Recently I was trying to understand what’s behind the mysterious condition

\sum_i [z_i]\wedge[1-z_i] = 0\in \Lambda^2 \mathbb C^\times

for elements z_i\in \mathbb C (z_i\neq 0, 1) to define an element \sum_i[z_i] in the Bloch group. It appears that the condition naturally appears if one studies hyperbolic 3-manifolds.

Let A be an abelian group. Let X be a triangulated oriented 3-dimensional manifold. Let \Delta=(a,b,c,d) be an oriented 3-dimensional simplex of the triangulation.

Definition
An angle structure on \Delta is a collection of 12 elements of A , denoted k_{xy} , one for each edge xy satisfying relations:

k_{xy}=k_{yx},\; \sum_{y} k_{xy}=0 \;\text{for any vertex}\; x.

We think of k_{xy} as the angle between the faces of \Delta meeting at xy.

We have

k_{ab}+k_{ac}+k_{ad}=0,\; k_{ab}+k_{bc}+k_{bd}=0,\\ \indent k_{ac}+k_{bc}+k_{cd}=0,\; k_{ad}+k_{bd}+k_{cd}=0.

This, in particular, implies

2(k_{ab}-k_{cd})=2(k_{ac}-k_{bd})=2(k_{ad}-k_{bc})=0.

Suppose each tetrahedron of the triangulation has an angle structure. The corresponding angles will be denoted k_{ab}^{cd}\in A , which means the angle between abc and abd, i.e. for a simplex \Delta=(a,b,c,d) we have 12 elements listed below:

k_{ab}^{cd}=k_{ba}^{dc}, k_{ac}^{db}=k_{ca}^{bd}, k_{ad}^{bc}=k_{da}^{cb}, k_{bc}^{ad}=k_{cb}^{da}, k_{bd}^{ca}=k_{db}^{ac}, k_{cd}^{ab}=k_{dc}^{ba}.

Let us define additionally

k_{ab}^{dc}=k_{ba}^{cd}=-k_{ab}^{cd}, \qquad \text{etc.}

Therefore the notation is invariant with respect to even permutations of vertices of \Delta.

Definition
For each edge ab of X let (ab x_i x_{i+1}) be the 3-simplices adjacent to ab with i=0,1,\ldots,n-1 , and x_n=x_0. Suppose the following condition is satisfied:

\sum_i k_{ab}^{x_i x_{i+1}}=0.

Then we call our manifold angled.

Suppose we have a finite set of tetrahedra with angle structure. Is it possible to glue them together and obtain an angled oriented manifold? We provide a necessary condition.

Theorem
In an angled oriented manifold the following condition is satisfied in the group \Lambda^2 A:

\sum_{abcd}(4 k_{ab}^{cd}\wedge k_{ac}^{db}+k_{ad}^{bc}\wedge k_{ad}^{bc}+k_{bc}^{ad}\wedge k_{bc}^{ad})=0,

the sum is over the 3-simplices which compose the fundamental class of the manifold.

Note that if 2 is invertible in A , then the terms k_{ad}^{bc}\wedge k_{ad}^{bc} and k_{bc}^{ad}\wedge k_{bc}^{ad} are {0}.

Remark
Let A={\mathbb C}^\times and each tetrahedron is realized as an ideal tetrahedron in the hyperbolic 3-space \mathfrak H with cross-ratio z. Then its angles are z, \frac{1}{1-z}, 1-\frac{1}{z}. To make the product 1 one can change angles to -z, \frac{1}{z-1}, \frac{1}{z}-1. Then we see that up to 2-torsion the sum of tensors z\wedge (1-z) is {0}, which is well known. In other words, hyperbolic 3-manifolds provide elements in the Bloch group. However our approach seems to be more general.

The rest of this text provides a proof of the theorem.

Suppose X is angled. Then we can construct elements h_{ab}^c\in A with the property

k_{ab}^{cd} = h_{ab}^d-h_{ab}^c

for each oriented simplex (a,b,c,d). If h'^c_{ab} is another such family then there is a family q_{ab} with

h'^c_{ab}-h_{ab}^c = q_{ab}.

Fix a vertex a. Let ab and ab' be edges. Join ab and ab' by a sequence of triangles (a b_i b_{i+1}) for i=0,1,\ldots,n-1 , b_0=b , b_n=b'. Put

m_a(b,b') = \sum_{i} (h_{a b_i}^{b_{i+1}} - h_{a b_{i+1}}^{b_i}).

