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  Why exactly do sometimes universal covers, and sometimes central extensions feature in the application of a symmetry group to quantum physics?

+ 7 like - 0 dislike

There seem to be two different things one must consider when representing a symmetry group in quantum mechanics:

  • The universal cover: For instance, when representing the rotation group $\mathrm{SO}(3)$, it turns out that one must allow also $\mathrm{SU}(2)$ representations, since the negative sign a "$2\pi$ rotation" induces in $\mathrm{SU}(2)$ is an overall phase that doesn't change the physics. Equivalently, all representations of the Lie algebra are what we seek. ($\mathfrak{so}(3) = \mathfrak{su}(2)$, but although every representation of the algebra is one of the universal cover, not every representation of the algebra is one of $\mathrm{SO}(3)$.)

  • Central extensions: In conformal field theory, one has classically the Witt algebra of infinitesimal conformal transformations. From the universal cover treatment one is used to in most other cases, one would expect nothing changes in the quantum case, since we are already seeking only representation of an algebra. Nevertheless, in the quantization process, a "central charge" appears, which is often interpreted to arise as an "ordering constant" for the now no longer commuting fields, and we have to consider the Virasoro algebra instead.

The question is: What is going on here? Is there a way to explain both the appearence of universal covers and central extensions in a unified way?

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user ACuriousMind
asked Sep 2, 2015 in Mathematics by ACuriousMind (910 points) [ no revision ]
If you enjoyed this question (and especially the answer), you will probably also enjoy the book on conformal field theory by Schottenloher (first linked book on the page), which covers this topic and much more.

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user Danu
This is explained very well (in my opinion) in the first chapter of Weinberg QFT I.

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user Peter Kravchuk

1 Answer

+ 7 like - 0 dislike

Yes. Both universal covers and central extensions incurred during quantization come from the same fundamental concept:

Projective representations

If $\mathcal{H}$ is our Hilbert space of states, then distinct physical states are not vectors $\psi\in\mathcal{H}$, but rays, since multiplication by a complex number does not change the expectation values given by the rule $$ \langle A\rangle_\psi = \frac{\langle \psi \vert A \vert \psi \rangle}{\langle \psi \vert \psi \rangle}$$ nor the transition probabilities $$ P(\lvert \psi \rangle \to \lvert \phi \rangle) = \frac{\lvert \langle \psi \vert \phi \rangle\rvert^2}{\langle \phi \vert \phi \rangle\langle \psi \vert \psi \rangle}$$ The proper space to consider, where every element of the space is indeed a distinct physical state, is the projective Hilbert space $$ \mathrm{P}\mathcal{H} := \mathcal{H} /\sim$$ $$ \lvert \psi \rangle \sim \lvert \phi \rangle :\Leftrightarrow \exists c\in\mathbb{C}: \lvert \psi \rangle = c\lvert\phi\rangle$$ which is just a fancy way to write that every complex ray has been shrunk to a point. By Wigner's theorem, every symmetry should have some, not necessarily unique, unitary representation $\rho : G \to \mathrm{U}(\mathcal{H})$. Since it has to descend to a well-defined ray transformation, the action of the symmetry is given by a group homomorphism into the projective unitary group $G \to \mathrm{PU}(\mathcal{H})$, which sits in an exact sequence $$ 1 \to \mathrm{U}(1) \to \mathrm{U}(\mathcal{H}) \to \mathrm{PU}(\mathcal{H}) \to 1$$ where $\mathrm{U}(1)$ represents the "group of phases" that is divided out when passing to the projective space. It is already important to notice that this means $\mathrm{U}(\mathcal{H})$ is a central extension of $\mathrm{PU}(\mathcal{H})$ by $\mathrm{U}(1)$.

