The n-Category Café

When I was making my way back into physics thanks to Henk van Dam and especially Tom Kephart, the language problem was a challenge. When Tom referred to the `little group’, it took me a while to realize it meant an isotropy subgroup. Even more challenging, the arrow went from the big group to the little group! because it was really denoting the restriction of irreps.
Hopefully more of us on both sides of the linguistic divide are becoming bilingual.

Unless you’re a dirty two-timing rat, SO(2,2) is not relevant to physics.

Given the revival in the last five years or so of twistor techniques and other ways of analytically continuing momenta to compute scattering amplitudes, a lot of physicists are now two-timing rats.

By the way, the splitting into an internal symmetry group and the Poincare group makes me think of Kaluza-Klein theories. Is there a neat way to formulate things in one big group where the distinction between spacetime and the internal gauge group disappears?

Yes, indeed, as you say yourself, this is the idea of Kaluza-Klein theory: the internal group is really the diffeomorphisms of the second factor $Y$ of a higher dimensional pseudo-Riemannian spacetime which locally looks like $M^4 \times Y$ with $Y$ a compact Riemannian manifold of very small Riemannian volume.

The basic idea is simple and works well, except for one little subtlety: in the full context of general relativity (as opposed to the special relativistic setup considered in the discussion here so far) the volume of the “internal space” $Y$ is itself dynamical, and hence apart from the desired Poincaré and internal group, the Kaluza-Klein mechanism always yields one more phenomenon: so-called “scalar moduli” which parametrize these scale degrees of freedom of the internal manifold.

As these moduli fields are not being observed, taken at face value this rules out KK-theories. On the other hand, since apart from the the KK-mechanism is so nice, one may start thinking about extending the idea slightly so as to take care of this problem. There are plenty of ideas on this and it becomes a long story. I’ll just mention this aspect:

In the string-theoretic context, after many years of being an open problem, the moduli were finally shown to be “fixed” i.e. achieve a constant dynamics which would, generally speaking, be consistent with these fields not being observed, by assuming that certain higher degree differential form fields (higher gauge theory fields) called “fluxes” in this context, are in a classical condensate state. So higher gauge theory comes to the rescue of the old KK-mechanism!

See also maybe the table at the very end of the entry Connes on spectral geometry of the standard model which lists which aspects of particle physics are encoded by which aspects of a Kaluza-Klein geometry.

Todd asked

[hey, how do you make that semidirect fish symbol?]

\ltimes gives $\ltimes$, \rtimes gives $\rtimes$.

(Thanks, internet!)

So how would a mathematician say it precisely, to make everything right?

I believe the way this is usually interpreted is with a rigged Hilbert space. The basic idea, as I understand it, is as follows. Given an unbounded normal operator $T$ on a separable Hilbert space $H$, it is possible to find a topological vector space $\Phi$ and a continuous embedding of $\Phi$ into $H$ such that the image of $\Phi$ is dense and contained in the domain of $T$, and such that the dual space $\Phi'$ contains a complete set of eigenvectors for $T$. In other words, there is a measure $\mu$ on the spectrum $\sigma(T)$ such that, for almost every $\lambda \in \sigma(T)$, there exists $\phi_{\lambda} \in \Phi'$ such that $\phi_{\lambda}(T x) = \lambda\phi_{\lambda}(x)$ for any $x \in \Phi$, and furthermore,

I believe this statement is more or less equivalent to the usual spectral theorem for unbounded normal operators.

As a simple example, we can recover the Fourier transform from the spectral theory of $\frac{d}{d x}$. We embed the Schwarz space $\mathcal{S}$ into $L^2(\mathbb{R})$; its dual is the space $\mathcal{S}'$ of tempered distributions. The eigenvectors of $\frac{d}{d x}$ lying in $\mathcal{S}'$ are the functions $e^{i k x}$ for $k \in \mathbb{R}$. (Note that none of these lie in $L^2(\mathbb{R})$ itself.) If we take $\mu$ to be a properly normalized Lebesgue measure on $\mathbb{R}$, then the completeness equation above is just a statement of Plancherel’s theorem.

As long as it’s a learning experience, I’ll throw in something about fundamental scalar fields which hasn’t been mentioned at all: they shouldn’t happen. Their existence violates certain “naturalness” considerations, which basically boil down to saying that our theories shouldn’t need exact cancellations between very large numbers to maintain small ones — unless, that is, such a cancellation is enforced at some “deeper” level by a symmetry. So actually the pions are very nice as scalar fields, because they are Goldstone bosons, so escape this naturalness problem. Gauge bosons are also protected, by having the gauge symmetry in the first place.

