0 \ee $$ \includegraphics[scale=.75]{pix12-m.pdf} $$ More precisely, the interior of a triangle is associated with a triplet $(c_1,c_2,c_3)/GL(1)$, with all ratios $c_a/c_b > 0$, so that the $c_a$ are either all positive or all negative, but here and in the generalizations that follow, we will abbreviate this by calling them all positive. Including the closure of the triangle replaces ``positivity'' with ``non-negativity'', but we will continue to refer to this as ``positivity'' for brevity. One obvious generalization of the triangle is to an $(n-1)$ dimensional simplex in a general projective space, a collection $(c_1, \dots, c_n)/GL(1)$, with $c_a > 0$. The $n$-tuple $(c_1, \dots, c_n)/GL(1)$ specifies a line in $n$ dimensions, or a point in $\mathbb{P}^{n-1}$. We can generalize this to the space of $k$-planes in $n$ dimensions --- the Grassmannian $G(k,n)$ --- which we can take to be a collection of $n$ $k-$dimensional vectors modulo $GL(k)$ transformations, grouped into a $k \times n$ matrix \be C = \left( \begin{array}{ccc} & & \\ c_1 & \dots & c_n \ \\ & & \end{array} \right)/GL(k) \ee Projective space is the special case of $G(1,n)$. The notion of positivity giving us the ``inside of a simplex'' in projective space can be generalized to the Grassmannian. The only possible $GL(k)$ invariant notion of positivity for the matrix $C$ requires us to fix a particular ordering of the columns, and demand that all minors in this ordering are positive: \be \langle c_{a_1} \dots c_{a_k} \rangle > 0 \, \, {\rm for} \, \, a_1<\dots

0 \quad {\rm for} \quad a_1 < a_2 < a_3 \ee Having arranged for this, the interior of the polygon is given by points of the form \be Y^I = c_1 Z_1^I + c_2 Z_2^I + \dots c_n Z_n^I \quad {\rm with}\quad c_a > 0 \ee Note that this can be thought of as an interesting pairing of two different positive spaces. We have \be (c_1, \dots, c_n) \subset G_+(1,n),\quad \left(Z_1, \dots, Z_n \right) \subset M_+(3,n) \ee If we jam them together to produce \be Y^I = c_a Z_a^I \ee for fixed $Z_a$, ranging over all $c_a$ gives us all the points on the inside of the polygon, living in $G(1,3)$. This object has a natural generalization to higher projective spaces; we can consider $n$ points $Z_a^I$ in $G(1,1+m)$, with $I = 1, \dots, 1+m$, which are positive \be \langle Z_{a_1} \dots Z_{a_{1 + m}} \rangle > 0 \ee Then, the analog of the ``inside of the polygon'' are points of the form \be Y^I = c_a Z_a^I, \quad {\rm with} \quad c_a > 0 \ee This space is very closely related to the ``cyclic polytope''~\cite{cyclic}, which is the convex hull of $n$ ordered points on the moment curve in $\mathbb{P}^{m}$, $Z_a = (1,t_a,t_a^2, \dots, t_a^{m})$, with $t_1 < t_2 \dots < t_n$. From our point of view, forcing the points to lie on the moment curve is overly restrictive; this is just one way of ensuring the positivity of the $Z_a$. \looseness=-1 We can further generalize this structure into the Grassmannian. We take positive external data as $(k + m)$ dimensional vectors $Z_a^I$ for $I =1, \dots, k+m$. It is natural to restrict $n \geq (k+m)$, so that the external $Z_a$ fill out the entire $(k+m)$ dimensional space. Consider the space of $k$-planes in this $(k+m)$ dimensional space, $Y \subset G(k,k+m)$, with co-ordinates \be Y_\alpha^I, \, \alpha = 1, \dots k, \, I = 1, \dots, k+m \ee We then consider a