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}
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1$\begingroup$ It's my understanding that the subscripts on these groups are merely labels to remind us of the objects they are acting on. So we write $SU(3)_C$ or $SU(3)_F$ depending on whether we're considering the group $SU(3)$ to be acting on the triplet of colour states of a quark, or the flavour triplet (up, down, strange). It's precisely the same group in both cases. Hence removing the labels is entirely legitimate. At least, I think. I would also say that I'm pretty sure the isomorphism is in fact $$ SU(2) \times U(1) = U(2) \times Z_2 $$ Perhaps somebody could explain why books often drop the $Z_2$? $\endgroup$– gj255Commented Mar 8, 2015 at 12:00
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$\begingroup$ Related math.SE question: math.stackexchange.com/q/1111766/11127 $\endgroup$– Qmechanic ♦Commented Mar 7, 2020 at 11:54
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$\begingroup$ @gj255 - it's not true that $\mathrm{SU}(2) \times \mathrm{U}(1)$ is isomorphic to $\mathrm{U}(2) \times \mathbb{Z}_2$. The easiest way to see this is to note that $\mathrm{SU}(2) \times \mathrm{U}(1)$ is connected, while $\mathrm{U}(2) \times \mathbb{Z}_2$ has two connected components. $\endgroup$– John BaezCommented Mar 18, 2021 at 19:40
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1 Answer
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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.
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 $k\in\mathbb{Z}$.
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!
answered Mar 8, 2015 at 11:27
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$\begingroup$ Notes for later: Define non-central(!) diagonal elements $D(\lambda):={\rm diag} (\underbrace{\lambda,~\ldots~, \lambda}_{n\text{ bosonic entries}}, \underbrace{\lambda^{-1},~\ldots~, \lambda^{-1}}_{m\text{ fermionic entries}} )$; Define semidirect product by $(\mu, g) \ltimes (\nu, h):= (\mu\nu, D(\nu)^{-1}gD(\nu) h)$; $\endgroup$– Qmechanic ♦Commented Feb 13, 2018 at 12:23
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$\begingroup$ Notes for later: $$U(p,q|m)~\cong~[U(1)\times SU(p,q|m)]/\mathbb{Z}_{|p+q-m|}\quad\text{if}\quad n\equiv p+q\neq m;$$ $$U(p,q|m)~\cong~[U(1)\ltimes SU(p,q|m)]/\mathbb{Z}_{p+q+m};$$ $$GL(n|m;\mathbb{F})~\cong~[\mathbb{F}^{\times} \times SL(n|m;\mathbb{F})]/\mathbb{Z}_{|n-m|}\quad\text{if}\quad n\neq m;$$ $$GL(n|m;\mathbb{F})~\cong~[\mathbb{F}^{\times} \ltimes SL(n|m;\mathbb{F})]/\mathbb{Z}_{n+m};$$ $\endgroup$– Qmechanic ♦Commented Feb 13, 2018 at 12:40
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$\begingroup$ Notes for later: $$U(p,q|m)~\ni~ g\quad\mapsto\quad $$ $$ \left(\sqrt[n+m]{{\rm sdet} g}, ~D\left(\sqrt[n+m]{{\rm sdet} g}\right)^{-1}g\right) \quad\sim\quad\left(\omega^k~\sqrt[n+m]{{\rm sdet} g}, ~D\left(\omega^k~\sqrt[n+m]{{\rm sdet} g}\right)^{-1}g\right) $$ $$ ~\in ~[U(1)\ltimes SU(p,q|m)]/\mathbb{Z}_{n+m};$$ $\endgroup$– Qmechanic ♦Commented Feb 14, 2018 at 14:17
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$\begingroup$ Similarly can I then write $$S(U(n) \times U(m))$$ $$ \backsimeq (U(1) \times SU(n) \times SU(m))/ \mathbb{Z}_{n+m}$$ ? $\endgroup$ Commented Oct 23, 2023 at 1:24