Is $SU(2)\times U(1) = U(2)$?

Question

In the many textbook of the Standard Model, I encounter the relation \begin{align} SU(2)_L \times U(1)_L = U(2)_L. \end{align} Here the subscript $L$ means the left-handness (i.e., the chirality of the fermions). Is the relation above true in the general case? That is, is \begin{align} SU(2) \times U(1) = U(2)\ ? \end{align}

Answer 1

1. The relevant Lie group isomorphism reads

   $$\begin{align} U(2)~\cong~&[U(1)\times SU(2)]/\mathbb{Z}_2, \cr Z(SU(2))~\cong~&\mathbb{Z}_2.\end{align}\tag{1a} $$

   In detail, the Lie group isomorphism (1a) is given by $$U(2)~\ni~ g\quad\mapsto\quad $$ $$ \left(\sqrt{\det g}, ~\frac{g}{\sqrt{\det g}}\right) ~\sim~ \left(-\sqrt{\det g}, ~-\frac{g}{\sqrt{\det g}}\right)$$ $$~\in ~[U(1)\times SU(2)]/\mathbb{Z}_2.\tag{1b}$$ Here the $\sim$ symbol denotes a $\mathbb{Z}_2$-equivalence relation. The $\mathbb{Z}_2$-action resolves the ambiguity in the definition of the double-valued square root.

2. It seems natural to mention that the Lie group isomorphism (1a) generalizes in a straightforward manner to generic (indefinite) unitary (super) groups

   $$\begin{align} U(p,q|m)~\cong~&[U(1)\times SU(p,q|m)]/\mathbb{Z}_{|n-m|}, \cr Z( SU(p,q|m))~\cong~&\mathbb{Z}_{|n-m|},\end{align}\tag{2a}$$

   where $$\begin{align} p,q,m~\in~& \mathbb{N}_0, \cr n~\equiv~p+q~\neq~&m, \cr n+m~\geq ~& 1,\end{align}\tag{2b}$$ are three integers. Note that the number $n$ of bosonic dimensions is assumed to be different from the number $m$ of fermionic dimensions. In detail, the Lie group isomorphism (2a) is given by $$U(p,q|m)~\ni~ g\quad\mapsto\quad $$ $$ \left(\sqrt[|n-m|]{{\rm sdet} g}, ~\frac{g}{\sqrt[|n-m|]{{\rm sdet} g}}\right) ~\sim~ \left(\omega^k~\sqrt[|n-m|]{{\rm sdet} g}, ~\frac{g}{\omega^k~\sqrt[|n-m|]{{\rm sdet} g}}\right)$$ $$ ~\in ~[U(1)\times SU(p,q|m)]/\mathbb{Z}_{|n-m|},\tag{2c}$$ where $$\omega~:=~\exp\left(\frac{2\pi i}{|n\!-\!m|}\right)\tag{2d}$$ is a $|n\!-\!m|$'th root of unity, and $k\in\mathbb{Z}$.

3. Interestingly, in the case with the same number of bosonic and fermionic dimensions $n=m$, the center $$ Z( SU(p,q|m))~\cong~U(1) \tag{3a}$$ becomes continuous! I.e. the $U(1)$-center of $U(p,q|m)$ has moved inside $SU(p,q|m)$, and formula (2a) no longer holds!