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March 25, 2009

The Algebra of Grand Unified Theories III

Posted by John Baez

If you’re a phenomenologist, here’s a question: is the nonsupersymmetric SO(10)SO(10) GUT still experimentally viable? I get the impression that it is.

If you’re a follower of Husserl or Heidegger, here’s a word of reassurance: I don’t mean that kind of ‘phenomenologist’. I mean an particle physicist who is not an experimentalist, but is nonetheless deeply concerned with experimental data.

And if you’re a physicist of a more theoretical or mathematical sort — the sort perhaps more likely to frequent the nn-Café — please ask your phenomenologist pals to stop by and pass on their wisdom!

Let me start with a few words for nonexperts before asking my question to the experts.

In trying to go beyond the Standard Model of particle physics, grand unified theories try to make sense of the messy pack of particles we see and find elegant patterns lurking in the noise. The SO(10)SO(10) theory does an astoundingly good job of this. There are 32 particles and antiparticles in each generation of the Standard Model, if we include the elusive right-handed neutrino and its antiparticle — and these fit neatly in the spinor representation of SO(10)SO(10)’s double cover Spin(10)Spin(10) on the exterior algebra ΛC 5\Lambda \C^5, which has dimension 2 5=322^5 = 32.

This is sort of amazing. In particular, the crazy-looking hypercharges of all 32 particles must be exactly what we see for this theory to make sense.

Let me explain this, in terms even a Ph.D. mathematician can understand.

Irreducible representations of the circle group, U(1)(1), are classified by integers. Physicists often call this integer the ‘charge’, because this is the modern explanation of why electric charge comes in discrete units. Any irreducible representation of the Standard Model group U(1)×SU(2)×SU(3)(1) \times SU(2) \times SU(3) thus has an associated ‘charge’. However, the U(1)(1) here is not associated to ordinary electric charge, so physicists use a different term in this case: ‘hypercharge’. They also divide this integer by 3, just to drive mathematicians crazy, but let’s not do that today.

Here are the 16 left-handed particles in the first generation of the Standard Model, and their hypercharges:

  • left-handed neutrino: -3
  • left-handed electron: -3
  • left-handed up quarks (red, green and blue): 1
  • left-handed down quarks (red, green and blue): 1
  • left-handed antineutrino: 0
  • left-handed positron: 6
  • left-handed up antiquarks (antired, antigreen and antiblue): -4
  • left-handed down antiquarks (antired, antigreen and antiblue): 2

It might seem hopeless to explain this list of numbers in an elegant way. But in fact Spin(10)Spin(10) has a U(1)(1) subgroup such that when we restrict the spinor representation to this subgroup, it splits up as a direct sum of irreducible representations with precisely these charges!

Of course, there is a vast amount of particle physics data that a successful grand unified theory must fit. I’ve just skimmed off a tiny bit of the cream, the numerological stuff that even a Ph.D. mathematician can understand. But in fact, the SO(10)SO(10) theory does pretty well on quite a wide front.

Could it be true?

I want to know the current conventional wisdom on this question! John Huerta and I are writing an expository paper for mathematicians on the algebra of grand unified theories. Our paper doesn’t touch serious issues like proton decay or running coupling constants… but still, we want to know: is SO(10)SO(10) without supersymmetry still a contender?

I get the impression that this theory is still viable with a number of symmetry-breaking schemes, including ones where at intermediate energies the unbroken symmetry group is the Pati–Salam group SU(2)×SU(2)×SU(4)SU(2) \times SU(2) \times SU(4). This makes me happy.

The following review article compares 12 symmetry-breaking schemes where the symmetry group breaks down from SO(10)SO(10) (or really its double cover) to U(1)×SU(2)×SU(3)(1) \times SU(2) \times SU(3) through two intermediate steps:

They use a notation for such schemes introduced here:

Is it true that all schemes with SU(5)SU(5) as an intermediate step are ruled out, while many models that pass through SU(2)×SU(2)×SU(4)SU(2) \times SU(2) \times SU(4) are viable?

Also: what might the Large Hadron Collider tell us about these issues? Suppose we find a Standard Model Higgs… then what else might we see, that would help confirm or disconfirm the SO(10)SO(10) GUT?

Posted at March 25, 2009 7:54 PM UTC

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right link

The right link seems to be Intermediate mass scales in the non-supersymmetric SO(10) grand unification: a reappraisal.

Posted by: RodM on March 26, 2009 12:12 AM | Permalink | Reply to this

Re: right link

Thanks! Fixed!

Posted by: John Baez on March 26, 2009 1:12 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

It’s probably worth saying that the hypercharges are highly constrained by anomaly cancellation conditions, so the coincidence is not quite as impressive as it might seem.

Posted by: Aaron Bergman on March 26, 2009 3:21 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Not “highly constrained.” Completely determined! If you demand that the gauge anomalies and the mixed gauge-gravitational anomaly for the U(1)U(1) cancel (ie, that the U(1)U(1) gauge symmetry is not violated in a curved background), then the U(1)U(1) hypercharge assignments (up to an overall normalization that John already complained about) are completely determined.

I don’t understand all this mystification surrounding the hypercharge assignments. Given the SU(3)SU(3) and SU(2)SU(2) representation content, the U(1)U(1) hypercharges could not possibly be anything other than what they are.

Posted by: Jacques Distler on March 26, 2009 4:37 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

If there were only one generation of fermions, or if the hypercharge assignments of each generation were constrained to be identical, it’s true the hypercharges would be determined up to a multiplicative factor. But a priori, why should the hypercharge assignments be the same for each generation?

Posted by: Water on March 27, 2009 1:16 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Water wrote:

….a priori, why should the hypercharge assignments be the same for each generation?

I consider this part of a larger but somehow simpler mystery: why should each generation of fermions be described by the same representation of SU(3)×SU(2)×U(1)SU(3) \times SU(2) \times U(1)? There are, after all, plenty of representations where the anomalies all cancel, asymptotic freedom is preserved, and no other nasty things happen.

Posted by: John Baez on March 29, 2009 12:08 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Well, OK, there is a second solution to the anomaly constraints.

Exercise: find and interpret that second solution.

Posted by: Jacques Distler on March 26, 2009 5:08 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Since I’ve been busy writing this expository paper that focuses purely on the finite-dimensional group representation theory, I sort of forgot about how far you can get just by demanding anomaly cancellation.

I’ll work out the constraints due to anomaly cancellation and solve Jacques’ exercise. I need to brush up on how that stuff works.

Posted by: John Baez on March 26, 2009 8:06 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Wait a minute—anomaly cancellation in the SM forces the particle content to look just like a breakdown of a Spin(10) irrep?

Isn’t that a bit … weird? Why would that happen?

Posted by: Tim Silverman on March 27, 2009 4:28 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Isn’t that a bit … weird?

Not really.

What’s remarkable is that the 5¯10\overline{5} \oplus 10 is an anomaly-safe representation of SU(5)SU(5) (or, for that matter, that the 1616 of Spin(10)Spin(10) is anomaly-safe).

Given that the groups embed, any anomaly-safe representation of G GUTG_{\text{GUT}} must decompose as an anomaly-safe representation of G SMG_{\text{SM}}. So one of the solutions of the anomaly constraints is guaranteed to be a representation of G SMG_{\text{SM}} that can be extended to a representation of G GUTG_{\text{GUT}}.

But, as I said, there’s a second solution …

Posted by: Jacques Distler on March 27, 2009 6:00 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Thanks, Jacques!

That is indeed a lot less weird than I imagined.

(A little learning and all that … )

Posted by: Tim Silverman on March 28, 2009 12:47 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Tim wrote:

Wait a minute — anomaly cancellation in the SM forces the particle content to look just like a breakdown of a Spin(10) irrep?

Sometimes I think we should all become script writers for science fiction TV shows. We could do such better jobs…

Sorry: just a random reaction to your exclamation, plus the fact that last night I watched that Star Trek episode where Barclay was zapped by aliens and suddenly got an IQ of about 1200. He was talking to Einstein on the holodeck about grand unification — but they were saying really silly stuff! I wish he’d said “Wait a minute — anomaly cancellation in the SM forces the particle content to look just like a breakdown of a Spin(10) irrep?

Anyway: I think your statement is considerably exaggerated — but it has a kernel of truth to it, and I’d like to know the precise statement.

I think this is what Jacques actually claimed: given how one generation of fermions transforms under SU(2)×SU(3)SU(2) \times SU(3), and knowing their handedness, the requirement that both gauge and gauge-gravitational anomalies cancel leaves only two choices for how they transform under U(1)(1).

One of these choices makes all the left-handed fermions into an irrep for Spin(10)Spin(10); the right-handed fermions form the dual irrep. The other choice… well, that’s the ‘exercise’ I’m supposed to do, so please: nobody give away the answer, unless I give up.

Anyway, as you can see, Jacques claim is considerably weaker than “anomaly cancellation forces the particle content to look just like a breakdown of a Spin(10) irrep”.

However, in my attempts to brush up anomaly cancellation, I peeked into this book:

  • Reinhold A. Bertlmann, Anomalies in Quantum Field Theory, Oxford U. Press, 1996.

and found this statement:

‘Safe groups’: There are no anomalies if the trace over the gauge group generators vanishes in a certain representation or even for all representations. The condition for it is

tr({T a,T b}T c)=0tr( \{T^a, T^b\} T^c) = 0

resulting from the triangle diagram; then all higher loop contributions vanish too.

There are ‘safe’ groups in 4 dimensions like SU(2),SO(2N+1)SU(2), SO(2N+1) and SO(4N)SO(4N) with N2N \ge 2, or exceptional groups like E 6,E 8E_6, E_8 where the above identity is valid for all representations; however, unfortunately note for SU(N)SU(N) with N3N \ge 3.

So, while anomaly cancellation doesn’t force the fermions to lie in a rep of SO(10)SO(10), lying in a rep of SO(10)SO(10) is enough for anomaly cancellation (at least the gauge anomalies; I don’t know about the gauge-gravitational ones).