Now for those who care a bit more:

When we write down a field theory, we are really encoding how spatially separate bits of the field interact with each other. We usually demand that the theory is “local”, which is to say that they only interact in their immediate (spacetime) neighbourhood. In theory, this is the mathematically ideal — infinitely small, infinitely close together. In practice however, our instruments only measure over an extended volume (in both space and time). So we usually incorporate this into theory as saying that the field has an ultraviolet cutoff — that is, fourier modes above some limit simply don’t exist. Concretely, we set integrals over momenta to have an upper limit (there are other ways, but they are technical improvements on the same idea).

But what if we have deliberately crude instruments (or if you live in a technologically inferior civilisation that can’t even build machines to probe the Planck scale)? Well, if we don’t excite too much the modes with length scale smaller than the instrument size, then we ought to be able to fudge in their effects by some statistical averaging. Concretely, you would take your Feynman sum over paths and perform the sums which are over the frequencies you now don’t care about. The actual effect of this is that you end up changing the dimensionless constants of your Lagrangian, even changing some of them from zero to non-zero. This coarse graining strictly loses you information, but that’s not a bad thing, as long as you end up focusing on the stuff you want. In fact, for physics, the ideal theory just talks about what you care about, and the rest somehow end up not mattering, or are just absorbed into some constants. So actually, there are good reasons to look at what happens in the limit of your instruments being macroscopically sized. This flowing of the constants is usually called renormalisation.

Now for some magic: most of the numbers will actually go downwards, and only a finite number will go up or stay the same. We name these irrelevant, relevant and marginal, respectively. Relevant terms which grow to much greater than unity are telling you that you should change your basic set of fields — because they are no longer weakly coupled together, and you will no longer be able to treat their interactions perturbatively. For instance, QCD goes from quarks and gluons to being confined, and it then makes more sense to talk about hadrons and mesons; in superconductors, you stop talking about electrons, and talk about Cooper pairs instead. Now we can make precise what I said earlier about naturalness: irrelevant terms will go to zero, relevant terms will necessitate a different set of fields to continue the renormalisation, and only the marginal ones should remain at the largest scales.

The mass term of a scalar field is relevant. This means that as we look at lower and lower energy phenomenon, the scalar field should actually drop out entirely of the set of fields we wish to consider, since it’s so massive that no process will actually create it. The only way for it to stay, is if somehow the various other numbers conspired to keep it small yet non-zero — but this essentially requires very large numbers to cancel off against each other, almost yet not quite. It will require, as a consequence, that we measure some non-zero, dimensionless numbers to absurd precision to be able to have any predictive power.

So what about the Higgs? The currently accepted method of introducing it is essentially chosen because of simplicity — it’s the simplest way to break electroweak symmetry and end up with things as we see it. However, because of the naturalness problem I’ve (tried, but probably failed) to explain, there are good reasons to believe that the picture fails just above the electroweak scale. This is essentially why we bothered to spend the billions to build the LHC — we’ve got actual theoretical reasons to believe that there are new, experimentally accessible phenomenon right above when we’re supposed to see the Higgs.

By the way, when I seem to get overheated and start scolding someone — say, for forgetting about the pions, or only caring about particles that are currently considered fundamental — it’s often because I’m preaching to the world at large […]

No problem at all! I wasn’t upset at all about your reply, certainly not.

I think I was just trying to say: oh, if you are not restricting yourself to particles which are fundamental, well then there is a large supply of spin-0 particles: just take any bound state with total spin vanishing. For instance I am a spin-0 particle to good approximation, for that matter!

Model builders therefore compute with compound spin-0 particles all the time. Not just with pions. For instance most computations of quantum fluctuations on cosmological backgrounds are done this way, for instance.

On top of that, it might be that even the spin-0 particles which are considered fundamental today, notably the Higgs, are secretly actually composite particles. There are various theories postulating this. None of them considered overly likely today, but certainly possible.

The people who’re actually daring to speak up in this conversation are just a tiny fraction of the people listening in.

True that.

Thank you, John! That’s very good to know.

