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**A flexible octahedron**

+Shantha Hulme pointed out this flexible octahedron on Wikipedia. It has the same number of vertices, edges and faces as a regular octahedron. The big difference is that the faces intersect each other. If you cut it in half along the dashed ellipse, you get not a square but an

**antiparallelogram**: a 4-sided shape where two sides cross each other!

This flexible octahedron was discovered by Raoul Bricard in 1897, along with others, like the weird one you'll see here:

https://en.wikipedia.org/wiki/Bricard_octahedron

It's impossible for a convex polyhedron with rigid faces to be flexible. Bricard was the first to discover flexible polyhedra with rigid faces, but his examples all had faces that intersect each other. In 1977, Robert Connelly found the first non-self-intersecting flexible polyhedron with the topology of a sphere. In 1978, Klaus Steffen found a simpler example, which turns out to be the simplest possible. He modified the Bricard octahedron by adding more faces, in order to move the self-crossing parts of the polyhedron away from each other while still allowing it to flex. He got something with 9 vertices and 14 triangular faces:

https://en.wikipedia.org/wiki/Steffen's_polyhedron

http://www.gregegan.net/SCIENCE/Steffen/Steffen.html

#geometry

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**Varignon's Theorem**

Draw any quadrilateral. Then the midpoints of the edges are the corners of a parallelogram!

Very beautiful. Very simple. So the big question is: why this wasn't known to the Greeks?

Well, maybe it was! But it's named after Pierre Varignon, who published it in 1731.

Maybe this is one of those results that's a lot easier to prove using modern methods. Over on Twitter, Guillermo Valle wrote approximately this:

*The average of H and F should be same as the average of G and E, as both are averages of two of the four original points. Therefore the diagonals of the new quadrilateral intersect at their midpoints, giving a parallelogram!*

That's very quick... but it uses facts about "averaging" points in Euclidean space, which work best if you know about vectors, or at least know enough analytic geometry to treat points in Euclidean space as lists of numbers, which can be averaged. I imagine Archimedes understood these facts, since he worked with the "center of mass" idea. But did earlier Greeks?

The key idea in Valle's proof, hidden in his words, is algebraic:

((A+B)/2 + (C+D)/2)/2 = ((A+D)/2 + (B+C)/2))/2

I'd say this is a law in the operad for convex linear combinations - but hey, I'm just that kind of guy.

My own slower proof also used ideas from analytic geometry, but I avoided the "lemma" saying that a quadrilateral is a parallelogram if its diagonals intersect at their midpoints. I just showed that the vector from E to H is the same as the vector from F to G:

H - E = (A+D)/2 - (A+B)/2 = (D - B)/2

G - F = (D+C)/2 - (C+B)/2 = (D - B)/2

Actually my mental proof was a bit different than this. I thought: there are two ways to go from B to D: using A as an intermediate step, or C. The first gives a vector which when halved forms one side of the would-be parallelogram, the second gives a vector which when halves forms the other side, so these must be equal.

As a category theorist I like this argument because it's about a commutative square! But when I write it out as equations it turns into this. The two ways to go from B to D give

(D - A) + (A - B) = (D - C) + (C - B)

so

(D - A)/2 + (A - B)/2 = (D - C)/2 + (C - B)/2

or

(D + A)/2 - (A - B)/2 = (D + C)/2 - (C + B)/2

or

H - E = G - F

In any case, it's an algebraic proof, like Valle's. And as Ian Agol pointed out, neither of these proofs relies on the original four points lying in the plane!

I heard the phrase "Varignon's Theorem" on the blog "My Favorite Theorems":

https://kpknudson.com/my-favorite-theorem/

but I don't like videos or podcasts, because they usually move too slow, so I just read about it on Wikipedia.

#geometry

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**The amplitudohedron**

What you usually read about the amplitudohedron is exciting but short on detail, like this:

**A Jewel at the Heart of Quantum Physics**

*Physicists have discovered a jewel-shaped geometric object that challenges the notion that space and time are fundamental constituents of nature.*

*Physicists have discovered a jewel-like geometric object that dramatically simplifies calculations of particle interactions and challenges the notion that space and time are fundamental components of reality.*

*“This is completely new and very much simpler than anything that has been done before,” said Andrew Hodges, a mathematical physicist at Oxford University who has been following the work.*

*The revelation that particle interactions, the most basic events in nature, may be consequences of geometry significantly advances a decades-long effort to reformulate quantum field theory, the body of laws describing elementary particles and their interactions. Interactions that were previously calculated with mathematical formulas thousands of terms long can now be described by computing the volume of the corresponding jewel-like “amplituhedron,” which yields an equivalent one-term expression.*

