15
$\begingroup$

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}

$\endgroup$
3
  • 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 $Z_2$? $\endgroup$
    – gj255
    Commented Mar 8, 2015 at 12:00
  • $\begingroup$ Related math.SE question: math.stackexchange.com/q/1111766/11127 $\endgroup$
    – Qmechanic
    Commented Mar 7, 2020 at 11:54
  • $\begingroup$ @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 Baez
    Commented Mar 18, 2021 at 19:40

1 Answer 1

21
$\begingroup$
  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!

$\endgroup$
6
  • $\begingroup$ 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
  • $\begingroup$ 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
  • $\begingroup$ 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
  • $\begingroup$ That was very useful $\endgroup$
    – R. Rankin
    Commented Oct 19, 2023 at 0:56
  • $\begingroup$ Similarly can I then write $$S(U(n) \times U(m))$$ $$ \backsimeq (U(1) \times SU(n) \times SU(m))/ \mathbb{Z}_{n+m}$$ ? $\endgroup$
    – R. Rankin
    Commented Oct 23, 2023 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.