subspace of $G(k,k+m)$ determined by taking all possible ``positive'' linear combinations of the external data, \be Y = C \cdot Z \ee or more explicitly \be Y_\alpha^I = C_{\alpha a} Z_a^I \ee where \be C_{\alpha a} \subset G_+(k,n), Z_a^I \subset M_+(k+m,n) \ee It is trivial to see that this space is cyclically invariant if $m$ is even: under the twisted cyclic symmetry, $Z_n \to (-1)^{k+m-1} Z_1$ and $c_n \to (-1)^{k-1} c_1$, and the product is invariant for even $m$. We call this space the generalized tree amplituhedron ${\cal A}_{n,k,m}(Z)$. The polygon is the simplest case with $k=1,m=2$. Another special case is $n = (k+m)$, where we can use $GL(k+m)$ transformations to set the external data to the identity matrix $Z_a^I = \delta_a^I$. In this case ${\cal A}_{k+m,k,m}$ is identical to the usual positive Grassmannian $G_+(k,k+m)$. \looseness=-1 The case of immediate relevance to physics is $m=4$, and we will refer to this as the tree amplituhedron ${\cal A}_{n,k}(Z)$. The tree amplituhedron lives in a $4k$ dimensional space and is not trivially visualizable. For $k=1$, it is a polytope, with inequalities determined by linear equations, while for $k>1$, it is not a polytope and is more ``curvy''. Just to have a picture, below we sketch a 3-dimensional face of the 4 dimensional amplituhedron for $n=8$, which turns out to be the space $Y = c_1 Z_1 +\dots c_7 Z_7$ for $Z_a$ positive external data in $\mathbb{P}^3$: $$ \includegraphics[scale=.6]{pix11-m.pdf} $$ ]]>

k+m$, there is always some linear combination of the $Z_a$ which sum to zero! We have to take care to avoid this happening, and in order to avoid ``0'' on the left hand side, we obviously need positivity properties on both the $Z$'s and the $C$'s. It is simple and instructive to see why positivity ensures that the $Y = C \cdot Z$ map is projectively well-defined. We will see this as a by-product of locating the co-dimension one boundaries of the generalized tree amplituhedron. Let us illustrate the idea already for the simplest case of the polygon with $k=1, m=2$, with $Y = c_1 Z_1 + \dots c_n Z_n$. In order to look at the boundaries of the space, let us compute $\langle Y Z_i Z_j \rangle$ for some $i,j$. If as we sweep through all the allowed $c$'s, $\langle Y Z_i Z_j \rangle$ changes sign from being positive to negative, then somewhere $\langle Y Z_i Z_j \rangle \to 0$ and $Y$ lies on the line $(Z_i Z_j)$ in the interior of the space, thus $(Z_i Z_j)$ should not be a boundary of the polygon. On the other hand, if $\langle Y Z_i Z_j \rangle$ everywhere has a uniform sign, then $(Z_i Z_j)$ is a boundary of the polygon: $$ \includegraphics[scale=.6]{pix22-m.pdf} $$ Of course for the polygon it is trivial to directly see that the co-dimension one boundaries are just the lines $(Z_i Z_{i+1})$, but we wish to see this more algebraically, in a way that will generalize to the amplituhedron where ``seeing'' is harder. So, we compute \be \langle Y Z_i Z_j \rangle =\sum_a c_a \langle Z_a Z_i Z_j \rangle \ee We can see why there is some hope for the positivity of this sum, since the $c_a > 0$, and also ordered minors of the $Z's$ are positive. It is however obvious that if $i,j$ are not consecutive, some of the terms in this sum will be positive, but some (where $a$ is stuck between $i,j$) will be negative. But precisely when $i,j$ are consecutive, we