In my post on Wigner's theorem, I explained the famous result that any symmetry transformation on quantum states can be realized as a unitary or antiunitary operator on Hilbert space. But when we study symmetries of quantum systems, we usually have in mind not a single symmetry but a full group of symmetries; Wigner's theorem tells us nothing about how the operators corresponding to different symmetries in the same group should compose with one another.
Suppose, for example, that a quantum system transforms under the symmetry group $G$, and that the unitary or antiunitary operator corresponding to the element $g \in G$ is denoted by $\hat{U}_g$. Because any two unitary operators related by a phase are physically equivalent, it may not be the case that $\hat{U}_{g_1}$ and $\hat{U}_{g_2}$ compose to $\hat{U}_{g_1 g_2}$; instead, we will have a relationship like
$$\hat{U}_{g_1} \hat{U}_{g_2} = e^{i \phi(g_1, g_2)} \hat{U}_{g_1 g_2}.$$
At first glance, it seems like we just made a mistake in our definition of $\hat{U}_{g_1 g_2}$; since the Wigner operator corresponding to a symmetry is only defined up to a phase, we could freely redefine $\hat{U}_{g_1 g_2} \mapsto e^{- i \phi(g_1, g_2)} \hat{U}_{g_1 g_2}$ and obtain the expected composition law $\hat{U}_{g_1} \hat{U}_{g_2} = \hat{U}_{g_1g_2}.$ But remember that, because $G$ is a group, there may be many ways of making the group element $g_1 \cdot g_2$; for example, there may exist group elements $h_1, h_2, h_3$ with $g_1 g_2 = h_1 h_2 h_3$. If we wanted to make all of the unitary compositions "phase-free," we would need to enforce $\hat{U}_{g_1} \hat{U}_{g_2} = \hat{U}_{h_1} \hat{U}_{h_2} \hat{U}_{h_3},$ $\hat{U}_{g_1} \hat{U}_{g_2} = \hat{U}_{h_1} \hat{U}_{h_2 h_3},$ and so on. It turns out that it is not always possible to redefine phases in the Wigner unitaries so that all of these compositions are free of phases. (I give an explicit example where rephasing is not possible at the end of section 1.1 of this post.)
The formal statement that comes out of this observation is that there exist group actions of certain groups $G$ on the space of quantum states $\mathbb{P}(\mathcal{H})$ that cannot be realized as representations of $G$ on the Hilbert space $\mathcal{H}.$ Instead, it is necessary to construct a central extension group $E_G$ whose representation on $\mathcal{H}$ is consistent with the action of $G$ on $\mathbb{P}(\mathcal{H}).$ Classifying the central extensions of a given symmetry group (or, sometimes, a symmetry algebra) is an essential problem in quantum physics; for example, one can show that the algebra of infinitesimal conformal symmetries in 2D Euclidean space has a unique central extension in the Virasoro algebra, which explains why two-dimensional Euclidean conformal field theories all have Virasoro symmetry.
One very important theorem about central extensions, due to Bargmann in his 1954 paper, is that there are certain special symmetry groups with no nontrivial (smooth) central extensions. These are — and I'll explain what all of these terms mean later in this post — simply connected, finite-dimensional Lie groups with semisimple Lie algebras. With some extra work — mainly in dealing with the parenthetical "(smooth)" in the first sentence of this paragraph — one can show that any action of such a group on quantum states can always be represented on Hilbert space without needing to worry about phases. Probably the most famous application of this theorem is in the theory of spinors. The local isotropy group of a signature-$(p, q)$ pseudo-Riemannian manifold, denoted $\operatorname{O}(p, q)$, is finite-dimensional and has a semisimple Lie algebra. Its universal cover (or, more precisely, the universal cover of its identity component; more on that later) is finite-dimensional and simply connected with a semisimple Lie algebra and thus satisfies the conditions of Bargmann's theorem. As a result, for any quantum system whose space of states is acted on by the identity component of the symmetry group $\operatorname{O}(p, q)$, we can realize that action as a representation of the universal cover on Hilbert space. Spinors are basically just fields that transform in representations of this universal covering group, with some extra consistency conditions enforced.
In this post, I'll explain and motivate central extensions of groups, and central extensions of Lie algebras. I'll also give an explanation of Bargmann's theorem and sketch the proof, though I'll rely on some lemmas about Lie theory that I won't prove here. I suspect this post will eventually serve as preliminary reading for two future posts on (i) the Virasoro algebra in 2D conformal field theory and (ii) spinor fields on arbitrary manifolds, but I don't know for sure.
I learned most of what I know about this subject from: (1) Chapter 13 of Wald's book on General Relativity, which defines spinor fields in four dimensions, and (2) Chapters 3 and 4 of Schottenloher's book on conformal field theory. Bargmann's 1954 paper is also an interesting read, but the proof of his theorem has been simplified considerably since its original formulation. The proof sketch I will give of Bargmann's theorem draws partially from these notes of Zinovy Reichstein, which give a nice direct proof of the fact that finite-dimensional semisimple Lie algebras have no nontrivial central extensions. (Traditionally, this statement is proven by showing that the set of inequivalent central extensions of a Lie algebra is in one-to-one correspondence with the second cohomology group of that Lie algebra, then invoking Whitehead's second lemma to show that the second cohomology of a semisimple Lie algebra vanishes. Reichstein's direct proof that there are no nontrivial central extensions actually gives an alternate proof of Whitehead's second lemma; I'll comment on this more in the section about central extensions of Lie algebras.)
Prerequisites: Group theory; basic familiarity with Lie groups and Lie algebras, including the definition of a Lie algebra representation. I'll make some side comments about cohomology and about continuous maps in the strong operator topology but they aren't essential to the main pedagogical flow. In section 2, I make some comments about the Lie algebra of an infinite-dimensional projective unitary group; the formalism needed to make sense of that idea is explained in detail in my post on Banach Lie groups and algebras, but if you're willing to accept me saying "it's the vector space of bounded antihermitian operators quotiented by multiples of the identity" on faith, then you can safely skip that background reading.
