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I'm reading Katz "Enumerative Geometry and String Theory". He shows that a degree $d$ hypersurface has cohomology class $dH $(where $H$ is the class of a hyperplane) by the following argument (p.80, I'm paraphrasing a bit):

...we can continuously deform a degree d hypersurface (given by F=0) to a union of d >hyperplanes by the equations: $$ G_t := tF(x) + (1-t) \Pi_{i=1}^{d} l_i (x) = 0 $$ where $ l_i $ are homogeneous linear forms.

I've studied the topic of intersection theory a bit already, so my question isn't about the result necessarily, but more about his argument. Specifically, what does he mean by continuous here (i.e. continuous in what topology)? For example, if we are in the the plane, and we take $d=2$, he is saying we can continuously deform a circle into a pair of lines. I'm pretty sure that deformation isn't continuous, though (or maybe it is in the complex plane?)

Adam
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1 Answers1

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When $t=0$, you see that $G_0$ is the union of $d$ hyperplanes, whereas when $t=1$, we have $G_1 = F$. In the case of $\mathbb{R}$ or $\mathbb{C}$, this is actually a continuous (in the usual topology) deformation as $t$ varies continuously from 0 to 1, if we work in the projective plane.

If it seems strange that a circle could be deformed into a pair of lines, try graphing this specific instance over the real numbers for particular values of $t$: $$ t (x^2 + y^2 - 1) + (1-t)(x^2-y^2) = 0 $$

When $t=1$ this is a circle $x^2 + y^2 = 1$. When $t=0$ this is a pair of lines ($y=\pm x$).

In between, when $1/2 < t < 1$, this is an ellipse. As $t$ approaches 1/2, the ellipse becomes more and more elongated. At $t=1/2$, we are left with a pair of distinct vertical lines ($x = \pm 1/\sqrt{2}$). The "top and bottom points" of the ellipse have gone off to infinity while the "left and right points" on the x-axis approach $\pm 1/\sqrt{2}$. Then for $0 < t < 1/2$, the two lines branch into a hyperbola. At $t=0$, the hyperbola has degenerated into a pair of distinct lines.

In the affine plane, we seem to have encountered a discontinuity at $t=1/2$. But in the projective plane, all these conics look the same. The "points at infinity" are actual points in the projective plane, and the two "branches" of the hyperbola join up twice at infinity (once for each of the two asymptotes) to make a single connected component, just like an ellipse.

Ted
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  • Thanks, that was really helpful. I wasn't thinking of the deformation correctly. As a follow-up question, how does one prove it is continuous in general? How do I know deforming the equation never results in discontinuities? – Adam Aug 12 '13 at 21:05
  • I made a few corrections in my original answer (at $t=1/2$ we get distinct vertical lines, not coincident). – Ted Aug 14 '13 at 08:04
  • For your follow-up question, I don't know the subject well enough to say. My impression is that in algebraic geometry one approaches this sort of continuity problem in an entirely different way, not by considering a deformation in a topology related to $\mathbb{R}$ or $\mathbb{C}$ as I did above, but by defining a deformation of a variety $V$ purely algebraically - a flat map $M \to A^1$ where the fiber over 0 is the original $V$ and all other fibers are the deformations. Then you prove that certain properties (like the cohomology class) are invariant under deformation. – Ted Aug 14 '13 at 08:19