
Any smooth cubic surface contains 27 lines - tokenadult
http://blogs.ams.org/visualinsight/2016/02/15/27-lines-on-a-cubic-surface/
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stephengillie
The Wikipedia article might be more accessible:
[https://en.m.wikipedia.org/wiki/Cubic_surface](https://en.m.wikipedia.org/wiki/Cubic_surface)

~~~
johnhattan
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.

~~~
manofcode
Haha, my thoughts exactly

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Cogito
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.

~~~
ttctciyf
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...](https://golem.ph.utexas.edu/category/2009/10/girih_by_egan.html)
where you can see the link to the applet goes to Egan's site at
[http://www.gregegan.net/](http://www.gregegan.net/) or on the paper
[http://arxiv.org/abs/gr-qc/0208010](http://arxiv.org/abs/gr-qc/0208010) co-
authored by Egan, Baez and J. Daniel Christensen (referenced in Egan's
wikipedia entry.)

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

~~~
user2994cb
[http://matthewarcus.github.io/polyjs/clebsch.html](http://matthewarcus.github.io/polyjs/clebsch.html)

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IngoBlechschmid
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/Mirror_symmetry_\(string_theory\))
[https://en.wikipedia.org/wiki/Homological_mirror_symmetry](https://en.wikipedia.org/wiki/Homological_mirror_symmetry)

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

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

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

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impendia
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)

~~~
thaumasiotes
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. :/

~~~
impendia
> 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](https://en.wikipedia.org/wiki/Bézout%27s_theorem)

~~~
thaumasiotes
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?

~~~
impendia
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.

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amelius
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.

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GFK_of_xmaspast
> 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 "x __2 + y __2
- 1 "

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DigitalJack
Can we get one in the shape of a kitten?

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ternbot
27 * 4 = 108

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jsprogrammer
> _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.

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omaranto
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.

~~~
amelius
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.

~~~
omaranto
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.

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ClayFerguson
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.

~~~
evanpw
That 26 already includes a dimension for time.

