This is the rest of the proof that the structure graph of a disc diagram bounded by a geodesic
triangle has one of the two shapes I described in class. The starting point is the diagram
E,
whose properties are recalled below.
The proof is a bit simpler than the one in class in fact, one doesn't need to check explicitly
that there are no internal cells in all of
E,
1
although this follows from the proof .
This proof is not examinable in its entirety it's too long but the techniques for
analysing disc diagrams are: applying the ladder theorem and Greendlinger theorem, doing
curvature computations and using Gauss-Bonnet.
1. The setting
X
is a
C 0 (1/6)
complex,
e
E → X
is a reduced disk diagram with the following properties
(established earlier):
• ∂p E is immersed;
• ∂p E = α10 α20 α30 , where each αi0 is a geodesic;
• every 0cell of E has curvature ≤ 0;
• every 2cell of E has curvature ≤ 0, with 3 exceptions:
• there are 3 shells R1 , R2 , R3 in E , with each ∂p Ri = Ii Oi , where Ii is internal (no 1cell
in common with ∂p E and Oi is a subpath of ∂p E . Moreover, |Ii | < |Oi |, as provided by
the Greendlinger theorem, so in particular Oi is not a subpath of α1 , α2 , or α3 and thus
contains the transition vertex between two of them.
We ruled out, earlier in the proof, the possibility that
E
is a ladder that led to the desired
conclusion. The situation is shown in Figure 1.
Figure 1. The diagram
2cells
that can surround
E,
R.
the internal
2cell R,
2. Structure of
Earlier, we saw that for each
2cell C
of
E
and some dierent types of
E
and each
j ∈ {1, 2, 3},
either
∂p C
disjoint, or their intersection in a single common subpath. So we can classify the
according to how many of the
1You'll
αj
and αj are
2cells of E
they intersect (and thus share exactly one subpath with).
see that this is checked, with much more ease, in a certain subdiagram
1
2
2.1. The case where there is a cell intersecting each
0
0
0
that intersects each of α1 , α2 , α3 , i.e.
0
the αj .
In particular,
i ∈ {1, 2, 3},
E−C
∂p E
has three components, one containing each of
there is a subdiagram
Ei
of
E
Figure 2. The case where a
C
is a
2cell
of
E
R1 , R2 , R3 .
For each
obtained from the closure of the complementary
E containing Ri by adding back C .
E1 ∪ E2 = E2 ∪ E3 = E3 ∪ E1 = C .
∂p E .
Suppose that
has a common (possibly trivial) subpath with each of
component of
and
αj0 .
2cell C
See Figure 2. Observe that
E = E1 ∪ E2 ∪ E3 ,
intersects each constituent geodesic of
The green path is the boundary path of the subdiagram
E1 .
We now prove:
If E has a 2cell intersecting each of α10 , α20 , α30 , then its structure graph is a
subdivided tripod.
Lemma 2.1.
Proof. The above discussion shows
E1 ; the same proof works for
that it suces to show that each
C0
Ei
is a ladder. We'll do
E1 with short
O be the outer path of C 0 and I the
0
0
0
0
inner path. Then either O is a subpath of α1 or α2 , or C = C , or C = R1 . But since |O| > |I|,
0
0
we can't have O a subpath of αj , because then we could shorten αj by replacing O by I . Hence
there are at most 2 shells with short inner paths, and hence ≤ 2 positively-curved shells. Thus,
by Greendlinger, E1 is a ladder. This completes the proof.
it for
the other two. Suppose that
inner paths (with respect to the boundary path of
E1 ).
is a shell in
Let
3 case. Now suppose that E has no 2cells that intersect every αj0 . Since
E is not a ladder, so R1 , R2 , R3 all exist, for each pair {i, j} of distinct elements of {1, 2, 3},
0
0
there is a 2cell Mij whose boundary path intersects αi , αj . Let Mij be chosen to be innermost
in the sense we described in class. Note that we could have e.g. M13 = R1 , but the Mij are all
2.2. Remaining case
distinct. See Figure 3.
Each
Mij
has a complementary component containing
Rk
k ∈ {1, 2, 3} − {i, j}, and the
00
subdiagram E . We thus have
for
Mij , is a
E = E 00 ∪ E1 ∪ E2 ∪ E3 , where:
• each Ek is, by the exact same argument as in the proof of Lemma 2.1, a ladder;
• E 00 ∩ E1 = M13 , E 00 ∩ E2 = M12 , E 00 ∩ E3 = M13 ;
• E1 − M13 is disjoint from E2 , E3 and same with indices permuted. See Figure 3.
