## March 21, 2021

### 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

(1)$i(h) = \left(h b^{-2}, \left(\begin{smallmatrix}b^{{\color{red}-}3}&0\\0&b^{{\color{red}+}3}\end{smallmatrix}\right), b^{\color{red}-1}\right)$

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

(2)$g(s)=\left(\left(\begin{smallmatrix}e^{2\pi i s/3}&0&0\\0&e^{2\pi i s/3}&0\\0&0&e^{-4\pi i s/3}\end{smallmatrix}\right),\begin{pmatrix}e^{i\pi s}&0\\0&e^{-i\pi s}\end{pmatrix},e^{2\pi i s/6}\right), \qquad s\in[0,1]$

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

(3)$h(s)= \begin{pmatrix}1&0&0\\0&1&0\\0&0&e^{{\color{red} -}2\pi i s}\end{pmatrix}, \qquad s\in[0,1]$

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

Posted by distler at March 21, 2021 12:27 AM

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### 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.

Posted by: John Baez on March 21, 2021 1:58 AM | Permalink | Reply to this

### Typo

Fixed. Thanks!

Posted by: Jacques Distler on March 21, 2021 2:08 AM | Permalink | PGP Sig | Reply to this

### Naïve

Note that if you did the completely naïve thing and assumed that the gauge group is $G'$, broken to $H'=SU(3)\times U(1)$, then (IIRC), you get the completely wrong answer that there are $\pi_1(G'/H')=\mathbb{Z}_2$ cosmic strings.

So the necessity of getting this much of the global structure right was clear back in 1982.

Posted by: Jacques Distler on March 21, 2021 2:44 AM | Permalink | PGP Sig | Reply to this

### Re: Cosmic Strings in the Standard Model

To make $i$ well-defined, the third component of $i(h)$ must be $b^{-1}$ instead of $b$. Accordingly, the third component of $g(s)$ should be $e^{-2\pi i s/6}$.

Such typos are easier to spot when we identify $G$ with $S(U(2)\times U(3))$ via $[B,A,z] \mapsto (z^3A,z^{-2}B)$. Then the corrected version of your $i\colon H=U(3) \to G=S(U(2)\times U(3))$ is

(1)$i(h) = \begin{pmatrix}\begin{pmatrix} 1 & \\ & \det(h)^{-1} \end{pmatrix}, h\end{pmatrix} ;$

whereas the other version is not well-defined because it involves $b^{-2-2}$, not just $b^6 = \det(h)$.

Posted by: Marc Nardmann on March 22, 2021 7:39 PM | Permalink | Reply to this

### Re: Cosmic Strings in the Standard Model

Thanks! I’ve corrected the typos (in $\color{red}\text{red}$).

It’s worthwhile repeating the calculation for other choices of global form for $H$ and $G$ (with the same Lie algebra embedding). Sometimes one gets $\pi_1(G/H)=0$ (hence no cosmic strings); sometimes one doesn’t.

Posted by: Jacques Distler on March 22, 2021 8:19 PM | Permalink | PGP Sig | Reply to this

### Re: Cosmic Strings in the Standard Model

You said: “It’s worthwhile repeating the calculation for other choices of global form for $H$ and $G$ (with the same Lie algebra). Sometimes one gets $\pi_1(G/H)=0$ (hence no cosmic strings); sometimes one doesn’t.”

It is even worthwhile considering the exact same $H=U(3)$ and $G=S(U(2)\times U(3))$ but with different embeddings. We have Lie group embeddings $i,i',j\colon H\to G$ given by

(1)\begin{aligned} i(h) &= \begin{pmatrix}\begin{pmatrix}\det(h)^{-1}&\\ &1\end{pmatrix}, h\end{pmatrix} ,\\ i'(h) &= \begin{pmatrix}\begin{pmatrix}1&\\ &\det(h)^{-1}\end{pmatrix}, h\end{pmatrix} ,\\ j(h) &= \Big(\det(h)\mathbb{1}, \det(h)^{-1}h\Big) . \end{aligned}

(The first option is your definition with the red correction.)

The first two are equivalent in the sense that there is a Lie group automorphism $\Phi$ of $G$ such that $i' = \Phi\circ i$: With

(2)$a = \begin{pmatrix}\begin{pmatrix}&i\\ i&\end{pmatrix}, \mathbb{1}\end{pmatrix} \in G ,$

we consider the inner automorphism $\Phi: g\mapsto a g a^{-1}$ and get $\Phi(i(h)) = i'(h)$ for all $h\in H$.

