How does one actually compute the amplituhedron?

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I was watching Nima's very popular talk (download if you're using chrome) (also mirrored at youtube here) about the "Amplituhedron", which has suddenly become very popular recently.

He talks all about how the amplituhedron computes the same result for the scattering amplitudes as ordinary peturbation theory in a simple and elegant way, but I fail to understand how one actually computes the amplituhedron for a certain scattering process anyway?

As per the recent TRF posts about amplituhedron and why they don't wear diapers, I can understand that one may calculate the scattering amplitudes by simply taking the volume of the amplituhedrons (ignoring constants, I guess), but how does one actually calculate the amplituhedron?

I'm especially stunned by the image (looks like a sort of a concrete example, don't know how they constructed the amplituhedron):

To summarise my question, how does one actually figure out, or construct, the amplituhedron based on the specific scattzering process?

edited Apr 25, 2014
Wait for the paper. It's a geometric object in some space whose dimension depends on the number of external particles, number of loops, and number of "helicity flips". The volume form, the integrand, is a simple form roughly scaling like $1/x$ where $x$ is the distance from a face, and the faces are given by inequalities of the type "determinants of a submatrix are zero". These inequalities depend on the external momenta and/or twistor variables, sort of linearly or simply. The scattering amplitude is the single simple integral of the volume form over the polytope.

This post imported from StackExchange Physics at 2014-03-07 16:33 (UCT), posted by SE-user Luboš Motl
@LubošMotl So does does the 'volume' of a single polytope represent the contribution of a single Feynman diagram or does it represent the exactly (non-perturbatively) obtained net scattering amplitude of the whole process? Or is it something else entirely?

This post imported from StackExchange Physics at 2015-11-25 11:20 (UTC), posted by SE-user dj_mummy
Maybe you will enjoy watching, "Scattering without space time". He elaborates on these ideas, in a 3 talk series

This post imported from StackExchange Physics at 2014-03-07 16:33 (UCT), posted by SE-user Prathyush

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To begin with, The Amplituhedron formalism only works for a specific theory, N=4 SYM in the planar limit (only planar Feynmann diagrams are considered). Because of supersymmetry, you can classify scattering processes with two parameters: $n$ and $k$. n is the number of particles involved, and k is, roughly speaking, the number of spin flips in the process.

In addition, there is also the number of loop $L$ at which you want to perform your calculation. Be aware that at loop level you are not computing the (super)amplitude, but rather the integrand of the amplitude. That is, thanks to the planar limit, you can uniquely define a function dependent on external data + virtual momenta that has to be integrated in order to obtain the amplitude. To clarify, is what you obtain if you swap $\sum_{diagrams} \int_{loops} = \int_{loops} \sum_{diagrams}$, you can do that with no ambiguity because the planarity allows to choose a scheme to fix the loop variables. This integrand is what is produced by the Amplituhedron at loop level.

Now, for any $n,k$ and $L$ and for fixed external data $Z$ (a $(k+4) \times n$ matrix which ultimately encodes the external momenta and supermomenta) you build an amplituhedron $A_{n,k,L}$ in a standard way. At tree level ($L=0$) it is the subset of $G_{+}(k,k+4)$ of points $Y$ written as $Y = C . Z$, for any $C$ in $G_{+}(k,n)$. At loop level is a little bit more complicate, but nothing terrible.

An important feature of the Amplituhedron is that you can "triangulate" it in a very interesting way. (By triangulate we mean divide it in zones that have in common only their boundary.) In fact, in a prievous work it was shown that the amplitude/integrand for $n,k,L$ is written as a sum over certain on-shell diagrams, which in turn label certain cells in the Grassmannian $G_{+}(k,n)$. (The BCFW rules are used to obtain this expression of the amplitudes.) If you consider the image of these cells under the map $Y = C . Z$ you obtain a triangulation for the Amplituhedron!

Now these Amplituhedra have codimension one "boundaries". These are zero-loci of non-negative functions defined on the Amplituhedra. A crucial point will be that these boundaries may be written as product of other Amplituhedra. You can define a (unique?) volume form on the Amplituhedron by the condition that it should have logarithmic singularities (i.e. like 1/x) at these codimension one boundaries.

A concrete way to do so is to consider a triangulation of the Amplituhedra, for example the one deriving from BCFW recursion relations as above. On every "triangle" you are already given a form that has log singularities at the face of the "triangle". This is so because the cells of $G_{+}(k,n)$ used in this triangulation are equipped with simple positive charts $(\alpha_1, ... , \alpha_d)$ that allows to reach a boundary of the cell by setting certain $\alpha$ to zero. Therefore the forms $\Pi d\alpha/\alpha$ have the desired property cell-wise. You just sum these forms over the cells: the "spurious" poles, associated to cell-boundaries that are not amplituhedron-boundaries cancels (they are shared by two cells) and you are left with the right log singularities.

But you can also obtain this volume form in other ways, following directing by the definition, as explained in the paper.

Finally, you evaluate the volume form obtained above in a particular point of the amplituhedra, and what you obtain is the amplitude/integrand.

The procedure seems a little bit complicate and abstract, but actually the idea is quite simple. The geometry of the amplituhedron captures the intricate factorization properties of amplitudes/integrands required by Locality and Unitarity: The boundaries of an amplituhedron factorizes in the corresponding smaller amplituhedra. Therefore a form with logarithmic singularities at these boundaries will have the right poles and factorizations as well.

This post imported from StackExchange Physics at 2015-11-25 11:20 (UTC), posted by SE-user giulio bullsaver
answered Nov 25, 2015 by (30 points)

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