The fact that physicists don’t come out and say this is not very surprising; it just doesn’t seem to be part of the prevailing culture to get involved with such details. Maybe the mathematicians do come out and say this, but I’ve not seen it (admittedly, I’m not all that well-read in the literature on the mathematical side). It’s the type of thing that I get hung up on, and I’ll bet there are hundreds or even thousands like me in this respect.

Just in case anyone is following this here and wondering: there is more discussion of this here on the nForum – a little bit right now and hopefully more in the near future.

I have just had one of the most insanely busy weeks of my life, with an article I am co-authoring, a book that I am co-editing and a bachelor thesis that I am advising by cosmic coincidence all wanting to finish right now… and I must go on vacation tomorrow morning.

But once I have a minute, I hope to join Todd on his quest to write out these things more clearly and more publicly on the $n$Lab…

Or because they are physicists talking to a restricted community who KNOW what’s going on or are presumed to?

Even if the background is fixed?

I vaguely remember in grad school, we would often assume “periodic boundary conditions”, which is equivalent (I think) to assuming space was a 3-torus. Then we let the size of the torus go to infinity.

The only reason I bring it up at all is that I seem to recall (from one of John’s lectures ages ago) that some of the mathematical difficulties with representations of the Poincare group have to do with the fact that $\mathbb{R}^4$ is infinite (or something). I was wondering if $\mathbb{R}^4$ might not be the simplest starting point. Is there a simpler group that becomes the Poincare group in some limit, e.g. a 4-torus in the limit as the radius in all 4 dimensions foes to infinity? I didn’t intend to introduce additional complexity and was actually hoping to start simpler.

Eric wrote:

… I seem to recall (from one of John’s lectures ages ago) some of the mathematical difficulties with representations of the Poincaré group have to do with the fact that $\mathbb{R}^4$ is infinite (or something).

Yes, the fact that $\mathbb{R}^4$ is noncompact makes life a bit difficult — but maybe even more important is the fact that the Poincaré group itself is noncompact. The representation theory of compact Lie groups is a piece of cake, by comparison, because all the irreps are finite-dimensional.

Is there a simpler group that becomes the Poincare group in some limit, e.g. a 4-torus in the limit as the radius in all 4 dimensions foes to infinity?

Unfortunately not. The Poincaré group is 10-dimensional: 4 translations, 3 rotations, 3 boosts. The isometry group of a 4-torus is just 4-dimensional: only translations.

The Poincare group is 10-dimensional, and 10 = $4(4+1)/2$. In general the maximum dimension of the symmetry group of an $n$-dimensional spacetime is $n(n+1)/2$, and the cases where the maximum is reached are known.

In dimension 4, if we’re talking about Lorentzian manifolds — meaning 1 time dimension plus 3 space — the options are these: Minkowski spacetime, DeSitter spacetime, and anti-DeSitter spacetime. For all of these, the isometry groups are noncompact and the relevant irreps are infinite-dimensional. The representation theory is probably easiest, or at least most familiar, in the Minkowski case. So, that’s what we’re talking about.

If you’re desperate to get a compact Lie group, to get a bunch of finite-dimensional reps, you can switch to Riemannian manifolds — where all 4 dimensions are spacelike. Then the maximally symmetric 4d ones are 4d Euclidean space, 4d hyperbolic space, and the 4-sphere. The isometries of the 4-sphere form the group $SO(5)$ — check that this is indeed 10-dimensional — and only in this case do we get a group that’s compact.

People know everything there is to know about the irreps of SO(5), and Todd probably knows plenty of this stuff, since he and Jim spent ages discussing Dynkin diagrams. There is a link between representations of SO(5) and representations of the Poincaré group, but it’s so indirect that without knowing the Poincaré reps ahead of time, it would be a major feat to see how this works! Starting from the 4-sphere, you have to take the limit as the radius goes to infinity and replace one dimension of space by one dimension of time, to reach Minkowski spacetime. Each procedure is separately fascinating, but I don’t have the energy to lead us through both of them.

So, I really urge that we dive in and directly attack Wigner’s classification of irreps of the Poincaré group. It’s incredibly beautiful math and incredibly important physics.

Eric wrote:

I vaguely remember in grad school, we would often assume “periodic boundary conditions”, which is equivalent (I think) to assuming space was a 3-torus. Then we let the size of the torus go to infinity.

[…]

Is there a simpler group that becomes the Poincare group in some limit, e.g. a 4-torus in the limit as the radius in all 4 dimensions goes to infinity?

