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Let $G \leq \mathrm{PGL}(2,\mathbb{C})$ be the subgroup of upper-triangular matrices. I am interested in the natural morphism on the Schur multiplier (i.e. group homology as discrete groups)

$H_{2}(G,\mathbb{Z}) \longrightarrow H_{2}(\mathrm{PGL}(2,\mathbb{C}),\mathbb{Z})$

It seems to me that this morphism is surjective but I am not able to prove it. I know that $G \cong \mathbb{C} \rtimes \mathbb{C}^{*}$ with the natural multiplication action. Using the Lyndon–Hochschild–Serre spectral sequence on $G$ I can compute the spectral sequence as follows: $$ \begin{matrix} q=2 & 0 & & &\\ q=1 & 0 & H_{1}(\mathbb{C}^{*},\mathbb{C}) & &\\ q=0 & \mathbb{Z} & \mathbb{C}^{*} & \mathbb{C}^{*} \wedge_{\mathbb{Z}} \mathbb{C}^{*} & (\bigwedge_{\mathbb{Z}}^{3}\mathbb{C}^{*})\oplus \mathbb{Q}/\mathbb{Z}\\ & p=0 & p=1 & p=2 & p=3 \end{matrix} $$ with the natural multiplication action. It's well known that $H_{2}(\mathrm{PGL}(2,\mathbb{C}),\mathbb{Z})$ is $K_{2}(\mathbb{C})\oplus \mathbb{Z}/2$ but I have little knowledge on this group.

EDIT: As pointed out by Dave Benson in comment my original computation of $H_{2}(G,\mathbb{Z})$ is wrong.

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    $\begingroup$ Are you computing homology as a discrete group or as a Lie group via the classifying space? They're very different. For example, what do you think $H_2(\mathbb{R},\mathbb{Z})$ is? $\endgroup$ Commented May 16 at 7:28
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    $\begingroup$ @DaveBenson They are discrete groups. I think otherwise they will not be called Schur multiplier? $\endgroup$
    – hyyyyy
    Commented May 16 at 13:32
  • $\begingroup$ I edited the question for clarity. $\endgroup$
    – hyyyyy
    Commented May 16 at 13:59
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    $\begingroup$ Then I don't understand your computation of $H_2(G,\mathbb{Z})$. Could you please explain. $\endgroup$ Commented May 16 at 17:02
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    $\begingroup$ I don't see why that makes $E_{2,0}$ and $E_{0,2}$ equal to zero. For example, isn't $H_2(\mathbb{C},\mathbb{Q})$ equal to $\Lambda^2_{\mathbb{Q}}H_1(\mathbb{C},\mathbb{Q})$, hence infinite dimensional? So with $\mathbb{Z}$ coefficients it won't be zero. $\endgroup$ Commented May 16 at 17:57

1 Answer 1

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This answer has been edited so that it not only reads better but also gives a better answer. I'll refer to Dupont and Sah, "Scissors congruences, II" as [DS] and the paper of Sah and Wagoner, "Second cohomology of Lie groups made discrete" as [SW] in what follows.

Let $F$ be an algebraically closed field of characteristic zero; the question asks about $F=\mathbb{C}$ but the answer is true in at least this generality. Then $PGL(2,F)\cong PSL(2,F)$ sits in an exact sequence $$1 \to \mathbb{Z}/2 \to SL(2,F) \to PGL(2,F) \to 1.$$ So $K_2(F)=H_2(SL(2,F),\mathbb{Z})$ and $H_2(PGL(2,F),\mathbb{Z})\cong K_2(F)\oplus \mathbb{Z}/2$ see [DS (A5)] and [SW].

Let $G$ be the upper triangular matrices in $PGL(2,F)$ and let $U$ be the subgroup of $G$ consisting of upper triangular matrices with ones on the diagonal. So there is a split exact sequence $1 \to U \to G \to F^\times \to 1$, and $U$ is isomorphic to the additive group of $F$.

It is shown in [DS (A11)] that the inclusion $F^\times \to G$ and the quotient map $G \to F^\times$ induce an isomorphism $H_*(F^\times,\mathbb{Z}) \cong H_*(G,\mathbb{Z})$. In particular, it follows that $H_1(F^\times,U)=0$ (in your case, this is $H_1(\mathbb{C}^\times,\mathbb{C})=0$). Thus $H_*(G,\mathbb{Z})\cong \Lambda^*_{\mathbb{Z}}F^\times$, the exterior algebra on $F^\times$ sitting in degree one, plus a copy of $\mathbb{Q}/\mathbb{Z}$ in each odd degree from $3$ onwards coming from the roots of unity [DS, (A10)].

Now Matsumoto's theorem says that $K_2(F)$ is the quotient of $F^\times \otimes_{\mathbb{Z}}F^\times$ by the subgroup generated by elements of the form $a \otimes (1-a)$. These relations imply that $u\otimes v$ is equivalent to $-v\otimes u$, so $K_2(F)$ is really a quotient of $\Lambda^2_{\mathbb{Z}}F^\times$. So there is a surjective symmetrising map from $\Lambda^2_{\mathbb{Z}} F^\times$ to $K_2(F)$ sending $u\wedge v$ to the Steinberg symbol $\{u, v\}$ represented by $u\otimes v$. Looking at [DS (A2)], this is the map in the Bloch-Wigner theorem. In particular, it comes from the inclusion of the split torus into $SL(2,F)$, see [DS, top of p189], so this is your map.

The conclusion is that your map surjects onto $K_2(F)$, but that $H_2(PGL(2,F),\mathbb{Z})$ contains an extra factor of two. Since $\Lambda^2_{\mathbb{Z}}F^\times$ is a $\mathbb{Q}$-vector space, I don't think the extra factor of two can be hit by $H_2(G,\mathbb{Z})$, but I admit that this confuses me.

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  • $\begingroup$ Thank you! That's exactly what I need. I think we only have $E_{3,0} \rightarrow E_{1,1}$ surjective? It seems to me that $E_{1,1}$ is not zero from computation. $\endgroup$
    – hyyyyy
    Commented May 17 at 18:09
  • $\begingroup$ Well, but the map out of $E_{3,0}$ has to be zero in order for the map $H_*(G,\mathbb{Z})$ to $H_*(F^\times,\mathbb{Z})$ to be an iso on $H_3$ doesn't it? $\endgroup$ Commented May 17 at 18:15

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