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In Weinberg's Classical Solutions of Quantum Field Theory, he states whilst introducing homotopy that groups, such as $SU(2)$, may be endowed with the structure of a smooth manifold after which they may be interpreted as Lie groups. My questions are:

  • If we formulate a quantum field theory on a manifold which is also a Lie group, does that quantum field theory inherit any special or useful properties?
  • Does a choice of metric exist for any Lie group?
  • Are there alternative interpretations of the significance of Killing vectors if they preserve a metric on a manifold which is also a Lie group?
user45389
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    Comments on question (v1). Note that the moment you use the term "Lie group," you must be certain that the object being considered is a smooth manifold. So I'd like to suggest the rephrasing "groups, such as $\mathrm{SU}(2)$, may be endowed with the structure of a smooth manifold after which they may be interpreted as Lie groups." Also, do you have a compelling reason to believe that a "natural" choice of metric exists for an arbitrary Lie group? Perhaps "does there exist a natural choice of metric..." would be a better phrasing? Great question! – joshphysics Apr 27 '14 at 16:26
  • @joshphysics: Thank you for the suggested edit, I will incorporate it into the question. I will add the question of the existence of a choice of metric for an arbitrary Lie group, but I would also like to know how to select one if possible, given the appropriate Lie group. – user45389 Apr 27 '14 at 16:33
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    This question is extremely similar to the following math.SE question. Coincidence? http://math.stackexchange.com/q/769080/ – joshphysics Apr 27 '14 at 16:38
  • @joshphysics: Thanks for the link, that answers part of my question. – user45389 Apr 27 '14 at 16:54
  • @joshphysics Pedantic, but interesting (at least I think so; You may already know this): Actually smooth is not required. You only need to guarantee a topological group that is also a topological (locally homeomorphic to $\mathbb{R}^N$) manifold with the no small subgroups property and then the Montgomery Zippin Gleason Yamabe results take over and guarantee a Lie group and therefore that the manifold is smooth (indeed $C^\omega$) automatically. See Hilbert's Fifth Problem Wiki Page – Selene Routley Mar 18 '15 at 00:06
  • @joshphysics Terry Tao does a great writeup of this at https://terrytao.wordpress.com/2011/06/17/hilberts-fifth-problem-and-gleason-metrics/ – Selene Routley Mar 18 '15 at 00:11
  • @WetSavannaAnimalakaRodVance Interesting, thanks for the link. – joshphysics Mar 18 '15 at 00:14
  • To make this question interesting, you would have to restrict to non-Abelian Lie-groups, since $\mathbb{R}^n$ is a Lie group by itself, so all usual QFTs already live on a Lie group. – ACuriousMind Mar 18 '15 at 00:43

2 Answers2

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What I'm about to say apply only to Lie groups with finite-dimensional Lie algebras. In this case, there can be a natural choice of metric for the Lie group under some assumptions. The relevant constructions are defined in the Lie algebra of the group.

In the Lie algebra $\mathfrak{g}$, we can define, for every element $X \in \mathfrak{g}$, the adjoint action on the algebra itself given by its Lie bracket $ad_X(.)=[X,.]$. Now, we can define a symmetric bilinear form on $\mathfrak{g}$, the Killing form, by $B(X,Y)=Tr(ad_X ad_Y)$, since $ad_X$ can be expressed in term of a matrix acting on the vectors of the Lie algebra. We now have that this form will also be non-degenerate iff $\mathfrak{g}$ is semi-simple (Cartan criterion). This form is now only defined on the tangent space over the identity of the group. We can define a metric by using the fact that any tangent space can be mapped, for such a Lie group, to the tangent space over the identity. This metric may not be positive-definite, but in certain case, as if the Lie group is compact, the Killing form is negative-definite and thus, we can form a positive-definite metric by using $-B(X,Y)$.

Now, the Killing form will be invariant under $ad$. Taking the Lie derivative with respect to $X \in \mathfrak{g}$ of the above-defined metric, we will obtain $B(ad_X(Y),Z) - B(Y,ad_X (Z)) = 0$. When extending this $X$ to a right invariant vector field on the Lie group, this property is maintained, meaning that this vector field is Killing, or composed of Killing vectors. This shows that there is a way to relate some Killing vectors to the underlying group.

Concerning your question about a quantum field theory over Lie group, I am not so sure. If you choose the above metric, than you also have symmetries generated by the Killing fields obtained via the Lie algebra structure. This then implies that your QFT fields need to be representations for the group of these symmetries.

G. Bergeron
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  • It would be well worth citing, as I'm sure you know, the theorem that for simple groups your Lie derivative formula, together with nondegeneracy, fully defines a symmetric billinear form to within a scaling constant (and that indeed it "almost" uniquely defines the form for semisimple groups, modulo separate scaling constants for each direct product member). I was always utterly mystified by the Killing form until I came across this theorem; there doesn't seem to be any other way of seeing why such an arbitrary looking formula should mean anything. BTW, great answer. – Selene Routley Dec 12 '16 at 09:37
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I'm not a field theorist but I can say that if you formulate a theory on a Lie group as opposed to a general manifold, there are most certainly properties inherited by that theory that it might not have in a general smooth manifold. Lie groups are extremely special manifolds: some of their properties that general manifolds do not have are:

  1. The fundamental group is Abelian, as it is for all topological groups. In contrast, for every finitely presented group, a smooth manifold with this group as its fundamental group can be found. The complement of a trefoil knot is a manifold with a non-Abelian braid group $B_3$ and thus a neat counterexample (see section B in chapter 6 beginning about p51 in Rolfsen's "Knots and Links" for a good discussion of this).

  2. The manifold is parallelizable: the left and right-invariant vector fields vanish nowhere. Informally, a Lie group's hair can always be combed everywhere flat. Thus there is a Lie group that is a torus and a Lie group that is a 3-sphere (as is your example $SU(2)$), but there is no Lie group that is also a 2-sphere, as this is ruled out by the Hairy Ball theorem.

  3. Through the operations of left / right translation, one can define a notion of parallel transport on a Lie group which is independent of path. So it can be given a flat connexion. Naturally, there is nonzero torsion here (and naturally, this is not the Levi-Civita connexion which can be defined, for compact groups, through the Killing form metric talked about in G. Bergeron's Answer).

Lastly, an interesting point is that one only needs to (i) show that a topological group is locally homeomorphic to $\mathbb{R}^N$ and (ii) show that it has the no small subgroups property and then the Montgomery-Zippin-Gleason-Yamabe results take over and guarantee a Lie group and therefore that the manifold is smooth (indeed $C^\omega$) automatically. See Hilbert's Fifth Problem Wiki Page. Terrence Tao also does a great writeup of this wonderful work in his article "Hilbert’s fifth problem and Gleason metrics" on his blog.

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Selene Routley
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