Table of Contents
- Projective Representations and Central Extensions of Groups
- Central Extensions of Lie Algebras
- Bargmann's Theorem
1. Projective Representations and Central Extensions of Groups
In my post on Wigner's theorem, I explained that the space of quantum states associated to a Hilbert space $\mathcal{H}$ is best thought of as the projective space $\mathbb{P}(\mathcal{H}).$ Since the only observables associated with states are the transition amplitudes
$$T(|\psi_1\rangle, |\psi_2\rangle) = \frac{|\langle \psi_1 | \psi_2 \rangle|^2}{\langle \psi_1 | \psi_1 \rangle \langle \psi_2 | \psi_2 \rangle},$$
any two vectors related by scalar multiplication (i.e., $|\psi'\rangle = \lambda |\psi\rangle$) are physically equivalent. The projective space $\mathbb{P}(\mathcal{H})$ is the quotient of $\mathcal{H}$ by this equivalence relation, i.e., the set of equivalence classes
$$[|\psi\rangle] = \{ \lambda |\psi \rangle | \lambda \in \mathbb{C}-\{0\}\}.$$
A symmetry transformation of a quantum system is a bijection $S : \mathbb{P}(\mathcal{H}) \rightarrow \mathbb{P}(\mathcal{H})$ that preserves all transition amplitudes. We will call the group of all such transformations $\operatorname{Aut}(\mathbb{P}(\mathcal{H})),$ since these are automorphisms with respect to the structure on $\mathbb{P}(\mathcal{H}).$ The statement of Wigner's theorem is that for any symmetry transformation $S$, there exists a unitary or antiunitary operator $\hat{S} : \mathcal{H} \rightarrow \mathcal{H}$ that agrees with $S$ on equivalence classes; i.e.,
$$S([|\psi\rangle]) = [\hat{S} |\psi \rangle ].$$
If there is a group $G$ of physical symmetry operations I can enact on a quantum system with Hilbert space $\mathcal{H}$, then each element of that group induces a symmetry transformation in $\operatorname{Aut}(\mathbb{P}(\mathcal{H})).$ Doing two physical symmetry operations in a row ought to correspond to composing the two corresponding elements of $\operatorname{Aut}(\mathbb{P}(\mathcal{H})).$ In other words, if a quantum system has symmetry group $G$ then there ought to be a group homomorphism $\rho : G \rightarrow \operatorname{Aut}(\mathbb{P}(\mathcal{H})).$ Such a map is called a projective representation of $G$ on $\mathcal{H}$. If $G$ has a topology, then we will usually assume that $\rho$ is continuous with respect to the topology on $G$. Making this assumption requires giving a topology to $\operatorname{Aut}(\mathbb{P}(\mathcal{H}))$; the natural choice is the strong operator topology of pointwise convergence; we have $S_n \rightarrow S$ in $\operatorname{Aut}(\mathbb{P}(\mathcal{H}))$ if and only if, for each $[|\psi \rangle] \in \mathbb{P}(\mathcal{H})$, we have
$$T(S_n([|\psi\rangle]),\, S([|\psi\rangle])) \rightarrow 1.$$
One goal of this post is to understand how to study the following question: for a given symmetry group $G$, what is the full list of ways that $G$ can act on a quantum system?
This amounts to classifying all of the possible projective
representations of $G$. In the next two subsections, we will explore two equivalent ways of classifying projective representations; the first in terms of cocycles, the second in terms of central extensions.
1.1 Cocycles and Phases
Suppose we have a projective representation $\rho : G \rightarrow \operatorname{Aut}(\mathbb{P}(\mathcal{H})).$ We will use the notation $S_g \equiv \rho(g)$ to denote the symmetry transformations corresponding to group elements $g$. By Wigner's theorem, we can choose unitary or antiunitary operators $\hat{S}_g$ on $\mathcal{H}$ that agree with $S_g$ on $\mathbb{P}(\mathcal{H}).$ Suppose some choice $\hat{S}_g$ has been made for each $g \in G$. Since $\rho$ is a group homomorphism into $\operatorname{Aut}(\mathbb{P}(\mathcal{H})),$ we have $S_{g_1} S_{g_2} = S_{g_1 g_2}$ for any $g_1, g_2 \in G.$ But since two unitary/antiunitary operators on $\mathcal{H}$ that agree on $\mathbb{P}(\mathcal{H})$ may differ on $\mathcal{H}$ by a global phase, we have
$$\hat{S}_{g_1} \hat{S}_{g_2} = e^{i \phi(g_1, g_2)} \hat{S}_{g_1 g_2}$$
for some real number $\phi(g_1, g_2).$
Thought of as a function on $G \times G$, the phase $e^{i \phi(g_1, g_2)}$ must obey a consistency condition due to associativity of operator multiplication:
$$\hat{S}_{g_1} (\hat{S}_{g_2} \hat{S}_{g_3}) = (\hat{S}_{g_1} \hat{S}_{g_2}) \hat{S}_{g_3},$$
which in terms of phases is
$$e^{i \phi(g_1, g_2 g_3) + i \phi(g_2, g_3)} = e^{i \phi(g_1, g_2) + i \phi(g_1 g_2, g_3)}. \tag{1}$$
It is also clear that, since we are free to redefine all our Wigner representatives $\hat{S}_{g}$ by phases $\hat{S}_{g} \mapsto e^{i \alpha(g)} \hat{S}_{g}$ without changing the underlying projective representation $S_g,$ we should call two phase functions $e^{i\phi(g_1, g_2)}$ and $e^{i \xi(g_1, g_2)}$ equivalent if there exists a function $\alpha : G \rightarrow \mathbb{R}$ so that
$$e^{i \phi(g_1, g_2) - i \xi(g_1, g_2)} = e^{i \alpha(g_1 g_2) - i \alpha(g_1) - i \alpha(g_2)}.\tag{2}$$
So to study all the possible ways $G$ can act on a quantum system, we will study all the phase functions $e^{i \phi(g_1, g_2)}$ satisfying equation (1), or more precisely the equivalence classes of such functions under the relation given by equation (2). (We won't need this, but one can actually show that for suitable groups $G$, there exist projective representations of $G$ corresponding to any phase function satisfying equation (1); see Bargmann section 3.) A phase function $e^{i \phi(g_1, g_2)}$ on $G$ satisfying (1) is sometimes called a 2-cocycle or just a cocycle on $G$. I'll use this terminology when it's convenient, but the actual reasoning behind calling it a "cocycle" is irrelevant for the purposes of this post. I will only mention, for readers with a taste for homological algebra, that such a function is indeed a cocycle of the cochain complex of $G$'s group cohomology with respect to the trivial $G$-module on $\operatorname{U}(1)$; the phase functions of the form $e^{i \alpha(g_1 g_2) - i \alpha(g_1) - i \alpha(g_2)}$, used to define the equivalence relation on cocycles, are the coboundaries in the second chain group. In this sense, the set of allowed phase functions is exactly the second cohomology group of $G$ in the sense of group cohomology.
