# Local Algebra: intersection multiplicity

1. Jan 3, 2016

### giulio_hep

I'm trying to understand why the intersection multiplicity of two singular subvarieties is not equal to the complex dimension of the local ring but it is instead the Euler characteristic.
Is it possible to find an intuitive explanation?

I think that the following concepts need some clarification:
1) when do smooth curves meet transversally?
2) why have we to focus on the group of invertible sheaves?
3) any simple examples of pull-back of line bundles?

2. Jan 7, 2016

### lavinia

I am not an Algebraic Geometer but I hope this helps. Perhaps you could ask a more specific question.

Line Bundles
On an algebraic curve, (actually any manifold), the complex line bundles form a group under tensor product of bundles and the trivial bundle is the identity element.
So each line bundle is invertible in this group. The inverse line bundle is the dual bundle,$L^{*}$, since $L⊗L^{*}$ is bundle isomorphic to $Hom(L,L)$ and the identity map is a global non-zero section.

Invertible Sheaves from Line Bundles
A line bundle is locally trivial which means that each point is surrounded by an open neighborhood where there is a bundle isomorphism $φ:π^{-1}(U) \rightarrow U$ x $C$. Since the line bundle is trivial over $U$, it makes sense on an algebraic curve to talk about regular sections of it. These are maps $s:U \rightarrow π^{-1}(U)$ such that $π(s) = id$ and the projection of $φ(s)$ onto the second factor of the local trivialization is holomorphic. These sections taken over all the sets $U$ determine an invertible sheaf. It turns out that this mapping from line bundles to invertible sheaves is a bijection so invertible sheaves and line bundles are in some sense the same. I wonder if invertible sheaves are called "invertible" because they correspond to line bundles.

Induced Bundles
Whenever one has a vector bundle $π:E \rightarrow M$ over a space $M$ and a continuous map $Φ: N \rightarrow M$ from a space $N$ into $M$ there is an induced bundle, $Φ^{*}(E)$, over $N$ which is defined as the subset of $N$ x $E$ of points $(x,v)$ where $Φ(x) = π(v)$. It is easy check that the induced bundle is a vector bundle. For holomorphic complex line bundles, if the mapping, $Φ$, is holomorphic then the induced bundle will be a holomorphic line bundle. For instance for any holomorphic mapping of the Riemann sphere into itself e.g. $z \rightarrow z^2$ the induced bundle will be a holomorphic line bundle.

Example
I am a little over my head here

If one starts with the dual to the tautological line bundle over the Riemann sphere, call it $L^{*}$, I am pretty sure that the induced bundle under $z \rightarrow z^2$ is the tangent bundle. Here is the argument. The induced map on $H^2(S^2;Z)$ is multiplication by 2(since $z \rightarrow z^2$ has mapping degree 2) so by naturally of the Chern class, the Chern class of the induced bundle is twice the Chern class of $L^{*}$. The Chern class of the $L^{*}$ integrates to 1 so the Chern class of the induced bundle integrates to 2. But the Chern class of the tangent bundle also integrates to 2 and since complex line bundles are classified uniquely by their Chern class, the induced bundle must be the tangent bundle.

Another Example

Consider the Gauss map from an embedded surface in 3 space to the unit sphere. The pull back of the tangent bundle of the sphere (the induced bundle) is the tangent bundle of the surface. This is easy to prove.

Transversal Intersection
Transversal intersection of smooth submanifolds means that the tangent spaces of the intersecting manifolds span the tangent space of the ambient manifold at the points of intersection. For instance, two smooth curves in the plane intersect transversally if their tangent vectors are linearly independent at the points of intersection.

Similarly, two algebraic curves in $CP^2$ that are transversal intersect in a finite number of oriented points.

I think in Algebraic Geometry this is called a "proper intersection".

Euler Class and Euler Characteristic

- Your question about Euler characteristic seems vague to me. Can you elaborate?

The Euler class of the normal bundle of an oriented submanifold is Poincare dual to the intersection of the submanifold with a section of the normal bundle that intersects the submanifold transversally. Such a section always exists. If the normal bundle is bundle-isomorphic to the tangent bundle of the submanifold, the transverse self-intersection is a finite set of oriented points whose oriented sum is the Euler characteristic of the manifold. More generally, the transverse intersection is a submanifold.

Last edited: Jan 7, 2016
3. Jan 8, 2016

### giulio_hep

A bit more context. I asked this when I started to read the first chapter of "Spectral Algebraic Geometry": at page 7 of this pdf (of 931 pages) you can find how the Euler characteristic is introduced/motivated. In the meantime I was searching some explanation and I bounced into this.
Do these two links above make my point about Euler characteristic less vague?

Thank you very much again.