*“The degree of efficiency is mind-boggling,” said Jacob Bourjaily, a theoretical physicist at Harvard University and one of the researchers who developed the new idea. “You can easily do, on paper, computations that were infeasible even with a computer before.”*

This is from the

*Quanta*article by Natalie Wolchover:

https://www.quantamagazine.org/physicists-discover-geometry-underlying-particle-physics-20130917/

It actually has a lot of good information... but it doesn't actually say what an amplitudohedron is. In this post here, +Refurio Anachro goes further. Maybe I can get him to go a bit further and say exactly what subset of the Grassmannian is the "positive Grassmannian". I don't think I quite got that.

Eventually I'll break down and read the original papers, or watch the videos he recommends. But right now I'm too engaged in my own quest!

#geometry

**Hunting for Amplituhedra in positive Grassmannian spaces**– Let's delve into a beautiful generalization of projective geometry: into the world of Grassmannian geometry. To get an idea of the territory we'll encounter a few handful of delightful elementary concepts which can be elegantly fused to inspect polytopes inside the positive section of Grassmannians, a region which is much easier to understand than the general beast. Come along, we're starting right now:

**Definition:**

*Gr(k,n)*is the space of all k-dimensional subspaces in R^n. Although other fields or manifolds might also work.

For

**k=1**we obtain the

*real projective space*RP^(n-1) or just P^(n-1). Gr(1,n) is the space of all lines in n-dimensional space which pass through the origin. We need the origin because otherwise we don't get a subspace. Gr(1,3) is the space of all such lines in 3-space, and that one looks like a half-sphere with a twisted

*crosscap*.

I've written about RP² here:

https://plus.google.com/+RefurioAnachro/posts/YbgFU8Xzbac

Remember? Any direction in R³ can be given by a 3-vector, and any line consists of all the points we can reach by stretching that vector while maintaing its direction. Since the length of the vector doesn't really matter we get a family of which any member describes a line: an element of, or a point in Gr(1,3).

p = t·v

p = t·(ax + by + cz)

p = [a:b:c] = [q·a:q·b:q·c]

This way to denote points on RP^n is called homogeneous coordinates. Note that we can also stretch [a:b:c] by a negative factor q and still end up describing the same point. Note also that there is no point deserving the name [0:0:0] so at least one coordinate has to be nonzero.

For

**k=2**it turns out we need two equations or vectors to describe a plane. This pattern holds and we can describe points in Gr(k,n) as a k×n matrix! For example:

(x1 y1 z2)

(x2 y2 z2)

Then Gr(2,3) should be the space of all planes in 3-space (containing the origin). But since every such plane can also be given by a 3-vector (its surface normal), we obtain the same space as Gr(1,3) = P². Which is a two-dimensional surface for which two coordinates seem to be enough, what's going on? Simple scaling won't rid us of the extra parameters...

Meet

*Plücker coordinates*, sometimes called Grassmann coordinates because it was actually

**Günther Graßmann**who generalized

**Julius Plücker**'s notation for arbitrary k. It seems Graßman helped Plücker out a little because it was Plücker who found Gr(2,4): the first Grassmannian that isn't just a projective space.

https://en.wikipedia.org/wiki/Pl%C3%BCcker_coordinates

Looking at our k×n matrix we can construct a square minor (matrix) of maximal dimension k by striking out the i'th column. Lets call the determinants of our minors delta Δ1 Δ2 Δ3. Remember, such a determinant gives the signed volume of a parallelpiped spanned by the vectors that ended up in our maximally sized minor.

Together with the scaling relation we recover the homogeneous coordinates for Gr(2,3). What about that smallest non-projective Gr(2,4)? It turns out that the scaling relations aren't enough to completely collapse families of coordinates pointing at the same element, we need more

*Plücker relations*!

https://en.wikipedia.org/wiki/Pl%C3%BCcker_embedding

Because of the scaling property we find that Δij = -Δji so we don't need both, and that means that if there were any Δjj they'd have to be zero. Playing some more you might find another relation:

Δ13Δ24 = Δ12Δ34 + Δ14Δ23

You can write this one graphically as a sum of pairs of chords in a circle:

(X) = (||) + (=)

For more on relations like these take a look at

*cluster algebras*:

https://en.wikipedia.org/wiki/Cluster_algebra

In the end it is not quite trivial to understand Grassmannians in full generality, but we can make things much simpler by restricting our interest to the

**positive Grassmannian**Gr+(k,n) which is a region or section of the full space that has the shape of a convex polytope!

