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Any smooth cubic surface contains 27 lines (ams.org)
80 points by tokenadult on Feb 16, 2016 | hide | past | favorite | 37 comments



The Wikipedia article might be more accessible: https://en.m.wikipedia.org/wiki/Cubic_surface


Ahh, thank you. My first reaction was "A surface of a cube is a square". I can define a square with way fewer than 27 lines.


Haha, my thoughts exactly



Is anyone aware of which Greg Egan is referenced in the article?

I couldn't see any links to the original work, but I might be blind.

I suspect that it is the Australian scifi author, as it is the sort of thing he has done before, but I haven't dug into it yet.


Yeah, it's the SF author - he's collaborated previously with John Baez, as for example at https://golem.ph.utexas.edu/category/2009/10/girih_by_egan.h... where you can see the link to the applet goes to Egan's site at http://www.gregegan.net/ or on the paper http://arxiv.org/abs/gr-qc/0208010 co-authored by Egan, Baez and J. Daniel Christensen (referenced in Egan's wikipedia entry.)


Someone needs to build a threeJS/WebGL scene of this. So we can drag it around.



Incidentally, this topic gave rise to a much larger theory, namely mirror symmetry (unrelated to reflection at everyday mirrors). Please take the following account of the history with some grain of salt, I'm not an expert on this topic.

After counting lines on a smooth cubic surface, you could also count lines on other manifolds, for instance three-dimensional ones given by a quintic polynomial (these are examples for "Calabi–Yau manifolds"). Also you could count curves of degree two, three, and so on instead of lines, which are curves of degree one. The calculations get increasingly harder: The case of degree two curves was only settled in 1986.

It therefore came as a surprise when a group of physicists (Philip Candelas, Xenia de la Ossa, Paul Green, and Linda Parkes) announced in 1991 a formula for calculating the result for curves of any degree. They did so by inventing a new technique, mirror symmetry, in which one relates the "complex geometry" (as in "complex numbers") of the manifold on which you're counting curves to the "symplectic geometry" of a certain other manifold, dubbed mirror of the original one.

Many aspects of mirror symmetry are still widely non-understood and many conjectures are motivated and made plausible by physical arguments.

https://en.wikipedia.org/wiki/Mirror_symmetry_(string_theory... https://en.wikipedia.org/wiki/Homological_mirror_symmetry


Is the converse statement also true -- i.e., do 27 distinct lines define a cubic surface? Will have to think about this.


It should be clear that 27 arbitrary distinct lines do not form a smooth cubic surface, as it is easy to create a counter example

eg. consider three lines through a single point and mutually orthogonal (the three axes say). There is no smooth surface that contains all three of these lines.

That said, you may be able to find conditions under which 27 lines define a smooth cubic surface, however I suspect you probably need as much information as you would need to construct the cubic surface in the first place.

A treatment of how you can parameterise cubic surfaces (which is what we are really talking about here) is at http://docs.lib.purdue.edu/cgi/viewcontent.cgi?article=2208&...


So what can we say about 27 lines? Is it enough to just exclude the counterexample you gave? Say, is it true that:

All sets of 27 distinct lines describe a smooth cubic surface as long as no subset of 3 of them both intersect at the same point and are mutually orthogonal.

What additional restrictions would it take to know that the lines describe a distinct smooth cubic surface? (if that's possible at all)


Take 27 lines that intersect in exactly one point (all coincide in the same). Would you tell that they define a smooth surface?


Certainly. Those lines could all lie in a plane, for instance -- that's very much a smooth surface.


It would not be cubic though (a plane is defined by a linear rather than a cubic equation).

(p.s. I was not the downvoter)


It's a little odd to say a polynomial "isn't cubic" just because the degree-3 coefficient is zero. Ordinarily what matters is that every coefficient of degree higher than three is zero.

Think of the result that you can fit a polynomial of degree n to go through n+1 points. It's easy to show that the polynomial indicated by that theorem which goes through a set of collinear points must be a line (degree-1 polynomial), no matter how many points are given[1]. But we don't say "as long as the points aren't collinear, this theorem holds"; we just accept that a line is a special case of cubic polynomial.

[1] Proof: the polynomial of degree n going through n+1 points is unique. The line going through n+1 collinear points goes through all of those points, and therefore must be the unique polynomial of degree n to do so.

I do see in the wikipedia article, though, that "cubic surface" appears to be defined to exclude polynomials which have any nonzero term of degree other than three. It's a weird world. :/


Well, it's defining it in projective space; in projective space, the polynomial has to be homogeneous to make sense. If you were working in affine space, you wouldn't have that requirement -- but then the theorem wouldn't be true without first passing to projective space, I imagine. Which would involve taking your polynomial and adding an extra variable so as to make it homogeneous.


> Proof: the polynomial of degree n going through n+1 points is unique.

Suppose you have nine points lined up in an exact 3x3 grid. What, according to your proposed definition, is the unique degree 8 polynomial going through them?

There are two obvious cubic polynomials I can think of: One, cover the nine points by three horizontal lines and multiply the equations of the lines. Two, do the same with the vertical lines.

