Why does the Cheeger--Chern--Simons class descend to H_3(G/B)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:37:00Z http://mathoverflow.net/feeds/question/72650 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72650/why-does-the-cheeger-chern-simons-class-descend-to-h-3g-b Why does the Cheeger--Chern--Simons class descend to H_3(G/B)? unknown (google) 2011-08-11T04:05:23Z 2011-08-11T05:50:46Z <p>Cheeger and Simons <a href="http://www.ams.org/mathscinet-getitem?mr=827262" rel="nofollow">here</a> 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).</p> <p>The usual way to calculate this class (c.f. Neumann and Yang <a href="http://www.ams.org/mathscinet-getitem?mr=1663915" rel="nofollow">here</a>, and Dupont and Zickert <a href="http://www.ams.org/mathscinet-getitem?mr=2255500" rel="nofollow">here</a>) 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 <code>$\mathbb C\setminus\{0,1\}$</code> 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 $\hat c_2$.</p> <p>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 <a href="http://mathoverflow.net/questions/66372/" rel="nofollow">question</a>), $H_\ast(G;\ker)$ 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.</p> <p>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 $H_3(SL(2,\mathbb C);\mathbb Z)$?</p> http://mathoverflow.net/questions/72650/why-does-the-cheeger-chern-simons-class-descend-to-h-3g-b/72656#72656 Answer by Agol for Why does the Cheeger--Chern--Simons class descend to H_3(G/B)? Agol 2011-08-11T05:50:46Z 2011-08-11T05:50:46Z <p>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. </p>