
This is the systolic argument that underlies distance bounds for homological codes [18, 14].
What is particular to this lattice is the local constant, and the fact that it is attained exactly.
The local ingredient comes first.
Lemma 1 (How big a shadow can be). Let π
ij
be the map that keeps only the coordinates x
i
and x
j
. Every triangle f of the honeycomb has area(π
ij
(f)) ≤
1
2
, and every 2-cell f
∗
of the dual
24-cell honeycomb has area(π
ij
(f
∗
)) ≤
1
4
. Both values are reached.
Proof. A triangle is {v, v + r, v + s} where r, s and r − s are all roots, so its shadow is a triangle
of area
1
2
|r
i
s
j
− r
j
s
i
|. A root has exactly two nonzero entries, each ±1, so |r
i
s
j
− r
j
s
i
| ≤ 2. The
value 2 needs r and s to both live on the coordinates i and j with opposite sign patterns, say
r = (1, 1, 0, 0) and s = (1, −1, 0, 0). But then r − s and r + s each have a single entry ±2, and
neither is a root, so there is no such triangle. Hence |r
i
s
j
− r
j
s
i
| ≤ 1 and the area is at most
1
2
.
It is reached, for instance by r = (1, 1, 0, 0) and s = (1, 0, 1, 0) shadowed onto (x
1
, x
2
).
A 2-cell of the dual honeycomb is the triangle joining the centres of the 3 16-cells that contain
a given triangle (Section 2). Running through the 96 kinds of triangle and all 6 coordinate pairs
gives a largest shadow of
1
4
. The computation is the published script projection.py (Data
Availability Statement); it is exact arithmetic and takes a second.
Theorem 2 (Lower bound). For even L ≥ 4, every X-type logical operator has weight at least
2L
2
, and every Z-type logical operator has weight at least 4L
2
.
Proof. Let z be an X-type logical operator, that is, a sheet that wraps the torus in a way no
stabilizer can undo. Its wrapping class lives in H
2
(T
4
; F
2
), which has a basis labelled by the six
pairs of coordinates; choose a pair (i, j) appearing in the class of z, and write π for the shadow
map onto (x
i
, x
j
).
Suppose some point p of the cross-section were missed by the shadow. Then z would lie entirely
inside the part of the torus lying over the punctured cross-section. A punctured two-dimensional
torus carries no wrapping sheets at all, so the class of z would have to be trivial in the (i, j)
direction , contradicting the choice of (i, j). So the shadow covers every point of the cross-section.
Areas can only add up, so
X
f∈supp z
area
π(f)
≥ area
cross-section
= L
2
.
By Lemma 1 each term is at most
1
2
, so z contains at least 2L
2
triangles.
A Z-type logical operator is the same kind of object in the dual honeycomb, where the qubits
correspond one-to-one to dual 2-cells. The same argument with the constant
1
4
gives at least
4L
2
.
Theorem 3 (A logical operator of exactly this size). Let L be even and c even, and let
Π
c
= {x : x
1
+ x
2
+ x
3
≡ c mod L, x
4
= 0}.
Then Π
c
contains exactly L
2
lattice points, and the 2L
2
triangles they span are triangles of the
honeycomb and form an X-type logical operator. Hence d
X
≤ 2L
2
.
Proof. Because L is even, fixing x
1
+ x
2
+ x
3
modulo L also fixes it modulo 2; for c even all L
2
solutions have even coordinate sum and so lie in D
4
. The roots that stay inside Π
c
are those with
r
4
= 0 and r
1
+ r
2
+ r
3
= 0, of which there are six, so the points of Π
c
form a triangular lattice,
5