What is the idea behind deformation to the normal cone and what are examples of its applications?
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8$\begingroup$ I'm not an expert, but the idea is more or less to imitate the tubular neighborhood theorem in differential topology. In the topological setting, given two manifolds $X \subset Y$, the embedding of $X$ in a neighborhood inside $Y$ is the same thing (diffeomorphic) as the embedding of $X$ as the zero section of $N_{X/Y}$. In the algebraic setting this is not true anymore, but the latter can be obtained from the former as a flat deformation, and this is enough for some purposes. $\endgroup$– Andrea FerrettiCommented Jun 2, 2010 at 12:55
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$\begingroup$ To add to Andrea's comment, the purposes that it's most suited for seem to be intersection theoretic, and the normal cone construction is very useful for constructing characteristic classes and intersection numbers. $\endgroup$– Charles SiegelCommented Jun 2, 2010 at 13:05
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3 Answers
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Here's a place to see the normal cone side-by-side with other familiar constructions, that I learned from Fulton's "Intersection Theory". Here $X \subset Y$.
Start with the space $Y \times {\mathbb P}^1$, thought of as a trivial family over ${\mathbb P}^1$. Blow up the subscheme $X \times \infty$. Now we still have a flat family over ${\mathbb P}^1$, in which all the fibers except the one over $\infty$ are still copies of $Y$. The fiber over $\infty$ is reducible: one piece $Z_1$ is the blowup of $Y$ along $X$, and the other is the projective completion $Z_2$ of the normal cone to $X$ inside $Y$. These intersect along the projectivization of the normal cone, which appears in $Z_1$ as the exceptional divisor, and in $Z_2$ as the stuff added in projective completion.
(An example, for people who like polytopal pictures of toric varieties: let $Y$ be ${\mathbb P}^2$, pictured as a triangle, and $X$ be a point, pictured as a vertex. Then $Y \times {\mathbb P}^1$ is pictured as a triangular prism, and its blowup by cutting a corner off that prism. The fiber over $\infty$ is then pictured as cutting that triangle into a trapezoid $Z_1$ union a small triangle $Z_2$, glued along an interval, matching the decomposition above.)
So, if perhaps you don't like degenerating $Y$ (which may be complete) to something noncomplete, you can complete it by including $Z_1$. The normal cone is $Z_2 \setminus Z_1$.
Two other comments. On $Y \times {\mathbb P}^1$ there is a circle action, dilating the ${\mathbb P}^1$. It acts trivially on the $0$ and $\infty$ fibers, moving the rest around. When we blow up $X \times \infty$, the circle action on the new $\infty$ fiber $Z_1 \cup Z_2$ is nontrivial on $Z_2$; it is the dilation of the cone fibers. This is one place to lay blame for the existence of this "cone" structure.
Finally, there's a conformally equivalent way to think about the $Y$ to $Z_1 \cup Z_2$ degeneration, at least when $X$ and $Y$ are smooth. Pick a small tubular neighborhood (nonalgebraic!) around $X$, with boundary $S$, a sphere bundle over $X$. Let the metric on $Y$ get very looong nearby $S$, in directions passing through $S$. You might say that $X$ is falling into a black hole, with $S$ the Schwarzschild boundary. In the limit, it gets infinitely long, and $X$ has bubbled off into its own universe.
answered Jun 2, 2010 at 14:32
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$\begingroup$ Some $X$s and $Y$s are switched, which is all the more confusing since you didn't say explicitly which of them is ambient space. $\endgroup$ Commented Jun 2, 2010 at 15:01
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$\begingroup$ There's also a purely algebraic way to think about deformation to the normal cone: A filtered ring admits a flat deformation "over \AA^1" to its associated graded. Off the top of my head, I don't remember where one can read about this - maybe someone has a reference. $\endgroup$– mdelandCommented Jun 2, 2010 at 15:23
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$\begingroup$ Incidentally, if you don't like $Y$ degenerating from a reduced scheme to a nonreduced one, I explain how to avoid that in my "Balanced normal cones" paper here: pamq.henu.edu.cn/downloadarticle.jsp?id=136 $\endgroup$ Commented Jun 2, 2010 at 17:44
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1$\begingroup$ @mdeland: I think S15 of Matsumura is a reference. Here is how Matsumura describes it. Given a ring $A$ (algebra analogue of $Y$) and ideal $I$ (algebra analogue of $X$) form $R_+ := \{ \sum c_n t^n \colon c_n \in I \}$ and then set $R = R_{+}[u]$. The ring $R$ is the desired ring. It satisfies $R/(1-u) = A$ and $R/u = \operatorname{Gr}_{I}(A)$. $\endgroup$– jlkCommented Jun 2, 2010 at 20:50
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an explanation of it in a very relaxed and friendly tone is in Ravi Vakils notes intersection theory notes, see https://math.stanford.edu/~vakil/245/245class14.pdf and just read section 3, which is really selfcontained and doesn't need the rest of the notes.
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I'm copying kai's answer but spelling it out in text here:
For $X \subset Y$ we want to emulate the tublar nbr from topology.
That's easy, one takes the normal bundle, and then takes its total space.
Why is that not good enough? Because the point of a tublar nbr is that it lets you move from $Y$ to the tublar nbr, not just abstractly construct the tublar nbr. In topology the tublar nbr is contained in the $Y$, so it's clear how to move between them. In contrast, it's not clear how to move data from $Y$ to the normal cone. That's what the deformation to the normal cone does.
In the link in kai's answer, vakil explains how to take a chow class in $Y$, and map it to one in the tbular nbr. Topologically, we'd intersect with the tublar nbr. In algebra, the tublar nbr doesn't lie as a subscheme of $Y$. Instead, deformation to the normal cone solves the problem as you see in the link.