To classify all possible quantumly allowed representations of a symmetry group $G$, we need to understand the allowed Lie group homomorphisms $\sigma : G\to\mathrm{PU}(\mathcal{H})$. Since linear representations are nicer to work with than these weird projective things, we will look at

Classifying projective representations by unitary linear representations

For any $g\in G$, choose a representative $\Sigma(g)\in\mathrm{U}(\mathcal{H})$ for every $\sigma(g)\in\mathrm{PU}(\mathcal{H})$. This choice is highly non-unique, and is essentially responsible for how the central extension appears. Now, since for any $g,h\in G$ we have $\sigma(g)\sigma(h) = \sigma(gh)$, the choices of representatives must fulfill $$ \Sigma(g)\Sigma(h) = C(g,h)\Sigma(gh)$$ for some $C : G\times G\to\mathrm{U}(1)$. Applying associativity to $\Sigma(g)\Sigma(h)\Sigma(k)$ gives the consistency requirement $$ C(g,hk)C(h,k) = C(g,h)C(gh,k)\tag{1}$$ which is also called the cocycle identity. For any other choice $\Sigma'$, we must have $$ \Sigma'(g) = f(g)\Sigma(g) $$ for some $f : G \to \mathrm{U}(1)$. $\Sigma'$ has an associated $C'$, and so we get $$ C'(g,h)\Sigma'(gh) = \Sigma'(g)\Sigma'(h) = f(g)f(h)C(g,h)f(gh)^{-1}\Sigma'(gh)$$ which yields the consistency requirement $$ C'(g,h)f(gh) = f(g)f(h)C(g,h)\tag{2}$$ Therefore, projective representations are classified giving the choice of unitary representatives $\Sigma$, but those that are related by $(2)$ give the same projective representation. Formally, the set $$ H^2(G,\mathrm{U}(1)) := \{C : G\times G \to \mathrm{U}(1)\mid C \text{ fulfills } (1)\} / \sim$$ $$ C \sim C' :\Leftrightarrow \exists f : (2) \text{ holds }$$ classifies the projective representations of $G$. We want to use it to construct a unitary representation of something that classifies the projective representation:

Define the semi-direct product $G_C := G \ltimes_C \mathrm{U}(1)$ for any representative $C$ of an element in $H^2(G,\mathrm{U}(1)$ by endowing the Cartesion product $G \times \mathrm{U}(1)$ with the multiplication $$ (g,\alpha)\cdot(h,\beta) := (gh,\alpha\beta C(g,h))$$ One may check that it is a central extension, i.e. the image of $\mathrm{U}(1)\to G \ltimes_C\mathrm{U}(1)$ is in the center of $G_C$, and $$ 1 \to \mathrm{U}(1) \to G_C \to G \to 1$$ is exact. For any projective representation $\sigma$, fix $\Sigma,C$ and define the linear representation $$ \sigma_C : G_C \to \mathrm{U}(\mathcal{H}), (g,\alpha) \mapsto \alpha\Sigma(g)$$ Conversely, every unitary representation $\rho$ of some $G_C$ gives a pair $\Sigma,C$ by $\Sigma(g) = \alpha^{-1}\rho(g,\alpha)$.

Therefore, projective representations are in bijection to linear representations of central extensions.

On the level of the Lie algebras, we have $\mathfrak{u}(\mathcal{H}) = \mathfrak{pu}(\mathcal{H})\oplus\mathbb{R}$, where the basis element $\mathrm{i}$ of $\mathbb{R}$ generates multiples of the identity $\mathrm{e}^{\mathrm{i}\phi}\mathrm{Id}$. We omit the $\mathrm{Id}$ in the following, whenever a real number is added to an element of the Lie algebra, it is implied to be multiplied by it.

