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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)×SU(2)×U(1))/N G = (SU(3)\times SU(2)\times U(1))/N where NN is the 6\mathbb{Z}_6 subgroup of the center of G=SU(3)×SU(2)×U(1)G'=SU(3)\times SU(2)\times U(1) generated by the element (e 2πi/3𝟙,𝟙,e 2πi/6)\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)Z(G)= U(1), rather than Z(G)=U(1)× 6Z(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, GG is broken to the subgroup H=U(3)H=U(3), where the embedding i:HGi\colon H\hookrightarrow G is given as follows. Let hHh\in H and let ddet(h)d\coloneqq \det(h). Choose a 6th root b=d 1/6 b = d^{1/6} Then

(1)i(h)=(hb 2,(b 3 0 0 b +3),b 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 bb leads precisely to an ambiguity in i(h)i(h) by multiplication by an element of NN. Thus (1) is ill-defined as a map to GG', but well-defined as a map to GG.

The (would-be) cosmic strings associated to the breaking of GG to HH are classified by π 1(G/H)\pi_1(G/H). Both π 1(H)\pi_1(H) and π 1(G)\pi_1(G) are equal to \mathbb{Z}. The long-exact sequence in homotopy yields 0π 1(H)π 1(G)π 1(G/H)0 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 π 1(H)\pi_1(H) in π 1(G)\pi_1(G). If the image is nn times the generator of π 1(G)\pi_1(G), then the quotient is nontrivial and we have n\mathbb{Z}_n cosmic strings.

π 1(G)\pi_1(G) is generated by the (homotopy class of) the loop

(2)g(s)=((e 2πis/3 0 0 0 e 2πis/3 0 0 0 e 4πis/3),(e iπs 0 0 e iπs),e 2πis/6),s[0,1]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]

π 1(H)\pi_1(H) is generated by the loop

(3)h(s)=(1 0 0 0 1 0 0 0 e 2πis),s[0,1]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)i(h(s))=g(s). Hence π 1(G/H)=0\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 GG' is U(1)× 6\mathrm{U}(1) \times \mathbb{Z}_6, not × 6\mathbb{Z} \times \mathbb{Z}_6. The center of GG 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 GG', broken to H=SU(3)×U(1)H'=SU(3)\times U(1), then (IIRC), you get the completely wrong answer that there are π 1(G/H)= 2\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 ii well-defined, the third component of i(h)i(h) must be b 1b^{-1} instead of bb. Accordingly, the third component of g(s)g(s) should be e 2πis/6e^{-2\pi i s/6}.

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

(1)i(h)=((1 det(h) 1),h); 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 22b^{-2-2}, not just b 6=det(h)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 red\color{red}\text{red}).

It’s worthwhile repeating the calculation for other choices of global form for HH and GG (with the same Lie algebra embedding). Sometimes one gets π 1(G/H)=0\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 HH and GG (with the same Lie algebra). Sometimes one gets π 1(G/H)=0\pi_1(G/H)=0 (hence no cosmic strings); sometimes one doesn’t.”

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

(1)i(h) =((det(h) 1 1),h), i(h) =((1 det(h) 1),h), j(h) =(det(h)𝟙,det(h) 1h). \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 Φ\Phi of GG such that i=Φii' = \Phi\circ i: With

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

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

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

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

(3)(A,B)j(h)(A,B) 1=(det(BhB 1)𝟙,det(BhB 1) 1BhB 1)j(H). (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 jj is not equivalent to ii (or ii'). Therefore it’s not guaranteed that π 1(G/i(H))=π 1(G/i(H))\pi_1(G/i(H)) = \pi_1(G/i'(H)) is isomorphic to π 1(G/j(H))\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, 𝔧:𝔥𝔤\mathfrak{j}: \mathfrak{h}\hookrightarrow \mathfrak{g} is not isomorphic to 𝔦\mathfrak{i} or 𝔦\mathfrak{i}'. That’s to say (in physicists’ terms) that the electric charges of the quarks and leptons are different for the embedding jj.

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

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

Changing the global form of GG (and/or HH) can, however, easily be achieved by some minor tweaks (e.g., adding some heavy vector-like quarks). That’s why this question about π 1(G/H)\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, jj is not equivalent to ii on the Lie algebra level. So, let us consider only Lie group embeddings Ξ:HG\Xi: H\to G whose induced Lie algebra embedding ξ:𝔥𝔤\xi\colon \mathfrak{h}\to \mathfrak{g} admits a Lie algebra automorphism ϕAut(𝔤)\phi\in Aut(\mathfrak{g}) such that ϕξ\phi\circ\xi is equal to the Lie algebra embedding 𝔦\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 Ξ\Xi could still yield nonisomorphic fundamental groups π 1(G/Ξ(H))\pi_1(G/\Xi(H)). The point is the following.

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

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

Maybe this phenomenon cannot happen for our given H,G,𝔦H,G,\mathfrak{i} (or for any H,GH',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

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