# The Sylvester Gallai Theorem and Sections of Varieties with “Simple Topology”.

The Sylvester-Gallai theorem asserts that for every collection of points in the plane, not all on a line, there is a line containing exactly two of the points.

One high dimensional extension asserts that for every collection of points not all on a hyperplane in a d-dimensional space there is a [d/2]-space L whose intersection with the collection is a spanning set of cardinality [d/2]+1

My question is: Given a k-dimensional real algebraic variety V [perhaps of a certain kind] embedded in n-dimensional affine space (whose image spans this space) can one find an affine r-dimensional space L so that $V \cap L$ spans L and is topologically "simple".

Remarks: 1) For smooth complex varieties the Lefschetz hyperplane theorem describes sort of the opposite phenomenon: the homology of the hyperplane section is (more or less) as complicated as the homology of the original manifold all the way to half the dimension.

2) There is an analog of the Sylvester-Gallai Theorem over the complex numbers (there, if the points are not colinear you can always find a line containing 2 or 3 points, and if the points are not coplanar you can find a line containing precisely two points. The later statement was a conjecture by Serre first proved by Kelly. See this post in Konard Swanepoel's Blog).

So the difference between real AG and complex AG is not the only issue at hand (another issue seems to be how reducible the manifold is), and we can ask for sections with simple topology for complex varieties as well. Are there any results known in this direction?

3) Note that the whole point in the Sylvester-Gallai theorem and the proposed generalization is about non-generic embeddings and about intersection with non-generic flats.

However, the behavior of "generic embedding" (regarding intersection with non-generic flats) is sort of a role for what we expect for non-generic embedding. One difficulty is that I am not aware of a definition of "generic embedding" of a real algebraic semi variety in a large Euclidean space. Is there such a definition?

Let me formulate an open problem (and a few variations) which takes into account the discussion so far. We say that an embedding of a real semi algebraic variety V into an Euclidean space is genuine if its image span the space.

Problem (special case): Let V be a 2-dimensional variety genuinly embedded into n-space. (n cannot be too small, but maybe n=4 will suffice.) Then there is a plane L whose intersection with V is 1-dimensional, $V \cap L$ affinely span V, and $V \cap L$ is of very restricted topological type. (Maybe belong to a finite list of homotopy types.)

Problem (general case): The same conclusion (V \cap L belongs to a small set of homotopy types) when V is k dimensional, L is r dimensional, the intersection between V and L is j-dimensional and the dimension of the ambient space n is sufficiently large (but perhaps being moderately large suffices.) Greg's example shows that n cannot be too small. (For k=1,r=3,j=1 we cannot take n=4.) We know also that for k=j=0 we need n>=2r.

Problem (special case) Consider the case where V is an arrangement of subspaces.

Problem (complex analog) Consider the case that V is a complex variety; (and the special case of an arrangement of subspaces).

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For example; suppose you have a 2 dimensional real algebraic variety V embedded in a high dimensional space (but not in any subspace). Can you find a plane L whose intersection with V affinely span V and belongs to one of a small number of homotopy type? –  Gil Kalai Nov 9 '09 at 13:22
If you claimed that Gil is short for Gilvester (which is a real first name although rare), then you could say that any of your results is the "Gilvester Kalai theorem". –  Greg Kuperberg Nov 24 '09 at 5:13

There are many cases of the question as stated that follow quickly from the standard Sylvester-Gallai theorem. If $V$ is an $r$-dimensional variety, then its intersection with a generic $(n-r)$-plane is a finite set of points. You can then apply the standard Sylvester-Gallai theorem, or the high-dimensional generalization stated here.

There are cases where nothing can be said for singular varieties. As a warm-up, let's consider a set which is not an algebraic variety but a union of line segments. Then it could be the union of all of the interior diagonals of a convex polytope $P$ with complicated facets. For instance you could take all of the interior diagonals of the Cartesian product of two $n$-gons. Any 3-plane that intersects $P$ 3-dimensionally has to intersect many of the edges.

A line segment is not a real algebraic variety. However, it can be replaced by a thin needle with cusps at the ends that is a real algebraic variety. You can replace all of the diagonals with these needles, as long as you skip the edges of $P$ itself, and the result will still lie in the convex hull of $P$.

A needle of this type can have a cross-section of any dimension and very complicated topology. If you asked for a hyperplane that specifically intersects in more than a finite set, then the diagonal-needle construction can force a lot of topology.

You could specifically look at non-singular varieties. I don't have a rigorous result here, but the smooth restriction would make it difficult to avoid hyperplanes that do something at the boundary of the convex hull of $V$. The Sylvester-Gallai theorem is more about things that have to happen in the interior if they do not happen at the boundary of the convex hull.

You could bound the degree of the variety $V$. Then a simple compactness argument bounds the complexity of intersects, and there are a lot of interesting bounds on the topology of $V$ itself. But that also goes against the spirit of Sylvester-Gallai, because the number of points in that result is not bounded.

Maybe a more interesting variation is to keep a finite intersection, but replace the hyperplane with a $V$ with bounded degree. However, that is no longer the question posted.

The question is a bit open-ended. I can think of several constructions that seem to ask for a less open-ended question, or a question which is open-ended in a different way.

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Excellent answer, Greg, thanks. –  Gil Kalai Nov 21 '09 at 10:29
This is an excellent answer and let me give some more detailes about it. First, it raises interesting special cases and generalizations of the problem. 1. Certainly a case which deserves to be singled out is the case where V comes from an arrangelemt of subspaces. I.e., when V is the union of affine subspaces of an n-dimensional space. (The original case deals with 0-subspaces.) 2. It suggests to study the problem (in the real case) for semi-algebraic varieties. Those include sets like skeleta of polytopes and skeleta with added diagonals. 3. To consider smooth varieties. –  Gil Kalai Nov 21 '09 at 16:24
I do not see yet how the example [or such an example] allows 1-dimensional intersection of a 3-planes [ot 2-plane] with V with arbitrary complicated topology, when the abbient space is of high dimension. It looks that indeed it shows that for 1-dimensional intersection with a 3-plane for a variety in 4 space, the topology cannot be as tight as in the original G-S thm. (Unbounded number of connected components seems anavidable.) I dont see if in this example the topology can be arbitrary complicated. I'll be happy to hear more on the last paragraph. (several constructions and open-ended probs) –  Gil Kalai Nov 21 '09 at 17:02
In any case Greg's answer pushed the problem enough to be accepted. More answers are nevertheless welcome. –  Gil Kalai Nov 21 '09 at 17:05
For semialgebraic sets, you can have same sort of construction using diagonal triangles or diagonal simplices, using a polytope with complicated enough k-faces with k near n or k near 0. But I concede that for singular algebraic varieties, there are some technical difficulties. I don't yet know a construction when the intersecting plane has codimension 2. –  Greg Kuperberg Nov 21 '09 at 17:48

Dear Gil,

Small point: the first sentence requires that the original set of points not all be on a line.

Best,

Joe

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Thanks, Joe. I corrected it now. –  Gil Kalai Nov 9 '09 at 6:17
In the future, these sorts of things should be posted as comments to the question, and not answers. –  Steven Sam Nov 22 '09 at 15:52
The perennial issue is that people don't always have enough reputation to comment. –  j.c. Nov 28 '09 at 18:49