Repeating the arguments above for the Lie algebras, we get that the projective representation $\sigma : G \to \mathrm{PU}(\mathcal{H})$ induces a representation of the Lie algebra $\phi : \mathfrak{g}\to\mathfrak{pu}(\mathcal{H})$. A choice of representatives $\Phi$ in $\mathfrak{u}(H)$ classifies such a projective representation together with an element $\theta$ in $$ H^2(\mathfrak{g},\mathbb{R}) := \{\theta : \mathfrak{g}\times\mathfrak{g} \to \mathbb{R}\mid \text{ fulfills } (1') \text{ and } \theta(u,v) = -\theta(v,u)\} / \sim$$ $$ \theta \sim \theta' :\Leftrightarrow \exists (b : \mathfrak{g}\to\mathbb{R}) :\theta'(u,v) = \theta(u,v) + b([u,v])$$ with consistency condition $$ \theta([u,v],w) + \theta ([w,u],v) + \theta([v,w],u) = 0 \tag{1'}$$ that $\theta$ respects the Jacobi identity, essentially.

Thus, a projective representation of $\mathfrak{g}$ is classified by $\Phi$ together with a $\theta\in H^2(\mathfrak{g},\mathbb{R})$. Here, the central extension is defined by $\mathfrak{g}_\theta := \mathfrak{g}\oplus\mathbb{R}$ with Lie bracket $$ [u\oplus y,v\oplus z] = [u,v]\oplus\theta(u,v)$$ and we get a linear representation of it into $\mathfrak{u}(\mathcal{H})$ by $$ \phi_\theta(u\oplus z) := \Phi(u) + a$$

Again, we obtain a bijection between projective representations of $\mathfrak{g}$ and those of its central extensions $\mathfrak{g}_\theta$.

Universal covers, central charges

We are finally in the position to decide which representations of $G$ we must allow quantumly. We distinguish three cases:

  1. There are no non-trivial central extensions of either $\mathfrak{g}$ or $G$. In this case, all projective representations of $G$ are already given by the linear representations of $G$. This is the case for e.g. $\mathrm{SU}(n)$.

  2. There are no non-trivial central extensions of $\mathfrak{g}$, but there are discrete central extensions of $G$ by $\mathbb{Z}_n$ instead of $\mathrm{U}(1)$. Those evidently also descend to projective representations of $G$. Central extensions of Lie groups by discrete groups are just covering groups of them, because the universal cover $\overline{G}$ gives the group $G$ as the quotient $\overline{G}/\Gamma$ by a discrete central subgroup $\Gamma$ isomorphic to the fundamental group of the covered group. Thus we get that all projective representations of $G$ are given by linear representations of the universal cover. No central charges occur. This is the case for e.g. $\mathrm{SO}(n)$.

  3. There are non-trivial central extensions of $\mathfrak{g}$, and consequently also of $G$. If the element $\theta\in H^2(\mathfrak{g},\mathbb{R})$ is not zero, there is a central charge - the generator of the $\oplus\mathbb{R}$ in $\mathfrak{g}_\theta$, or equivalently the conserved charge belonging to the central subgroup $\mathrm{U}(1)\subset G_C$. This happens for the Witt algebra, where inequivalent $\theta(L_m,L_n) = \frac{c}{12}(m^3 - m)\delta_{m,-n}$ are classified by real numbers $c\in \mathbb{R}$.

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user ACuriousMind
answered Sep 2, 2015 by ACuriousMind (910 points) [ no revision ]
A question on your choice of language: when you say highly nonunique to describe the choice $\sigma\mapsto\Sigma$, are you emphasizing the in general uncountable choice in contrast with the at most countable choice that arises for covers of a finite dimensional Lie group (whose algebra of course doesn't have a central charge)?

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
@WetSavannaAnimalakaRodVance: I say "highly" because the $\Sigma$ together with its $C$ represents a cohomology class (the notation $H^2$ is not an accident, although I didn't explain the connection). I tend to imagine (co)homology classes as "very large" (e.g. in the singular case, the chain groups are indeed absurdly large), but I didn't have a specific cardinality for it in mind.

This post imported from StackExchange Physics at 2015-11-01 18:02 (UTC), posted by SE-user ACuriousMind

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