get a manifestly positive sum: \be \langle Y Z_i Z_{i+1} \rangle = \sum_a c_a \langle Z_a Z_i Z_{i+1} \rangle > 0 \ee Since $\langle Z_a Z_i Z_{i+1} \rangle > 0$ for $a \neq i, i+1$, this is manifestly positive. Thus the boundaries are lines $(Z_i Z_{i+1})$ as expected. \looseness=-1 This also tells us that the map $Y = C \cdot Z$ is projectively well-defined. There is no way to get $Y \to 0$, since this would make the left hand side identically zero, which is impossible without making all the $c_a$ vanish, which is not permitted as we we mod out by $GL(1)$ on the $c_a$. We can extend this logic to higher $k,m$. Let's look at the case $m=4$ already for $k=1$. We can investigate whether the plane $(Z_i Z_j Z_k Z_l)$ is a boundary by computing \be \langle Y Z_i Z_j Z_k Z_l \rangle = \sum_a c_a \langle Z_a Z_i Z_j Z_k Z_l \rangle \ee Again, this is not in general positive. Only for $(i,j,k,l)$ of the form $(i,i+1,j, j+1)$, we have \be \langle Y Z_i Z_{i+1} Z_j Z_{j+1} \rangle = \sum_a c_a \langle Z_a Z_i Z_{ i+1} Z_j Z_{j+1} \rangle > 0 \ee For general even the $m$, the boundaries are when $Y$ is on the plane\\ $(Z_{i} Z_{i+1} \dots Z_{i_{m/2 - 1}} Z_{i_{m/2}})$. This again shows that the $Y = C \cdot Z$ is projectively well-defined. The result extends trivially to general $k$, provided the positivity of $C$ is respected. For $m = 4$ the boundaries are again when the $k$-plane $(Y_1 \cdots Y_k)$ is on $(Z_i Z_{i+1} Z_j Z_{j+1})$, as follows from \be \langle Y_1 \dots Y_k Z_i Z_{i+1} Z_j Z_{j+1} \rangle = \sum_{a_1< \dots < a_k} \langle c_{a_1} \dots c_{a_k} \rangle \langle Z_{a_1} \dots Z_{a_k} Z_i Z_{i+1} Z_j Z_{j+1} \rangle > 0 \ee which also shows that $Y$ is always a full rank $k$-plane in $k+4$ dimensions. The emergence of boundaries on the plane $(Z_{i} Z_{i+1} Z_j Z_{j+1})$ is a simple and striking consequence of positivity. We will shortly understand that the location of these boundaries are the ``positive origin'' of locality from the geometry of the amplituhedron. ]]>

0 \quad {\rm for} \quad a_1 < \dots < a_{4 + k} \ee The amplituhedron lives in $G(k,k+4;L)$: the space of $k$ planes $Y$ in $(k+4)$ dimensions, together with $L$ 2-planes ${\cal L}_{(i)}$ in the 4 dimensional complement of $Y$, schematically $$ \includegraphics[scale=.65]{pix28-m.pdf} $$ We will denote the collection of $({\cal L}_{(i)}, Y)$ as ${\cal Y}$. The amplituhedron ${\cal A}_{n,k,L}(Z)$ is the subspace of $G(k,k+4;L)$ consisting of all ${\cal Y}$'s which are a positive linear combination of the external data, \be {\cal Y} = {\cal C} \cdot Z \ee More explicitly in components, the $k$-plane is $Y_{\alpha}^I$, and the 2-planes are ${\cal L}_{\gamma (i)}^I$, where $\gamma = 1,2$ and $i = 1, \dots, L$ . The amplituhedron is the space of all $Y, {\cal L}_{(i)}$ of the form \be Y_{\alpha}^I = C_{\alpha a} Z_a^I, \, \, {\cal L}_{\gamma (i)}^{I} = D_{\gamma a (i)} Z_a^I \ee \looseness=-1 where as in the previous section the $C_{\alpha a}$ specifies a $k$-plane in $n$-dimensions, and the $D_{\gamma a (i)}$ are $L$ 2-planes living in the $(n-k)$ dimensional complement of $C$, with the positivity property that for any $0 \leq l \leq L$, all the ordered $(k + 2l) \times (k+2l)$ minors of the $(k + 2 l) \times n$ matrix \be \left(\begin{array}{ccc} & D_{(i_1)} & \\ \hdashline & \vdots & \\ \hdashline & D_{(i_l)} & \\ \hdashline & C \end{array} \right) \ee are positive. The notion of cells, cell decomposition and canonical form can be extended to the full amplituhedron. A cell $\Gamma$ is associated with a set of positive co-ordinates $\alpha^\Gamma = (\alpha^\Gamma_1, \dots, \alpha^\Gamma_{4(k + L)})$, rational in the ${\cal C}$, such that for $\alpha$'s positive, ${\cal C}(\alpha) = (D_{(i)}(\alpha), C(\alpha))$ is in $G_+(k,n;L)$. A cell decomposition is a collection $T$ of non-intersecting cells $\Gamma$ whose images under ${\cal Y} = {\cal C} \cdot Z$ cover the entire amplituhedron. The rational form $\Omega_{n,k,L}({\cal Y}; Z)$ is defined by having the property that \begin{center} $\Omega_{n,k,L}(Y;Z)$ has logarithmic singularities on all the boundaries of ${\cal A}_{n,k,L}(Z)$ \end{center} A concrete formula follows from a cell decomposition as \be \Omega_{n,k,L}({\cal Y}; Z) = \sum_{\Gamma \subset T} \prod_{i = 1}^{4(k + L)} \frac{d \alpha_i^\Gamma}{\alpha_i^\Gamma} \ee Of course any cell decomposition gives the same form $\Omega_{n,k,L}$. ]]>

0, \quad (x_i - x_j)(z_i - z_j) + (y_i - y_j)(w_i - w_j) < 0 \ee We can rephrase this problem in a simple, purely geometrical way by defining two dimensional vectors $\vec{a}_i = (x_i,y_i), \vec{b}_i = (z_i, w_i)$. The points are in the upper quadrant of the plane. The mutual positivity condition is just $(\vec{a}_i - \vec{a}_j) \cdot (\vec{b}_i - \vec{b}_j) < 0$. Geometrically this just means that the $\vec{a}, \vec{b}$ must be arranged so that for every pair $i,j$, the line directed from $\vec{a}_i \to \vec{a}_j$ is pointed in the opposite direction as the one directed from $\vec{b}_i \to \vec{b}_j$. An example of an allowed configuration of such points for $L=3$ is $$ \includegraphics[scale=.6]{pix26-m.pdf} $$ Finding a cell decomposition of this $4L$ dimensional space directly gives us the integrand for the four-particle amplitude at $L$-loops. Now, we know that the final form can be expressed as a sum over local, planar diagrams. This makes it all the more remarkable that no-where in the definition of our geometry problem do we reference to diagrams of any sort, planar or not! Nonetheless, this property is one of many that emerges from positivity. As we will describe at greater length in~\cite{Into}, it is easy to find a cell decomposition for the full space ``manually'' at low-loop orders. We suspect there is a more systematic approach to understanding the geometry that might crack the problem at all loop order. As an interesting warmup to the full problem, we can investigate lower-dimensional ``faces'' of the four-particle amplituhedron. Cellulations of these faces corresponds to computing certain cuts of the integrand, at all loop orders. We will discuss many of these faces and cuts systematically in~\cite{Into}. Here we will content ourselves by presenting some especially simple but not completely trivial examples. Let us start by considering an extremely simple boundary of the space, where all the $w_i \to 0$. This corresponds to having all the lines intersect $(Z_1 Z_2)$. The positivity conditions then simply become \be (x_i - x_j)(z_i - z_j) < 0 \ee which is trivial to triangulate. Whatever configuration of $x$'s we have are ordered in some way, say $x_1 < \dots < x_L$. Then we must have $z_1 > \dots > z_L$. The $y_i$ just have to be positive. The associated form is then trivially (we omit the measure $\prod_i dx_i dz_i d y_i$): \be \frac{1}{y_1} \dots \frac{1}{y_L} \frac{1}{x_1} \frac{1}{x_2 - x_1} \dots \frac{1}{x_L - x_{L-1}} \frac{1}{z_L} \frac{1}{z_{L-1} - z_L} \dots \frac{1}{z_1 - z_2} + {\rm perm.