I wish I understood the above equation

tr(T aT bT c)+tr(T bT aT c)=0tr( T^a T^b T^c) + tr(T^b T^a T^c) = 0

at a more conceptual level. Here the guys T aT^a are a basis of Lie algebra elements as represented as linear operators on some rep of that Lie algebra. Of course you can write this out diagrammatically, and then it says that this sort of triangle, plus its ‘flipped version’, vanishes:

Here the wiggly lines stand for the Lie algebra, while the straight line stands for a given representation of the Lie algebra. The fact that this diagram minus its ‘flipped version’ vanishes is what makes the ‘triangle anomaly’ vanish.

But still, this isn’t exactly a conceptual explanation of what it means for a representation to satisfy that identity: it’s diagrammatic.

Posted by: John Baez on March 27, 2009 6:18 PM | Permalink | Reply to this

Anomaly-safe Groups

The condition for it is tr([T a,T b]T c)=0tr([T^a,T^b]T^c)=0

I think you mean the totally-symmetrized trace, which is usually written with brace, rather than square brackets:

tr({T a,T b}T c)=0tr(\{T^a,T^b\}T^c)=0

And your list of anomaly-safe gauge theories is somewhat abbreviated. There are the groups with only self-conjugate representations: Spin(2n+1),Spin(4n),Sp(n),G 2,F 4,E 7,E 8 Spin(2n+1),\, Spin(4n),\, Sp(n),\, G_2,\, F_4,\, E_7,\, E_8

These are unsuitable for physics, because the gauge theories are non-chiral.

E 6,Spin(4n+2),n2 E_6,\, Spin(4n+2),\, n\geq 2 are anomaly-safe, but have complex (i.e., non-self-conjugate) representations.

…at least the gauge anomalies; I don’t know about the gauge-gravitational ones.

Those vanish for any semi-simple group. For the Standard Model, I needed to impose the vanishing of the gauge-gravitational anomaly as a separate (but, physically, eminently reasonable) condition.

Posted by: Jacques Distler on March 27, 2009 7:19 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Jacques wrote:

John wrote:

The condition for it is

tr([T a,T b]T c)=0tr([T^a,T^b]T^c)=0

I think you mean the totally-symmetrized trace, which is usually written with brace, rather than square brackets:

tr({T a,T b}T c)=0tr(\{T^a,T^b\}T^c)=0

Whoops! Right. I’m going to zip back in time and fix that mistake — not to appear smarter than I really am, but just to reduce the amount of misinformation floating around the universe.

So, you’re hinting that this condition holds for any representation that’s self-conjugate; that should be easy for me to check. Hmm, maybe something like this.

Let’s say we have a unitary representation of a Lie group, which we can assume since we’re talking about compact Lie groups here. Then the generators T aT^a are skew-adjoint in this representation, so

T a=(T a) *=(T a¯) tT^a = -(T^a)^* = -(\overline{T^a})^t

where tt denotes the transpose matrix and overline denotes the conjugate matrix.

But if our representation is also self-conjugate we have

T a=T a¯T^a = \overline{T^a}

so

T a=(T a) t T^a = -(T^a)^t

so

tr(T aT bT c)=tr((T aT bT c) t)=tr((T c) t(T b) t(T a) t)=tr(T cT bT a) tr(T^a T^b T^c) = tr((T^a T^b T^c)^t) = tr((T^c)^t (T^b)^t (T^a)^t) = - tr(T^c T^b T^a)

so

tr({T a,T b}T c)=0.tr(\{T^a,T^b\}T^c)=0.

There must be a shorter way, but nobody is paying me to find it.

But you’re also saying it holds for other representations too — which are actually more interesting, because they describe chiral theories.

Okay, some things are starting to make sense. I’ll have to keep on studying this. Any good comprehensive references?

And while I have your attention: is what Arvind Rajaraman writes here correct?

In four dimensions, the gauge and gravitational anomalies come from triangle diagrams which involve either one or three chiral currents. The gauge anomalies involve the U(1) hypercharge and the SU(2) gauge bosons. The nontrivial anomalies are the ones with U(1)-SU(2)-SU(2), U(1)-U(1)-U(1), and SU(2)-SU(2)-SU(2). The last vanishes always. The first two vanish if the sum of hypercharges in a generation is zero, and the sum of the cubes is also zero. These happen to be satisfied.

and then later

I should make the above statements precise. Actually, the first vanishes if the sum of hypercharges of the left handed particles is zero, and the second if the sum of the cubes of the left handed particles minus that of the right handed particles is zero. So the gravitational anomaly is actually new, it says that the sum of all the hypercharges in a generation is zero.

This is from an old conversation I saw back when I was a moderator of sci.physics.research.

Posted by: John Baez on March 27, 2009 7:58 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Just a note for the unwary: what I mean by “self-conjugate” is that RR¯R\simeq \overline{R}. So, that really means that T a=MT¯ aM 1 T^a = M \overline{T}^a M^{-1} (where MM is the isomorphism) rather than T a=T¯ aT^a = \overline{T}^a. This subtlety doesn’t matter when you take the trace.

And while I have your attention: is what Arvind Ramarajan writes here correct?

I would have said; U(1) 3{U(1)}^3, U(1)SU(2) 2U(1) {SU(2)}^2, U(1)SU(3) 2U(1) {SU(3)}^2 and SU(3) 3{SU(3)}^3 are the possible gauge anomalies. The SU(2) 3{SU(2)}^3 anomaly automatically vanishes by the considerations above.

So the gravitational anomaly is actually new, it says that the sum of all the hypercharges in a generation is zero.

That’s correct (and, as you can see, would be automatic in the semisimple case).

Posted by: Jacques Distler on March 27, 2009 8:15 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

What is the exact definition of symmetrized trace? The definitions I found imply it’s of the form χ 3=tr(T aT bT c) \chi_3 = \sum \tr(T^a T^b T^c) where T aT^a are the generators of the group in the representation under consideration. I’m not sure what the sum is over; my guess is that it’s over the product of 3-tuples of generators. Is it? The other part that’s not clear is that whether this is over the generators of the group itself or its algebra. The generators of the algebra are skew adjoint (T a) *=T a (T^a)^* = - T^a not those of the group
Posted by: rntsai on March 29, 2009 10:31 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

rntsai writes:

What is the exact definition of symmetrized trace? The definitions I found imply it’s of the form χ 3=tr(T aT bT c) \chi_3 = \sum \tr(T^a T^b T^c) where T aT^a are the generators of the group in the representation under consideration. I’m not sure what the sum is over; my guess is that it’s over the product of 3-tuples of generators. Is it?

I don’t know what definitions you’re reading, but the quantity that matters for anomalies is

tr({T a,T b}T c) tr( \{ T^a, T^b \} T^c)

This is shorthand for

tr(T aT bT c+T bT aT c) tr(T^a T^b T^c + T^b T^a T^c)

where T aT^a ranges over a basis of the Lie algebra LL of some Lie group GG as a=1,,na = 1,\dots, n, and we are implicitly taking the trace in a given representation of this Lie algebra.

A more fussy notation would specify the representation ρ:LEnd(V)\rho : L \to End(V) and make it visible, like this:

tr(ρ(T a)ρ(T b)ρ(T c)+ρ(T b)ρ(T a)ρ(T c)) tr(\rho(T^a) \rho(T^b) \rho(T^c) + \rho(T^b) \rho(T^a) \rho(T^c))

Note: wouldn’t be useful sense to sum this over basis elements a,b,ca,b,c, since the result would be basis-dependent in a nasty way. If we avoid summing over basis elements, we can think of

tr(ρ(T a)ρ(T b)ρ(T c)+ρ(T b)ρ(T a)ρ(T c)) tr(\rho(T^a) \rho(T^b) \rho(T^c) + \rho(T^b) \rho(T^a) \rho(T^c))

as the index-ridden notation for the trilinear map

χ:L×L×L \chi : L \times L \times L \to \mathbb{R}

given by

χ(x,y,z)=tr(ρ(x)ρ(y)ρ(z)+ρ(y)ρ(x)ρ(z)) \chi(x,y,z) = tr(\rho(x) \rho(y) \rho(z) + \rho(y) \rho(x) \rho(z))

for x,y,zLx,y,z \in L. As such, χ\chi is a completely symmetric trilinear map from our Lie algebra to the real numbers.

I don’t know if this χ\chi is secretly the ‘χ 3\chi_3’ you’ve been reading about, but I would bet you a doughnut that it is.

The other part that’s not clear is that whether this is over the generators of the group itself or its algebra.

Yikes!

The generators of the algebra are skew adjoint (T a) *=T a (T^a)^* = - T^a not those of the group.

Right. But let’s get this cleared up right away: when physicists say ‘generators of a Lie group’, they always mean a basis for its Lie algebra! If they feel like being careful, they call these ‘infinitesimal generators’ of the Lie group. But again: they just mean a basis for the Lie algebra.

Of course every group has a set of ‘generators’, meaning elements such that any element of the group is a product of these elements and their inverses. But for a Lie group, a set of generators in this sense must typically be uncountable.

On the other hand, if T aT^a is a basis for the Lie algebra of your Lie group, and your Lie group is connected, every element of the group is a finite product of elements of the form

exp(tT a)exp(t T^a)

as tt ranges over real numbers.

This is why physicists are quite reasonable to call basis vectors of a Lie algebra ‘infinitesimal generators for the Lie group’: if you exponentiate them, you get generators for the group.

Posted by: John Baez on March 29, 2009 11:09 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

The definition in “Consistency Conditions for Kaluza-Klein Anomalies”, P.H.Frampton and T.W.Kephart, Physcial Review Letters Volume 50, Number 18, 2 May 1983 for example has that form; which also looks easier to generalize. Thanks for the clarification about the infinitesimal generators of the group…although that “Yikes” wasn’t too polite

Posted by: rntsai on March 29, 2009 11:35 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I’ll take a peek at that paper when I get some time. I’m due for a delivery of time any day now.

I’m sorry about the “Yikes! ” I just suddenly realized that you might be a mathematician adrift in physics terminology that sounds like familiar math terminology but means something quite different… and the idea of taking an uncountable sum over triples of generators — in the mathematician’s sense — for a Lie group struck terror into my heart.