The positivity condition we want is that all Plücker symbols must be positive. Well, as I said earlier, switching the indices like Δij = -Δji also flips the sign, so we need to pick a canonical order of the indices and demand their symbols to be positive. With the pyhsicist's short-hand for a determinant we might state this as:

<ViVj...Vk> > 0 for i<j<...<k

And call that that all

*cyclic maximal minors*should be positive. There's a subtilety if the minor wraps around, as for example in Δ14 = -Δ41. For (xxx look it up) you get an extra minus sign when that happens.

The advantage of the positive region is that we can stop worrying about cross-caps and other weird topological features. Our first example, the real projective plane P² is completely symmetric: any of its points is equivalent to all others, and if you cut it into pieces the crosscap is gone! So the positive part of Gr(1,3) is exactly an octant of a sphere!

But there is more. We will be interested in high-dimensional Grassmannians, and that means that we will get a polytope made of of points, lines, faces, hyperfaces up to (n-k)-hypervolume. Like polyhedra the positive Grassmannians have a discrete and combinatorial structure. And as it is with platonic solids, when we understand this structure we have also understood something about the symmetries of the space they live in!

Now let's take a look at Gr^+(2,4) shall we?

Example:

**Gr^+(2,4) looks like an octahedron!**There are six ways to strike out two of the four columns of a 2×4 matrix, so we get six Plücker symbols.

Δ13

Δ12 Δ23 Δ34 Δ14

Δ24

Here, let me show you in binary which indices are mentioned in each symbol

1010

1100 0110 0011 1001

0101

There is a cycle of four where the ones and zeros are adjacent, and two symbols have no such pairs. It looks like an octahedron! There are two more internal faces, one is the square 1010-1100-0101-0011, the other the square 1010-0110-0101-1001, both are suspended vertically from the pair-free top and bottom vertices.

There are also four more 3-cells (volumes), I remember those been called tetrahedra, but I'm not quite sure.

This is part 1 of a series of three blog posts to explain in detail how the Amplituhedron simplifies the calculation of scattering amplitudes (Feynman diagrams and stuff). I should have everything, but I still need to clean up my notes a bit so they are more fun to read. Here are the titles, I'll add the links here when they become available:

Permutation 2:

**Locating the Amplituhedron, decompositions and projections**

Part 3:

**Shooting and bagging Amplituhedra for profit**

**Links:**

This post relates to about the first third to half of what we'll need from

**Alexander Postnikov**'s paper

*"Total positivity, Grassmannians, and networks"*:

https://arxiv.org/abs/math/0609764

That is, if you need more detail or can't wait for the next post. Or try this fun 4-part lecture series by

**Alexander Postnikov**,

*"Combinatorics of the Grassmanian"*:

https://youtube.com/watch?v=5m6j_yiepFM

https://youtube.com/watch?v=gIHF8N6QjAI

https://youtube.com/watch?v=Qi9GgAuzWPY

https://youtube.com/watch?v=qd-Tro6ibEM

https://ncatlab.org/nlab/show/positive+Grassmannian

https://en.wikipedia.org/wiki/Grassmannian

Once again, hat tip to +John Baez for making me want to write about this in detail!

https://plus.google.com/+johncbaez999/posts/A8sQSPmETHk

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**Penrose triangle**

This looks like an image first created by the Swedish artist Oscar Reutersvärd in 1934. The psychiatrist Lionel Penrose and his mathematician son Roger Penrose independently discovered it in the 1950s, calling it "impossibility in its purest form".

It's possible!

But when you walk around it, you see.... interesting things. Watch the video.

For more on this shape read:

https://en.wikipedia.org/wiki/Penrose_triangle

#geometry

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**Möbius strip**

*Mathematics is not a book confined within a cover and bound between brazen clasps, whose contents it needs only patience to ransack. It is not a mine, whose treasures may take long to reduce into possession, but which fill only a limited number of veins and lodes. It is not a soil, whose fertility can be exhausted by the yield of successive harvests. It is not a continent or an ocean, whose area can be mapped out and its contour defined.*

*It is limitless as that space which it finds too narrow for its aspirations. Its possibilities are as infinite as the worlds which are forever crowding in an multiplying among the astronomer’s gaze. It is as incapable of being restricted within assigned boundaries or being reduced to definitions of permanent validity, as the consciousness, the life, which seems to slumber in each monad, in every atom of matter, in each leaf and bud and cell, and is forever ready to burst forth into new forms of vegetable and animal existence.*

That was the famous mathematician James Joseph Sylvester, from a speech at Johns Hopkins. Heady stuff! It's true. And believe it or not, it was originally one sentence, with lots of semicolons. I broke it into pieces for you impatient modern tweeters.

