Any smooth cubic surface contains 27 lines
blogs.ams.org
blogs.ams.org
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.
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
(p.s. I was not the downvoter)
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. :/
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:
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?
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.
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&...
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)
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.
This is not true with multiple variables, consider for example "x2 + y2 - 1"
Is there a proof of this?
Also, AMS: adding additional text to the text I selected to copy is not cool.
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?
Of course, such proofs tend to be longer than most proofs that are considered non-trivial.
...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."
"I am aware of several proofs of this theorem. None of them is trivial."