0
Now, we can write ∂p M13 = O13 S1 I13 S3 , where Sj is a subpath of αj and O13 is a part of the
00
boundary path of E1 , and I13 is the innerpath of M13 within the subdiagram E .
intersection of these components, together with the
that
3
Figure 3. Finding
Hence
I13 ,
I13
curvature
I13
The green-bounded subdiagrams are the ladders
Ei .
2cell N whose boundary path
α30 , contradicting that M13 is innermost.
contains at least 1 piece-transition, so that κ(M13 ) ≤ 2π/3 (here we are computing
00
00
in the diagram E , using the boundary path of E ). Similarly, κ(M23 ), κ(M12 ) ≤
Suppose that
contains
E 00 .
is a single piece. Then there would be some
and thus
α10
N
would intersect both
E 00
cannot have any more shells with short inner paths, since such a shell
and
2π/3.
On the other hand,
would have outer path on a geodesic, leading to an impossible shortening. Thus (see curvature
only
E 00
are Mij , and their
at most 2π . But combinatorial Gauss-Bonnet wants the sum of the curvatures
E 00 contains no negatively-curved cells !
computation in Greendlinger theorem), the
positively-curved cells in
curvatures sum to
to be
2π ,
so
Some observations:
(1) Each
Mij
has exactly one piece-transition on its inner path: if it had
2
or more, the
≤ π/3, not enough to satisfy combinatorial Gauss-Bonnet.
2cells, since these have curvature ≤ −π/3 by C 0 (1/6) (we did
00
this computation in a lecture). By construction/innermostness of the Mij , E has no
0
00
2cell intersecting more than one of the αj , so every 2cell of E is a shell.
00
Each shell in E other than Mij has exactly 4 pieces in its inner path: any fewer, and
curvature would be
(2)
(3)
E 00
contains no internal
it would give an illegal shortening, and any more, and it would contribute impossible
negative curvature.
(4) There is no degree
> 4
internal vertex in
E 00 ,
since it would contribute impossible
negative curvature.
Let P be a shortest (hence
v12 . The subdiagram bounded by P , M13 , M23 ,
00
and the subtended part of the boundary path of E is a ladder, by Greendlinger. There are
two other ladders of this type; see Figure 4 at left. The subdiagram formed by P and the other
00
two paths joining the vij must have area 0, since any 2cell inside it would be internal in E ,
0
00
which was ruled out above. Letting P, P , P denote the sides of this tripod, we see that, say,
P is a concatenation of P ∩ P 0 and P ∩ P 00 . Now, as shown in Figure 4, we can't identify the
0
0
piece-transitions in P with those in P , else they wouldn't be piece-transitions, unless we have a
vertex which has degree 3 and lies in three dierent cells, serving as a piece-transition in each one.
Let
vij
be the vertex of
Iij
where the piece-transition occurs.
embedded) combinatorial path in
E 00
from
v13
to
There is at most one such vertex (called the centre), and, as shown in Figure 4, there can't be
more than
3 2cells
there, for otherwise we would see
> 4,
contradicting the above observation.
On the other hand, if there is a centre, then, as shown, the other
2cells
can't be staggered in
the appropriate way there would be disallowed identications of piece-transitions.
4
E 00 , and the constituent ladders. The triangular subdiagram
0 00
bounded by P P P , and has area 0. The coloured 2cells are
Figure 4. At left is
in the middle is
shells in the three ladders, and the coloured vertices are piece-transitions that
must occur in their boundary paths. In the middle are the possible ways the cells
come together along the tripod
P P 0 P 00 .
Cells that stay away from the centre
of the tripod are staggered the couloured piece-transition in one cannot
coincide with the coloured piece-transition in the other, since otherwise each
would lie in the middle of a single piece, and not be a transition. But, if there
are three cells at the centre, their piece-transitions are
at right, there are at most
So, there are
3
3
not
staggered. As shown
cells at the centre.
2cells (possible including some of the Mij ) and no non-central ones.
TE 00 is either a 3cycle or a 3cycle with a single edge dangling o each
central
It's now easy to see that
vertex. Various cases (two possible and the impossible ones) are shown in Figure 5.
3 central cells. There are
nonMij central cells there are.
Figure 5. The situation once we know there are
several cases to consider according to how many
But, in several cases, there is a problem: some cell has too few pieces in its inner
path. In each allowable case, the structure graph is as claimed.
Attaching the structure graphs of the
described in the lecture.
Ei
ladders shows that
TE
is of one of the two types I