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:

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

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

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.