Posted by: John Baez on March 30, 2009 12:19 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

The two definitions are then actually the same. The one in the paper would give the symmetric trace as : χ 3(x,y,z)=tr(xyz+zxy+yzx+yxz+zyx+xzy)/6=tr(xyz+yxz)/2\chi_3(x,y,z)=\tr(xyz+zxy+yzx+yxz+zyx+xzy)/6=\tr(xyz+yxz)/2 for higher order versions, this would involve taking all permutations of x,y,z,… so there would be 24 terms in χ 4(x,y,z,w)\chi_4(x,y,z,w),….for skew adjoint reps (x *=x)x^*=-x), χ 1=0\chi_1=0 and χ 3=0\chi_3=0.
Posted by: rntsai on March 30, 2009 2:13 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I’ll take a peek at that paper when I get some time. I’m due for a delivery of time any day now.

I’m sorry about the “Yikes!” I just suddenly realized that you might be a mathematician adrift in physics terminology that sounds like familiar math terminology but means something quite different… and the idea of taking an uncountable sum over triples of generators — in the mathematician’s sense — for a Lie group struck terror into my heart.

Posted by: John Baez on March 30, 2009 7:13 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

But still, this isn’t exactly a conceptual explanation of what it means for a representation to satisfy that identity: it’s diagrammatic.

The conceptual explanation is the Families Index Theorem (for the Dirac operator coupled to a vector bundle VV) of Atiyah and Singer.

The relevant bit of index formula is Ch 3(V)Ch_3(V), where VV is associated, via some representation, RR, to a GG principal bundle.

When you compute Ch 3(V)Ch_3(V), you find that it is proportional to tr R({T a,T b}T c) tr_R(\{T^a,T^b\} T^c) and so vanishes when that particular trace vanishes.

Posted by: Jacques Distler on March 27, 2009 8:24 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

“The conceptual explanation is the Families Index Theorem (for the Dirac operator coupled to a vector bundle V) of Atiyah and Singer.”

Even though I’m hopelessly behind in this discussion, let me see if I have a vague idea of what this means. Some kind of functional integral associated to the theory leads you to

Det(D A)Det(D_A)

where D AD_A is a twisted Dirac operator acting on the sections of your bundle VV. When the connection form AA is hit with a gauge transformation gg, one gets a formula like

Det(D A g)=Det(D A)c(A,g)Det(D_{A^g})=Det(D_A)c(A,g)

for a function c(A,g)c(A,g) that should be a cocycle in some sense for the group of gauge transformations acting on the space of connections. Now, the function c(A,g)c(A,g) can be computed using the families index theorem, and one sees that it is proportional to

tr R({T a,T b}T c),tr_R(\{T^a,T^b\}T^c),

where I guess we should take gg to be somehow built out of the T aT^a. Thus, the determinant is invariant under gauge transformations when that trace vanishes.

Is this roughly the line of reasoning? (I’m vaguely recalling something I read twenty years ago. So I fear it may be complete nonsense.)

Posted by: Minhyong Kim on March 30, 2009 10:41 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Sorta.

If you consider the chiral Dirac operator which I helpfully defined for Garrett Lisi, here, the fermion path integral gives Det(D A)Det(D_A),

Unfortunaqtely, as should be apparent from the definition, Det(D A)Det(D_A) is not naturally a number, but rather an element of a complex line. If we consider a family of such operators (parametrized by a family of connections modulo gauge transformation, say), Det(D A)Det(D_A) is not naturally a function on the parameter space, but rather a section of a line bundle (called, appropriately enough, the Determinant Line Bundle).

The Index Theorem for Families computes the first Chern class of that line bundle. (In fact, in a more sophisticated, geometrical form, it computes the actual curvature of a certain connection on that line bundle.)

This is the obstruction to trivializing the Determinant Line Bundle, and hence to treating Det(D A)Det(D_A) as a function on the parameter space.

Posted by: Jacques Distler on March 31, 2009 1:43 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

I see. Because you have a chiral Dirac operator, it’s a priori not a number.

Now this business is coming back to me. But since I looked at these constructions only from a geometric point of view, there were a few things about the physics that always puzzled me. For example, at the end of the day, I presume one does need to assign a number to the path integral. So is there a physically natural trivialization one chooses, when the anomaly vanishes?

Also, are there applications of the refined differential form version of the index theorem in this context?

Posted by: Minhyong Kim on March 31, 2009 3:19 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

So is there a physically natural trivialization one chooses, when the anomaly vanishes?

Indeed, there’s a natural connection on the Determinant Line Bundle, called the Bismut-Freed connection (though I think the physicists got there first) which is flat when the anomaly vanishes.

[Actually, this might be a fine time to point out that the anomaly cancellation conditions we have been discussing are sufficient to ensure the flatness of the Bismut-Freed connection. But that’s not quite enough to ensure that the Determinant Line Bundle is trivial. Even if c 1(DET)=0c_1(DET) =0 rationally, it could still be nonzero (torsion). A flat connection could have holonomy around nontrivial cycles in the parameter space. This actually happens for G=SU(2)G= SU(2) (where Tr({T a,T b}T c)0Tr(\{T^a,T^b\}T^c)\equiv 0) with an odd number of doublets. Note that the SU(2)SU(2) in the Standard Model has an even number of doublets.]

A flat connection, with vanishing holonomies gives you a trivialization.

Posted by: Jacques Distler on March 31, 2009 5:43 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Aha, so one does use all the fine information of the Bismut-Freed connection. And, if I understood you correctly, there is still the question of holonomy, which one needs to deal with somehow.

However, there is still one small point, even in the trivial holonomy case: one still needs to choose a trivialization of some particular fiber of the line bundle, which can then be transported to the other fibers. Is there a canonical way to do this?

I can see a way to do this in the index zero case, because then, I suppose the generic operator D AD_A will have zero kernel and cokernel as well. But then, there is a canonical isomorphism Det(D A)C.Det(D_A)\simeq C. [Now I’m using the det symbol for the determinant line as you suggest.]

Even this is rather unsatifactory, because it seems to arbitrarily set the path integral for some particular D AD_A to be 1.

Perhaps there is a better way?

By the way, I really appreciate your explanations. These are questions I’ve had in the back of my mind for years.

Posted by: Minhyong Kim on March 31, 2009 10:28 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

However, there is still one small point, even in the trivial holonomy case: one still needs to choose a trivialization of some particular fiber of the line bundle, which can then be transported to the other fibers. Is there a canonical way to do this?

This is the question of “What is the overall normalization of the Feynman Path Integral?”

Normally, mathematicians would consider the Feynman Path Integral so hopelessly ill-defined that they would find that question ludicrous.

Here, I think we can do better.

Let’s assume that the anomaly vanishes, a trivialization exists, and the Bismut-Freed connection supplies that trivialization up to multiplication by a nonzero complex number. Thus Det(D A)Det(D_A) is a function on the parameter space, up to this overall rescaling.

We can fix the magnitude of that complex number by the following observation.

Let me assume that the index=0, so that, generically, the kernel and cokernel vanish.

D A D AD_A^\dagger D_A is a self-adjoint elliptic operator and we can define Det(D A D A)Det(D_A^\dagger D_A), without any such ambiguity, by ζ\zeta-function regularization. This gives us a definition of the magnitude |Det(D A)||Det(D_A)| (which is, moreover, compatible with the Bismut-Freed connection).

Defining the overall phase of Det(D A)Det(D_A) is a much more subtle business. People, who are smarter than I, know how to do it.

Posted by: Jacques Distler on March 31, 2009 11:31 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Thanks. That gives me at least a rough idea of what peope do.

Here is a more technical question: The only mathematical treatment of determinant line bundles, Bismut-Freed connection, etc., I know of starts with a compact Riemannian manifold or a family of them. But I assume physicists are interested in the Dirac operator on Minkowski space. Even assuming that one applies some kind of ‘Wick rotation,’ which I don’t really understand, how does one deal with the non-compactness?

Posted by: Minhyong Kim on April 1, 2009 12:09 AM | Permalink | Reply to this

Euclidean

The Minkowski signature Feynman Path Integral is just not well-defined at all (even by physicists’ standards). It is given meaning by an analytic continuation from Euclidean signature.

And, even there, you need to introduce some sort of infrared cutoff (at least as an intermediate step) to make sense of the calculation (which is to say, I don’t understand John’s crack about hoping no mathematicians are watching). Particularly in nonabelian gauge theories, that means working in finite volume, rather than in noncompact 4\mathbb{R}^4.

The lattice gauge theorists work on T 4T^4. Many other people prefer to work on the one-point compactification of 4\mathbb{R}^4, namely S 4S^4.

The principle of locality in field theory really says that the details of this choice should drop out when you take the VV\to\infty limit at the end of the day. So, really, you should be able to use any (compact, oriented, spin) 4-manifold for your computations.

Posted by: Jacques Distler on April 1, 2009 4:34 PM | Permalink | PGP Sig | Reply to this

Re: Euclidean

So if I combine your two comments, John and Jacques, I should conclude that the pure mathematician and the physicist are in league.

But the assertion does seem to be an astounding one. Let me try to formulate it in a way I can understand. Given a compact, oriented, spin, Riemannian four-manifold, MM, one has a path integral that looks something like P(M)=e L(A,ψ)dψdAP(M)=\int e^{L(A,\psi)}d\psi dA =Det(D A)dA=\int Det(D_A) dA whose evaluation itself will involve some discrete approximations (which would be called the ultraviolet cut-off?) that become finer. Am I to understand that any limit lim V(M)P(M)\lim_{V(M)\rightarrow \infty} P(M) should exist, and give the same value?

Somehow, this seems strange if I consider, say, the 1+1 dimensional situation where I can at least conceive of some concrete limits. For example, if I take a sequence M gM_g of surfaces of increasing genus gg as well as increasing volume, then it seems strange to expect lim graP(M g)\lim_{g\ra \infty}P(M_g) to be the same as a limit over larger and larger spheres.

Perhaps you would like to take a family of spaces MM that converge to R 4R^4 in the some suitable sense, say that of Gromov and Haudorff?

If this is truly what you expect, it seems to be an excellent mathematical problem to prove or disprove it, say at least in the 1+1-dim case. Note that in resolving this issue, we will genuinely need to nail down the natural normalization, since we want to compare two numbers at the end.

Oh, should I be normalizing the terms as well before taking the limit, say by dividing each P(M)P(M) by the volume of MM?