The condition that the cocycle $e^{i \phi(g_1, g_2)}$ is equivalent to $1$ is exactly the condition under which the projective representation of $G$ on $\mathcal{H}$ can be lifted to a true representation of $G$ on $\mathcal{H}$; this is the case where all of the Wigner operators can be re-phased so their composition laws are phase-free. Before proceeding to section 1.2, I will briefly show that there do exist nontrivial projective representations, i.e., nontrivial cocycle equivalence classes $[e^{i \phi(g_1, g_2)}]$. Let $\operatorname{W}$ be the Pauli group (I'm using $\operatorname{W}$ here for Wolfgang Pauli, since I'll need P for projective space), i.e., a set of $2 \times 2$ unitary matrices $\{\hat{1}, \hat{X}, \hat{Y}, \hat{Z}\}$ satisfying $\hat{X}^2 = \hat{Y}^2 = \hat{Z}^2 = \hat{1}$ and $\hat{X}\hat{Y} = i \hat{Z} = - \hat{Y} \hat{X}, \hat{Y} \hat{Z} = i \hat{X} = - \hat{Z} \hat{Y}, \hat{Z} \hat{X} = i \hat{Y} = - \hat{X} \hat{Z}.$ This is a subgroup of the unitary group on $\mathbb{C}^2$, i.e. $\operatorname{W} \leq \operatorname{U}(\mathbb{C}^2)$, so if we quotient by phases, we obtain the projective Pauli group $\operatorname{PW} = \{1, X, Y, Z\} \leq \operatorname{Aut}(\mathbb{P}(\mathbb{C}^2)).$
The projective Pauli group has multiplication laws $X^2 = Y^2 = Z^2 = 1, XY = Z = YX, YZ = X= Z Y, ZX = Y = X Z.$ But there is no way to rephase the Pauli operators $\hat{X}, \hat{Y}, \hat{Z}$ to satisfy this phase-free multiplication law! In particular, the symmetry transformations $X$ and $Y$ commute, while the operators $\hat{X}$ and $\hat{Y}$ do not; there is no way to multiply $\hat{X}$ and $\hat{Y}$ by phases to make them commute. So the Pauli subgroup $\operatorname{PW}$ of the projective unitary group $\operatorname{PU}(\mathbb{C}^2)$ has a nontrivial projective representation on $\mathbb{C}^2$.1.2 Central Extensions
1.2 Central Extensions
So far, we've come up with one way of thinking about the allowable ways for a $G$-action on $\mathbb{P}(\mathcal{H})$ to be represented on $\mathcal{H},$ in terms of $2$-cocycles $e^{i \phi(g_1, g_2)}.$ We will now introduce an equivalent way, in terms of the central extensions of $G$. As usual, different ways of thinking about the same underlying structure will give us different insights; we will try to mine these insights by moving back and forth smoothly between cocycles and central extensions.
The idea of a central extension is as follows. We are handed an action of $G$ on $\mathbb{P}(\mathcal{H}),$ i.e., a projective representation $\rho : G \rightarrow \operatorname{Aut}(\mathbb{P}(\mathcal{H})).$ We would like to find a way to lift this projective representation to a true representation, i.e., a group homomorphism $\rho : G \rightarrow \mathcal{I}(\mathcal{H})$, with $\mathcal{I}(\mathcal{H})$ the group of unitary and antiunitary operators on $\mathcal{H}$. (The symbol $\mathcal{I}$ here stands for "isometry.") This is not always possible, as evidenced by the construction of a nontrivial cocycle at the end of the previous subsection. What we can hope to do instead, is to extend $G$, i.e., to find some larger group $E_G$ with a representation on $\mathcal{H}$ that descends to the action of $G$ on $\mathbb{P}(\mathcal{H})$ under appropriate quotients. The group $E_G$ will be called a central extension of $G$ by $\operatorname{U}(1)$.
The sense in which $E_G$ is "larger" than $G$ is a bit subtle. It won't end up being the case that $G$ is a subgroup of $E_G$, but rather that $E_G$ comes furnished with a projection onto $G$. To understand this, let's start by constructing a central extension corresponding to a cocycle $e^{i\phi(g_1, g_2)},$ then give the abstract definition once this construction has been understood.
Let the underlying set of $E_G$ be $\operatorname{U}(1) \times G$, i.e., the set of pairs $(e^{i \alpha}, g)$ with $g \in G.$ Define the group multiplication law by
$$(e^{i \alpha_1}, g_1) \cdot (e^{i \alpha_2}, g_2) = (e^{i \alpha_1 + i \alpha_2 + i \phi(g_1, g_2)}, g_1 g_2).$$
We need to check that this is a group. The cocycle identity
$$e^{i \phi(g_1, g_2 g_3) + i \phi(g_2, g_3)} = e^{i \phi(g_1, g_2) + i \phi(g_1 g_2, g_3)} \tag{3}$$
with $g_2 = g_3 = 1$ or $g_1 = g_2 = 1$ gives
$$e^{i \phi(1, g)} = e^{i \phi(1, 1)} = e^{i \phi (g, 1)}.$$
From this, we may check that $(e^{- i \phi(1, 1)}, 1)$ is an identity element for $E_G.$ Setting $g_1 = g, g_2 = g^{-1},$ and $g_3 = g$ in (3) gives
$$e^{i \phi(g, g^{-1})} = e^{i \phi(g^{-1}, g)},$$
from which one can check that
$$(e^{- i \phi(1, 1) - i \phi(g, g^{-1}) - i \alpha}, g^{-1})$$
is both a left and right inverse for the element $(e^{i \alpha}, g).$ Finally, with some fairly trivial manipulations, associativity of group multiplication can be seen to follow from (3).
