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Cheeger and Simons here defined a family of characteristic classes for principal $G$-bundles ($G$ a Lie group). I'm interested in the special case of $\hat c_2:H_3(SL(2,\mathbb C);\mathbb Z)\to\mathbb C/\mathbb Z$ (depending on normalization, the target might also be $\mathbb C/2\pi i\mathbb Z$). Here $H_3(SL(2,\mathbb C);\mathbb Z)$ means group homology where the group is given the discrete topology (i.e. group homology as an abstract group). Thought of another way, this map is basically calculating the "complex volume" $\operatorname{vol}+i\operatorname{cs}$ (volume of a hyperbolic 3-manifold plus $i$ times its Chern--Simons invariant).

The usual way to calculate this class (c.f. Neumann and Yang here, and Dupont and Zickert here) is via the pre-Bloch group and a certain dilogarithm formula. It is known that (up to 2-torsion or 3-torsion), the following sequence is exact: $$ 0\to H_3(SL(2,\mathbb C);\mathbb Z)\to\mathcal P(\mathbb C)\to\wedge^2\mathbb C^\times\to 0 $$ where $\mathcal P(\mathbb C)$ is the free group generated by $\mathbb C\setminus\{0,1\}$ modulo a "five-term" relation (which is related to dividing an ideal dipyramid in hyperbolic space into two ideal simplices and into three ideal simplices). The Rogers dilogarithm gives a function $R:\mathcal P(\mathbb C)\to\mathbb C$ explicitly as an analytic function of the generators $[z]$ ($R$ satisfies the five-term relation and so is defined on $\mathcal P(\mathbb C)$). Restricting $R$ to $H_3(SL(2,\mathbb C),\mathbb Z)$ gives exactly $\hat c_2$.

Now let us recall that the exact sequence above can be understood as follows. Let $G=SL(2,\mathbb C)$, and let $B$ be the Borel subgroup of upper triangular matrices. Then $G/B$ is $\mathbb CP^1$. We have an exact sequence of left $\mathbb Z[G]$-modules: $$ 0\to\ker\to\mathbb Z[G/B]\to\mathbb Z\to 0 $$ This gives rise to a long exact sequence of group homology over $G$: $$ \cdots\to H_\ast(G;\ker)\to H_\ast(G;\mathbb Z[G/B])\to H_\ast(G;\mathbb Z)\to\cdots $$ By Shapiro's Lemma, $H_\ast(G;\mathbb Z[G/B])=H_\ast(B;\mathbb Z)$, so perhaps this sequence is best written: $$ \cdots\to H_\ast(G;\ker)\to H_\ast(B;\mathbb Z)\to H_\ast(G;\mathbb Z)\to\cdots $$ Now (as per the answer to the author's earlier question), $H_\ast(G;\ker)$ can be thought of as the relative group homology $H_{\ast+1}(G,B;\mathbb Z)$. Thus our long exact sequence can be written: $$ \cdots\to H_\ast(B,\mathbb Z)\to H_\ast(G,\mathbb Z)\to H_\ast(G,B;\mathbb Z)\to\cdots $$ Looking around the $\ast=3$ part, we get exactly the presentation of $H_3(SL(2,\mathbb C);\mathbb Z)$ given above in terms of the pre-Bloch group.

Now, one way of interpreting this is to say that $\hat c_2$ descends to a map on $H_3(G,B;\mathbb Z)$, and furthermore, this map is computable (in terms of the Rogers dilogarithm). Is there a nice explanation for this fact? I guess maybe it's easy to see that $\hat c_2$ annihlates the image of $H_3(B;\mathbb Z)$ (this just involves analyzing the restriction to $BB$ of the tautological principal $G$ bundle over $BG$), but why should this exact sequence give such a nice presentation of $H_3(G;\mathbb Z)$, and why does it give essentially the only known way to calculate $\hat c_2$ of a class in $H_3(SL(2,\mathbb C);\mathbb Z)$?

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The way I view this is that $H_3(SL(2,\mathbb{C});\mathbb{Z})$ can be realized by maps of 3-manifolds into a classifying space for $SL(2,\mathbb{C})$ thought of as a discrete group (any 3-dimensional homology class may be represented by a map of a 3-manifold). This group, of course, acts on $\mathbb{H}^3$, and you obtain an induced representation of the 3-manifold group. Triangulate the 3-manifold, then map the universal cover of this triangulation equivariantly to $\mathbb{H}^3$, and straighten the tetrahedra so that they have geodesic edges. Then you can compute the hyperbolic volume by summing the signed volume of these tetrahedra (the Chern-Simons invariant is a bit more subtle). Now, "spin" the vertex to infinity in hyperbolic space, and you get ideal tetrahedra. Thus, the homology class is realized by a signed sum of ideal hyperbolic tetrahedra, which gives an element in the Bloch group. These spun ideal triangulations go back to Thurston.

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There are only countably many compact $3$-manifolds [or more precisely, there are only countably many pairs consisting of a compact $3$-manifold and a connected component of $\operatorname{Hom}(\pi_1(M),SL(2,\mathbb C))$], so your statement "any 3-dimensional homology class may be represented by a map of a 3-manifold" would imply that $H_3(SL(2,\mathbb C);\mathbb Z)$ is countable. This is known as the Bloch Rigidity Conjecture, which I thought was still unsolved. –  John Pardon Aug 12 '11 at 0:34
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@ unknown: in fact, the issue with the Bloch conjecture, is that the same 3-manifold may represent many different homology classes, by deforming the representation. If you could prove that the homology classes of curves of representations give the same homology class, then this would imply the conjecture. –  Ian Agol Aug 12 '11 at 5:13

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