4. Jan 8, 2016

### lavinia

The "this" link covers the case where the normal bundle is isomorphic to the tangent bundle. If you embed a manifold,$C$, as the diagonal of the Cartesian product, $C$ x $C$, then its normal bundle in $C$ x $C$ is isomorphic to its tangent bundle. The map that sends a tangent vector $v$ to the tangent vector, $(v,-v)$, in the tangent bundle of $C$ x $C$ is an explicit isomorphism. One sees that $(v,-v)$ is in the normal bundle to the diagonal by selecting a Riemannian metric on $C$ then taking the product metric on $C$ x $C$. Under this metric, $(v,-v)$ is orthogonal to $(v,v)$. (see Milnor Characteristic Classes pg. 121)

Last edited: Jan 8, 2016
5. Jan 8, 2016

### giulio_hep

Thank you! Really, I was slightly more interested in the point of view of homological algebra: the fact that (in page 7 of the first link) the intersection multiplicity is given by the Serre's formula and can not be determined using the tensor product functor, since it is not exact... I don't know if there is something else to say about it, to help me in understanding the concept... For example, I've found also this answer, saying that Tor measures the failure of exactness of the tensor product (but it is still a bit difficult for me)

6. Jan 8, 2016

### lavinia

I found this Wikipedia link https://en.wikipedia.org/wiki/Serre's_multiplicity_conjectures

A projective resolution $... \rightarrow F_{n} \rightarrow F_{n-1} \rightarrow ... F_1 \rightarrow F_0 \rightarrow A \rightarrow 0$ of an $R$-module $A$ is an exact sequence of $R$-modules (possibly infinitely long) where each $F_{n}$ is projective. If one takes the tensor product of each term with an $R$-module $B$ one gets another sequence $... \rightarrow F_{n}⊗B \rightarrow F_{n-1}⊗B \rightarrow ... F_2⊗B \rightarrow F_1⊗B \rightarrow A⊗B \rightarrow 0$

While the last map on the right is still surjective, the rest of the sequence may no longer be exact. One is left with a chain complex, that is, the image of the map, $f_{n+1}$, on $F_{n+1}$ is still in the kernel of the next map, $f_{n}$, but may not equal the entire kernel. In this case one can "measure" the failure of exactness by the quotients,

kernel$(f_{n})/$image$(f_{n+1})$ and this is the n'th torsion group. These are similar to homology groups.

In your case, if $R$ is a regular local ring and I suppose if $A⊗B$ satisfies the right conditions, then by analogy with Euler characteristic in homology, the alternating sums of the lengths of these torsion modules is called the Euler characteristic of $A⊗B$

Example

The sequence, $0 \rightarrow Z \rightarrow Z \rightarrow Z_2 \rightarrow 0$ is a free resolution of $Z_2$. Tensoring with $Z_2$ gives the sequence,
$0 \rightarrow Z _2\rightarrow Z_2 \rightarrow Z_2 \rightarrow 0$ which is not exact since the left copy of $Z_2$ is mapped identically to $0$. So the torsion group is $Z_2$. If one tensors with $Z_4$ one gets the sequence ,$0 \rightarrow Z _4\rightarrow Z_4 \rightarrow Z_2 \rightarrow 0$ which is also not exact. Again the torsion is $Z_2$. This example generalizes to $Z_n$.

Last edited: Jan 13, 2016
7. Jan 9, 2016

### giulio_hep

Sorry for the trivial questions.

Is the idea that the alternating sum of the Betti numbers gives the Euler characteristic, thus genralizing the historical V−E+F=2? In order to see the correspondence between Betti numbers and Tor, should I refer to this math q&a?

But I fail to understand how the Euler characteristic generalizes the dimension of the tensor product
$O$$X$,p$O$CP2,p $O$$Y$,p
as a definition of the multiplicity of an intersection. Maybe I need to study (here and here?) why the scheme-theoretic intersection of two lines is the spectrum of the ring... but my intuitive understanding is that the ordinary tensor product is something like the first (lowest) term of $R$-modules Tor homologies... I think that this is the main open point of my question (sorry if you already explained it and I've misunderstood your reply)

How are the maps defined in the sequences of your example? Does "Tensoring with $Z_2$" mean "Tensoring with $Z_2$ = $Z$/$2Z$"? Are you making the point that tensor product is right exact but not necessary left exact? I would appreciate if you could expand all the steps in your example... in particular if it helps to clarify the above said "unanswered part" of my question.

Edit

I guess that content of this thesis in "Homological algebra : Tor functors, Betti numbers, and free resolutions." contains most of the answers and in particular I recognize your example, completed with the maps, at the end of pag 5, so that clarifies a lot...