Given a set of Wigner operators $\hat{S}_{g}$ for $G$, satisfying $\hat{S}_{g_1} \hat{S}_{g_2} = e^{i \phi(g_1, g_2)} \hat{S}_{g_1 g_2},$ the map $R : E_G \rightarrow \mathcal{I}(\mathcal{H})$ given by $(e^{i \alpha}, g) \mapsto e^{i \alpha} \hat{S}_g$ is easily checked to be a homomorphism. So we have succeeded in our goal of representing $E_G$ on $\mathcal{H}$ in such a way that the subset $\{(1, g) \in E_G\, |\, g \in G\}$ maps to $\{\hat{S}_{g}\}.$ An
interesting fact that we won't prove here, but which is explained
nicely in theorems 3.10-3.12 of Schottenloher's book, is that if $G$ has a topology then with a slightly different construction of $E_G$ we can guarantee that $R$ is a continuous map with respect to the strong operator topology on $\mathcal{I}(\mathcal{H})$ and a natural topology on $E_G.$
So in what sense is $E_G$ an extension of $G$ by $\operatorname{U}(1)$? Well, $E_G$ contains a subset $\{(1, g)\}$ that is in bijection with $G$, and moreover there is a homomorphism $\pi : E_G \rightarrow G$ given by $(e^{i \alpha}, g) \mapsto g$ that maps $E_G$ onto $G$. $E_G$ also contains a subgroup $\{\{e^{i \alpha} e^{- i \phi(1, 1)}, 1)\}$ that is isomorphic to $\operatorname{U}(1)$ and that lies in the center of $E_G.$ Finally, the subgroup isomorphic to $\operatorname{U}(1)$ is exactly the kernel of $\pi$.
These properties form the definition of a central extension of $G$ by $\operatorname{U}(1)$. We require a subgroup $\operatorname{U}(1) \leq E_G$ that lives in the center $Z(E_G)$, a surjective homomorphism $\pi : E_G \rightarrow G$, and the condition $\operatorname{ker}(\pi) = \operatorname{U}(1).$ Formally, this data can be written compactly in the form of a short exact sequence
$$1 \rightarrow \operatorname{U}(1) \rightarrow E_G \rightarrow G \rightarrow 1.$$
For readers not familiar with short exact sequences, all this expression means is that there exists a homomorphic injection of $\operatorname{U}(1)$ into $E_G$ and a homomorphic surjection of $E_G$ onto $G$ such that the kernel of the surjection is the image of the injection. The fact that the image of $\operatorname{U}(1)$ lies in the center of $E_G$ is not encoded in this notation, but will be assumed going forward.
We will say that two central extensions $E_G \rightarrow^{\pi} G$ and $E_G' \rightarrow^{\pi'} G$ are equivalent if there exists an isomorphism $\psi : E_G \rightarrow E_G'$ that respects the structure; i.e., it maps the $\operatorname{U}(1)$ subgroups onto each other and satisfies $\pi' \circ \psi = \pi.$ By repeating the construction of the past few paragraphs with an equivalent cocycle $e^{i \phi(g_1, g_2) + i \alpha(g_1, g_2) - i \alpha(g_1) - i \alpha(g_2)},$ one can easily check that the corresponding central extension is equivalent to the one that was constructed using the cocycle $e^{i \phi(g_1, g_2)}$ under the isomorphism
$$(e^{i \theta}, g) \mapsto (e^{i \theta} e^{i \alpha(g)}, g).$$
Conversely, given any central extension $E_G \rightarrow^{\pi} G$, we can construct a cocycle on $G$ by picking an injective map $\tau : G \rightarrow E_G$ with $\pi \circ \tau = \operatorname{id}_G$, and defining
$$e^{i \phi(g_1, g_2)} = \tau(g_1) \tau(g_2) \tau(g_1 g_2)^{-1}.$$
The condition $\pi \circ \tau = \operatorname{id}_G$ guarantees that the right-hand side of this equation is in the kernel of $\pi$, which guarantees that it is in the $\operatorname{U}(1)$-subgroup of $E_G$. This definition of $e^{i \phi(g_1, g_2)}$ can be checked to satisfy the cocycle identity by straightforward manipulations:
$$e^{i \phi (g_1, g_2)} e^{i \phi(g_1 g_2, g_3)} = \tau(g_1) \tau(g_2) \tau(g_3) \tau(g_1 g_2 g_3)^{-1}$$
$$= \tau(g_1) \tau(g_2) \tau(g_3) \tau(g_2 g_3)^{-1} \tau(g_2 g_3) \tau(g_1 g_2 g_3)^{-1}$$
$$ = e^{i \phi(g_2, g_3)} \tau(g_1) \tau(g_2 g_3) \tau(g_1 g_2 g_3)^{-1} = e^{i \phi(g_2, g_3)} e^{i \phi(g_1, g_2 g_3)}.$$
For any other injective map $\sigma : G \rightarrow E_G$ with $\pi \circ \sigma = \operatorname{id}_{G}$, the associated cocycle $e^{i \xi(g_1, g_2)}$ satisfies
$$e^{i \xi(g_1, g_2) - i \phi(g_1, g_2)} = \sigma_{g_1} \sigma_{g_2} \sigma_{g_1 g_2}^{-1} \tau_{g_1 g_2} \tau_{g_2}^{-1} \tau_{g_1}^{-1}.$$
Since $\pi$ is a homomorphism, $\sigma{g_1 g_2}^{-1} \tau_{g_1 g_2}$ is in the kernel of $\pi$, and is thus of the form $e^{i \alpha(g_1 g_2)}.$ This gives
$$e^{i
\xi(g_1, g_2) - i \phi(g_1, g_2)} = e^{i \alpha(g_1 g_2)} \sigma_{g_1} \sigma_{g_2} \tau_{g_2}^{-1} \tau_{g_1}^{-1}.$$
Repeating the same argument for $\sigma_{g_2} \tau_{g_2}^{-1}$ and $\sigma_{g_1} \tau_{g_1}^{-1}$, we obtain
$$e^{i
\xi(g_1, g_2) - i \phi(g_1, g_2)} = e^{i \alpha(g_1 g_2) - i \alpha(g_1) - i \alpha(g_2)}.$$This is exactly the condition for $e^{i \xi}$ and $e^{i \phi}$ to be equivalent as cocycles.
