What is the amplituhedron? 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.
 A: 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.
