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The global conformal group in 2D is $SL(2,\mathbb{C})$. It consists of the fractional linear transforms that map the Riemann sphere into itself bijectively and is finite dimensional.

However, when studying $CFT_2$ people always use the full Virasoro algebra, not just the $L_{0,\pm1}$ which actually exponentiate to invertible transformations. I would like to know why people consider the other $L_n$'s to be symmetries of the theory.

I am aware that Ward identities are local statements, and that I can consider a coordinate patch where the additional conformal transformations are bijective in order to derive relations between correlation functions in this patch. I am also familiar with the representations of the Virasoro algebra and how constraining the symmetry is.

However, we are doing quantum mechanics, and a symmetry of the theory should take me from one physical state to another. In addition, the symmetry should have an inverse which undoes this transformation. This means that the physical Hilbert space should organize itself into representations of the symmetries of the theory. However, the local conformal transformations cannot have inverses, and so they do not form a group as far as I know. So why is it assumed that the states of a $CFT_2$ should organize themselves into representations of the Virasoro algebra? (I am aware $L_{n \leq -2}|0\rangle\neq 0$, $\langle 0|L_{n\geq 2} \neq 0$ so all but the $L_{0,\pm 1}$ are "spontaneously broken" on the in/out vacuum, but this is not relevant since the states of the theory are still assumed to assemble into Verma modules since it is assumed that the Virasoro was a symmetry of the theory which is just violated by the vacuum).

My question basically boils down to: How can I have symmetries of a theory which are not invertible? I'd appreciate any comment's that clear up my confusion.

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The Virasoro algebra is a centrally extended algebra. This means that in every representation, its central element must be represented by the unit operator. Thus (for a non-vanishing central charge) it cannot be fully implemented at the quantum level as a symmetry of the vacuum, otherwise one can get a contradiction of the type $1 |0\rangle = 0 |0\rangle $.

The algebra $\frak{su}(1,1)$ generated by $L_{0, \pm1}$ is the largest subalgebra not containing the central element, thus the largest subalgebra which can be implemented as a symmetry of the vacuum.

The correct way is to view the (centrally extended) group $ G = \widetilde{Diff(S^1)}$ as the dynamical group of the theory (Please see Souriau's book page 100). (Although this book treats only finite dimensional groups). This means that this group can be implemented classically by means of a canonical transformation. The Hamiltonian will be an element of universal enveloping algebra of the group. Both these conditions are valid in our case.

The importance of this construction is that upon quantization, the action of the dynamical group can be (optimistically) lifted to a unitary representation on the quantum Hilbert space. This representation is induced from a representation of the small group (vacuum preserving subgroup for example a subgroup $H$ corresponding to the Lie algebra $\frak{su}(1,1)$).

In the modern terminology, this construction is called a quantization of the coadjoint orbit $G/H$. The representation of the small group actually fixes a symplectic structure on the coadjoint orbit. Please see the article by: Gay-Balmaz on the Virasoro group coadjoint orbits and references therein (The article is available online in the following page).

(It should be mentionioned that the theory of coadjoint orbits in the finite dimensional case is much more straightforward, please see for example the following review by Kirillov).

This is not the whole story in our case because of the infinite dimensionality of the group and the representations. As mentioned in the question, the broken generators in this case cannot be unitarily implemented because they drive the vacuum outside the quantum Hilbert space.

This situation was considered by Bowick and Rajeev and also by Kirillov and Yuriev, please see the following review by Sergeev. They constructed the Hilbetrt space and then acted on it by an arbitrary element of $Diff(S^1)$ shifting the vacuum state. The vacuua generate a line bundle over $Diff(S^1)$ with a Hermitian connection. They found that the condition that this connection becomes flat is exactly the when the quantum theory becomes $Diff(S^1)$ invariant at the quantum level (for example in the flat case of the bosonic string this happens when $d=26$). The quantization they performed is a Kähler quantization. They found that the coadjoint orbit canonical bundle contribution to the anomaly is just equal to the contribution of the ghosts (This is understandable because the ghost term originates from the volume measure of the path integral). The flatness condition of the vacuum bundle connection can be interpreted as a unitary equivalence of the quantization Hilbert spaces. In this case the Virasoro algebra can be exponentiated since it does not have a net central charge.

This type of quantization was used by Hitchin and later by Axelrod, della Pietra, and Witten in the quantization of the Chern Simons theory.

Thus the solution of the implementability of the full Virasoro group at the quantum level is to take the vacuum as a tensor product of the matter Fock vacuum and the ghost sectors such that its full central charge vanishes.

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    $\begingroup$ Do I understand the logic of your first paragraph right? Suppose we want all $L_n$ to be symmetries. This means for all $n$ the vacuum is invariant: $\rho(L_n) |0>=0$ for all $n$. On the other hand the representation of a center is always proportional to the unity $\rho(Z)=c \times 1$. Now $[\rho(L_n),\rho(L_{-n})]|0>=0|0>=2n \rho(L_{0})|0>+c \times 1|0>=c \times 1|0>$ so we have $0=c \times 1|0>$ eventhough we assmumed $c \neq 0$. So the full Virasoro algebra can not be implemented at quantum level as a symmetry of the vacuum. $\endgroup$ – ungerade Apr 14 '16 at 19:59
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    $\begingroup$ @ungerade you are correct $\endgroup$ – David Bar Moshe Apr 15 '16 at 5:31
  • $\begingroup$ I'm also a little bit puzzled by: "the broken generators in this case cannot be unitarily implemented because they drive the vacuum outside the quantum Hilbert space." To my (probably wrong) understanding I thought that $\rho(L_n)|0>$ is either $0$ (for $n>-2$) or a new state in the Hilbert space. How plays this into the discussion? (Of course if this question is too long to answer, I'm happy to ask a separate stackexchange question) $\endgroup$ – ungerade Apr 15 '16 at 9:38
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    $\begingroup$ in quantum field theory (in contrast to quantum mechanics), spontaneously broken generators drive the vacuum outside the Hilbert space of the theory since we can prove that minimum energy vectors of the type $e^{i\alpha G}|vac\rangle$ where $G$ is a spontaneously broken generator are orthogonal to the vacuum for all $\alpha \ne 0$ to the vacuum, thus cannot be represented by a unitary operator (the infinitesimal action $ G|vac\rangle$, however, exists as it is the Nambu-Goldstone mode.) $\endgroup$ – David Bar Moshe Apr 15 '16 at 20:39
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    $\begingroup$ This means that the symmetry group in contrast to the symmetry algebra will not be unitarily representable on the Hilbert space, and this is not the case that we need, since as mentioned above, we need to gauge this symmetry out. $\endgroup$ – David Bar Moshe Apr 15 '16 at 20:42

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