I can't really think of any reasonable way to distinguish any one of these (or another) polynomial over other candidates.

In contrast, with the definition that a cubic polynomial requires its coefficients to be of degree 3, you get things like Bezout's theorem:

https://en.wikipedia.org/wiki/Bézout%27s_theorem


I was bothered as I was writing my comment by an issue related to this one. A curve covering a horizontal/vertical 3x3 grid of points cannot be a polynomial, as, if it is interpreted as a function, it has the range {true,false}, but some other domain, probably ℝ^2. The result I mention should be applicable to a set of points which can be represented as a function from ℝ to ℝ; no curve covering two different points which are vertically aligned can be represented that way (well, you could transpose the axes, but for a 3x3 grid that won't work either).

I do have some questions, because I don't fully understand what you're saying:

Suppose the 9 points are {-1,0,1} × {-1,0,1}. The horizontal line equations are y = -1, y = 0, and y = 1. If I multiply those together, I get y^3 = 0, which is a single horizontal line. I feel I must have done something wrong there.

If I instead do (y + 1)(y - 0)(y - 1) = 0 I get y^3 - y = 0, which at least has y = -1, y = 0, and y = 1 as solutions. Since a cubic equation can't have four roots, this graph consists exactly of three horizontal lines. It's not a polynomial, since it has the " = 0" constraint embedded. (Using the definition on wikipedia, "[i]n mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents".) Is this what you meant?

I know I've seen a diagram of exactly the example you're talking about somewhere, as something in the spirit of "math fun facts". But I can't find it; do you know of a writeup you could point me toward?


I don't know of a good writeup, but y^3 - y = 0, or alternatively x^3 - x = 0, among other polynomials, is what you want.

In "ordinary" algebra, typically we consider polynomials of one variable, which we write f(x). The graph of the polynomial represents the set of solutions to y = f(x), or alternatively y - f(x) = 0.

In algebraic geometry, the solution set to y - f(x) = 0 is indeed a plane curve, but we think of y - f(x) as a special case of a polynomial in the two variables x and y. All such polynomials (except constant polynomials) also define plane curves. This is the concept that generalizes.

In particular: y^3 - y = 0 and x^3 - x = 0 are both plane curves, they are both polynomials in x and y (i.e. we can think of y^3 - y = 0 as a polynomial in not only y but also x, even though x doesn't appear), and the intersection of these two curves is the set of nine points in question.

Hope this helps.


I don't know why you're getting downvotes. Thanks for the example.


I would like to know the reason for the downvotes as well.


Good point.


Any polynomial of degree n has n roots, but some of them may be degenerate, and some of them may be complex.

So, how would that translate to the cubic-surface situation? Are there cubic-surfaces that are "degenerate", and thus have actually less than 27 lines?

Are all the lines expressible in real numbers? Or do we need complex numbers, like it is the case for polynomial roots?

Just some questions that popped into my mind.


> Any polynomial of degree n has n roots, but some of them may be degenerate, and some of them may be complex

This is not true with multiple variables, consider for example "x2 + y2 - 1"


Can we get one in the shape of a kitten?


27 * 4 = 108


>No matter how one goes about it, the proof of this theorem is nontrivial.

Is there a proof of this?

Also, AMS: adding additional text to the text I selected to copy is not cool.


Are you asking for a proof of the "fact" that all proofs of the theorem about cubic surfaces having 27 lines are non-trivial? If so, then no, there is not even a definition of non-trivial proof, let alone a proof that any specific proof is non-trivial or a proof that there exists a fact all of whose proofs are non-trivial.


My definition of a trivial proof would be a proof that is mechanically verifiable. I.e., it is verifiable by even something as "dumb" as a computer.

Of course, such proofs tend to be longer than most proofs that are considered non-trivial.


Maybe years ago that would have been a plausible definition, but by now many proofs have been verified by computer, including several which probably no-one would feel comfortable calling trivial.


Yes, that is what I was asking for.

What does the sentence mean then?

One particular proof (of which we do not know if it is exclusive) is non-trivial (whatever that means)? Or, perhaps, all known (to the author) proofs are non-trivial?


Baez didn't mean that as a mathematical statement rather as a statement in the field of history and human psychology: "So far nobody has been able to devise a proof that is widely regarded by people who study it as being trivial (where triviality of a proof is a subjective judgment for which there is (1) no formal criterion, but also (2) a surprising amount of agreement)."


>No matter how one goes about it, the proof of this theorem is nontrivial.

...I'm going to guess he really means:

"No matter how one goes about it, this proof of the theorem is nontrivial."

...and not...

"No matter how one goes about it, all proofs of this theorem are nontrivial."


More like:

"I am aware of several proofs of this theorem. None of them is trivial."


If you look up "-1/12" on Wikipedia page about; 1+2+3+4...=-1/12 it has a mention that: "Bosonic string theory not consistent in dimensions other than 26". But if you add one dimension for time you get 27. I wonder if there is a relation between all these pointing to the number 27.


That 26 already includes a dimension for time.




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