Since $i(H) \neq i'(H)$, this shows also that the subgroups $i(H)$ and $i'(H)$ of $G$ are not normal.

In contrast, $j(H)$ is a normal subgroup of $G$: for all $h\in H$ and $(A,B)\in G$, we have

(3)$(A,B)j(h)(A,B)^{-1} = \Big(\det(B h B^{-1})\mathbb{1}, \det(B h B^{-1})^{-1} B h B^{-1}\Big) \in j(H).$

In particular, the embedding $j$ is not equivalent to $i$ (or $i'$). Therefore it’s not guaranteed that $\pi_1(G/i(H)) = \pi_1(G/i'(H))$ is isomorphic to $\pi_1(G/j(H))$ (which I haven’t computed yet).

Posted by: Marc Nardmann on March 22, 2021 10:10 PM | Permalink | Reply to this

### Different H↪G

If I understand correctly, the embedding induced on the Lie algebra, $\mathfrak{j}: \mathfrak{h}\hookrightarrow \mathfrak{g}$ is not isomorphic to $\mathfrak{i}$ or $\mathfrak{i}'$. That’s to say (in physicists’ terms) that the electric charges of the quarks and leptons are different for the embedding $j$.

For physicists, $j$ is the breaking pattern where ${\mathfrak{u}(1)}_{\text{EM}}$ is identified with $\mathfrak{u}(1)_Y$, rather than with the diagonal subalgebra of the Cartan of $\mathfrak{su}(2)$ and $\mathfrak{u}(1)_Y$.

That’s not some minor tweak to the Standard Model.

Changing the global form of $G$ (and/or $H$) can, however, easily be achieved by some minor tweaks (e.g., adding some heavy vector-like quarks). That’s why this question about $\pi_1(G/H)$ was interesting to me back in the day.

Posted by: Jacques Distler on March 22, 2021 11:25 PM | Permalink | PGP Sig | Reply to this

### Re: Different H↪G

Indeed, $j$ is not equivalent to $i$ on the Lie algebra level. So, let us consider only Lie group embeddings $\Xi: H\to G$ whose induced Lie algebra embedding $\xi\colon \mathfrak{h}\to \mathfrak{g}$ admits a Lie algebra automorphism $\phi\in Aut(\mathfrak{g})$ such that $\phi\circ\xi$ is equal to the Lie algebra embedding $\mathfrak{i}$ which describes the Standard Model symmetry breaking. A priori (i.e., for all I know without having thought seriously about the issue), different embeddings $\Xi$ could still yield nonisomorphic fundamental groups $\pi_1(G/\Xi(H))$. The point is the following.

The map $Aut(G)\to Aut(\mathfrak{g})$ that assigns to each Lie group automorphism the induced Lie algebra automorphism is (injective because $G$ is connected but) probably not surjective. (If we consider $SU(3)\times SU(2)\times U(1)$ instead of $S(U(2)\times U(3))$, then the nonsurjectivity is obvious, because already $Aut(U(1))\to Aut(\mathfrak{u}(1))$ is not surjective: $Aut(U(1))\cong\mathbb{Z}_2$, $Aut(\mathfrak{u}(1))\cong GL(1,\mathbb{R})$.)

We can take a Lie algebra automorphism $\phi\in Aut(\mathfrak{g})$ that does not lie in the image, and consider $\xi = \phi^{-1}\circ\mathfrak{i}$. The question is whether there exists such a $\phi$ that also admits a Lie group embedding $\Xi\colon H\to G$ with induced Lie algebra embedding $\xi$. If so, then it is a priori not clear whether $\pi_1(G/\Xi(H))$ is isomorphic to $\pi_1(G/i(H))$: the subgroups $\Xi(H)$ and $i(H)$ might “sit differently” in $G$, even though the Lie subalgebras $\xi(\mathfrak{h})$ and $\mathfrak{i}(\mathfrak{h})$ “sit similarly” in $\mathfrak{g}$.

Maybe this phenomenon cannot happen for our given $H,G,\mathfrak{i}$ (or for any $H',G'$ with the same Lie algebras). But I would be surprised if one could not construct some Lie groups and Lie algebra embedding where this occurs.

Posted by: Marc Nardmann on March 24, 2021 9:42 AM | Permalink | Reply to this

### Re: Cosmic Strings in the Standard Model

“… the presence or absence of various topological defects (in particular, cosmic strings) …” What does the amplituhedron predict about cosmic strings?

Posted by: David Brown on June 17, 2021 1:46 AM | Permalink | Reply to this

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