Nima Arkani-Hamed had a series of talks at JHU roughly 6 months ago which I attended related to this topic. He discussed it at Stony Brook a little bit over a week ago (pointed out by Emilio Pisanty in the comments above in which he used the term "amplituhedron", but my understanding of this comes mostly from his earlier talks.

*Update:* I managed to track Nima down today and get him to explain the details. Surprisingly almost everything was correct, but the loop level description has some changes. I've also made some aesthetic changes, attempting to follow the notation in Trnka's slides as closely as possible while being readable by mathematicians. Also note that Nima claimed that their paper on this would appear "very soon".

*Update 2:* Their paper has appeared on the arXiv: http://arxiv.org/abs/1312.2007. At first glance everything seems consistent with what I've written below. I think it should be readable to mathematicians simply by skipping the few sections which require knowledge of physics. They do find some combinatorial results therein which may be of interest, but as this answer is already rather large I'll just direct those interested to the paper directly.

Everything here is over $\mathbb R$. We know that the quotient of the subset $\tilde{M}(k,n) \subset M(k,n)$ of matrices with full rank by the $GL(k)$ left action gives the Grassmannian $G(k,n)$, in which each matrix is mapped to its row space. Define $M_+ (k,n)$ to be the subset of $M(k,n)$ in which all $k \times k$ minors are positive (ordering the rows in the process). The image of $M_+(k,n)$ under this quotient is the *positive Grassmannian* $G_+(k,n)$.

Fix a matrix $Z \in M_+(k+m,n)$ as input data, which comes from physics, but won't really affect the combinatorial structure* at all.

Then there is a map $Y_{n,k,m}: G_+(k,n) \rightarrow G(k,k+m)$ given by $([C],Z) \mapsto [CZ^T]$, where brackets denote equivalence classes under the $GL(k)$ action (the requirement that $CZ^T$ has full rank is automatically satisfied if both are in their respective positive pieces). Its image $\mathcal P_{n,k,m}$ is called the tree-level amplituhedron. The combinatorial structure does depend on $n,k,$ and $m$. This case is apparently fairly well understood thanks to their work with Alexander Postnikov (according to Nima).

(One of the interesting aspects of this positive-real story is that while the map $Y_{n,k,m}$ is only a rational map of algebraic varieties, its base locus doesn't intersect $G_+(k,n)$.)

*I've seen it claimed that $Z$ can be taken to live in $G_+(k+m,n)$. I must admit this doesn't make sense to me because the map described above is not invariant under the right action by $GL(k+m)$. If I'm missing something obvious though than feel free to correct me. At the very least, I'm pretty sure that the above description works, but it might be a bit more redundant than necessary.

As I alluded to above, $\mathcal P_{n,k,m}$ is not the full amplituhedron. Rather, it's just the tree-level case. The full amplituhedron has another nonnegative integer parameter $l$ which determines the loop order. This gives additional coordinates to points in the amplituhedron. The subregion of $M(k+2l, k+m)$ of interest which the $C$ vary through satisfies somewhat more stringent positivity constraints than those of the ordinary positive Grassmannian that we had above in the case $l=0$. I will call this $l$-positivity, though this is my own terminology. $C' = \left( \begin{matrix} C \\ C^{(1)} \\ \vdots \\ C^{(l)} \end{matrix} \right)$ with $C$ as $k \times n$ and each $C^{(i)}$ as $2 \times n$ is $l$-positive iff for any $I = \{i_1 , \ldots, i_r\} \subseteq \{1, \ldots, l\}$ (with $i_1 < i_2 < \cdots < i_r$), the submatrix $\left( \begin{matrix} C \\ C^{(i_1)}\\ \vdots \\ C^{(i_r)} \end{matrix} \right) \in M_+(k+2r,n)$ (including the case $I=\phi$), *and* each $C^{(i)}$ is only well-defined up to addition of elements of $C$.

For convenience of notation let $\mathcal A_{n,m} = Y_{n,2,m}$. The $l$-loop amplituhedron is then the image of $\{[C'] | C'\text{ is }l\text{-positive}\} \times \{Z\}$ in $G(k,k+m) \times (G(2,k+m))^l$ by just applying $Y_{n,k,m}$ to $([C],Z)$ and $\mathcal A_{n,m}$ to each $([C^{(i)}],Z)$ (all with the same $Z$). This is called $\mathcal P_{n,k,l,m}$ (Trnka drops the $m$, presumably since $m=4$ for physics). The space that this is embedded in has no significance either combinatorially or physically. We could take it to be in the $l$-fold product of the $G(2,m)$ bundle over $G(k,k+m)$ such that the fiber at each point is those $2$-planes orthogonal to that $k$-plane.

It is important to realize that the amplituhedron itself is not so much the object of interest: rather, that is a meromorphic volume form defined on it (and on the Grassmannian, or bundle over Grassmannian, in which the amplituhedron is Zariski-dense). The principal job of the amplituhedron is to help nail down this form: the form is required to be well-defined on the interior of the amplituhedron. It seems very hard to make this statement mean anything without talking about positive real parts of varieties.

The parameters are important for physics, so I'll list them, but of course if you're not interested in the physics you can set them to be whatever you like. $k$ is the order in perturbation theory. $l$, the loop order. $m=4$ is the case for physics, but in principle $m$ can be any even positive integer ($m=2$ makes the loop part trivial, so $m=4$ is in some sense the first interesting case). $n$ is the number of momenta in the scattering process. $Z$ is a positive matrix that represents all of the momenta, but at least for the purpose of combinatorics the structure doesn't really depend on the choice of $Z$. There are of course cases in which the construction does not make sense; these are irrelevant for physics (e.g. $n-k < m$ is unphysical). Also, while I'm talking about physics, to get physical predictions out of the amplituhedron, there is a particular volume form which is simply integrated over the region. This volume gives the amplitude for the process.

anypreprint on the amplituhedron proper has been released yet, so as a partial response to your question, even if such a physics-free way of "appreciating" this exists, certainly no such reference exists at the current time. But you might browse this earlier paper arxiv.org/abs/1212.5605 and these slides staff.science.uu.nl/~tonge105/igst13/Trnka.pdf . $\endgroup$ – j.c. Sep 22 '13 at 1:136more comments