We now have a recipe for mapping cocycles to central extensions that maps equivalent cocycles to equivalent extensions, and a recipe for mapping central extensions to cocycles that maps equivalent central extensions to equivalent cocycles. We conclude that the two concepts are exactly the same; equivalence classes of cocycles are in bijection with equivalence classes of central extensions $E_G.$
As a final comment, we will ask: why does a central extension need to have this finicky property of projecting onto $G$? Why couldn't $G$ be a subgroup of $E_G$ in the first place? The answer is that if $G$ is realized as a subgroup of $E_G$, then the inclusion map $\tau : G \rightarrow E_G$ satisfies $\pi \circ \tau = \operatorname{id}_{G}$ and is a homomorphism; this means that the associated cocycle $\tau(g_1) \tau(g_2) \tau(g_1 g_2)^{-1}$ is trivial. Because we know from section 1.1 that there are cocycles not equivalent to trivial ones, we know there are central extensions for which no such $\tau$ can exist. A homomorphism $\tau : G \rightarrow E_G$ with $\pi \circ \tau = \operatorname{id}_G$ is called a splitting; a central extension has a splitting if and only if it is equivalent to a trivial extension.
We now know how to study the question posed at the beginning of this section: for a given symmetry group $G$, what is the full list of ways that $G$ can act on a quantum system? We classify all possible central extensions of $G$, then classify all possible representations of those central extensions. This sounds like a monumental task, but it isn't actually as hard as it seems; for many groups of interest, the list of central extensions is finite or trivial. We will return to that point in section 3; for now, though, let's turn to Lie algebras.
2. Central Extensions of Lie Algebras
In the previous section, we learned that the inequivalent ways a group $G$ can act symmetrically on a quantum system are classified by the central extensions of $G$, or equivalently by its cocycles. What if, instead of assuming our quantum system has symmetry group $G,$ we assume it has a symmetry algebra $\mathfrak{g}$? There are two ways to think about this assumption.
- We imagine that we have some physical system that, at the classical level, is symmetric under a Lie group $G$. Is every element of $G$ a symmetry transformation of the quantization of that system? The answer appears to be no, in general; parity violation, for example, indicates that the Standard Model is not invariant under the full 4D Lorentz group $\operatorname{O}(1, 3),$ since the weak interaction is not symmetric under spatial reflections. But it seems reasonable enough to assume that there is a neighborhood of the identity in $G$ that is included among the quantum symmetries of the system — these symmetry transformations are so small, goes the reasoning, that they couldn't possibly break down under quantization. Because any neighborhood of the identity in a Lie group generates the entire connected component $G_e$ containing the identity, it is generally assumed that a system classically symmetric under $G$ is quantumly symmetric under $G_e$. This in turn should induce an action of the Lie algebra $\mathfrak{g}$ on our quantum system in terms of "infinitesimal symmetries," and if we understand how those infinitesimal symmetries work then we can understand how the full symmetry group $G_e$ works.
- We forget about Lie groups and just assume, by hand, that a particular Lie algebra $\mathfrak{g}$ is included among the infinitesimal symmetries of our quantum system. This is often done on physical grounds — for example, it is assumed that the algebra of two-dimensional conformal killing vector fields acts as an infinitesimal symmetry algebra on the Hilbert space of any 2D CFT, without thinking of it as the Lie algebra of any particular group. From this point of view, all we need to do is define what we mean by an "infinitesimal symmetry," and we can start working. At the end of the day, this isn't such a wild thing to do; doing physics is about making a set of assumptions about how nature is represented in mathematical structures, making predictions from those structures, and trying to verify those predictions experimentally. The assumption that 2D CFTs carry an infinitesimal symmetry action of the 2D conformal algebra, for example, holds up extremely well in experimental studies of critical spin chains. So it seems like a good assumption to make; do we really need to worry about justifying it post-facto?
So let's say what it should mean for a Lie algebra $\mathfrak{g}$ to be be among the infinitesimal symmetries of a quantum system. It's pretty clear what an infinitesimal symmetry should be; the group of symmetries of a quantum system is $\operatorname{Aut}(\mathbb{P}(\mathcal{H})),$ so the algebra of infinitesimal symmetries ought to be the Lie algebra of that group. To make sense of what we mean by the Lie algebra of $\operatorname{Aut}(\mathbb{P}(\mathcal{H})),$ we'll start by observing that due to Wigner's theorem, every element of $\operatorname{Aut}(\mathbb{P}(\mathcal{H}))$ is realizable as the $\operatorname{U}(1)$ quotient of either a unitary or antiunitary operator on $\mathcal{H}$. We will call the space of unitary operators $\operatorname{U}(\mathcal{H})$ and the space of antunitary operators $\operatorname{AU}(\mathcal{H}).$ The corresponding $\operatorname{U}(1)$ quotients, which are spaces of symmetry transformations, will be denoted $\operatorname{PU}(\mathcal{H})$ and $\operatorname{PAU}(\mathcal{H}).$ Wigner's theorem is
$$\operatorname{Aut}(\mathbb{P}(\mathcal{H})) = \operatorname{PU}(\mathcal{H}) \cup \operatorname{PAU}(\mathcal{H}).$$
$$\operatorname{Aut}(\mathbb{P}(\mathcal{H})) = \operatorname{PU}(\mathcal{H}) \cup \operatorname{PAU}(\mathcal{H}).$$
In any reasonable topology on operators — either the strong operator topology discussed in the previous section, or the topology induced by the operator norm — the sets $\operatorname{U}(\mathcal{H})$ and $\operatorname{AU}(\mathcal{H})$ are disconnected from one another, as are their projective versions. Since $\operatorname{PU}(\mathcal{H})$ contains the identity transformation, what we mean by the Lie algebra of $\operatorname{Aut}(\mathbb{P}(\mathcal{H}))$ ought to be the same thing as the Lie algebra of $\operatorname{PU}(\mathcal{H}).$ We have to be a little careful in defining what this means; when $\mathcal{H}$ is infinite dimensional, $\operatorname{PU}(\mathcal{H})$ is infinite dimensional, so we have to define a notion of infinite dimensional Lie groups to proceed. It turns out — see this blog post for details — that the projective unitary group is an example of what's called a Banach Lie group with respect to a smooth structure induced by the operator norm. Its Lie algebra, $\mathfrak{pu}(\mathcal{H}),$ is the set of equivalence classes of bounded antihermitian operators on $\mathcal{H}$, where two operators are equivalent if their difference is a multiple of the identity. The Lie bracket on this algebra is given by the usual commutator of operators. We'll adopt the usual physics convention of multiplying every element of $\mathfrak{pu}(\mathcal{H})$ by $i$ so we can think of it as the space of bounded Hermitian operators on $\mathcal{H}$ up to addition by multiples of the identity.
