Claude St-Louis
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Dvorák -----------------------------------Cello Concerto in B minor, Op. 104 - II. Adagio

Cello: Truls Mørk
Conductor: Jonathan Nott
Orchestra: Bamberg Symphony
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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.

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 Hermann 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) = (||) + (=)

The general form for this Grassmann-Plücker relation looks like this:

Δ_i1,…ik·Δ_j1,…jk = sum_s=1^k Δ_js,i2,…ik·Δ_j1,…j(s-1),is,j(s+1)…jk

And here is the sign-flip for any k:

Δ_i1,…ik = (-1)^sign(w) Δ_iw(1),…iw(k)

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

Or say that all cyclic maximal minors should be positive. As is true for any determinant, if you cyclically shift or "roll" the vectors, and k is odd, it changes sign. When k is even nothing happens, such determinants are cyclically invariant.

The advantage of the positive region is that we can stop worrying about crosscaps 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

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

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

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We will rock you! If you like this you can pay a visit to collection of Music, Art and Culture. It's always interesting!
Because even in alternate universes Freddie Mercury is awesome.
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Let's dance!
Rockabilly!
Brian Setzer is going to show you what a guitar is for--throwing down some classic Billy Riley.﻿
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The world's most important number.

One you probably haven't heard of. And the man behind it.

https://www.bloomberg.com/news/features/2016-11-29/the-man-who-invented-libor-iw3fpmed