4
$\begingroup$

My question is from P.98 of the book by Di Francesco on Conformal Field theory. He gives the six non-vanishing commutation relations between the elements $P_{\mu}, D, L_{\mu \nu}$ and $K_{\mu}$ comprising the four dimensional space of generators of the conformal group. These are then rewritten by defining a set of four more generators: $$J_{\mu \nu} = L_{\mu \nu}\,\,\,,\,\,\,J_{-1, \mu} = \frac{1}{2}(P_{\mu} - K_{\mu})\,\,\,,\,\,\,J_{-1,0} = D\,\,\,,\,\,\,J_{0,\mu} = \frac{1}{2}(P_{\mu} + K_{\mu}),$$ and I believe the motivation for doing so is so that the six commutation relations mentioned above can be eloquently recast in a single one line commutation relation $$[J_{ab}, J_{cd}] = i(\eta_{ad}J_{bc} + \eta_{bc}J_{ad} - \eta_{ac}J_{bd} - \eta_{bd}J_{ac})$$ It also says that $a,b \in \left\{-1,0,1,\dots,d\right\}$ and that the new generators obey the $SO(d+1,1)$ commutation relations (which I think is the one-line equation above).

My question is: What do the indices $a,b$ represent and what does the notation $SO(d+1,1)$ mean? I think there are $d$ spatial dimensions but I can't see what the significance of the $-1$ and $0$ elements are.

Many thanks.

$\endgroup$
  • $\begingroup$ Since no one has said anything yet I will make a total guess. See if it makes sense. There are $d$ spatial dimensions. The index 0 is for time (I'm pretty sure of this). The index -1 has something to do with the $D$ operator, which I am guessing are dilations. $SO(d+1,1)$ is just the group of lorentz transformations in minkowski space with $d+1$ spatial dimensions instead of $d$. The only confusing thing to me is what $\eta_{-1-1}$ is supposed to be be. I'd guess $\eta_{-1-1}$ is supposed to have the opposite sign of the other diagonal elements of $\eta$, but maybe $\eta_{00}$ is the odd one. $\endgroup$ – Brian Moths Jun 11 '14 at 22:50
3
$\begingroup$

The indices $a$ and $b$ are chosen in such a way that inserting different combinations of values from $-1$ to $d$ gives just the original six commutation relations for the generators of conformal symmetry transformations. As you have correctly suggested, the point of this is to rewrite the six relations in compact form.

The latter shows that the conformal group is actually given by $\text{SO}(d+1,1)$, which is the group of special orthogonal transformations in $d+1$ spacelike and $1$ timelike dimensions, where $d$ is the number of spatial space-time dimensions. $0$ represents the time-component, while $-1$ is spacelike and appears because we have dilatations and special conformal transformations. Note that that $-1$ is not a spatial dimension with respect to spacetime but rather an additional dimension with respect to the group action.

$\endgroup$
  • $\begingroup$ Hi Frederic, many thanks for your response. So is it the case that an additional spacelike component and a timelike component is added to make the isomorphism between the conformal group and SO(d+1,1) explicit? In some sense these extra components are fictitious dimensional quantities that are added to make the connection to SO(d+1,1)? Could you also explain how the $-1$ appears because we gave dilatations and special conformal transformations? Thanks again. $\endgroup$ – CAF Jun 12 '14 at 7:37
  • 1
    $\begingroup$ @CAF: Nobody adds an additional timelike component, it is already there. The group of Lorentz transformations in d space-like dimensions is given by SO(d,1), there is only one time component. Adding dilatations and special conformal transformations results in SO(d+1,1). But yes, you can interpret the additional dimension as a tool to make SO(d+1,1) manifest. We call that additional component $-1$ and assign them to the generators in such a way that the nature of the object is preserved. $\endgroup$ – Frederic Brünner Jun 12 '14 at 8:38
  • $\begingroup$ @CAF Tensors turn into tensors, vectors into vectors and scalars remain scalars. This way, one can assign the indices uniquely. $\endgroup$ – Frederic Brünner Jun 12 '14 at 8:38
2
$\begingroup$

Just following up on Frederic's answer. I wouldn't get too hung up on thinking about a metric that includes $\eta_{-1-1}$. The metric really refers to spacetime and there is no new spacetime dimension we've introduced. It's just a way of labeling the generators - their indices don't necessarily refer to spacetime dimensions, though they do for most of the $J$ here.

You should separate the idea of the symmetry group from the spacetime somewhat. What I mean is, you seem to be hung up on thinking of SO(3) as the group of isometries of 3-space and maybe SO(3,1) as the isometries of Minkowski space (maybe you didn't know that last part but it's true), and therefore thinking you need a larger spacetime to have SO(d+1,1).

However, it is not necessarily the case that the symmetry group of a theory in d spatial dimensions and n timelike dimensions is SO(d,n). It happens to be true for Euclidean space and Minkowski space without conformal invariance, so that prejudices us. But really, it's just a prejudice to assume that the (maximal) symmetry group is SO(d,n). I mean, it is true in most of the QFTs we first learn. And furthermore, in theories with conformal invariance, SO(d,n) is contained as a subgroup, and we didn't need to worry about whether it was part of a larger group when we were learning QFT.

I think there is one mistake in Frederic's answer. The conformal group in d spatial and n timelike dimensions is SO(d+1,n+1). So a CFT in 4-D Euclidean space is a rep of SO(5,1) and in 4-D Minkowski space is a rep of SO(4,2).

One very interesting thing about all this: you might ask, what is a spacetime where SO(4,2) really is just generalized rotations (as opposed to rotations + SCTs + dilatations)? Well, $AdS_5$ is one! This might be your first clue towards the existence of the AdS-CFT correspondence! A CFT in 3+1-dimensional spacetime obeys the same algebra as the isometries of $AdS_5$. See "ANTI-DE SITTER SPACE" by Ingemar Bengtsson - pages 1-5 give a nice concise introduction to AdS spacetime and it's isometries.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.