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The Serre intersection formula, as an alternating sum of contributions from Tor-groups, is something that combines a lot of ingredients that I'm interested in, but I've never really felt that I have a "grip" on it. One of the reasons for this is that, despite making attempts on a couple of occasions, I never seem to have been able to concoct an example where I get contributions from higher Tor-terms that's geometrically satisfying.

This is, unfortunately, a vague description of what I'm looking for. When you describe intersection multiplicity, it's relatively easy to give examples of double and triple intersection of planar curves, give a proto-definition in terms of degree of tangency, show how this is captured by their scheme-theoretic intersection, and talk about what happens when the curves are moved slightly. This is the kind of thing I miss having.

I should also clarify that I'm looking for something slightly more complicated than intersecting a planar curve with a point.

Are there examples of the Serre intersection formula in this vein?

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The standard example is intersecting a 2-plane $Y$ with $X$, a union of 2-planes meeting at a point. Here the Tor formula gives $i(X,Y,p)=2$, which is the 'correct' multiplicity since $Y$ meets each of the planes transversely. However, the length of $R/(I_X+I_Y)$ is not 2. –  J.C. Ottem Nov 2 '11 at 14:13
Assuming the subvarieties intersect properly (i.e. the dimensions add up correctly), then the higher Tor are non-zero if and only if one of the subvariety is not Cohen-Macaulay. So Serre formula in some sense corrects the non-Cohen-Macaulayness. That may explain why a satisfying geometric example is not obvious, as CM is not a very geometrically obvious condition. –  Hailong Dao Nov 2 '11 at 15:59
To build on Hailong Dao's comment, many geometric incidence conditions are CM. For example, Schubert varieties are CM, so you will never see a higher Tor term in a Schubert calculus example. –  David Speyer Nov 6 '11 at 13:08
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1 Answer

up vote 8 down vote accepted

Consider a flat morphism $f:X\to Y$ of smooth connected varieties. For instance let $X=Y\times F$ with $X,Y,F$ all smooth. Further let $Z\subset X$ be a generically reduced subvariety such that $f|_Z:Z\to Y$ is a finite morphism, which is not flat. For any $y\in Y$ let $X_y\subset X$ denote the fiber of the original $f$.

As any two points on $Y$ are equivalent, the intersection product $Z\cdot_X X_y$. is independent of $y\in Y$.

Since $Y$ is smooth and $Z$ is generically reduced, most fibers of $f|_Z$ are smooth, so the intersection product $Z\cdot_XX_{y}$ for a general $y\in Y$ is given by the length of $\mathscr O_{X_{y}}$, in other words, there is no $\mathrm{Tor}$ contribution there.

Now take a point $y\in Y$ such that $f_Z$ is not flat in any neighbourhood of $y$ in $Y$. It follows that the length of $\mathscr O_{X_y}$ (which is also the Hilbert polynomial of the fiber) has to be different from the value that we get from the smooth fibers. Therefore there has to be $\mathrm{Tor}$ contribution there.

Perhaps this example shows why it is $\mathrm{Tor}$ that comes in: $\mathscr O_{Z_y}$ is not flat on $Z$, so it does not give the right length and thus has to be corrected. It is arguably intuitive that the alternating sum of the length of the $\mathrm{Tor}$'s will be constant as $y$ runs through the (closed) points of $Y$.

I don't know if you find this geometrically satisfying. I would paraphrase what Hailong said and say that anything involving $\mathrm{Tor}$ should not be geometrically obvious, on the other hand $\mathrm{Tor}$ measures the failure of exactness of the tensor product so the failure of flatness is an obvious condition to look at. Then again, all of this just reiterates the fact that flatness is a difficult notion to truly comprehend (and I don't claim I do).

The examples/comments given by JC and Hailong fit into this example:

  • Since $Y$ is smooth and $f|_Z$ is finite, it is flat if and only if $Z$ is Cohen-Macaulay, which gives us Hailong's comment, and

  • a concrete example is given by $X=\mathbb A^4$, $Y=\mathbb A^2$, $f$ the obvious projection, and $Z$ the union of two planes meeting in a single point both of which project onto $Y$ isomorphically, which gives us JC's example.

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