### Cosmic Strings in the Standard Model

Over at the n-Category Café, John Baez is making a big deal of the fact that the global form of the Standard Model gauge group is $G = (SU(3)\times SU(2)\times U(1))/N$ where $N$ is the $\mathbb{Z}_6$ subgroup of the center of $G'=SU(3)\times SU(2)\times U(1)$ generated by the element $\left(e^{2\pi i/3}\mathbb{1},-\mathbb{1},e^{2\pi i/6}\right)$.

The global form of the gauge group has various interesting topological effects. For instance, the fact that the center of the gauge group is $Z(G)= U(1)$, rather than $Z(G')=U(1)\times \mathbb{Z}_6$, determines the global 1-form symmetry of the theory. It also determines the presence or absence of various topological defects (in particular, cosmic strings). I pointed this out, but a proper explanation deserved a post of its own.

None of this is new. I’m pretty sure I spent a sunny afternoon in the summer of 1982 on the terrace of Café Pamplona doing this calculation. (As any incoming graduate student should do, I spent many a sunny afternoon at a café doing this and similar calculations.)

At low energies, $G$ is broken to the subgroup $H=U(3)$, where the embedding $i\colon H\hookrightarrow G$ is given as follows. Let $h\in H$ and let $d\coloneqq \det(h)$. Choose a 6th root $b = d^{1/6}$ Then

The ambiguity in defining $b$ leads precisely to an ambiguity in $i(h)$ by multiplication by an element of $N$. Thus (1) is ill-defined as a map to $G'$, but well-defined as a map to $G$.

The (would-be) cosmic strings associated to the breaking of $G$ to $H$ are classified by $\pi_1(G/H)$. Both $\pi_1(H)$ and $\pi_1(G)$ are equal to $\mathbb{Z}$. The long-exact sequence in homotopy yields $0\to \pi_1(H)\to \pi_1(G) \to \pi_1(G/H)\to 0$ So what we need to do is compute the image of the generator of $\pi_1(H)$ in $\pi_1(G)$. If the image is $n$ times the generator of $\pi_1(G)$, then the quotient is nontrivial and we have $\mathbb{Z}_n$ cosmic strings.

$\pi_1(G)$ is generated by the (homotopy class of) the loop

$\pi_1(H)$ is generated by the loop

Plugging (3) into (1), we see that $i(h(s))=g(s)$. Hence $\pi_1(G/H)=0$ and there are no cosmic strings.

## Re: Cosmic Strings in the Standard Model

Nice! Before I really read this, a little correction: the center of $G'$ is $\mathrm{U}(1) \times \mathbb{Z}_6$, not $\mathbb{Z} \times \mathbb{Z}_6$. The center of $G$ needs fixing too.