Posted by: Minhyong Kim on April 1, 2009 7:55 PM | Permalink | Reply to this

Re: Euclidean

Minhyong wrote:

Am I to understand that any limit

lim v(M)P(M)lim_{v(M) \to \infty} P(M)

should exist, and give the same value?

That’s a bit overoptimistic, for two reasons.

First, what we can measure through particle physics experiments is not the partition function P(M)P(M), but instead expectation values of observables, which are ratios

O(A,ψ)e S(A,ψ)dψdAe S(A,ψdψdA \frac{\int O(A,\psi) e^{-S(A,\psi)} d \psi d A}{\int e^{-S(A,\psi} d \psi d A}

where SS is the action (you wrote LL) and OO is some ‘observable’ — that is, some gauge-invariant function of AA and ψ\psi.

In this formula the partition function only serves as a ‘normalization factor’. That’s good: it means that if we multiply the mysterious measure dψdAd \psi d A by 42, nothing changes.

So, we don’t really need

lim v(M)P(M)lim_{v(M) \to \infty} P(M)

to exist. It’s enough for

lim v(M)O(A,ψ)e S(A,ψ)dψdAe S(A,ψ)dψdA lim_{v(M) \to \infty} \frac{\int O(A, \psi) e^{-S(A,\psi)} d \psi d A}{\int e^{-S(A,\psi)} d \psi d A}

to exist for a nice class of observables OO.

Second, I doubt this limit exists except for sequences of Riemannian manifolds that ‘locally approach flat Euclidean space’ in some sense. As you note, there are sequences of manifolds with v(M)v(M) \to \infty whose topology keeps changing as MM gets big… but the real problem comes from sequences where the local geometry keeps wiggling around as MM gets big! That is, these are problematic when the observable OO is ‘local’ in a suitable sense.

Posted by: John Baez on April 2, 2009 1:04 AM | Permalink | Reply to this

Re: Euclidean

Hmm, so one can’t work just with the partition function? A difficulty I have with the formulation you’ve given is to come up with the observables O(A,ψ)O(A,\psi) that make sense ‘uniformly’ in a family of manifolds. I suppose if we had just a single scalar field ϕ\phi instead of our gauge/fermion set-up, then we could consider families of tuples (M,x 1,,x n)(M,x_1,\ldots, x_n) that converge to (R 4,p 1,,p n)(R^4,p_1, \ldots, p_n). Then we could investigate the independence of the limit limϕ(x 1)ϕ(x n)e S(ϕ)dϕe S(ϕ)dϕ.\lim \frac{\int \phi(x_1)\cdots \phi(x_n) e^{S(\phi)}d\phi}{\int e^{S(\phi)}d\phi}. If there were only a gauge field, then perhaps we should consider ‘loop functionals’ limTr l 1(A)Tr l n(A)e S(A)dAe S(A)dA\lim \frac{\int Tr_{l_1}(A)\cdots Tr_{l_n}(A) e^{S(A)}dA}{\int e^{S(A)}dA} as we go over families (M,l 1,,l n)(M,l_1,\ldots,l_n) converging to something. Does this make sense? I just realized that in our setting, we actually have three fields: the chiral fermion fields, the gauge fields, and then the fields coming from the representation VV of the gauge group. I suppose you folks know how to describe a ‘complete set of correlation functions’ with all that data…

Posted by: Minhyong Kim on April 2, 2009 3:44 AM | Permalink | Reply to this

Re: Euclidean

I suddenly realized that there is also the choice of a spin structure that goes into all this, so we need to consider something like convergence of oriented spin manifolds. Perhaps this is a good reason to stick to things like spheres, where the spin structure is unique.

On the other hand, I can imagine an average of any quantity over the finitely many spin structures also converging to the right thing, if anything converges at all.

Posted by: Minhyong Kim on April 2, 2009 3:56 AM | Permalink | Reply to this

Re: Euclidean

Heh. I amuse myself looking back at the thread. I started out the questioning sincerely intending to emulate the physicists’ way of thinking, whatever that is, but, as one scrolls down, there is a gradual and natural degeneration into mathematical pedantry. At least maybe it’s amusing for some other people as well.

Once again, confirming the wisdom of the decision to change fields.

Posted by: Minhyong Kim on April 2, 2009 11:22 AM | Permalink | Reply to this

Re: Euclidean

Even worse, come to think of it. A family of choices of principal GG-bundles.

Posted by: Minhyong Kim on April 2, 2009 8:53 AM | Permalink | Reply to this

Re: Euclidean

It is fairly well-known that one must sum over choices (of equivalence classes) of GG-bundle, in addition to integrating over (equivalence classes of) connections on each given GG-bundle.

Try Googling the phrase “cluster decomposition.”

Posted by: Jacques Distler on April 2, 2009 2:05 PM | Permalink | PGP Sig | Reply to this

Re: Euclidean

Ah, that makes sense. On the other hand, perhaps one can just choose spin structures and principal GG-bundles for each space and hope that they will converge to the unique ones on R 4R^4.

In any case, could I trouble you to give a precise formulation of the kind of limit you would regard as allowable when we want to calculate interesting quantities for R 4R^4? Of course including, say, one non-trivial example of the kind of quantities of which I should be considering the limit.

Thanks in advance.

Posted by: Minhyong Kim on April 2, 2009 6:27 PM | Permalink | Reply to this

Re: Euclidean

Minhyong wrote:

Hmm, so one can’t work just with the partition function?

As Jacques noted, the partition function

Z(M)=O(A,ψ)e S(A,ψ)dψdAZ(M) = \int O(A,\psi) e^{-S(A,\psi)} d \psi d A

is usually subject to ambiguities depending on our normalization of the ‘measure’ dAdψd A d \psi — though as he also noted, we can sometimes eliminate those ambiguities.

If we can eliminate those ambiguities, it might be very fun to study how the partition function Z(M)Z(M) varies as we vary the spacetime — that is, the compact Riemannian manifold MM.

However, Z(M)Z(M) is still just a single number, so it’s not very informative. If Z(M)Z(M) converges as MM converges to Euclidean space in some sense, that doesn’t mean everything we’re actually interested in converges.

Conversely, because multiplying the ‘measure’ by a constant doesn’t change anything physically observable, Z(M)Z(M) might fail to converge even if everything we’re actually interested in does converge.

A difficulty I have with the formulation you’ve given is to come up with the observables O(A,ψ)O(A,\psi) that make sense ‘uniformly’ in a family of manifolds.

True!

However, instead of tackling these questions in full generality, it’s probably more productive to start with a special case or two. That way, all sorts of distracting nuances become less important, and we can smash more quickly into the brick wall of real difficulties.

For example: instead of wondering what it means for an arbitrary sequence of compact Riemannian manifolds to approach Euclidean 4\mathbb{R}^4, it’s probably more productive to study the limits

lim nO(A,ψ)e S(A,ψ)dψdAe S(A,ψ)dψdA lim_{n \to \infty} \frac{\int O(A, \psi) e^{-S(A,\psi)} d \psi d A}{\int e^{-S(A,\psi)} d \psi d A}

for one sequence of compact Riemannian manifolds M nM_n that obviously approaches 4\mathbb{R}^4, and for some set of observables that obviously make sense uniformly in nn.

For example: let M nM_n be the torus S 1××S 1S^1 \times \cdots \times S^1 where each circle has radius nn. Let O(A,ψ)O(A,\psi) be an open Wilson network with fixed geometry (more on that later). If you can prove the limit exists in this case, then your grad students — and their grad students — can surely keep generalizing your proof, publishing endless refinements of the basic argument, writing each other letters of recommendation, etc. You know how it goes.

But don’t count your chickens before they’re hatched! What observables O(A,ψ)O(A,\psi) should you use? As you note, Wilson loops are an obvious choice for any sort of gauge theory. More generally you could use ‘Wilson networks’: graphs with edges labelled by representations and vertices labelled by intertwining operators. If our theory only has gauge fields, these may be enough. But if we also have charged fermions, we can use networks with ‘loose ends’ labelled by those fermions. For example, in QCD we can use a ‘Wilson line’ with a quark at one end and an antiquark at the other.

In this situation the observable O(A,ψ)O(A,\psi) depends not just on the combinatorics of the labelled graph, but also on its geometry. However, it’s easy to understand what we mean by ‘keeping the geometry of the graph fixed’ as we expand our torus S 1××S 1S^1 \times \cdots \times S^1 — since a small portion of a torus is isometric to Euclidean space.

In practice many people also discretize the torus: that’s ‘lattice gauge theory’.

I just realized that in our setting, we actually have three fields: the chiral fermion fields, the gauge fields, and then the fields coming from the representation VV of the gauge group.

I’m not sure what you mean by that. In the theories we’ve been discussing, the chiral fermion fields are the field transforming in a certain representation VV of the gauge group.

Posted by: John Baez on April 2, 2009 8:09 PM | Permalink | Reply to this

Re: Euclidean

I’ve been rather ill since arriving at IHES, so I’m tending to stay in my bungalow with the computer on my lap. So while I ponder your points, here are a few quick replies.

“If you can prove the limit exists in this case, then your grad students — and their grad students — can surely keep generalizing your proof, publishing endless refinements of the basic argument, writing each other letters of recommendation, etc. You know how it goes.”

Actually, I really don’t know how this goes, even for myself! This is one of the reasons it’s so frustrating to be my student.

“I’m not sure what you mean by that. In the theories we’ve been discussing, the chiral fermion fields are the field transforming in a certain representation V of the gauge group.”

Am I completely confused? I thought chiral fermion fields referred to sections of even and odd bundles of spinors Δ ±\Delta^{\pm} while we are actually considering a twisted Dirac operator DI+I A:Δ +VΔ VD\otimes I+I\otimes \nabla_A: \Delta^+\otimes V\rightarrow \Delta^-\otimes V (sorry I can’t make one of those fancy Dirac operator symbols) where the connection A\nabla_A acts on a bundle VV associated to a representation of the gauge group. My formulation was strange, but I thought of the vector valued fields as being (sums of) ψs\psi\otimes s where ψ\psi is a spinor and ss is this auxiliary field.

By the way, the spinors are not associated to a representation of the standard model gauge group, are they? Am I writing nonsense?

