Consider a projective representation of $\mathbb Z^2$ with $U(1)$ coefficients. I would like to find the covering group corresponding to this representation. For this, one needs to find the appropriate central extension by $\mathbb Z^2$ of the Schur multiplier $H^2(\mathbb Z^2, C^\times)$. I have some confusion in calculating this object. If I use the definition of the Schur multiplier given in terms of the homology group $H_2(\mathbb Z^2,\mathbb Z)$, I obtain the group $\mathbb Z$. This is in contradiction to the result $H^2(\mathbb Z^2, C^\times) = U(1)$ that I have seen quoted elsewhere, and which I expect to obtain on certain physical grounds (there is a separate physical interpretation of this problem). I would like to know which result is correct.
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$\begingroup$ People use "Schur multiplier" to refer to both $H_2(G, \mathbb{Z})$ and $H^2(G, \mathbb{C}^{\times})$ for $G$ a finite group because they are (noncanonically) isomorphic in that case, which they aren't in general. Also, the universal central extension is an extension of $G$ by $H_2(G, \mathbb{Z})$ and it only exists in general if $G$ is perfect, which $\mathbb{Z}^2$ isn't. $\endgroup$– Qiaochu YuanCommented Jun 26, 2018 at 23:19
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$\begingroup$ @QiaochuYuan Thanks. I didn't know the finiteness condition. I am trying to interpret 'central extension' as the group of symmetries on a two-dimensional lattice in the presence of a magnetic field (for a fixed field, the translation operators determine a projective representation of $\mathbb Z^2$). So different extensions would correspond to different magnetic fields, and I would not expect there to be a universal central extension for this problem. $\endgroup$– Naren ManjunathCommented Jun 27, 2018 at 7:29
1 Answer
I think that we have
$$H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z}) \simeq \mathbf{Z}, \quad H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})\simeq \mathbf{C}^{\times}.$$
In general, by the Universal Coefficient Theorem there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$
The isomorphism $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}) \simeq \mathbf{C}^{\times}$ is explicitly given by associating to every element $h \in \mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times})$ its value in $1$, namely $h \mapsto h(1)$.