This does not depend on the choice of the sequence since for any oriented 3-simplex (a b c d) we have

(h_{a b}^{c} - h_{a c}^b) + (h_{ac}^d-h_{ad}^c) + (h_{ad}^b-h_{ab}^d)=-k_{ab}^{cd}-k_{ac}^{db}-k_{ad}^{bc}=0.

Therefore there is a family q_{ab}\in A with the property

m_a(b,b')=q_{ab'}-q_{ab}.

If q'_{ab} is another such family, there exists a family p_a\in A with

q'_{ab}-q_{ab}=p_a.

In particular for any 2-simplex (abc) we have

h_{a b}^c-h_{ac}^b = q_{ac}-q_{ab}.

We see that we can replace h_{a b}^c with h_{a b}^c+q_{ab} to make h satisfying

h_{ab}^c=h_{ac}^b\;\text{for any}\; 2-\text{simplex}\; (abc).

If h'^c_{ab} is another family with this condition then there is a family p_a\in A with the property

h'^c_{ab}-h^c_{ab}=p_a.

Put h'^c_{ab}=-h^c_{ba}. Then for any oriented 3-simplex (abcd)

k_{ab}^{cd} = k_{ba}^{dc}=h_{ba}^c-h_{ba}^d=h'^d_{ab}-h'^c_{ab}.

Therefore there exists a family q_{ab} (this q is different from the one used before) with property

h^c_{ab}-h'^c_{ab} = q_{ab}.

This means that for any 2-simplex (abc) we have

h^c_{ab}+h^c_{ba} = q_{ab}.

Let us summarize the properties of h^c_{ab}:

h_{ab}^c=h_{ac}^b,\;h^c_{ab}+h^c_{ba} = q_{ab},\;k_{ab}^{cd} = h_{ab}^d-h_{ab}^c.

Let (abc) be a 2-simplex. Then

q_{ab}+q_{ac}-q_{bc}=h_{ab}^c+h_{ba}^c+h_{ac}^b+h_{ca}^b-h_{bc}^a-h_{cb}^a = 2 h_{ab}^c.

Consider the following element in \Lambda^2 A:

\phi_{abc}=h_{ab}^c\wedge h_{bc}^a + h_{bc}^a\wedge h_{ca}^b + h_{ca}^b\wedge h_{ab}^c.

This element is invariant under cyclic permutations:

\phi_{abc}=\phi_{bca}=\phi_{cab}.

Moreover,

\phi_{acb}=h_{ac}^b\wedge h_{cb}^a + h_{cb}^a\wedge h_{ba}^c + h_{ba}^c\wedge h_{ac}^b \\ \indent =h_{ab}^c\wedge h_{ca}^b + h_{ca}^b\wedge h_{bc}^a + h_{bc}^a\wedge h_{ab}^c=-\phi_{abc}.

Since h_{bc}^a=q_{ab}-h_{ab}^c and h_{ca}^b=q_{ac}-h_{ab}^c , we can also write \phi_{abc} as

\phi_{abc}=2 h_{ab}^c\wedge(q_{ab}-q_{ac}) +q_{ab}\wedge q_{ac} - h_{ab}^c\wedge h_{ab}^c \\ \indent =q_{ac}\wedge q_{ab} +q_{ab}\wedge q_{bc} +q_{bc}\wedge q_{ac} \\ \indent + q_{ab}\wedge q_{ab} + q_{ac}\wedge q_{ac} + h_{ab}^c\wedge h_{ab}^c\\ \indent =q_{ac}\wedge q_{ab} +q_{ab}\wedge q_{bc} +q_{bc}\wedge q_{ac} + q_{bc}\wedge q_{bc} + h_{ab}^c\wedge h_{ab}^c.

Let

\phi_{abc}^0=q_{ac}\wedge q_{ab} +q_{ab}\wedge q_{bc} +q_{bc}\wedge q_{ac},\\ \indent \phi_{abc}^1=q_{bc}\wedge q_{bc} + h_{ab}^c\wedge h_{ab}^c.

For any oriented 3-simplex (abcd) put

\psi^*_{abcd} =\phi^*_{bcd}-\phi^*_{acd}+\phi^*_{abd}-\phi^*_{abc}.