Posted by: Minhyong Kim on April 2, 2009 8:52 PM | Permalink | Reply to this

Re: Euclidean

Am I completely confused? I thought chiral fermion fields referred to sections of even and odd bundles of spinors

Everything is right. It’s just that the language is used a bit differently than you are imagining here: what is called a “fermion” is the entire section of Δ +V\Delta^+ \otimes V, not just the tensor factors in summands in Δ +\Delta^+.

You can tell this from the way these fermiuons are denoted in physics text, namely as symbols like

ψ α a \psi^a_\alpha

which is, as I suppose you know well, the physicist way of saying that ψ\psi is a section of a tensor product of two bundles, one tensor factor indicated by a{-}^a (this is your VV) and the other indicated by α{-}_\alpha (this is your Δ +\Delta^+).

And to be precise, such a section is not “the fermion”, strictly spealing, but one field configuration of that fermion field.

But that’s all just language, of course.

Posted by: Urs Schreiber on April 2, 2009 9:35 PM | Permalink | Reply to this

Re: Euclidean

I understand. Thanks. Is it also correct that Δ ±V\Delta^{\pm}\otimes V are not associated to representations of the gauge group GG? It seems to me rather that they come from the group SL 2(C)×G.SL_2(C)\times G.

Posted by: Minhyong Kim on April 2, 2009 9:43 PM | Permalink | Reply to this

Re: Euclidean

Minhyong wrote:

I’ve been rather ill since arriving at IHES, so I’m tending to stay in my bungalow with the computer on my lap.

I hope you get better! But then you’ll become too busy to grace us with your presence.

John wrote:

If you can prove the limit exists in this case, then your grad students — and their grad students — can surely keep generalizing your proof, publishing endless refinements of the basic argument, writing each other letters of recommendation, etc. You know how it goes.

Minhyong wrote:

Actually, I really don’t know how this goes, even for myself!

What I meant is: you know how the academic game goes. Someone really smart does something really interesting, and then other people keep on generalizing it. The problem we’re talking about is extremely hard, except perhaps in 2 dimensions, where it’s just ‘quite hard’. It’s not much use figuring out how to formulate generalizations of a difficult result before it’s been proved in a single special case.

Of course, even if it’s not much ‘use’, it could still be fun… but having battered my head against constructive quantum field theory in grad school, I find it more frustrating than fun to state highly general theorems that should be true in this subject.

Am I completely confused?

No, not yet — but I’m working on it.

I thought chiral fermion fields referred to sections of even and odd bundles of spinors Δ ±\Delta^{\pm}, while we are actually considering a twisted Dirac operator… where the connection acts on a bundle VV associated to a representation of the gauge group.

That’s indeed what we’re considering. But most folks typically feel free to use the term ‘chiral fermion’ to refer to a section of a chiral spinor bundle Δ ±\Delta^{\pm} tensored with a bundle VV associated to a representation of the gauge group. After all, a fermion is a kind of particle, and a particle is the whole package.

For example, the left-handed neutrino is a chiral fermion: it’s a section of Δ +\Delta^+ tensored with a bundle VV associated to a representation of U(1)×SU(2)×SU(3)U(1) \times SU(2) \times SU(3). More generally, all the particles I listed in this blog entry are chiral fermions.

So, the problem was you made it sound like the Standard Model involved three kinds of fields:

I just realized that in our setting, we actually have three fields: the chiral fermion fields, the gauge fields, and then the fields coming from the representation VV of the gauge group.

In fact it just has two: gauge fields and what I’m calling chiral fermions — which you might prefer to call VV-valued chiral spinor fields.

(And then of course there’s the Higgs, but we’re not talking about that now.)

Posted by: John Baez on April 2, 2009 9:44 PM | Permalink | Reply to this

Re: Euclidean

Yes, I know what you mean by the academic process, but I’m also quite serious that I’m personally very bad at it. I envy people who have systematic programmes that can be chipped away at in the manner you describe. Instead, I tend to ensnare impressionable young minds with dreamy allusions that sound nice at the outset but are quite hard to turn into real mathematics.

Be that as it may, it certainly would be nice to work this out for some field theories in two dimensions. But it seems to me that this limiting idea can’t possibly work for certain field theories in two-dimensions, that is, *the conformally invariant ones.* If I understand the term correctly, the correlation functions will remain unchanged when we scale the metric, so there’s no need to take an infinite volume limit. On the other hand, I’m too lazy and incapacitated to look it up, but a bunch of string theorists in the 80’s were computing correlation functions for conformal field theories on Riemann surfaces and coming up with functions that definitely depended on the complex structure. That is to say, two tori corresponding to inequivalent fundamental domains in CC will give you different correlation functions, limit or no limit. (I’m assuming we can somehow deal with the question of comparing correlation functions on different surfaces.)

So for which field theories is this idea supposed to work? I’m sure I’m still missing something very basic.

Posted by: Minhyong Kim on April 2, 2009 11:40 PM | Permalink | Reply to this

Re: Euclidean

I think I figured out one point just as I clicked on the ‘post’ button. It must be that to converge to R 2R^2, the family of tori must have square domain. That seems to make sense. But it does show that we certainly shouldn’t take *any* infinite volume limit, even with a plausible topology.

Posted by: Minhyong Kim on April 2, 2009 11:48 PM | Permalink | Reply to this

Re: Euclidean

John wrote:

If we can eliminate those ambiguities, it might be very fun to study how the partition function Z(M)Z(M) varies as we vary the spacetime — that is, the compact Riemannian manifold MM.

However, Z(M)Z(M) is still just a single number, so it’s not very informative.

It’s not just a number. It’s a function of the Riemannian metric, gg. As such, it contains quite a lot of information. For instance, varying it with respect to gg gives the correlation functions of the stress energy tensor of the QFT.

Of course, that’s only one operator, of the multitude whose correlations we might wish to compute. But, if we also allow ourselves to study Z(M)Z(M) for different topology of MM, we can learn more.

Posted by: Jacques Distler on April 3, 2009 1:13 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Minhyong wrote:

But I assume physicists are interested in the Dirac operator on Minkowski space. Even assuming that one applies some kind of ‘Wick rotation,’ which I don’t really understand, how does one deal with the non-compactness?

The usual method is to work on a Riemannian S 4S^4 and hope that no mathematicians are watching.

Or, more precisely, no mathematical physicists — since ‘pure’ mathematicians will let you use whatever manifold you like!

Posted by: John Baez on April 1, 2009 1:23 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I guess the S 4S^4 approach corresponds to allowing only certain kinds of connections. In fact, I seem to recall theorems to the effect that for some class of AA on R nR^n, D AD_A has no discrete spectrum at all.

But joking aside, there must be some justification given, if only loosely, for the S 4S^4 model…

Posted by: Minhyong Kim on April 1, 2009 11:09 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

The conceptual explanation is the Families Index Theorem (for the Dirac operator coupled to a vector bundle V) of Atiyah and Singer…When you compute Ch 3(V)Ch_3(V), you find that it is proportional to tr R({T a,T b}Tc)\tr_R(\{T_a,T_b\}T c) and so vanishes when that particular trace vanishes.

Hey this sounds great. I’m a latecomer to this thread, and I much appreciate the lifeline to learn something about the Standard Model!

I’d very much like to learn more about this anomaly cancellation business for group representations. Is there a friendly reference somewhere that discusses this tr R({T a,T b}T c)\tr_R (\{T_a, T_b\}T_c) term, and the connection to the families index theorem?

Associated to each representation RR, we have a number tr R({T a,T b}T c)\tr_R (\{T_a, T_b\}T_c). What does it measure? I guess the right answer is “it measures Ch 3Ch_3” but I haven’t got my head around what that means.

Posted by: Bruce Bartlett on March 30, 2009 11:40 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I also started out thinking that tr R({T a,T b}T c)\tr_R(\{T_a,T_b\}T_c) is a number; it’s actually a (trilinear) function L 3RealsL^3 \to Reals
Posted by: rntsai on March 30, 2009 11:58 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I guess in the end you really can associate just a single number with a given representation; since these forms are multiples of each other. χ 3(ρ 1)(x,y,z)=αχ 3(ρ 2)(x,y,z) \chi_3(\rho_1)(x,y,z)=\alpha \chi_3(\rho_2)(x,y,z) for example, for SU(5)SU(5), rep 55 can be given index 1,then 1010 would get 1,5 *5^* would get -1, and 10 *10^* would get -1. So none of these would be anomaly free on their own, but 510 *5\oplus 10^* would be.
Posted by: rntsai on March 31, 2009 7:42 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Associated to each representation R, we have a number tr R({T a,T b}T c)tr_R(\{T_a,T_b\}T_c)

A priori, t’s not a number; it’s a symmetric trilinear form on the Lie algebra.

For some groups, there is no candidate for such a symmetric trilinear form; hence the trace must vanish. For other groups (e.g. SU(n),n3SU(n),\, n\geq 3), there is, up to scale, a unique such form, and thus the trace in representation RR gives you a number times that form (defined, say, as the trace in the fundamental representation).

I guess the right answer is “it measures Ch 3Ch_3

If you write the formula for Ch 3Ch_3 in terms of the characteristic polynomial in the curvature, Ch 3(V)=i48π 3Tr(F 3) Ch_3(V) = -\tfrac{i}{48\pi^3}Tr (F^3) you see that you are evaluating exactly such a symmetrized trace. For some choices of structure group, GG, and representation, RR, Ch 3Ch_3 will just vanish identically (ie, independent of which representative connection you choose).

Posted by: Jacques Distler on March 31, 2009 1:22 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Bruce wrote:

Is there a friendly reference somewhere that discusses this tr R(T a,T bT c)tr_R({T_a,T_b}T_c) term, and the connection to the families index theorem?

This is pretty friendly:

Reinhold A. Bertlmann, Anomalies in Quantum Field Theory, Oxford U. Press, 1996.

Posted by: John Baez on March 31, 2009 7:21 AM | Permalink | Reply to this

anomaly cancellation

Bruce wrote:

I’m a latecomer to this thread, […]

I’d very much like to learn more about this anomaly cancellation business

I once wrote a series of blog entries on this. Relevant bits might be here:

Charges and twisted bundles III: Anomalies

and here

Charges and twisted bundles IV: Anomaly cancellation

I still have an email of you, Bruce, dated Feb 27, 2008, where you wrote:

I wish someone had told me that earlier! Now finally I know what an “anomaly” is.

