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.

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 $h \mapsto h(1)$.

finitegroup 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 Yuan Jun 26 '18 at 23:19