} \ee Now, this cut is particularly simple to understand from the point of view of the familiar ``local'' expansions of the integrand --- there is only only local diagram that can possibly contribute to this cut: the ``ladder'' diagram. The corresponding cut is precisely what we have above from positivity. $$ \includegraphics[scale=.8]{pix19-m.pdf} $$ We can continue along these lines to explore faces of the amplituhedron which determine cuts to all loop orders that are difficult (if not impossible) to derive in any other way. For instance, suppose that some of the lines intersect $(Z_1 Z_2)$, so that the $w_i \to 0$ for $i = 1, \dots, L_1$ and others intersects $(Z_3 Z_4)$, so that $y_I \to 0$ for $I =L_1+1, \dots, L$. To pick a concrete interesting example, let choose $L-2$ lines to intersect $(12)$ and 2 lines to intersect $(34)$. We can further specialize the geometry and take more cuts by making the $L$'th line pass through the point 3 --- this corresponds to sending $z_{L} \to 0$. Let us also take the $(L-1)$'st line to pass through the point 4 --- this corresponds to sending $z_{L-1}, w_{L-1} \to \infty$ with $w_{L-1}/z_{L-1} \equiv W_{L-1}$ fixed. We can again label the $x_i; x_I$ so they are in increasing order; then the positivity conditions become \be x_1 < \dots < x_{L-2}, z_1 > \dots > z_{L-2}; \, x_{L-1} < x_{L} \ee and \be W_{L-1} y_i > (x_{L-1} - x_i), \, \, w_L y_i > z_i (x_i - x_L) \ee This space is also trivial to triangulate, but the corresponding form is more interesting. The ordering for the $z$'s is associated with the form $$ \frac{1}{z_{L\mi2}(z_{L\mi3}-z_{L\mi2})(z_{L\mi4}-z_{L\mi3})\dots (z_1-z_2)} $$ The interesting part of the space involves $x_i,y_i$. Note that if $x_i < x_{L-1}$, the second inequality on $y_i$ is trivially satisfied for positive $y_i$, and the only constraint on $y_i$ is just $y_i > (x_{L-1} - x_i)/W_{L-1}$. If $x_{L-1} < x_i < x_{L}$, then both inequalities are satisfied and we just have $y_i > 0$. Finally if $x_i > x_L$, the first inequality is trivially satisfied and we just have $y_i > z_i (x_i - x_L)/w_L$. Thus, given any ordering for all the $x's$, there is an associated set of inequalities on the $y$'s, and the corresponding form in $x,y$ space is trivially obtained. For instance, consider the case $L=5$, and an ordering for the $x$'s where $x_1 < x_4 < x_2 < x_5 < x_3$. The corresponding form in $(x,y)$ space is just \begin{align} \frac{1}{x_1 (x_4 - x_1) (x_2 - x_4) (x_5 - x_2)(x_3 - x_5)} \frac{1}{y_1 - (x_4 - x_1)/W_4} \frac{1}{y_2} \frac{1}{y_3 - z_3(x_3 - x_5)/w_5} \end{align} By summing over all the possible orderings $x$'s, we get the final form. For general $L$, we can simply express the result (again omitting the measure) as a sum over permutations $\sigma$: \begin{align} &\prod_{l=1}^{L-2}\frac{1}{(z_l - z_{l+1})}\quad\times\hspace{-0.7cm}\sum_{\sigma; \sigma_1 < \dots < \sigma_{L-2}; \sigma_{L-1} < \sigma_L} \frac{1}{w_L W_{L-1}} \prod_{l=1}^{L} \frac{1}{(x_{\sigma^{-1}_l} - x_{\sigma^{-1}_{l-1}})}\\ &\hspace{4cm}\times\prod_{i=1}^{L-2} \left\{ \begin{array}{cc} (y_i - (x_{L-1} - x_i)/W_{L-1})^{-1} & \sigma_i < \sigma_{L-1} \\ y_i^{-1} & \sigma_{L-1} < \sigma_i < \sigma_{L} \\ (y_i - (x_i - x_L) z_i/w_L)^{-1} & \sigma_L < \sigma_i \end{array}\right\} \nonumber \end{align} where we define for convenience $z_{L-1} = x_{\sigma^{-1}_0} = 0$. This gives us non-trivial all-loop order information about the four-particle integrand. The expression has a feature familiar from BCFW recursion relation expressions for tree and loop level amplitudes. Each term has certain ``spurious'' poles, which cancel in the sum. This result can be checked against the cuts of the corresponding amplitudes that are available in ``local form''. The diagrams that contribute are of the type $$ \includegraphics[scale=.7]{pix20-m.pdf} $$ but now there are non-trivial numerator factors that don't trivially follow from the structure of propagators. The full integrand is available through to seven loops in the literature~\cite{Bern-ml-2005iz,Bern-ml-2006ew,Bern-ml-2007ct,Bourjaily-ml-2011hi,Eden-ml-2012tu}. The inspection of the available local expansions on this cut does not indicate an obvious all-loop generalization, nor does it betray any hint that that the final result can be expressed in the one-line form given above. For instance just at 5 loops, the local form of the cut is given as a sum over diagrams, $$ \includegraphics[scale=.8]{pix29-m.pdf} $$ \looseness=-1 with intricate numerator factors. If all terms are combined with a common denominator of all physical propagators, the numerator has 347 terms. Needless to say, the complicated expression obtained in this way perfectly matches the amplituhedron computation of the cut. ]]>

n, k^\prime > k, L^\prime > L$. The objects needed to compute scattering amplitudes for any number of particles to all loop orders are thus contained in a ``master amplituhedron'' with $n, k , L \to \infty$. In this vein it may also be worth considering natural mathematical generalizations of the amplituhedron. We have already seen that the generalized tree amplituhedron ${\cal A}_{n,k,m}$ lives in $G(k,k+m)$ and can be defined for any even $m$. It is obvious that the amplituhedron with $m=4$, of relevance to physics, is contained amongst the faces of the object defined for higher $m$. If we consider general even $m$, we can also generalize the notion of ``hiding particles'' in an obvious way: adjacent particles can be hidden in even numbers. This leads us to a bigger space in which to embed the generalized loop amplituhedron. Instead of just considering $G(k,k+4;L)$ of ($k-$ planes) $Y$ together with $L$ ($2-$planes) in $m=4$ dimensional complement of $Y$, we can consider a space $G(k,k+m;L_2,L_4, \dots, L_{m-2})$, of $k$-planes $Y$ in $(k+m)$ dimensions, together with $L_2$ (2-planes), $L_4$ (4-planes), $\dots L_{m-2}$ ($(m-2)$-planes) in the $m$ dimensional complement of $Y$, schematically: $$ \includegraphics[scale=.6]{pix10-m.pdf} $$ Once again we have ${\cal Y} = {\cal C} \cdot Z$, with the obvious extension of the loop positivity conditions on ${\cal C}$ to $G(k,n;L_2, L_4, \dots, L_{m-2})$. We can call this space ${\cal A}_{n,k;m,L_2, \dots, L_{m-2}}(Z)$. The $m=4$ amplituhedron is again just a particular face of this object. It would be interesting to see whether this larger space has any interesting role to play in understanding the $m=4$ geometry relevant to physics. ]]>