:-)

Posted by: Urs Schreiber on March 31, 2009 9:01 AM | Permalink | Reply to this

Re: anomaly cancellation

That is cruel :D

I know I’ve done the same thing on MANY occasions :)

Posted by: Eric on March 31, 2009 4:44 PM | Permalink | Reply to this

Re: anomaly cancellation

Just to reduce the potential misunderstanding I am causing to a minimum:

I just meant to say: “Hi Bruce, nice to see you interested in anomalies, I recall that we once chatted about that; maybe you remember those links.”

By the way, the only good discussion of anomalies truly undertandable by a mathematician that I have seen is the remarkable article

Dan Freed, Dirac charge quantization and generalized differential cohomology.

This discusses anomalies in abelian gauge theories only (but in higher abelian gauge theories, too) since Freed relies on the tool of differential abelian cohomology, but it gives the cleanest description of anomalies I have seen.

Does any comparable discussion exist for nonabelian gauge theory?

Posted by: Urs Schreiber on March 31, 2009 5:48 PM | Permalink | Reply to this

Re: anomaly cancellation

Now finally I know what an `anomaly’ is.

Doh! I get a big insight one day, only to have to re-learn it again a few months later.

I think my mind was in a spin after seeing John’s ‘triangle’ diagram above:

pic

I love those sorts of diagrams. So it’s really nifty that this esoteric concept of an “anomaly” is related to these diagrams, in this setting.

I now have a shining copy of Anomalies in Quantum Field Theory by Reinhold Bertleman (thanks) from the library, together with the statement he makes on page 245.

When I was writing up my thesis, I did some stuff about ‘fusion categories’, looking at the work of Etingof, Nikshych and Ostrik. Basically, in a semisimple rigid linear monoidal category, associated to any triple of simple objects R i,R j,R kR_i, R_j, R_k there is a weird (to me at least) map

(1)T ijk:Hom(R iR jR k,1)Hom(R iR jR k,1) T_{ijk} : \Hom(R_i \otimes R_j \otimes R_k, 1) \rightarrow \Hom(R_i \otimes R_j \otimes R_k, 1)

given by ‘rotating the simple objects around a full revolution’:

pic

You can give a nice graphical proof that T ijkT_{ijk} is an involution (i.e. it squares to the identity), based on an idea of Hagge and Hong. It turns out that in order for the fusion category to admit a pivotal structure (meaning in some sense: the duality operation on the category is “anomaly-free”), it is necessary that all the T ijkT_{ijk} maps be plus or minus the identity map. Don’t ask me what is the deeper significance of that.

Anyhow, I guess there isn’t a connection with John’s diagram above. This pivotal category business is trivial for unitary stuff, like unitary representations of compact groups. To understand it, I think I will have to understand planar algebras.

I also started out thinking that tr R({T a,T b}T c)\tr_R(\{T_a,T_b\}T_c) is a number; it’s actually a (trilinear) function L 3RealsL^3 \rightarrow Reals

Ok, thanks for clearing up my confusion.

… remarkable article Dan Freed, Dirac charge quantization and generalized differential cohomology.

Yes, I should have read this many times now.

Posted by: Bruce Bartlett on March 31, 2009 8:43 PM | Permalink | Reply to this

Re: anomaly cancellation

Thanks for the link to Bertleman
How recent is that book?

Pardon my failing neurons but I dimly remember there were good anomalies as well as bad ones, e.g. ones that accounted for some physical data - some kind of muons or pions?

Posted by: jim stasheff on April 3, 2009 1:42 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

How would Science Officer Spock approach this problem?

Posted by: Jonathan Vos Post on March 28, 2009 3:46 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Jacques wrote:

If you demand that the gauge anomalies and the mixed gauge-gravitational anomaly for the U(1)U(1) cancel (i.e., that the U(1)U(1) gauge symmetry is not violated in a curved background), then the U(1)U(1) hypercharge assignments (up to an overall normalization that John already complained about) are completely determined.

[…]

Given the SU(3)SU(3) and SU(2)SU(2) representation content, the U(1)U(1) hypercharges could not possibly be anything other than what they are.

and then:

Well, OK, there is a second solution to the anomaly constraints.

Exercise: find and interpret that second solution.

Okay. Okay. I admit that I got stuck and peeked in the back of the book — just to figure out what some of the constraints actually are!

Let’s ignore the right-handed neutrino and introduce this notation:

  • Y(e L)Y(e_L) is the hypercharge of the left-handed electron and neutrino (a total of 2 particles),
  • Y(q L)Y(q_L) is the hypercharge of all colors of left-handed up and down quarks (a total of 6 particles)
  • Y(e¯ L)Y(\overline{e}_L) is the hypercharge of the left-handed positron (a total of 1 particle),
  • Y(u¯ L)Y(\overline{u}_L) is the hypercharge of all colors of left-handed anti-up quark (a total of 3 particles),
  • Y(d¯ L)Y(\overline{d}_L) is the hypercharge of all colors of left-handed anti-down quark (a total of 3 particles).

Let’s continue to use conventions where these hypercharges take integral values. Here are the constraints:

  1. No U(1) ×\times gravitational anomaly: the hypercharges of all the left-handed fermions and antifermions must add up to zero: 2Y(e L)+6Y(q L)+Y(e¯ L)+3Y(u¯ L)+3Y(d¯ L)=0 2Y(e_L) + 6Y(q_L) + Y(\overline{e}_L) + 3Y(\overline{u}_L) + 3Y(\overline{d}_L) = 0
  2. No U(1) 3{}^3 anomaly: the cubes of the hypercharges of all the left-handed fermions and antifermions add up to zero: 2Y(e L) 3+6Y(q L) 3+Y(e¯ L) 3+3Y(u¯ L) 3+3Y(d¯ L) 3=0 2Y(e_L)^3 + 6Y(q_L)^3 + Y(\overline{e}_L)^3 + 3Y(\overline{u}_L)^3 + 3Y(\overline{d}_L)^3 = 0
  3. No U(1)×SU(2) 2(1) \times SU(2)^2 anomaly: the hypercharges of all the left-handed fermions must add up to zero: 2Y(e L)+6Y(q L)=0 2Y(e_L) + 6Y(q_L) = 0
  4. No U(1)×SU(3) 2(1) \times SU(3)^2 anomaly: the hypercharges of all the left-handed quarks and antiquarks must add up to zero: 6Y(q L)+3Y(u¯ L)+3Y(d¯ L)=0 6Y(q_L) + 3Y(\overline{u}_L) + 3Y(\overline{d}_L) = 0

So, we have 4 Diophantine equations in 4 variables. But the equations are homogeneous: given one solution, we can rescale it to get a 1-parameter family of solutions. So, we naively expect a finite set of 1-parameter families of solutions.

I see two 1-parameter families of solutions:

  • The solution chosen by the real world: Y(e L)=3,Y(q L)=1,Y(e¯ L)=6,Y(u¯ L)=4,Y(d¯ L)=2Y(e_L) = -3, Y(q_L) = 1, Y(\overline{e}_L) = 6, Y(\overline{u}_L) = -4, Y(\overline{d}_L) = 2 or any integer multiple of this.
  • The solution where only left-handed antiquarks have nonzero hypercharge: Y(e L)=Y(q L)=Y(e¯ L)=0,Y(u¯ L)=1,Y(d¯ L)=1Y(e_L) = Y(q_L) = Y(\overline{e}_L) = 0, Y(\overline{u}_L) = 1, Y(\overline{d}_L) = -1 or any integer multiple of this.

I never would have guessed the first if I hadn’t known it from the start. The second jumped out at me even before I knew all the constraints!

(Note that we can take all the YY’s to be zero, which is technically a special case of both families, but is physically quite distinct.)

The back of the book says these are the only solutions, but let’s see why.

The 3 linear equations can be written like this:

Y(e L)=3Y(q L) Y(e_L) = -3 Y(q_L)

Y(e¯ L)=6Y(q L) Y(\overline{e}_L) = 6Y(q_L)

Y(u¯ L)+Y(d¯ L)=2Y(q L) Y(\overline{u}_L) + Y(\overline{d}_L) = -2 Y(q_L)

These let us express Y(e L)Y(e_L) and Y(e¯ L)Y(\overline{e}_L) in terms of Y(q L)Y(q_L), and then express that in terms of Y(u¯ L)Y(\overline{u}_L) and Y(d¯ L)Y(\overline{d}_L). Doing this to our cubic equation:

2Y(e L) 3+6Y(q L) 3+Y(e¯ L) 3+3Y(u¯ L) 3+3Y(d¯ L) 3=0 2Y(e_L)^3 + 6Y(q_L)^3 + Y(\overline{e}_L)^3 + 3Y(\overline{u}_L)^3 + 3Y(\overline{d}_L)^3 = 0

we get first this:

54Y(q L) 3+6Y(q L) 3+216Y(q L) 3+3Y(u¯ L) 3+3Y(d¯ L) 3=0 -54 Y(q_L)^3 + 6 Y(q_L)^3 + 216 Y(q_L)^3 + 3Y(\overline{u}_L)^3 + 3Y(\overline{d}_L)^3 = 0

or this:

3Y(u¯ L) 3+3Y(d¯ L) 3=168Y(q L) 3 3Y(\overline{u}_L)^3 + 3Y(\overline{d}_L)^3 = -168 Y(q_L)^3

and then this:

Y(u¯ L) 3+Y(d¯ L) 3=56Y(q L) 3 Y(\overline{u}_L)^3 + Y(\overline{d}_L)^3 = -56 Y(q_L)^3

This is getting spooky! I did a whole issue of This Week’s Finds on the number 168.

