Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

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

TrackBack URL for this Entry:

8 Comments & 0 Trackbacks

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


Fixed. Thanks!

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


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

Post a New Comment