If we take the first, "explicit" point of view described above, and assume that we have a projective representation $\rho : G_e \rightarrow \operatorname{Aut}(\mathbb{P}(\mathcal{H})),$ then connectedness of $G_e$ — paired with the assumption mentioned in section 1 that $\rho$ is continuous in the strong operator topology — implies that $\rho$ is a projective unitary representation $\rho : G_e \rightarrow \operatorname{PU}(\mathcal{H}).$ If we further assume that $\rho$ is differentiable as a map between Banach manifolds, then its differential automatically furnishes us with a Lie algebra homomorphism
$$\rho : \mathfrak{g} \rightarrow \mathfrak{pu}(\mathcal{H}). \tag{4}$$
If we take the second, "abstract" point of view, then we take equation (4) as the definition of a projective representation of the algebra $\mathfrak{g}$ on $\mathcal{H}.$
We can now repeat all the machinery of section 1 for projective Lie algebra representations $\rho : \mathfrak{g} \rightarrow \mathfrak{pu}(\mathcal{H}).$ The ideas are all basically the same, so we will move quite a lot faster. Remember that $\mathfrak{pu}(\mathcal{H})$ is the space of equivalence classes
$$[\hat{H}] = \{ \hat{H} + \alpha\, |\ \alpha \in \mathbb{R}\},$$
where each $\hat{H}$ is Hermitian. If we pick a representative operator $\hat{H}_{X}$ in each equivalence class $\rho(X)$ for $X \in \mathfrak{g}$, then the commutators must satisfy a relationship of the form
$$[\hat{H}_{X_1}, \hat{H}_{X_2}] = \hat{H}_{[X_1, X_2]} + \phi(X_1, X_2).$$
Consistency of this equation requires that $\phi(X_1, X_2)$ is bilinear and antisymmetric, and the Jacobi identity of the commutator enforces the condition
$$\phi([X_1, X_2], X_3) + \phi([X_2, X_3], X_1) + \phi([X_3, X_1], X_2) = 0. \tag{5}$$
Any bilinear, antisymmetric map $\phi : \mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{R}$ satisfying equation (5) will be called a cocycle of $\mathfrak{g}$; as in the case of groups, this terminology comes from the fact that $\phi$ is a 2-cocycle in the Chevalley-Eilenberg cochain complex used to define the Lie algebra cohomology of $\mathfrak{g}.$ Because the translation $\hat{H}_{X} \mapsto \hat{H}_{X} + \alpha(X)$ for $\alpha$ linear does not affect the projective representation $\rho : \mathfrak{g} \rightarrow \mathfrak{pu}(\mathcal{H}),$ we will say that two cocycles $\phi$ and $\xi$ are equivalent if there exists some real, linear function $\alpha$ on $\mathfrak{g}$ satisfying
$$\phi(X_1, X_2) - \xi(X_1, X_2) = \alpha([X_1, X_2]).$$
(As an aside, the condition $\phi(X_1, X_2) = \alpha([X_1, X_2])$ is exactly the condition for $\phi$ to be a 2-coboundary in the Chevalley-Eilenberg cochain complex.)
(As an aside, the condition $\phi(X_1, X_2) = \alpha([X_1, X_2])$ is exactly the condition for $\phi$ to be a 2-coboundary in the Chevalley-Eilenberg cochain complex.)
As in the case of groups, we can define an extended Lie algebra $\mathfrak{e}_{\mathfrak{g}}$ on the set $\mathbb{R} \oplus \mathfrak{g}$ with Lie bracket
$$[(\alpha, X), (\beta, Y)] = (\phi(X, Y), [X, Y]).$$
Bilinearity and antisymmetry of $\phi$ guarantee bilinearity and antisymmetry of this Lie bracket, while the cocycle identity (5) can be shown to guarantee that this bracket satisfies the Jacobi identity. So $\mathfrak{e}_{\mathfrak{g}}$ is a genuine Lie algebra, and the representation of $\mathfrak{e}_{\mathfrak{g}}$ given by $(\alpha, X) \mapsto \alpha + \hat{H}_{X}$ coincides with the projective representation $\rho : \mathfrak{g} \rightarrow \mathfrak{pu}(\mathcal{H})$ on the set $\{(0, X)\}$. The Lie algebra $\mathfrak{e}_{\mathfrak{g}}$ has an abelian subalgebra $\mathbb{R} \cong \{(\alpha, 0)\}$ in the center of $\mathfrak{e}_{\mathfrak{g}},$ a surjective homomorphism $\pi : \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g}$ given by $(\alpha, X) \mapsto X$, and the kernel of that homomorphism is exactly the abelian subalgebra $\mathbb{R}$. In other words, there is a short exact sequence
$$0 \rightarrow \mathbb{R} \rightarrow \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g} \rightarrow 0$$
such that the image of $\mathbb{R}$ is in the center of $\mathfrak{e}_{\mathfrak{g}}.$ Any $\mathfrak{e}_{\mathfrak{g}}$ with these properties is called a central extension of $\mathfrak{g}$ by $\mathbb{R}$. For any linear map $L : \mathfrak{g} \rightarrow \mathfrak{e}_{\mathfrak{g}}$ with $\pi \circ L = \operatorname{id}_{\mathfrak{g}}$, the function
$$\phi(X_1, X_2) = [L(X_1), L(X_2)]$$
is a cocycle that reproduces $\mathfrak{e}_{\mathfrak{g}}$ by following the construction of this paragraph. As in the case of groups, there is a natural definition of isomorphisms between central extensions of Lie algebras, and one can show that the central extensions corresponding to equivalent cocycles are equivalent as central extensions, and that cocycles induced by equivalent central extensions are equivalent as cocycles. If there exists a map $L : \mathfrak{g} \rightarrow \mathfrak{e}_{\mathfrak{g}}$ with $\pi \circ L = \operatorname{id}_{\mathfrak{g}}$ that, in addition to being linear, is also a Lie algebra homomorphism, then the associated cocycle is equivalent to the trivial cocycle. $L$ is then called a splitting, and the existence of a splitting is equivalent to triviality of a central extension.