Anyway, expressing everything in terms of Y(u¯ L)Y(\overline{u}_L) and Y(d¯ L)Y(\overline{d}_L) we get

Y(u¯ L) 3+Y(d¯ L) 3=7(Y(u¯ L)+Y(d¯ L)) 3 Y(\overline{u}_L)^3 + Y(\overline{d}_L)^3 = 7(Y(\overline{u}_L) + Y(\overline{d}_L))^3

Or, in less complicated notation:

U 3+D 3=7(U+D) 3 U^3 + D^3 = 7(U+D)^3

It’s easy to see that U=1,D=1U = 1, D = -1 is a solution. So is the real world solution U=4,D=2U = -4, D = 2, thanks to this charming fact:

(4) 3+2 3=7(2) 3 (-4)^3 + 2^3 = 7 \cdot (-2)^3

And so, of course, is U=2,D=4U = 2, D = -4, but that’s not really different. I am too exhausted to check that these solutions and their integer multiples are the only ones! Is there any fan of cubic Diophantine equations out there who wants to finish off the job? Minhyong, maybe?

My main reaction to this calculation is that only increases my feeling that the hypercharges in the Standard Model are mysterious. Okay, they’re almost completely determined by anomaly cancellation. But still, they’re pretty weird.

Posted by: John Baez on March 29, 2009 4:07 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

By the way, I should note that any integer multiple of (2,1)(-2,1) or (1,2)(1,-2) is also a solution of U 3+V 3=7(U+V) 3U^3 + V^3 = 7(U + V)^3. However, we don’t just need UU and VV to be integers in this game! We also need (U+V)/2(U+V)/2 to be an integer!

Posted by: John Baez on March 29, 2009 4:14 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I’m not sure why you are so stuck on integers.

If you set that aside for the moment, one notes that we have 4 homogeneous equations in 5 variables, which is enough to determine them up to an overall rescaling.

More specifically, we have three linear equations and one cubic equation, which means that (again, up to an overall rescaling) we expect three solutions.

But, as you discovered, two of the solutions are really “the same” (merely swapping Y(u¯ L)Y(d¯ L)Y(\overline{u}_L) \leftrightarrow Y(\overline{d}_L)).

The remaining solution is the “new” one.

In fact, if you were willing to work a little harder …

U 3+D 3=(U+D)(U 2UD+D 2) U^3+D^3 = (U+D)(U^2-U D+D^2) so your “final” equation U 3+D 3=7(U+D) 3 U^3 +D^3 = 7 (U+D)^3 can be written in simpler form 0 =3(U+D)(2U 2+5UD+2D 2) =3(U+D)(2U+D)(U+2D) \begin{aligned} 0 &= 3(U+D) ( 2 U^2 +5 U D +2 D^2)\\ &= 3 (U+D) (2U +D) (U + 2D) \end{aligned}

This makes the three solutions (and the fact that two of them differ merely by UDU\leftrightarrow D) manifest.

Now, as to normalization…

There are two conventional normalizations for the generator of U(1) YU(1)_Y.

The most natural one is to normalize it in the same way that one normalizes the rest of the generators of Spin(10)Spin(10). The conventional choice is to normalize these generators such that Tr 16T 2=2 Tr_{16} T^2 = 2 In this normalization (with the standard embedding of G SMSpin(10)G_{\text{SM}}\subset Spin(10)), the eigenvalues of the SU(2)SU(2) generators in the 2-dimensional defining representation are ±1/2\pm 1/2, which is the normalization we are all familiar with from elementary QM class.

Anyway, in that normalization, Y(q L)=1215 Y(q_L) = \frac{1}{2\sqrt{15}}

The other natural normalization is the one in which the generator of the unbroken U(1) EM{U(1)}_{\text{EM}} is just the sum of the Cartan generator of SU(2)SU(2) and the generator of U(1) Y{U(1)}_Y.

That normalization differs from the “GUT” normalization, above, by a factor of 5/3\sqrt{5/3}: Y(q L)=16 Y(q_L) = \frac{1}{6}

Your normalization is neither natural from the point of view of GUTs, nor from the point of view of how U(1) EM{U(1)}_{\text{EM}} sits inside of G SMG_{\text{SM}} (you’re not going to make that pesky factor of 1/61/6 go away; you’re just going to displace it into the relation between the generators of U(1) Y{U(1)}_Y and U(1) EM{U(1)}_{\text{EM}}).

[I’ll grant you one thing: since we managed to factorize the cubic equation with integer coefficients, the solutions for all five unknowns are rational multiples of each other. Which means that, by an overall rescaling, we can make them all integers. It was not 100% guaranteed that the cubic would factorize over the rationals. The fact that it had to be symmetric under UDU\leftrightarrow D meant that it could be written as α(U+D)(U+βD)(βU+D)\alpha (U+D)(U + \beta D) (\beta U + D). Because the coefficients were integers, β=(p±p 24q 2)/2q \beta = \left( p \pm \sqrt{p^2 -4 q^2}\right)/2q for some coprime integers p,qp,q. This, I’ll confess, did not have to be rational, though it turned out to be.]

Posted by: Jacques Distler on March 29, 2009 6:54 AM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Okay, after a nice dinner I’ve recovered my steam:

Since the equation is homogeneous, every solution (U,D)(U,D) of

U 3+D 3=7(U+D) 3U^3 + D^3 = 7(U + D)^3

is a constant times a solution with D=1D = 1. So, we need to solve

U 3+1=7(U+1) 3U^3 + 1 = 7(U + 1)^3

But, we’ve already seen that U=1U = -1 is a solution, and also U=2U = -2, and also (if you paid attention) U=1/2U = -1/2. And that’s all the solutions this cubic can have.

So, given how the fermions and antifermions in one generation transform under SU(2)×SU(3)SU(2) \times SU(3), anomaly cancellation implies that the only allowed hypercharges are these:

  1. The solution chosen by the real world: Y(e L)=3,Y(q L)=1,Y(e¯ L)=6,Y(u¯ L)=4,Y(d¯ L)=2Y(e_L) = -3, Y(q_L) = 1, Y(\overline{e}_L) = 6, Y(\overline{u}_L) = -4, Y(\overline{d}_L) = 2 or any integer multiple of this.
  2. The solution where only left-handed antiquarks have nonzero hypercharge: Y(e L)=Y(q L)=Y(e¯ L)=0,Y(u¯ L)=1,Y(d¯ L)=1Y(e_L) = Y(q_L) = Y(\overline{e}_L) = 0, Y(\overline{u}_L) = 1, Y(\overline{d}_L) = -1 or any integer multiple of this.
Posted by: John Baez on March 29, 2009 7:43 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Jacques wrote:

I’m not sure why you are so stuck on integers.

I’m a mathematician. I like integers. I prefer using integers to parametrize irreps of U(1)U(1) than using integers divided by 3 or 6. But that’s no big deal.

More importantly, as you note, we have to start by worrying that not every solution of the constraints can be rescaled to be an integer solution. But as it turns out, all 3 roots of the dehomogenized equation U 3+1=7(U+1) 3U^3+1=7(U+1)^3 are rational, so that’s not a problem.

Posted by: John Baez on March 29, 2009 7:50 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

More importantly, as you note, we have to start by worrying that not every solution of the constraints can be rescaled to be an integer solution

As far as I can see, that’s the only thing we had to worry about. If there’s any mystery at all about U(1) Y{U(1)}_Y charge assignments, it’s that the anomaly constraints yielded only rational solutions.

The embedding in a (semi)simple group guaranteed this, but working only from the anomaly constraint equations themselves, it was not obvious that this would work out.

I’m a mathematician. I like integers. I prefer using integers to parametrize irreps of U(1) than using integers divided by 3 or 6.

Yes, but if you’re trying to explain GUTs, you should use the same normalization for the generator of U(1) Y{U(1)}_Y as for the other generators. The definition of the U(1) Y{U(1)}_Y gauge coupling depends on your choice of normalization. And, when one says “the gauge couplings unify at the GUT scale,” that’s only true when one uses the GUT normalization for U(1) Y{U(1)}_Y.

Posted by: Jacques Distler on March 29, 2009 4:36 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

In email communication, John expressed the hope of getting elliptic curves. This is what would (most likely) have happened if the anomaly constraint eventually resulted in a homogeneous cubic equation of three variables rather than two. I agree that it would be interesting to come up with natural examples of this sort.

But there was such a paper:

Bremner, Murray R.(1-MSRI) Modular invariant Virasoro modules and elliptic curves. Lett. Math. Phys. 20 (1990), no. 2, 113–123.

I’ve never read it, but I suspect that the context was quite close to the anomaly cancellation conditions of the sort we’re interested in. At least that’s what I remember from his talk 20 years ago. Murray worked with George Seligman at Yale around the same time I was working with Igor Frenkel. That’s the reason I happen to know about this.

Posted by: Minhyong Kim on April 2, 2009 7:36 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

It may not be too relevant here, but I get the 16 dimensional rep of so(10) giving hypercharges (using above convention)
-3,-3,1,1,0,6,-4,2
(the last 4 are inverted). Are both
combinations possible?

Posted by: rntsai on March 26, 2009 8:11 AM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

That’s because John, in an attempt to further obfuscate matters, refers to the hypercharge assignments, not of the left-handed fermions (which fit into a 16 of Spin(10)Spin(10)), but to the hypercharges of some mixture of left-handed and right-handed fermions.

Since the right-handed fermions transform in the 16¯\overline{16}, you should flip the hypercharge assignments that John wrote down for them, to obtain the hypercharges of the corresponding left-handed fermions.

Note, too, that one of those fermions is not like the others. One of those fermions has a large (nearly GUT-scale) mass, and does not appear in the low-energy theory.

Posted by: Jacques Distler on March 26, 2009 2:04 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Indeed one of those fermions, *and two of those bosons*, are not like the others. Both of them get mass from the electroweak scale. So one could expect that in the limit where the W and Z are massless, so is the top.

Incidentally, the massless and infinitely massive limits live, respectively, in 7 and 5 extra dimensions from the point of view of Kaluza Klein, do they?

Posted by: Alejandro Rivero on April 6, 2009 2:35 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Rivero wrote

Indeed one of those fermions, *and two of those bosons*, are not like the others. Both of them get mass from the electroweak scale.

No.

  1. No bosons were mentioned in that comment.
  2. The physics I was talking about, when I said “one of those fermions is not like the others” has nothing to do with the electroweak scale.

and

So one could expect that in the limit where the W and Z are massless, so is the top.

