kind of like a hyperboloid of one sheet, where the exceptional divisor is a plane through the narrowest circle on the hyperboloid. "Blowing up" should be physically reminding you of inflating a balloon at the origin and things separate a bit after you inflate.

Answer 2

If you think of a blow-up as replacing your blown-up point by a copy of $\mathbb P^2$, the strict transform (even its real picture analog) will not really be a cylinder because of the topology of $\mathbb P^2$~; informally speaking, a line passing through the point you blew up will have both its ends identified, so in topological terms, the real picture is a cylinder in which you took a circle (which would be the intersection of your cylinder with a plane passing through the origin) and you glued points on that circle lying on opposite ends, as in the classical construction of $\mathbb{RP}^2$.

In the complex picture, since you blew up a cone given by one equation, the strict transform (i.e. the blow-up of your cone $C$ at the origin) is very easy to compute : it's the equation itself! In other words, the strict transform looks like this : $$ \mathrm{Bl}_0(C) = \{ ((x,y,z),[p_x,p_y,p_z]) \in \mathbb A^3 \times \mathbb P^2 \mid x^2+y^2=z^2 \text{ and } p_x^2+p_y^2=p_z^2 \}. $$ So the blow-up is obtained by replacing the origin by the set of lines through the origin, and because your equation is given by homogeneous polynomials, the strict transform is given exactly by the same equations ; this is because if you think of it as blowing up $\mathbb P^3$ at the point $[0:0:0:1]$ instead of blowing-up $\mathbb A^3$ at the origin, the definition of the blow-up tells you that the coordinates of the point $([x,y,z,w],[p_x,p_y,p_z])$ have to satisfy equations like $$ p_x y = p_y x, \quad p_x z = p_z x, \quad (\cdots) $$ which essentially means that $[p_x,p_y,p_z]$ and $(x,y,z)$ have to satisfy the same homogeneous polynomial equations.

Perhaps it is now easier to see that the intersection of $\mathrm{Bl}_0(C)$ with $0 \times \mathbb P^2$ is just the projectivization of your cone, or in other words, a smooth conic in $\mathbb P^2$. It is not quite enlightening to think of the complex variety $C$ as a cone, but even though the real picture of that equation looks like a cone, its blow-up definitely is not a cylinder.

As for $E$, I assume you are referring to the exceptional divisor as in Hartshorne. It is not complicated to determine~: by definition, it is just $0 \times \mathbb P^2 \subseteq \mathrm{Bl}_0(\mathbb A^3)$. The smooth conic we obtained earlier was just $\mathrm{Bl}_0(C) \cap E$.

Hope that helps,