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The groups $U(N)$ and $SU(N)$ are the most important Lie groups in quantum field theory. The most popular are the $U(1),SU(2),SU(3)$ groups (these gauge groups form the Standard model). But is there mentioned a $SU(\infty)$ gauge theory in physics literature?

An example of such a theory could be the following: May be $g \in SU(\infty)$ smooth and for a function $f(x,y)$ with spacetime coordinate $x$ and the new $SU(\infty)$ degree of freedom $y$ it holds $gf(x,y) = \int d^4y (g(x,y,y')f(x,y'))$. Now it is straighforward to define a gauge connection and the gauge field strength.

In more non-theoretical words: Some quantum states have degeneracies and these degeneracies are based on a special symmetry (operator) that exists in a quantum system. If now the degeneracy symmetry operator is unitary and local symmetry one can define a gauge theory. Was this concept used in quantum mechanics or does such a concept makes sense?

Another interesting case is this: One can perform the following switch of coordinates $g(x,y,y') = g(x,x-y,x-y')$ and hence the generators $T_a(y,y')$ defined by $g(x,y,y') = \sum_a g_a(x)T_a(y,y')$ become dependent on the spacetime coordinate. Another question: Is it possible to define spacetime dependent generators of a Lie algebra?

kryomaxim
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    How do you propose to define $SU(\infty)$? – Danu Aug 28 '15 at 21:02
  • Here, the internal variables $y \in \mathbb R^4$ represent an infinite-dimensional vector space; this vector space is a Hilbert space. – kryomaxim Aug 28 '15 at 21:12
  • $SU(N)$ is not even a vector space. – Danu Aug 28 '15 at 21:35
  • Also, your sentence doesn't make much sense to me in general. Could you try to rephrase it? – Danu Aug 28 '15 at 21:57
  • I am assuming that the $g$ is an unitary operator acting on the internal degree of freedom $y$. Because linear operators can be considered as infinite dimensional matrices while vectors are functions I talked about the matrix representations of $SU(\infty)$. – kryomaxim Aug 28 '15 at 22:08
  • Before you define any representation, can you define $SU(\infty)$? – Danu Aug 28 '15 at 22:09
  • The space $SU(\infty)$ is defined as the set of all operators $exp(i \Lambda)$ where $\Lambda$ is an operator on the Hilbert space and it holds $tr(\Lambda) = \int d^4y \Lambda(y,y)= 0$ – kryomaxim Aug 28 '15 at 22:21
  • I am pretty sure $SU\left( \infty \right)$ is not a thing, but in conformal field theory you deal with things called Kac-Moody (Affine Lie) algebras which are infinite dimensional extensions of your usual algebras. Generally written like $\widehat U\left( 1 \right)$, $\widehat {SU}\left( 2 \right)$, etc. –  Aug 28 '15 at 23:29

2 Answers2

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Comments to the question (v2):

  1. The idea to consider the planar large $N_c\to \infty$ limit in $SU(N_c)$ QCD goes back to Ref. 1.

  2. In light-cone membrane theory, pioneered in Ref. 2, the group $SU(\infty)$ is naturally identified with area-preserving diffeomorphisms ${\rm SDiff}_0(T^2)$ on the torus $T^2$ connected to the identity.

  3. Concretely, OP's proposal resembles a Fourier series expansion of an extra (compact) spacetime dimension. Such exercises are customary in string theory.

References:

  1. G. 't Hooft, A planar diagram theory for strong interactions, Nucl. Phys. B72 (1974) 461.

  2. J. Goldstone, unpublished; J. Hoppe, MIT Ph.D. Thesis, 1982.

Qmechanic
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There are apparently several thousand references to "SU(\infty)" on arxiv.org, and some of them are definitely talking about gauge fields or Yang-Mills.

I suspect that some of the time, this will just be a way of talking about the large N limit of SU(N), i.e., not referring to a literal SU(∞) field theory, but rather the N→∞ limit of some quantity in SU(N) field theory.

Mitchell Porter
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  • Thanks for your answer. I have asked whether it is possible that the generators of the $\lim_{N \mapsto \infty} SU(N)$ group are explicitely spacetime dependent. Generators of Yang-Mills theories like QCD are constant matrices (Gell-Mann matrices), but it is possible for a field theory that generators are explicitely dependent on the spacetime coordinate? – kryomaxim Aug 29 '15 at 09:43
  • @kryomaxim I think the fact that you choose a basis for a finite group is because it is finite. You would have have consider whatever theory you work with no consideration of basis. If everything works out fine, I suppose it works. I'm making this analogy with the way mathematicians deal infinite dimensional vector spaces. –  Oct 27 '15 at 04:38
  • Even for the Lorentz group, with which I gather you want your generalized SU(N) to commute, the matrices themselves do not depend on coordinates, even though, of course, they represent coordinate-dependent generators. I wonder if your question stems for the fact that infinite N normally demarcates infinite sets of integers which, when Fourier transformed, amount to continuous phase-space type surfaces cf. , as Qmechanic's answer addresses in 2 & 3; but such surfaces are ancillary world-sheet constructs, and not our spacetime. – Cosmas Zachos Apr 05 '16 at 14:24