All of the remaining 15 fermions are massless in that limit.

Incidentally, the massless and infinitely massive limits live, respectively, in 7 and 5 extra dimensions from the point of view of Kaluza Klein, do they?

I don’t understand.

Posted by: Jacques Distler on April 6, 2009 3:37 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

I apologize. Second reading, I suppose you refer to the neutrino and its GUT mass.

In any case, the observation about the mass of gauge bosons extends to GUT: the neutrino is different because there is a scale where the symmetry is broken. If the symmetry were unbroken, the neutrino should stand in the same foot that the rest of the particles. Thus for a discussion of the algebra of GUT not including the mechanism for symmetry breaking, I do not see if it is relevant to argue about a different fermion. Of course it could be very different :) if there were a mechanism where this fermion is the cause of the breaking.

Posted by: Alejandro Rivero on April 6, 2009 4:13 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

About the extra-dimensional off topic. The point was that the two limits of the ElectroWeak model are unbroken gauge theories SU(3)xSU(2)xU(1) when M WM_W, M Z0M_Z \to 0 and SU(3)xU(1) when M W,M ZM_W, M_Z \to \infty.

Following classical arguments, these groups can be obtained from Kaluza Klein theories with 7 and 5 extra dimensions, ie in spacetime of dimension 11 or 9, respectively. These KK theories - at least for dimension 11- do not admit chiral fermions, and then they are a no go. But I find amusing that the old good electroweak model happens to jump across the same dimensions that string dualities jump.

Posted by: Alejandro Rivero on April 6, 2009 4:31 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

I’ll fix this. By the way, Jacques: on this blog, even when people fail to explain things well, the other participants don’t accuse them of ‘attempting to obfuscate’ the issue. They simply point out the mistake. That way they come out looking helpful instead of rude. We’re actually incredibly polite around here, and I like that.

What I’d really hope is that you could say a bit about my question. You must know a lot about this sort of thing.

Posted by: John Baez on March 26, 2009 6:35 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

for so(10) one of the 16 reps gives
-3,-3,1,1,0,6,-4,2
and the other 16 gives
3,3,-1,-1,0,-6,4,-2
(besides hypercharge, these reps can give
the correct weak isospin and charge)

if you move to so(12), each of the 32 reps
give the union of the above. Interestingly
enough the adjoint rep (66) has a third copy (as a subset)

Posted by: rntsai on March 26, 2009 8:00 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

rntsai wrote:

for so(10) one of the 16 reps gives

-3,-3,1,1,0,6,-4,2

and the other 16 gives

3,3,-1,-1,0,-6,4,-2

Right, because these reps are dual to each other. In our paper, John Huerta and I focus on the direct sum of these, which is just the exterior algebra Λ 5\Lambda \mathbb{C}^5. These two summands are the even and odd parts.

if you move to so(12), each of the 32 reps give the union of the above.

Hmm! Good point! Yeah, that’s because the above 16-dimensional reps are the left-handed and right-handed complex spinor reps of so(10), and quite generally:

  • taking the left-handed or right-handed complex spinor rep of so(2n), and restricting it to so(2n-1), we get the unique complex spinor rep of so(2n-1), and
  • taking the unique complex spinor rep of so(2n-1), and restricting it to so(2n-2),we get the direct sum of the left-handed and right-handed spinor reps of so(2n-2).

Has anyone tried extending the SO(10)SO(10) GUT to an SO(12)SO(12) GUT by this method? (Surely yes; everything in this realm has been tried.) Has anything intriguing emerged?

Interestingly enough the adjoint rep (66) has a third copy (as a subset)

Hmm, now that’s a lot less obvious. Unlike everything so far, this must be special to the particular dimensions 10 and 12. Could this coincidence be used in supergravity or something? Getting 11d spinors to sit inside so(12)?

Posted by: John Baez on March 26, 2009 8:24 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

more concisely: this is how
so(12) irreps decompose into su(5) irreps
(branching rules) :

32 -> (1+5+10+10’+5’+1)
32’-> (1+5+10+10’+5’+1)
66 -> (1+5+10+10’+5’+1) + 5+5’+24

where what’s between the brackets can be
identified with the 32 pacticles

Posted by: rntsai on March 26, 2009 10:17 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

More succinctly, the irreducible spinor representations of Spin(12)Spin(12) are pseudo-real (quaternionic). Under Spin(10)Spin(10), they decompose as 32 =1616¯ 32 =1616¯ \begin{aligned} 32 &= 16\oplus \overline{16}\\ 32' &= 16\oplus \overline{16} \end{aligned} That is, starting with Spin(12)Spin(12), you can never get a chiral gauge theory (like the Standard Model).

Posted by: Jacques Distler on March 26, 2009 10:41 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

My apologies.

I’ve become overly sensistive when it seems that people (I am thinking, in particular, of Chamseddine, Connes and Marcolli, but there are others) go out of their way to make the Standard Model seem much more complicated than it really is.

I know that wasn’t your intention.

It is complicated, but the complicated parts aren’t the the bits you’ve been talking about.

What I’d really hope is that you could say a bit about my question. You must know a lot about this sort of thing.

I don’t know much about the current experimental status of non-supersymmetric Spin(10)Spin(10) GUTs.

Spin(10)Spin(10) is attractive because, now that we have definitive evidence for neutrino masses, Spin(10)Spin(10) provides a natural realization of the seesaw mechanism: ie, a mechanism for generating the dimension-5 operator (which schematically looks like h 2ψψh^2 \psi\psi, where ψ\psi is the lepton doublet, hh is the Higgs doublet, and the SU(2)SU(2) and spinor indices are contracted in the obvious way) responsible for those neutrino masses, with a coefficient suppressed by a very high mass scale (nearly the GUT scale).

You can obviously implement such a mechanism in other theories, but in Spin(10)Spin(10), it comes for free.

(Well, sorta for free, having e.g., a Higgs in the 126 — the self-dual part of 5V\wedge^5 V, where VV is the defining representation of 𝔰𝔬(10)\mathfrak{so}(10) — hardly feels like “free.”)

Posted by: Jacques Distler on March 26, 2009 10:33 PM | Permalink | PGP Sig | Reply to this

Re: The Algebra of Grand Unified Theories III

Have physicists listed all reductive groups intermediate between the standard model gauge group and Spin(10)Spin(10)?

For example, does the following chain make sense?

SU(3)×SU(2)×U(1)SU(3)\times SU(2)\times U(1), G 2×SU(2)×U(1)G_2\times SU(2)\times U(1), Spin(7)×Spin(3)×U(1)Spin(7)\times Spin(3)\times U(1), Spin(10)Spin(10).

I am not convinced about the U(1)U(1).

Posted by: Bruce Westbury on March 26, 2009 8:41 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Bruce wrote:

Have physicists listed all reductive groups intermediate between the standard model gauge group and Spin(10)?

If they haven’t, someone is being overpaid! It’s pretty important, and probably not all that hard.

It would be good to understand the whole partially ordered set of reductive subgroups between Spin(10)Spin(10) and the Standard Model subgroup ((U(1)×SU(2)×SU(3))/ 6(1) \times SU(2) \times SU(3))/\mathbb{Z}_6 — though physicists may have already discarded a bunch of these as unpromising. And not just the isomorphism classes of subgroups, but the inclusion ordering, because ‘chains’ of subgroups are important. I guess the papers I cited claim to list all chains with two intermediate steps.

Allen Knutson keeps telling me about the recipe for finding maximal-rank Lie subalgebras of semisimple Lie algebras, and I keep forgetting it…

Hmm, now I see that Allen helped Jacques Distler and Eric Sharpe do this for E 8E_8 in Appendix A of their paper Heterotic compactifications with principal bundles for general groups and general levels. I’m not sure if this method gets at reductive subgroups, or just semisimple ones.

For example, does the following chain make sense?

SU(3)×SU(2)×U(1),G 2×SU(2)×U(1),Spin(7)×Spin(3)×U(1),Spin(10). SU(3) \times SU(2) \times U(1), G_2 \times SU(2) \times U(1), Spin(7)\times Spin(3) \times U(1), Spin(10).

I am not convinced about the U(1).

Yeah, I don’t see how you fit an extra U(1)U(1) in Spin(10)Spin(10) that commutes with everything in Spin(7)×Spin(3)Spin(7) \times Spin(3). It’s easy to fit U(1)×SU(n)×SU(m)U(1) \times SU(n) \times SU(m) into SU(n+m)SU(n+m), but that extra U(1)U(1) consists of diagonal matrices with unit complex numbers as entries. We can’t use that particular trick to similarly stuff U(1)×SO(n)×SO(m)U(1) \times SO(n) \times SO(m) inside SO(n+m)SO(n+m), or U(1)×Spin(n)×Spin(m)U(1) \times Spin(n) \times Spin(m) inside Spin(n+m)Spin(n+m).

(Note, I’m following you here in not distinguishing between groups that have the same Lie algebra. We don’t really have Spin(7)×Spin(3)Spin(7) \times Spin(3) as a subgroup of Spin(10)Spin(10), just (Spin(7)×Spin(3))/ 2(Spin(7) \times Spin(3))/\mathbb{Z}_2. Similarly, we don’t really have U(1)×SU(n)×SU(m)U(1) \times SU(n) \times SU(m) inside SU(n+m)SU(n+m), just a quotient of U(1)×SU(n)×SU(m)U(1) \times SU(n) \times SU(m) by a finite normal subgroup. But such nuances get a bit tiresome at times — and one can worry about them later, after understanding things at the Lie algebra level.)

Posted by: John Baez on March 26, 2009 11:11 PM | Permalink | Reply to this

Re: The Algebra of Grand Unified Theories III

Regarding the phenomenology of non Susy SO(10):

There are ‘viable’ models that exist that have not been ruled out, including some that mimic some of the SuSY version successes. Otoh quite generally they do require an intermediate scale or two (like the PS route)) and some finetuning.

A lot of the industry will be greatly elucidated the second the LHC gets results back. Particularly relevant is whether or not we need an axion as a DM particle or not, b/c a lot of the remaining calculationally viable models like having a PQ U(1) floating around, so not entirely minimal.

Posted by: Haelfix on April 3, 2009 10:01 AM | Permalink | Reply to this

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