Last edited: Jan 9, 2016
8. Jan 10, 2016

### lavinia

Yes. The correspondence comes from a theorem of homology that says that the Euler characteristic is the alternating sum of the ranks (betti numbers) of the homology groups with $Z$ coefficients. If one uses $R$ or $Q$ instead of $Z$, the it is just the alternating sum of the dimensions of the homology groups.
As described above, $Tor_{n}(A,B)$ is the n'th homology group of the projective resolution of $A$ tensored with $B$. In your case $A$ is $O$$X$,p and $B$ is $O$$Y$,p. These are modules over the ring, $O$CP2,p

Here is an important example which shows how projective resolutions can be used to compute homology.

Let $G$ be any group. The group ring of $G$, denoted $ZG$, is the free abelian group of all formal sums of symbols $m[g]$ where $m$ is an integer and $[g]$ is the symbol corresponding to the group element,$g$. So a typical element of $ZG$ looks like $Σ_{i}m_{i}[g_{i}]$ and multiplication looks like $(Σ_{i}m_{i}[g_{i}])⋅(Σ_{j}n_{j}[h_{j}]) = Σ_{i,j}m_{i}n_{j}[g_{i}h_{j}]$. This is a ring with unit but will not be commutative if $G$ is not abelian. For simplicity assume $G$ is abelian (The tensor product of modules over non-commutative rings is well defined but a little different.).

For groups rings there is the notion of a "trivial $ZG$ module". This is a module where $[g]⋅x = x$ for all $x$ and $[g]$.

Now take a projective resolution of the integers,$Z$, considered as a trivial $ZG$ module. Then the homology of this resolution tensored over $ZG$ with a $ZG$ module, $M$, is called the homology of $G$ with coefficients in $M$. This illustrates how Tor groups are like homology groups. In this case, they are the same. A good exercise is to show that the homology of the abelian group,$Z$, with coefficients in $Z$ is $Z$ in dimensions 1 and 0 and zero for all other dimensions.Topologically, this is the same as computing the homology of the circle.

In general, the homology/cohomology of a group corresponds to the ordinary homology of a topological space. For instance, the cohomology of $Z(ZxZ)$ is the cohomology of a torus. For $Z_2$ it is the infinite dimensional real projective space.

The Example from Mathoverflow

This example was given as an example in the link to Mathoverflow that you posted above.

"The standard example is intersecting a 2-plane YY with XX, a union of 2-planes meeting at a point. Here the Tor formula gives i(X,Y,p)=2iXYp2, which is the 'correct' multiplicity since YY meets each of the planes transversely. However, the length of R/(IX+IY)RIXIY is not 2"

For $XX$ take the union of two orthogonal planes in affine 4 space, $R = A^4$, that intersect only at the origin. For $YY$ take a plane that intersects $XX$ only at the origin. Since $YY$ intersects each plane in $XX$ exactly once, the intersection multiplicity is 2. But the dimension of $R/(IX+IY)RIXIY$ is 3. So for this case the intersection multiplicity formula is wrong.

For instance, take $XX$ to be the the two planes defined by the polynomials, $xz,xw,yz,yw$ and $YY$ the plane, $(x-z,y-w)$.
Then the ideal generated by $(xz,xw,yz,yw, y-w,x-z)$ is minimally generated by $(xy, x^2,y^2)$ so the dimension of the local ring is three.

Computing the Euler characteristic from a free resolution of $O$$YY$,0, one finds that all of the $Tors$ are zero except $Tor_1$ which has length 1. So the Euler characteristic is 2, ($^{-}1 + 3$), which is what you want. We can work this out together if you like.

Last edited: Jan 10, 2016
9. Jan 10, 2016

### lavinia

Another thing to prove is that the $Tor$ groups are all zero above dimension zero in the case of a proper intersection. So the Euler characteristic formula reduces to the usual formula for intersection multiplicity.

Last edited: Jan 10, 2016
10. Jan 11, 2016

### giulio_hep

I think it's important to put this in the K-theory perspective (as suggested by David Speyer's answer:)
Also because I'd like to have a grasp of the next development of K-theory, the central extension (Virasoro algebra and its application in physics, like the critical dimensions in CFT, etc...)

Last edited: Jan 11, 2016
11. Jan 11, 2016

### lavinia

For proper intersections,
Can you explain David Spencer's Answer?

12. Jan 11, 2016

### giulio_hep

This is the idea: "Serre’s intersection multiplicity is a special case of the product in K-theory with supports". Henri Gillet explains it in details in his handbook, in part 2: see K-Theory and Intersection Theory, by Henri Gillet, 235-294. In the initial introduction (pag. 237) he writes that there are two approaches to intersection theory, one method is to reduce the problem of defining the intersection product of arbitrary cycles to intersections with divisors. The other method is to use the product in K-theory to define the product of cycles. The second construction is Serre’s “tor formula”, which is equivalent to taking the product, in K-theory with supports, of the classes [$O$$Y$ ] ∈ $K$$Y$ 0 (X) and [$O$$Z$] ∈ $K$Z 0 (X). This definition works for an arbitrary regular scheme X, since it does not involve reduction to the diagonal. Of course I would need to study a lot to fully understand it!

