Visualizing a projective variety

Question

What does the variety $V(x_0^2+x_1^2+x_2^2)\subset \mathbb{P}^2$ look like? It seems to me like a single point...

In general, are there any good ways/tips/tricks to visualize projective varieties?

Answer 1

It is not a point--it's a curve.

There are many ways to see this. One possible approach is to know that the coordinate ring of $V:=V(x_0^2+x_1^2+x_2^2)$ is $R=k[x_0,x_1,x_2]/(x_0^2+x_1^2+x_2)^2$. Thus,

$$\dim V=\dim R-1=2-1=1$$

so that, in fact we do see that $V$ is really a curve.

Another way to see this dehomogenizing at $x_2$ gives that $U_0\cap V$ is in bijection with the curve $x_0^2+x_1^2+1=0$ in $\mathbb{A}^2$ which is, up to a change of coordinates just a circle.

In fact, a good way to visualize projective varieties is to think about how they look in each affine coordinate chart. For your curve dehomogenizing with respect to the three coordinate charts gives rise to three different circles.

Think about this as the analogy of "to understand a submanifold, see how it looks in various coordinate charts". Of course, the real geometry comes from how you glue these coordinate charts together. The same is true here. The real geometry of $V$ comes from how precisely we are gluing these three circles along their intersections.

I leave it to you, as an exercise if you will, to identify precisely what the gluings are between the various affine coordinate charts.

Answer 2

Since the variety is in $\mathbf{P}^2$, you can look at a cone in $\mathbf{A}^3$. You've discovered that the real picture can be misleading, but we can change coordinates to get $z^2 - x^2 - y^2$ and that's something with an interesting set of real points.

It would also be good to write down an isomorphism of this curve with $\mathbf{P}^1$. There might be a more geometric way to do this but here's what I have now: change coordinates so that the conic is $xz - y^2$ and remember the Veronese map $\mathbf{P}^1 \to \mathbf{P}^2$.