Then

\psi^0_{abcd}=q_{bc}\wedge q_{cd}+q_{cd}\wedge q_{bd}+q_{bd}\wedge q_{bc}-q_{ac}\wedge q_{cd}  -q_{cd}\wedge q_{ad}   \\ -q_{ad}\wedge q_{ac} + q_{ab}\wedge q_{bd} + q_{bd}\wedge q_{ad} + q_{ad}\wedge q_{ab} - q_{ab}\wedge q_{bc} \\ -q_{bc}\wedge q_{ac} - q_{ac}\wedge q_{ab} \\ \indent = q_{ab}\wedge (q_{bd}-q_{ad}-q_{bc}+q_{ac}) + q_{cd}\wedge (-q_{bc}+q_{bd}+q_{ac}-q_{ad}) \\ +(q_{bd}+q_{ac})\wedge(q_{bc}+q_{ad}) \\ \indent =(q_{bc}+q_{ad})\wedge(q_{ab}+q_{cd}) + (q_{ab}+q_{cd})\wedge(q_{bd}+q_{ac}) \\ + (q_{bd}+q_{ac})\wedge(q_{bc}+q_{ad}).

We may rewrite

q_{bc}+q_{ad}  =h_{bc}^a+h_{cb}^a+h_{ad}^c+h_{da}^c = h_{bc}^a+h_{dc}^a + q_{ac}-k_{ca}^{bd},\\ \indent q_{ab}+q_{cd} =h_{ab}^c+h_{ba}^c+h_{cd}^a+h_{dc}^a = h_{bc}^a+h_{dc}^a +q_{ac}+k_{ca}^{bd}.

Therefore

\psi^0_{abcd}=2 k_{ca}^{bd}\wedge(q_{bd}+q_{ac})-2 k_{ca}^{bd}\wedge(h_{bc}^a+h_{dc}^a + q_{ac}) \\ +k_{ca}^{bd}\wedge k_{ca}^{bd}+h_{bc}^a\wedge h_{bc}^a+h_{dc}^a\wedge h_{dc}^a+q_{ac}\wedge q_{ac}\\ \indent =2k_{ca}^{bd}\wedge(q_{bd}-h_{bc}^a-h_{dc}^a) + q_{bd}\wedge q_{bd} + q_{ac}\wedge q_{ac}.

Taking into account that

q_{bd}-h_{bc}^a-h_{dc}^a = h_{bd}^a+h_{db}^a-h_{bc}^a-h_{dc}^a =h_{ba}^d+h_{da}^b-h_{ba}^c-h_{da}^c \\ \indent =-k_{ba}^{dc}+k_{da}^{cb},

we obtain

\psi^0_{abcd}=2 k_{ca}^{bd}\wedge(k_{da}^{cb}-k_{ba}^{dc})+q_{bd}\wedge q_{bd} + q_{ac}\wedge q_{ac}\\ \indent =4 k_{ab}^{cd}\wedge k_{ac}^{db}+q_{bd}\wedge q_{bd} + q_{ac}\wedge q_{ac}.

Now we turn to \psi^1_{abcd}.

\psi^1_{abcd}=q_{cd}\wedge q_{cd} + h_{bc}^d\wedge h_{bc}^d + q_{cd}\wedge q_{cd}+h_{ac}^d\wedge h_{ac}^d \\ + q_{ab}\wedge q_{ab} + h_{da}^b\wedge h_{da}^b+q_{ab}\wedge q_{ab}+h_{ca}^b\wedge h_{ca}^b\\ \indent =q_{ad}\wedge q_{ad}+q_{bc}\wedge q_{bc} + k_{da}^{cb}\wedge k_{da}^{cb}+k_{cb}^{da}\wedge k_{cb}^{da}.

Therefore

\psi_{abcd}=4 k_{ab}^{cd}\wedge k_{ac}^{db}+k_{ad}^{bc}\wedge k_{ad}^{bc}+k_{bc}^{ad}\wedge k_{bc}^{ad}.

We see that \psi_{abcd} depends only on the angles and its sum over the manifold is zero. If 2 is invertible in A then

\frac{1}{4}\psi_{abcd}=k_{ab}\wedge k_{ac}.

This specializes to z\wedge(1-z) in the case of ideal hyperbolic tertrahedron.

Advertisements