13. Jan 11, 2016

### mathwonk

a few comments. warning: I am not an expert on this deep topic! but i will make a few general comments.

invertible sheaves are equivalent to line bundles, and they are called invertible because a sheaf (O-module) L is invertible if and only if there is another sheaf M such that LtensorM ≈ O, i.e. if and only if L has an inverse for the operation of tensor product. This is true precisely for line bundles. (And the inverse is the dual line bundle.)

Line bundles and invertible sheaves (the sheaves of sections of line bundles) also correspond precisely to ("linear") equivalence classes of divisors, which are subvarieties of codimension one. Essentially the line bundle associated to a divisor is the normal bundle of the divisor, which has fiber dimension one because the divisor has codimension one. Since a divisor in a smooth variety is always defined by one equation, it is easy to define intersection with a divisor, just by restricting the equation. Similarly you can pull back a line bundle via a map, by pulling back its divisor, simply by composing the map with the equation for the divisor. So divisors are good guys in intersection theory.

Thus it is nice to deal with subvarieties that are intersections of the right number of divisors, or "complete intersections" at least locally. Such subvarieties are "Cohen Macaulay" and that's why it is nicer to deal with such subvarieties, and also why the Tor terms do not appear in those cases. The normal sheaf of a local complete intersection is also a vector bundle and that is a nice property too.

A subvariety of codimension d is thus nice if it can be defined by d equations, i.e. is the complete intersection of d divisors. Then we can intersect such a subvariety with anything just by successively intersecting with those divisors. Thus plane curves are especially easy to intersect since they are both divisors, hence both cut out by one equation. But a space curve of degree 3 for example cannot by Bezout's theorem be a complete intersection of two divisors, i.e. of two surfaces, since its degree is prime. Similarly, a space curve that is not even locally a complete intersection is going to be more challenging to intersect with another surface. oh no it isn't since that surface is itself a divisor. but if i go up in dimension i guess a pair of surfaces in 4 space that are not local complete intersections will be harder to intersect. So that is why the standard bad example is an intersection of a plane in 4 space with a union of two planes that meet in one point. interesting: we still have one of the varieties is cohen macaulay (the plane) but not the other one i guess.

An intersection number should be unchanged by certain motions. think about a poyhedron and counting its number of edges say. this is highly variable. but the alternating sum vertices - edges+faces, is not so sensitive, it does not change even when you change the number of edges, it is entirely determined by the topology of the polyhedral surface. so euler characteristics are easier things to compute and they are more likely to have good properties of topological invariance.

If you have studied algebraic topology you know that computing cohomology from homology, or changing coefficients is not always so simple. it is not always dual to homology or done just by tensoring with the new coefficient ring. Rather to get nice exactness properties, one has to consider higher order contributions from exts and tors. the tensor product is just Tor^0. the same holds true for intersection theory. tensor is just the first or 0th order term of a formula that needs to behave well under variation.

I am getting above my pay grade, and will drop out. The standard authoritative work on intersection theory is the book by William Fulton, but there is also a new book by David Eisenbud and Joe Harris fromn a less sophisticated point of view. See pages 20-23 and 193-196 of their free book:

http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf [Broken]

unless you need to do otherwise, perhaps the moral is that it is prudent to do intersection theory in the restricted setting of Cohen Macaulay varieties. Indeed Serre's theory is not even settled yet. It is an open problem that his definition always yields a positive number in the case of two general intersecting objects of complementary dimensions (positivity conjecture in the wikipedia article)! There is a nice short summary of intersection theory in Hartshorne's Algebraic Geometry, Appendix A.1.

Thanks for this great question! I think this is what you have been waiting for Lavinia. I.e. these wonderful questions stimulate us to learn something new.

Last edited by a moderator: May 7, 2017
14. Jan 12, 2016

### lavinia

Indeed!

15. Jan 12, 2016

### mathwonk

Another interesting reference may be Mumford's book Algebraic Geometry I, Complex projective varieties, the appendix to chapter 6, especially formulas A.8, A.10, A.14, and A.17 and the discussion on page 125. He uses the Hilbert- Samuel polynomial instead of the Serre Euler characteristic, but he gives also a formula of Weil, in A.8 which has only a tensor product, but a tensor product with an analytic ingredient, suggesting perhaps that the Euler characteristic may be needed only to get a completely algebraic definition?