Sylvester loved poetry: he read and translated works from the original French, German, Italian, Latin and Greek, and many of his mathematical papers contain quotes from classical poetry. His love of language helped him invent many important mathematical terms, including "matrix" and "graph" (in the sense of a thing with vertices and edges).

He left England and became a professor of mathematics at the brand-new Johns Hopkins University in Maryland in 1876. His salary was $5,000, a good amount for the time, and

*he demanded to be paid in gold*. However, after negotiation, he dropped this demand.

The animated gif here shows a Möbius strip, which illustrates how mathematics is endless and always has another side... which is really the

*same*side, just seen from another view. It was made by "Dave" at

*Bees and Bombs*, one of many excellent images he's created:

https://beesandbombs.tumblr.com/image/163256181074

#geometry

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**Collapsible**

Check out +Gerard Westendorp's posts if you like pictures of cool math.

This structure is based on his observation that a 'hinged tiling' made of squares can be collapsed down to a single square. For an animation of that, go here:

https://tinyurl.com/westendorp-tile-2

He's also created a larger structure of this sort:

https://preview.tinyurl.com/westendorp-tile-3

Unfortunately these structures don't look like they'd be practical folding tables - not without some change in design, anyway! Can anyone figure out how to fix that?

#geometry

**Collapsible hinged tessellation in action.**

There might be a relation to number theory:

Each square has 2 numbers (i,j). Let these correspond to the coordinates of the horizontal plane that the centre of the square has at the most expanded state. We can normalise them to the integers. The centres of the squares move as (sin(alfa)+cos(alfa)) * (i,j), where (alfa, -alfa) are the alternating rotation of the squares.

So if a square centre exactly overlaps an integer point in is movement back to the origin, then (i,j) are not relative primes.

So this thing is a relative prime detection machine.

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**Chopping a square into 7 triangles with almost the same area**

It's impossible to divide a square into an odd number of triangles with the same area. This amazing result was proved by Paul Monsky in 1970, but I only heard about it today, from +David Eppstein, here on G+.

So it's impossible. But you can still try your best! These are the two best tries with 7 triangles and at most 8 vertices. These were discovered last year by Jean-Philippe Labbé, Günter Rote, and Günter Ziegler. They used a lot of clever math and programming to do this.

What do they mean by "best"? They mean that the standard deviation of the area of the triangles is as small as possible. For these two it's 0.0008051.

They couldn't check all the ways that used more than 8 vertices - it took too much time!

This was just the start of what Labbé, Rote, and Ziegler did. +David Eppstein summarized the rest:

*According to Monsky's theorem (*

*https://en.wikipedia.org/wiki/Monsky's_theorem*

*), it is impossible to divide a square into an odd number of triangles, all the same area as each other. But if it can't be done exactly, how close to equal-area can you get? The answer is, at least superpolynomially close. The construction uses the Thue–Morse sequence (*

*https://en.wikipedia.org/wiki/Thue-Morse_sequence*

*) to balance out the inequalities in area.*

**Superpolynomially close**means that for any number a, with n triangles you can get the difference in areas to be less than 1/nª if n is big enough.

**Puzzle.**Show that if n is even, you can divide a square into n triangles all having the same area.

This puzzle should be quite easy for most of you out there. How many seconds did it take you?

Here's the actual paper:

• Jean-Philippe Labbé, Günter Rote, Günter M. Ziegler, Area difference bounds for dissections of a square into an odd number of triangles. Available at https://arxiv.org/abs/1708.02891.

It's full of cool techniques... a great example of how even "useless" and "unimportant" problems in math can inspire interesting thoughts.

Here's the sketch of Monsky's proof, from Wikipedia:

1. Take the square to be the unit square with vertices at (0,0), (0,1), (1,0) and (1,1). If there is a dissection into n triangles of equal area then the area of each triangle is 1/n.

2. Color each point in the square with one of three colors, depending on the 2-adic valuation of its coordinates.

3. Show that a straight line can contain points of only two colors.

4. Use Sperner's Lemma to show that every triangulation of the square into triangles meeting edge-to-edge must contain at least one triangle whose vertices have three different colors.

5. Conclude from the straight-line property that a tricolored triangle must also exist in every dissection of the square into triangles, not necessarily meeting edge-to-edge.