As mentioned, the answer to this question is no. But to be fair, I should say that it is quite interesting to ponder particles on spacetimes like $\mathbb{R} \times T^3$ and $\mathbb{R} \times S^3$.

When space is the 3-torus $T^3$, the isometry group is just 4-dimensional: space and time translations. This means we get concepts of momentum and energy, but no angular momentum – bummer! On on the other hand $T^3$ is a compact abelian group so we can take full advantage of Fourier series: that’s why people like ‘periodic boundary conditions’.

When space is the 3-sphere $S^3$, the isometry group is 7-dimensional: 6 dimensions for the rotational symmetries of the 3-sphere, plus time translation. So, we get 6 conserved quantities reminiscent of momentum and angular momentum, together with energy. Note that momentum and angular momentum are ‘unified’, since there’s no real difference between a ‘translation’ of the 3-sphere and a ‘rotation’. But, we still don’t get anything like Lorentz boosts.

My whole thesis was actually about quantum field theory on $\mathbb{R} \times S^3$!

OK, since everybody else has gone quiet, I’ll step up. I’ve been cheating and looking at Shlomo Sternberg’s delightful Group Theory and Physics, so I think I can more or less explain the derivation of the Klein-Gordon equation.

As Bruce explains, the group $G = SL_2(\mathbb{C}) \ltimes \mathbb{R}^4$ acts on the space $\widehat{N}$ of irreducible representations of $N = \mathbb{R}^4$. The representation of $N$ corresponding to the point $p = (E, p_x, p_y, p_z)$ will be the unitary character

where $g(p, X)$ is shorthand for $E t - p_x x - p_y y - p_z z$, where $X = (t, x, y, z)$.

Next, we pick a point $p$ with $g(p, p) = m^2 \gt 0$ and $E \gt 0$. As John points out, such a point corresponds to a massive particle. The stabilizer of the point $p$ is isomorphic to $H = SU(2) \ltimes \mathbb{R}^4$. Since we are considering “spin-zero” representations, we want $SU(2)$ to act trivially. So the point $p$ corresponds to a representation of $H$ where $SU(2)$ acts trivially, and the $\mathbb{R}^4$ part acts like $\chi_p$. We’ll abuse notation and continue to call this representation $\chi_p$.

Now, we want to turn this representation of $H$ into a representation of $G$. We do this by constructing a certain vector bundle over $G / H$ and taking its $L^2$ sections (where integrability is taking with respect to an invariant measure). Specifically, we are looking at the vector bundle $E = G \times \mathbb{C} / H$, where $H$ acts on $G \times \mathbb{C}$ by $h \cdot (g, s) \mapsto (g h, \chi_p(h^{-1})s)$. $G$ acts on this vector bundle by left-multiplication on the left coordinate, and this action gives us our representation of $G$ on global sections $\Gamma(E)$.

I claim that this bundle is actually trivial. Define a function $S: G \to \mathbb{C}$ by $S(A X) = \chi_p(-X)$, where $A \in SL_2(\mathbb{C})$, $X \in N$. This function is equivariant with respect to the right action of $H$, so it defines a function $\sigma: G / H \to E$ by $\sigma(g H) = (g, S(g))$. We then get a bundle isomorphism from $E$ to the trivial bundle by sending a section $\gamma \in \Gamma(E)$ to the function $\gamma / \sigma$.

We have thus constructed a unitary representation of $G$ on $L^2(G / H)$. But $H$ was the stabilizer of the point $p$ in $\widehat{N}$, so what we have is a representation of $G$ on $L^2$ functions on the $E \gt 0$ branch of the hyperboloid $g(p, p) = m^2$ in $\widehat{N}$. We can then think of a function $f \in L^2(G / H)$ as a certain distribution on $\widehat{N}$ supported on this hyperboloid, i.e., a distribution of the form

where we assume that $f$ vanishes along the negative energy branch of the hyperboloid and is square integrable with respect to the invariant measure on the positive energy branch.

A distribution will be of the form (2) if and only if $(g(p, p) - m^2)u = 0$. Taking the inverse Fourier transform of this equation turns the multiplication operators $E^2$, $p_x^2$, $p_y^2$, and $p_z^2$ into the differentiation operators $-\frac{\partial^2}{\partial t^2}$, $-\frac{\partial^2}{\partial x^2}$, $-\frac{\partial^2}{\partial y^2}$, and$-\frac{\partial^2}{\partial z^2}$, respectively, so we recover the Klein-Gordon equation

We have thus realized our representation of $G$ on the space of solutions of the Klein-Gordon equation whose Fourier transforms satisfy certain constraints (namely, positive energy and integrability).