So, to classify the projective representations $\rho : \mathfrak{g} \rightarrow \mathfrak{pu}(\mathcal{H})$, it suffices to classify the the central extensions of $\mathfrak{g}$ by $\mathbb{R},$ and the representations of those extensions on $\mathcal{H}$. This classification is of great physical importance; in the next section, for example, we will see that when $\mathfrak{g}$ is semisimple, there are no nontrivial central extensions of $\mathfrak{g}.$ We will also likely see in some future post that when $\mathfrak{g}$ is the algebra of infinitesimal conformal transformations in two Euclidean dimensions, the only nontrivial central extension of $\mathfrak{g}$ is the Virasoro algebra.
3. Bargmann's Theorem
Now, a big result. Let $G$ be a connected Lie group with a projective representation $\rho : G \rightarrow \operatorname{PU}(\mathcal{H}).$ Let $\widetilde{G}$ be its universal cover, with projection map $\Pi : \widetilde{G} \rightarrow G.$ Because $\rho$ and $\Pi$ are both continuous homomorphisms, the lift $\widetilde{\rho} = \rho \circ \Pi : \widetilde{G} \rightarrow \operatorname{PU}(\mathcal{H})$ is a continuous homomorphism, i.e.,a projective representation of $\widetilde{G}$ on $\mathcal{H}.$ So, since every projective representation of $G$ induces a projective representation of $\widetilde{G},$ we can classify the projective representations of $G$ by classifying those of $\widetilde{G}.$
This is advantageous because structurally, $\widetilde{G}$ is a good deal simpler than $G$; in particular, it is simply connected. We will now sketch the proof of the first part of Bargmann's theorem, which is that when $G$ is finite-dimensional and semisimple, $\widetilde{G}$ has no nontrivial central extensions into Lie groups. The second part of Bargmann's theorem, which I will comment on at the end of the post, says that every projective representation of $\widetilde{G}$ can be realized as a true representation of a central extension into a Lie group. The two parts of Bargmann's theorem together imply a beautiful result: every projective representation of a finite-dimensional, connected, semisimple $G$ is induced by a true representation of $\widetilde{G}.$ This is an extremely powerful result; it's at the origin of the theory of spinors, which are representations of the universal cover of the connected component of the Lorentz group.
Before sketching the proof, let us be precise about what it means for $G$ to be semisimple. This is a condition on the Lie algebra of $\mathfrak{g}.$ A Lie algebra $\mathfrak{g}$ is called simple if it has no nontrivial ideals. Recall that an ideal is a Lie subalgebra $\mathfrak{h} \subset \mathfrak{g}$ that is "absorbing" under the Lie bracket: it satisfies
$$[X, Y] \in \mathfrak{h} \text{ for all } X \in \mathfrak{h}, Y \in \mathfrak{g}.$$
The trivial subalgebras $\{0\}$ and $\mathfrak{g}$ are obviously ideals; a Lie algebra is called simple if thoes are the only ideals. A Lie algebra is called semisimple if it can be written as a direct sum of simple Lie algebras:
$$\mathfrak{g} = \mathfrak{g}_{1} \oplus \dots \oplus \mathfrak{g}_{n}.$$
A Lie group is called semisimple if its Lie algebra is semisimple; this property holds, for example, for the Lorentz group in arbitrary signature and dimension.
The key result we will need on semisimple Lie algebras is the Weyl reducibility theorem, which states the following. If $\rho : \mathfrak{g} \rightarrow \mathfrak{gl}(V)$ is a finite-dimensional representation of a semisimple Lie algebra $\mathfrak{g}$ on the vector space $V$, and $W \subset V$ is a $\mathfrak{g}$-invariant subspace, then there exists a complementary $\mathfrak{g}$-invariant subspace $W' \subset V$ with $V = W \oplus W'.$ The use of this lemma, for our purposes, is that it implies that a semisimple Lie algebra $\mathfrak{g}$ has no nontrivial central extensions by $\mathbb{R}.$ To see this, let
$$0 \rightarrow \mathbb{R} \rightarrow \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g} \rightarrow 0$$
be a central extension of $\mathfrak{g}$ with projection $\pi : \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g}.$ Let $\tau : \mathfrak{g} \rightarrow \mathfrak{e}_{\mathfrak{g}}$ be a linear map, not necessarily a Lie algebra homomorphism, with
$$\pi \circ \tau = \operatorname{id}_{\mathfrak{g}}.$$ We will use this map to define a representation of $\mathfrak{g}$ on $\mathfrak{e}_{g}$:
$$X \cdot v = [\tau(X), v].$$
To see that this is a representation, we must verify the identity
$$[\tau([X, Y]), v] = [[\tau(X), \tau(Y)], v].$$
This does not necessarily require $\tau$ to be a Lie algebra homomorphism; it only requires that $\tau([X, Y]) - [\tau(X), \tau(Y)]$ is in the center of $\mathfrak{e}_{g}$. But because $\pi$ is a Lie algebra homomorphism, $\tau([X, Y]) - [\tau(X), \tau(Y)]$ is in the kernel of $\pi$, meaning it is in the center of $\mathfrak{e}_{g}.$ The subspace $\mathbb{R} \subset \mathfrak{e}_{\mathfrak{g}}$ is sent to zero by our $\mathfrak{g}$-action, so it is clearly an invariant subrepresentation of $\mathfrak{g}.$ The Weyl reducibility theorem therefore implies the identity