6. Use Cartesian geometry to show that the 2-adic valuation of the area of a triangle whose vertices have three different colors is greater than 1. So every dissection of the square into triangles must contain at least one triangle whose area has a 2-adic valuation greater than 1.

7. If n is odd then the 2-adic valuation of 1/n is 1, so it is impossible to dissect the square into triangles all of which have area 1/n.

Thinking about this this will force you to understand p-adic numbers and Sperner's Lemma, both of which are more important and interesting than this particular theorem!

https://en.wikipedia.org/wiki/Sperner's_lemma

https://en.wikipedia.org/wiki/P-adic_order

The Thue-Morse sequence is also lots of fun:

https://en.wikipedia.org/wiki/Thue-Morse_sequence

#geometry

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**Hinged tilings**

Al Grant has a great interactive page of tilings that move on "hinges". Check it out:

http://algrant.ca/projects/hinged-tessellations/

And Al is short for "Alexander", not "Artificial intelligence".

#geometry

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**Kepler, the Hydrogen Atom and the Fourth Dimension**

Kepler loved Platonic solids. He even made a theory of the planets' orbits where they moved on spheres between a nested octahedron, icosahedron, dodecahedron, tetrahedron, and cube. It fit the data pretty well... but not perfectly. He eventually realized that the orbits aren't circles: they're ellipses!

We now know that the force of gravity drops off as the inverse square of the distance. That's why planets move in elliptical orbits. Platonic solids have nothing to do with it.

But last week, +Greg Egan, +Layra Idarani and I found a solid mathematical connection between the Platonic solids and the inverse square force law. It involves quantum mechanics, and a detour into the 4th dimension!

The force between the electron and proton in a hydrogen atom also obeys an inverse square law... but we need to use quantum mechanics to understand it. Instead of having a definite position, the electron has a

**wavefunction**saying how likely it is that you'll find it in any location.

Amazingly, the wavefunction for the electron in a hydrogen atom can also be described as a function on a

**3-sphere**: a sphere in 4 dimensions. We can rotate a sphere in 4 dimensions and turn one wavefunction into another with the same energy.

This is a 'hidden symmetry' of the hydrogen atom. It's sort of obvious how 3-dimensional rotations act on a hydrogen atom. The amazing part is that you can do 4-dimensional rotations.

This picture, made by Egan, shows a wavefunction on the 3-sphere. It's positive in the blue regions, negative in the yellow regions, and almost zero where it's black. We’re seeing a moving slice of the 3-sphere, which is an ordinary sphere. When the image fades to black, our moving slice is passing through a sphere where the function is zero.

This particular function describes a very symmetrical state of a hydrogen atom, where it has a definite energy and 7200 symmetries, coming from the

*symmetries of the group of symmetries of a dodecahedron*.

That's quite a mouthful! Luckily, I've already written a nice gentle explanation of this stuff with lots of pictures, including nice pictures of Kepler's original Platonic solid theory, start here:

https://johncarlosbaez.wordpress.com/2018/01/07/the-kepler-problem-and-the-600-cell/

Make sure to follow the links to Egan's explanation:

https://tinyurl.com/egan-600

We worked out a lot of this math with +Layra Idarani, so if you're a mathematician you'll also enjoy his work:

https://tinyurl.com/idarani-SG

It generalizes a lot of this stuff to even higher dimensions!

#geometry

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**The golden ratio - yet again**

In math, all sufficiently beautiful entities are connected to all others. Here's another example. A regular octahedron has 12 edges. A regular icosahedron has 12 corners. So there could be a way to draw the icosahedron with its corners on the edges of the octahedron. And yes - there, is!

But as a final twist of the knife, you don't put the corners at the middle of the edges. That wouldn't work. Instead, each of the corners divides each of the edges

*according to the golden ratio!*

Math just

*had*to do that.

It seems this image was created by Jen-chung Chuan using Cabri 3D:

http://steiner.math.nthu.edu.tw/d3/d2/quick-and-dirty/

**Puzzle 1.**What shape has 12 corners, with one located exactly in the center of each edge of the regular octahedron?

**Puzzle 2.**Can you make or find an animated gif of that shape morphing into a regular icosahedron as its corners move from the midpoints of the octahedron edges to the points shown here?

**Puzzle 3.**How many ways are there to create a regular icosahedron whose corners lie on the edges of a regular octahedron?

You see, there are two choices of how to chop each octahedron edge in two parts, one longer than the other by a factor of the golden ratio. So, in principle, you have to check 2^12 different possibilities and see which ones give regular icosahedra. But only a few of these actually work.

#geometry

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