I am reading Di Francesco's "Conformal Field Theory" and in page 95 he defines a conformal transformation as a mapping $x \mapsto x'$ such that the metric is invariant up to scale:

$$g'_{\mu \nu}(x') = \Lambda(x) g_{\mu \nu} (x).$$

On the other hand we know from GR that under any coordinate transformation the metric changes as

$$ g_{\mu \nu} (x) \mapsto g'_{\mu \nu}(x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} .$$

I feel like there is a notation problem (inconsistency) in these formulas, or maybe I am mixing active and passive coordinate transformations. For instance, if we consider a simple rotation (which is of course a conformal transformation with no rescaling, i.e. $\Lambda(x)=1$) then from the first formula we see that $g'_{\mu \nu}(x') = g_{\mu \nu} (x)$, whereas from the second formula we get something more complicated. Where is the flaw?

In the "String theory" lecture notes by David Tong the same definition of conformal transformation is given. Then he says:

A transformation of the form (4.1) has a diferent interpretation depending on whether we are considering a fixed background metric $g_{\mu \nu}$, or a dynamical background metric. When the metric is dynamical, the transformation is a diffeomorphism; this is a gauge symmetry. When the background is fixed, the transformation should be thought of as an honest, physical symmetry, taking the point $x$ to point $x'$. This is now a global symmetry with the corresponding conserved currents.

I think it has to do with my question, but I don't fully understand it...

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    $\begingroup$ Related: physics.stackexchange.com/q/38138/2451 , physics.stackexchange.com/q/226464/2451 and links therein. $\endgroup$ – Qmechanic Mar 28 '19 at 17:48
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    $\begingroup$ I feel my question is not fully answered there. $\endgroup$ – MBolin Mar 28 '19 at 17:54
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    $\begingroup$ The actual definition of a conformal transformation is a mess. By my count, the usual references use at least 5 totally different definitions interchangeably, so you're right to be annoyed. I thought the top answer to the second question Qmechanic linked was good. $\endgroup$ – knzhou Mar 28 '19 at 17:56

OK I think I know what is going on. It's all about primes. Consider an active spacetime transformation:

$$ x^{\mu} \mapsto x'^{\mu}(x) \, ,$$

$$g_{\mu \nu} (x) \mapsto g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} \, .$$

(the transformation of the metric tensor follows from the fact that it is a rank 2 tensor). With this notation both Di Francesco and David Tong are wrong (as far as I understand). The GR book by Zee on the other hand writes it properly. First of all consider an isometry. This is an spacetime transformation as before that leaves the metric invariant, meaning

$$ g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} = g_{\mu \nu} (x') \, .$$

(watch the primes). On the other hand a conformal transformation is a transformation that satisfies a weaker condition: it leaves the metric invariant up to scale, meaning

$$ g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} = \Omega^2(x')g_{\mu \nu} (x') \, .$$

Now there should be no inconsistency. Di Francesco's definition was wrong (according to this convention/notation/understanding) because it compared the metric before and after the transformation at different points, and you have to compare them at the same point.

  • $\begingroup$ Did you check what equation do you get by imposing invariance under a small transformation? I mean, if you use your definition do you get the covariantized Killing equation $\nabla_{(a}v_{b)} = \nabla^cv_c\,g_{ab}$? $\endgroup$ – MannyC Mar 28 '19 at 23:10
  • $\begingroup$ Yes, I get $\nabla_{\mu} v_{\nu} + \nabla_{\nu} v_{\mu} = \frac{2}{d} (\nabla \cdot v ) g_{\mu \nu}$. $\endgroup$ – MBolin Mar 28 '19 at 23:32
  • $\begingroup$ Instead if you use the other one there are some pieces missing I imagine? This is very weird... $\endgroup$ – MannyC Mar 28 '19 at 23:38
  • $\begingroup$ Yes, there are missing pieces and it is logical. Try it. From Di Francesco's definition, for a change of coordinates to be a conformal transformation you should have something like $\frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} = \Omega^2(x) \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu}$, which doesn't seem like a conformal transformation in general. Take a look at Zee. $\endgroup$ – MBolin Mar 28 '19 at 23:40
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    $\begingroup$ Damn it, I thought the second block-quoted expression for diffeomorphism in your question to be from Tong which is what I was defending (while thinking that you had quoted it from Tong because I had misread). Yes, now I have no disagreements. Thanks for the responses and for providing a nice addition to my PSE bookmarks ;) $\endgroup$ – Dvij D.C. Nov 17 '20 at 13:12

I'm a mathematician, not a physicist, so I learned all of these ideas with different notation, but I think I understand what might be confusing you.

Conformal transformations are indeed a special kind of diffeomorphism, and a rotation (say in the plane with the usual metric) is indeed conformal, so the two formulas you listed had better agree in this case.

But in fact, if your manifold is $\mathbb{R}^2$, your metric is the usual one ($g_{\mu\nu}$ is the identity matrix at every $x$), and your coordinate change is a rotation, the second formula you listed will show you that the metric looks unchanged in the new coordinates. (This is not a coincidence: preserving this metric is exactly the property that makes rotations special in the first place!) That is, there is no conflict between the two formulas here, it's just that seeing it involves a bit of computation.

Working this out is a very good exercise and I don't think you'd gain much from me typing it all out here. A hint that might help you get oriented is that, since rotations are linear in the coordinate system we've chosen, the Jacobian matrix at every point is the same as the matrix for the rotation itself.

  • $\begingroup$ You are right, but I think that is not saying anything. Notice that for a simple metric like Euclidean metric, which doesn't depend on position, the formula for the transformation of the metric $g'_{\mu \nu}(x') = \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} g_{\alpha \beta} (x)$ and the equation for an isometry $g'_{\mu \nu}(x') = g_{\mu \nu}(x') $ look the same. $\endgroup$ – MBolin Mar 28 '19 at 18:42
  • $\begingroup$ Your answer indeed gives one of the five definitions of the phrase “conformal transformation”. But it isn’t the one that CFT is actually about. Your definition makes conformal invariance just a subset of diffeomorphism invariance. $\endgroup$ – knzhou Mar 28 '19 at 18:43
  • $\begingroup$ This is definitely our (i.e. physicists’) fault. We stole a word from math and used it to mean several different things, and the notation conventionally used is so ambiguous one can never tell between them. $\endgroup$ – knzhou Mar 28 '19 at 18:44
  • $\begingroup$ @knzhou: That's unfortunate! But it does seem like the definition I had in mind writing this is the one that the original question used in the first formula. Can you explain the discrepancy more clearly? I certainly don't want to contribute further to the confusion, so this answer ought to be edited if it's currently doing that. $\endgroup$ – Nicolas Ford Mar 28 '19 at 19:28
  • $\begingroup$ @MBolin: As I said in the last comment there may be a difference in the definitions that I'm not aware of, but it's worth pointing out that I think what you just said is not strictly true. There are certainly even linear transformations on the plane that are not conformal (and therefore not isometries). An example is multiplying the $x$ coordinate by 2, that is, $\begin{pmatrix}2&0\\0&1\end{pmatrix}$. $\endgroup$ – Nicolas Ford Mar 28 '19 at 19:31

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