jcreed blog > Orienting Simplices

Orienting Simplices

Suppose I have nn points in Rn1\R^{n-1}. For example, 3 points in R2\R^2.

If I think of these as objects in a 2-category, and I think of the xx-direction of R2\R^2 as representing 'the flow of time as far as the 1-cells are concerned', and the yy-direction as 'the flow of time as far as the 2-cells are concerned', then there's two possible sorts of 2-cell that these points (together with the lines and triangle that lie between them) can represent. One where two morphisms 'compose' across the 2-cell to yield a third, and one where one 'decomposes' into two morphisms.

More precisely, I want to observe the following things: There is a categorical sense in which every cell lies either in the domain or codomain of the higher-dimensional cells it's part of. For example, aa is in the domain of hh, bb is in the codomain of gg, hh is in the codomain of α\alpha, ff and gg are both in the domain of α\alpha, kk is in the codomain of β\beta, etc.

But there is also a geometric sense in which we can take any cell and ask whether it looks like it's in the domain or codomain of a higher dimensional cell. For example, to ask whether hh is in the codomain of α\alpha, we ask whether det(axay1cxcy1bxby1)<0\det\left(\begin{matrix}a_x & a_y & 1 \\ c_x & c_y & 1 \\ b_x & b_y & 1 \end{matrix}\right) < 0 at least assuming the points are laid out as depicted, with cxaxc_x - a_x.

In general, if we have points q,p1,,pnq, p^1, \ldots, p^n in Rn\R^{n} then det(p11p21pn11p12p22pn21p1np2npnn1q1q2qn1)det(p11p21pn111p12p22pn121p1np2npn1n1)\det\left(\begin{matrix} p^1_1 & p^1_2 & \cdots & p^1_{n} & 1\\ p^2_1 & p^2_2 & \cdots & p^2_{n} & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p^n_1 & p^n_2 & \cdots & p^n_{n} & 1\\ q_1 & q_2 & \cdots & q_{n} & 1 \end{matrix}\right) \over \det\left(\begin{matrix} p^1_1 & p^1_2 & \cdots & p^1_{n-1} & 1\\ p^2_1 & p^2_2 & \cdots & p^2_{n-1} & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p^n_1 & p^n_2 & \cdots & p^n_{n-1} & 1 \end{matrix}\right) is how far qq is — in the nn-coordinate — from the hyperplane determined by p1,,pnp^1, \ldots, p^n. Thus, this will be positive if it's on the `positive nn-coordinate side' of the hyperplane, and negative for the other.

So if we have a large sequence of points S=(S1,,Sm)(R)mS = (S^1, \ldots, S^m) \in (\R^\infty)^m, then we can extract some coarse information about how they are shaped, as some data of type Πn<m.ΠK:Pn(S).Πq(SK).{dom,cod}\Pi n < m. \Pi K : \mathcal P_n(S). \Pi q \in (S \setminus K) . \{\dom, \cod\} That is, for any n<mn < m and any nn-sized subset p1,pnSp^1, \ldots p^n \subseteq S and some qSp1,,pnq \in S \setminus p^1, \ldots, p^n not already in that subset, we can project each of the p1,,pn,qp^1, \ldots, p^n, q from R\R^\infty down to just Rn\R^n by taking their first nn coordinates, and ask whether the sign of the expression is above is positive or negative. We interpret this as telling us whether the (n1)(n-1)-cell determined by p1,pnp^1, \ldots p^n is in the domain or codomain of the nn-cell determined by q,p1,pnq, p^1, \ldots p^n.

Question

For each mm, how many distinct such functions can actually be attained by a suitable choice of mm points SS? Call this ϕ(m)\phi(m).

I know ϕ(1)=1\phi(1) = 1 and ϕ(2)=2\phi(2) = 2 and ϕ(3)=12\phi(3) = 12, (because of the two triangle diagrams above, plus a factor of 3!3! for permutations of merely in which order the three vertices appear in SS) and I think ϕ(4)=576=4!226\phi(4) = 576 = 4! \cdot 2 \cdot 2 \cdot 6, because of a choice of permutation of the 4 vertices, a choice of how to flip all the 2-cells, a choice of how to flip the 3-cell, and the fact that 6 of the following 8 diagrams

seem 'consistent' with a downward-pointing 2-cell going 'out of the page' from the top 1-cell of each little diagram to the bottom 1-cell. Where 'consistency' means that there is a 'least' and 'greatest' 1-path. But I'm less confident about that.