PENTAGONS
N. P. STRICKLAND
1. Introduction
The Stasheff operad involves certain spaces K(n). A construction due to Loday embeds K(n) as a convex
polytope of dimension n − 2 in Rn−1 . In particular, K(2) is a single point. Each face of K(n) admits a
natural piecewise-linear identification with K(i) × K(j) for some i and j with i + j = n + 1. In particular,
it can happen that j = 2, so some faces of K(n) are piecewise-linearly identified with K(n − 1). It is
interesting to ask whether these identifications can be improved to be globally linear, either with the Loday
embedding or some other embedding. We can also weaken this slightly. The Loday polytopes do not contain
the origin, so we can project onto the sphere S n−2 . It is convenient here to regard S n−2 as the space of
rays in Rn−1 , which gives us an action of GLn−1 (R)/R+ and allows us to talk about projective geometry.
In this note we will consider a certain face F of K(5), which is combinatorially identified with K(4) (and so
is combinatorially a pentagon). We will classify pentagons up to projective equivalence, and thus show that
the Loday embeddings of F and K(4) are not equivalent in this sense.
2. Classification
A pentagon means a system (V ; (Li )i∈Z/5 ) where
• V is a 3-dimensional vector space
• Each Li is a ray in V
• For all i, the rays Li+2 , Li+3 and Li+4 lie on the same side of the plane spanned by Li and Li+1 ,
and do not meet that plane.
We write P for the category of pentagons. We want to understand π0 P. (It is easy to see that for any
pentagon P we have Aut(P ) = R+ .)
Put
u0 = (1, 1, 1)
u1 = (−1, 1, 1)
u2 = (−1, −1, 1)
u3 = (1, −1, 1).
Given x > 1 and −1 < y < 1 we also put
Q(x, y) = (R3 ; R+ .u0 , R+ .u1 , R+ .u3 , R+ .u4 , R+ .(x, y, 1)),
so Q(x, y) ∈ P.
Lemma 2.1. If Q(u, v) ' Q(x, y) then (u, v) = (x, y).
Proof. Let g : R3 → R3 carry Q(u, v) to Q(x, y). Then the vectors ui are all eigenvectors for g (with positive
eigenvalue). As any three of these four vectors are linearly independent, it follows that all the eigenvalues
are the same, say λ, so g is λ times the identity. We also have g(u, v, 1) ∈ R+ .(x, y, 1), and it follows that
(u, v) = (x, y)
Proposition 2.2. For any pentagon P = (V ; L), there exists a unique pair (x, y) such that P ' Q(x, y).
Proof. Let A be the plane spanned by L0 and L2 , and let B be the plane spanned by L1 and L3 . The
definitions imply that A 6= B, so the space M = A ∩ B is a line. Choose m ∈ M \ {0}. There are then unique
elements ai ∈ Li and i ∈ {±1} such that m = 0 a0 + 2 a2 = 1 a1 + 3 a3 . The conditions are supposed
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to imply that all the i have the same sign. After replacing m by −m if necessary, we may assume that
m = a0 + a2 = a1 + a3 . Now put e0 = (a0 − a1 )/2 and e1 = (a0 − a3 )/2 and e2 = m/2, so
a 0 = e0 + e1 + e2
a1 = −e0 + e1 + e2
a2 = −e0 − e1 + e2
a 3 = e0 − e1 + e2 .
Now choose a point a4 = xe0 + ye1 + ze2 ∈ L4 \ {0}. This must lie on the same side of R{a0 , a1 } as a3 , and
on the same side of R{a2 , a3 } as a0 . It follows from this that z > |y| ≥ 0. We may replace a4 by a4 /z and
thus assume that z = 1 and |y| < 1. Let C be the plane spanned by a4 and a0 , and let D be spanned by a4
and a3 . These must not separate the other ai ’s, and it follows that x > 1. Thus Q(x, y) is defined, and it is
clear that P ' Q(x, y).
Remark 2.3. If V = R3 , we can compute as follows. Choose any generator ui for Li , then put di =
ui+1 .(ui+2 ×ui+3 ) and ai = di ui , where indices are interpreted mod 4. It then turns out that a0 +a2 = a1 +a3
as above. I think that also di > 0, but I have not checked this.
3. Loday polytopes
Example 3.1. Let K(4) be the Loday model of the 4-point associahedron. This has vertices
u0 = (1, 2, 3)
u1 = (2, 1, 3)
u2 = (3, 1, 2)
u3 = (3, 2, 1)
u4 = (1, 4, 1).
The corresponding vectors ai and ei are as follows:
a0 = (1, 2, 3)
a1 = (4, 2, 6)
a2 = (6, 2, 4)
a3 = (3, 2, 1)
a4 = (1/2, 2, 1/2) = 2e0 + e2
e0 = (−3/2, 0, −3/2)
e1 = (−1, 0, 1)
e2 = (7/2, 2, 7/2)
We thus have x = 2 and y = 0.
Example 3.2. The faces of K(5) are indexed by the integer intervals J ⊂ [0, 4] with |J| ≥ 2. The face F
corresponding to the interval {0, 1} has vertices as follows:
u0 = (1, 2, 3, 4)
u1 = (1, 4, 1, 4)
u2 = (1, 6, 1, 2)
u3 = (1, 6, 2, 1)
u4 = (1, 2, 6, 1).
These lie in the 3-dimensional space
V = {(w, x, y, z) ∈ R4 | 9w = x + y + z}.
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The corresponding vectors ai and ei are as follows:
a0 = (1, 2, 3, 4)
a1 = (2, 8, 2, 8)
a2 = (3, 18, 3, 6)
a3 = (2, 12, 4, 2)
a4 = (2/3, 4/3, 4, 2/3) = 7e0 /3 + e1 /3 + e2
e0 = (−1/2, −3, 1/2, −2)
e1 = (−1/2, −5, −1/2, 1)
e2 = (2, 10, 3, 5).
We thus have x = 7/3 and y = 1/3. There is a natural combinatorial equivalence between F and K(4),
under which the vertices ui in Example 3.1 correspond to the vertices ui in this example. Our calculation of
x and y implies that this cannot be improved to a projective equivalence between F and K(4) as cones.
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