$$\mathfrak{e}_{\mathfrak{g}} = \mathbb{R} \oplus \mathfrak{h}$$
for some subrepresentation $\mathfrak{h}.$ For any vector $v \in \mathfrak{e}_{\mathfrak{g}},$ we denote by $v_{\mathbb{R}}$ and $v_{\mathfrak{h}}$ its projections onto the direct summands $\mathbb{R}$ and $\mathfrak{h}$, respectively. We now define a new linear map $\sigma : \mathfrak{g} \rightarrow \mathfrak{e}_{\mathfrak{g}}$ by
$$\sigma(X) = \tau(X)_{\mathfrak{h}}.$$
This map satisfies $\pi \circ \sigma = \operatorname{id}_{\mathfrak{g}},$ because $\mathbb{R}$ is the kernel of $\pi$: we have
$$X = (\pi \circ \tau)(X) = \pi(\tau(X)_{\mathfrak{h}}) + \pi(\tau(X)_{\mathbb{R}}) = \pi(\sigma(X)).$$
Furthermore, it is a homomorphism: we already know that $\sigma([X, Y]) - [\sigma(X), \sigma(Y)]$ is in $\mathbb{R}$, because it is in the kernel of $\pi.$ But because $\mathbb{R}$ is in the center of $\mathfrak{e}_{\mathfrak{g}},$ we have
$$[\sigma(X), \sigma(Y)] = [\tau(X) - \tau(X)_{\mathbb{R}}, \sigma(Y)] = [\tau(X), \sigma(Y)].$$
Because $\mathfrak{h}$ is invariant under the $\mathfrak{g}$-action $X \cdot v = [\tau(X), v]$, we conclude that $[\sigma(X), \sigma(Y)]$ is in $\mathfrak{h}$. So $\sigma([X, Y]) - [\sigma(X), \sigma(Y)]$ is in both $\mathbb{R}$ and $\mathfrak{h},$ i.e., it is the zero vector and $\sigma$ is a Lie algebra homomorphism. We conclude that $\sigma$ is a splitting of $\mathfrak{e}_{\mathfrak{g}},$ and so it is a trivial central extension.
In passing, I will note that since inequivalent central extensions of Lie algebras are in one-to-one correspondence with elements of their second cohomology group, the proof of the preceding paragraph implies that the cohomology group $H^{2}(\mathfrak{g}, \mathbb{R})$ vanishes, which is a special case of Whitehead's second lemma. With some mild generalizations, this same proof can be used to show the full version of Whitehead's second lemma (which states that the cohomology group $H^{2}(\mathfrak{g}, V)$ vanishes for any finite-dimensional $\mathfrak{g}$-representation $V$). This is given as corrollary 5.13 in Reichstein's notes.
Because semisimpleness of a Lie group is a property of its Lie algebra, $G$ being semisimple implies that $\widetilde{G}$ is semisimple. So what we really want to prove is that a simply connected, semisimple, finite-dimensional Lie group has no nontrivial central extensions into Lie groups. Suppose $E_{\widetilde{G}}$ is a central extension of $\widetilde{G}$ by $\operatorname{U}(1)$ such that (i) $E_{\widetilde{G}}$ is a Lie group, and (ii) the embedding of $\operatorname{U}(1)$ into $E_{\widetilde{G}}$ and the projection $\pi : E_{\widetilde{G}} \rightarrow \widetilde{G}$ are both smooth. By taking differentials of the embedding map and the projection map, we obtain a central extension of Lie algebras: from
$$1 \rightarrow \operatorname{U}(1) \rightarrow E_{\widetilde{G}} \rightarrow \widetilde{G} \rightarrow 1$$
to
$$0 \rightarrow \mathbb{R} \rightarrow \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g} \rightarrow 0.$$
Because we know that there are no nontrivial central extensions of semisimple Lie algebras, we know there is a splitting homomorphism $\sigma : \mathfrak{g} \rightarrow \mathfrak{e}_{\mathfrak{g}}.$ We now appeal to one of the most important and fundamental theorems about the Lie group/Lie algebra correspondence: if a Lie group $\widetilde{G}$ is simply connected, then any homomorphism $\mathfrak{g} \rightarrow \mathfrak{h}$ of Lie algebras lifts to a homomorphism of Lie groups $\widetilde{G} \rightarrow H.$ So because $\widetilde{G}$ is simply connected, the splitting of the exact sequence
$$0 \rightarrow \mathbb{R} \rightarrow \mathfrak{e}_{\mathfrak{g}} \rightarrow \mathfrak{g} \rightarrow 0$$
lifts to a splitting of the exact sequence
$$1 \rightarrow \operatorname{U}(1) \rightarrow E_{\widetilde{G}} \rightarrow \widetilde{G} \rightarrow 1,$$
which implies that $E_G$ is a trivial central extension of $\widetilde{G}.$
We have proven the first part of Bargmann's theorem: that there are no nontrivial central extensions of semisimple, simply connected Lie groups into Lie groups. The second part of Bargmann's theorem is that if $\rho : \widetilde{G} \rightarrow \operatorname{PU}(\mathcal{H})$ is a projective representation, i.e. is a homomorphism that is continuous in the strong operator topology, then there exists a central extension $E_{\widetilde{G}}$ and a representation $R : E_{\widetilde{G}} \rightarrow \operatorname{U}(\mathcal{H})$ such that (i) $E_{\widetilde{G}}$ is a Lie group, (ii) $R$ is continuous in the strong operator topology, and (iii) the projective and true representations agree, i.e. $\rho \circ \pi = P \circ R,$ where $P$ is the quotient map from $\operatorname{U}(\mathcal{H})$ to $\operatorname{PU}(\mathcal{H}).$ The proof is technical, and I don't find it particularly informative; for a clear proof, see theorem 4.8 and lemma 4.9 of Schottenloher. Taken together, the two parts of Bargmann's theorem imply that any projective representation of a semisimple, connected Lie group $G$ can be realized as a true representation of its universal cover $\widetilde{G}.$
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