Anyway, I may have gotten a few details wrong, so please feel free to correct me as needed. A similar story holds for representations with nonzero spin, but with the additional complication that our vector bundle is nontrivial. But we should probably work the kinks out of the “easy” spin-zero case before moving on.

The $X$ here is the spectrum of $A$? Which can be identified with a subspace inclusion $i: X \hookrightarrow \mathbb{R}$ (with the measurable function $a: X \to \mathbb{R}$ being this very inclusion)?

Todd wrote:

The $X$ here is the spectrum of $A$? Which can be identified with a subspace inclusion $i: X \hookrightarrow \mathbb{R}$ (with the measurable function $a: X \to \mathbb{R}$ being this very inclusion)?

Wow, you’re asking a question about something I said two years ago! Let me try to remember what we were talking about…

Initially, I said:

The spectral theorem says:

Suppose $A$ is a (possibly unbounded) self-adjoint operator on a Hilbert space. Then this Hilbert space is isomorphic to $L^2(X)$ for some measure space $X$, and making use of this isomorphism, $A$ becomes a multiplication operator:

$$(A \psi)(x) = a(x) \psi(x)$$

where

$$a : X \to \mathbb{R}$$

is some measurable function. And in this situation,

$$(exp(- i t A) \psi)(x) = e^{-i t a(x)} \psi(x)$$

Here $X$ is not the spectrum of $A$, except when the ‘multiplicity’ of each point in the spectrum is 1.

To see what I mean, let $A$ be an $n \times n$ self-adjoint matrix. If each eigenvalue of $A$ has multiplicity 1, then we can take $X$ to be the spectrum of $A$. Otherwise, we can’t. Consider for example:

$$A = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$$

The spectrum of $A$ contains one point, but we need $X$ to contain two points.

This is why I later said:

And now we see a flaw in how I stated the spectral theorem. It tells us that the space $X$ and the function $a$ exist, but it doesn’t say what they are.

You can fix this flaw, but it’s not as easy as simply taking what I said and adding the remark that $X$ is the spectrum of $A$ and $a: X \to \mathbb{R}$ is the inclusion. That’s why I was, and am still, too lazy to fix the flaw. But I could be persuaded to fix it if I thought someone really needed to know how. In general $X$ is some kind of ‘bundle’ of measurable spaces over the spectrum of $A$, where the cardinality of the fiber, the ‘multiplicity’, can vary from point to point.

Greg wrote:

I wonder if anyone would like to say a bit about the representation for non-zero spins…

Hi! I’d been a bit annoyed at how this thread had gone dead. Now, looking back at it, I see there was plenty left to say about Evan Jenkin’s post — I’d just forgotten to do it. And now you’ve joined in! So, there’s plenty for me to talk about… but my laptop died a while back, and this computer in the engineering building at Glasgow isn’t loaded with the necessary fonts, so it’s tough to read what people have written. So, I’ll just say a little bit now.

The ‘induced bundle’ trick is indeed crucial to understanding representations of the Poincaré group. In generality, we can start with a Lie group $G$ acting smoothly and transitively on a smooth manifold $M$. The stabilizer of your favorite point $x \in M$ will be a Lie subgroup $H \subseteq G$, and we have

$$M \cong G/H$$

Starting from this, there’s a process that takes any representation $s$ of $H$ on a vector space $V$ and turns it into a vector bundle $E$ over $M$ — the so-called ‘induced bundle’. Even better, the group $G$ acts on this bundle, and the projection

$$\pi : E \to M$$

gets along with the action of $G$:

$$\pi(g e) = g \pi(e) .$$

So, we say the $E$ is a $G$-equivariant vector bundles over $M$.

You’ve described how this works, so let me just gild the lily a bit.

First, the ‘process’ I described is actually a functor.

There’s a category

$$Rep(H)$$

of linear representations of $H$, and a category

$$Vect(M,G)$$

of $G$-equivariant vector bundles over $M$. The induced bundle construction gives a functor

$$L: Rep(H) \to Vect(M,G)$$

But, if you think about it, you’ll notice there’s also a functor going back the other way:

$$R: Vect(M,G) \to Rep(H)$$