What is the amplituhedron?

Question

I would like to know what exactly this object is without any nasty physics stuff. The wikipedia says that it "generalizes the idea of a simplex in projective space." It's easy for me to see how the idea of a simplex could be generalized to any vector space, but not for a projective space, let alone a Grassmannian.

Answer 1

To establish notation, let's begin with the Grassmannian $Gr(k,n)$ of $k$-dimensional subspaces of $\mathbb{C}^n$. Every $V \in Gr(k,n)$ can be represented by a matrix $$ \begin{bmatrix} \leftarrow & \vec{v}_1 & \to \\ & \vdots & \\ \leftarrow & \vec{v}_k & \to \end{bmatrix} = \begin{bmatrix} \uparrow & & \uparrow \\ \vec{c}_1 & \dots & \vec{c}_n \\ \downarrow & & \downarrow \end{bmatrix} $$ where $\vec{v}_1, \dots, \vec{v}_k$ form a basis of $V$. The totally non-negative Grassmannian $Gr_{\geq}(k,n)$ is the subset of $Gr(k,n)$ so that every maximal minor is non-negative. This means for appropriate matrix representation and every $\{i_1,\dots,i_k\} \subset \{1,2,\dots,n\}$ we have $\det [\vec{c}_{i_1} \dots \vec{c}_{i_k}] \geq 0$.

Now take $Z$ a $(k{+}m) \times n$ matrix that is totally positive, i.e. every maximal minor is strictly positive. With some work, one can show that $$ \dim \langle Z \vec{v}_1, \dots, Z \vec{v}_k \rangle = k, $$ so $Z$ induces a map $\tilde{Z}: Gr_\geq(k,n) \to Gr(k,k{+}m)$. The (tree) amplituhedron $\mathcal{A}_{n,k,m}(Z)$ is the image $\tilde{Z}(Gr_\geq(k,n))$ of this map.

Physicists are most interested in the case where $m = 4$, since this should correspond to $\mathcal{N} = 4$ supersymmetric Yang-Mills theory, whatever that is. It is known that $\mathcal{A}_{n,1,m}(Z)$ is a cyclic polytope in projective space, which I believe is what Wikipedia is referring to. Additionally, when $k+m = n$ we have $\mathcal{A}_{n,k,m}(Z) \cong Gr_\geq(k,n)$, so it also in some sense generalizes the totally non-negative Grassmannian.

From a mathematician's perspective, the conjecture I have heard referenced most often is that the choice of $Z$ does not matter - $\mathcal{A}_{n,k,m}(Z) \cong \mathcal{A}_{n,k,m}(Z')$ for any totally positive choices $Z$ and $Z'$. There's been a significant push in the algebraic combinatorics community to study this object. Many of the experts on the non-negative Grassmannian are working on this, for instance Thomas Lam, Alex Postnikov, Hugh Thomas and Lauren Williams.