# Classical conformal invariance

So I am trying to understand classical conformal invariance.

So we move gently from general coordinate invariance to Weyl invariance to conformal invariance, and now they start out with this thing called Weyl invariance $$g_{\mu \nu}(x) \rightarrow \Omega(x)g_{\mu \nu}(x),$$ and also mention the infintesimal form $$g_{\mu \nu}(x) \rightarrow g_{\mu \nu} + \omega(x)g_{\mu \nu}(x)$$ I then read the following statement which seems to make some sense:

"So then a cordinate transformation which acts on the metric as a Weyl transformation is a conformal transformation."

Now I don't understand what this last part quite means. Are we just making rotations? Can someone walk with with me in similar spirit but showing clearly what these transformations are to arrive at classical conformal invariance? I am following this text because it is more approachable than some of the things I've seen.

• 1. Please proofread your posts before submitting and don't use excessive and superfluous punctuation. 2. What do you mean by "Are we just making rotations?" - why would rotations in particular come into play here? 3. The text you use defines a conformal transformation as a coordinate transformation that induces a Weyl transformation on the metric. What is your question about that? 4. If you want to know what explicitly a conformal transformation is (rotation, dilatation, whatever), then this depends on whether you are in two or in higher dimensions, and you should just read on. Aug 23, 2015 at 22:21
• So I seem to understand what that means conceptually, but in 1.1.3 they right after they demeonstrate the effect of the general coordinate transformation on the metric, they mention rotations and translations do not change the metric and all . . . I do not see where this comes in the conformal transformation. Is the transformation rotating the metric?
– user37343
Aug 23, 2015 at 22:28
• Do you know how coordinate transformations, in general, act on the metric? (It's equation (1.8)) All translations trivially don't change the metric, just plug them in. Rotations only don't change the metric if the metric is the standard Euclidean metric, the statement doesn't hold in general. Again, just plug them in. Aug 23, 2015 at 22:38

I'm going to answer this as a non- conformal field theorist, but I have been thinking about this stuff a bit lately. But I do believe I can answer the simplest of your questions: I'm sure an expert will put the following straight if there are mistakes.

"So then a cordinate transformation which acts on the metric as a Weyl transformation is a conformal transformation." Now I don't understand what this last part quite means.

It's simply a definition. But that's probably not a very satisfying answer to you: you want motivation for this definition; I believe this comes as an answer to your question:

Are we just making rotations?

You are almost correct: what is true is that locally, the transformation preserves all angles; this means that all inner products of vectors in the tangent space at any point to the manifold transform by $\langle X,\,Y\rangle \mapsto \Omega\langle X,\,Y\rangle$ where $\Omega$ is a real scaling constant that in general depends on the position in the manifold but not on the vectors $X$ and $Y$ ($\Omega$ is fixed for a given tangent space). An equivalent statement is that $\frac{\langle X,\,Y\rangle}{\|X\|\,\|Y\|}$ is left invariant at each tangent space. In turn, this is equivalent to the statement that tangent vectors to the manifold are transformed by proper Lorentz transformations multiplied by a position dependent real scale factor $\sqrt{\Omega}$. In more formal language, the pushforward of the transformation lives in $(\mathbb{R}^+,*)\times SO(1,\,3)$ at all points in the manifold (sometimes we might restrict these statements to an open subset of the manifold).

The situation is wholly analogous to a holomorphic function. One way of stating the Cauchy-Riemann relationships is as follows: "the pushforward of a conformal transformation on $\mathbb{C}$ lives in $(\mathbb{R}^+,*)\times SO(2)$ at all points of conformalness", i.e. if the holomorphic function is written as $(x,\,y)\mapsto (u,\,v)$ with $x,\,y,\,u,\,v\in\mathbb{R}$ then the matrix:

$$\left(\begin{array}{cc}\frac{\partial\,u}{\partial\,x}&\frac{\partial\,u}{\partial\,y}\\\frac{\partial\,v}{\partial\,x}&\frac{\partial\,v}{\partial\,y}\end{array}\right) = \left(\begin{array}{cc}\alpha&-\beta\\\beta&\alpha\end{array}\right) = \sqrt{\Omega} \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)\tag{1}$$

is a scalar multiple of a rotation matrix (here the scale factor $\sqrt{\Omega}=\sqrt{\alpha^2+\beta^2}$ and $\tan\theta = \beta/\alpha$) (if you've not done this before, check that the Cauchy-Riemann relationships are equivalent to (1)). So the local transformation is a rotation composed with a uniform dilation (or the other way around; clearly they commute).

Your situation is simply the above with the notion of "rotation matrix in $SO(2)$" replaced by "(proper) Lorentz isometry in $SO(1,\,3)$", each being proper, homogeneous isometries of the tangent space in question.

It seems that the above was not altogether clear: rather than rewriting the answer I am capturing a conversation between me and user ACuriousMind that illustrates my lack of clarity and the answer I gave to try to make amends and clear my description up.

ACuriousMind says:

Nope. While the Lorentz group is part of the conformal group in 4D (with signature (1,3)), the group of conformal transformations is larger - it also has dilatations (the scale factor), but also "special conformal transformations". The group of conformal transformations in 2D is in fact infinite-dimensional (its algebra is the Witt algebra). Conformal transformations are more than rotations or Lorentz transformations (otherwise they would be quite boring!), in fact the conformal algebra in $d>2$ with signature $(p,q)$ is isomorphic to $so(p+1,q+1)$.

Are you speaking globally? Because here I'm speaking about the pushforward of the transformation: so that (in 2D), yes, the algebra of vector fields that exponentiate to e.g. level curves of $Re$ and $Im$ of meromorphic functions is indeed infinite dimensional, but locally a conformal transformation is an isometry and dilation. I'm well aware that the group of globally conformal transformations of $\mathbb{R}^{p+q}$ is bigger than the group of global isometries composed with dilations (but, astoundingly, not much bigger when $p+q>2$, as you are aware).

And this seemed to clear up the confusion I had caused as ACuriousMind answers:

Ah, I misunderstood what you meant by "locally" - reading it again, it is clear and correct.

I hope the situation is now clearer to other readers!

• Nope. While the Lorentz group is part of the conformal group in 4D (with signature $(1,3)$), the group of conformal transformations is larger - it also has dilatations (the scale factor), but also "special conformal transformations". The group of conformal transformations in 2D is in fact infinite-dimensional (its algebra is the Witt algebra). Conformal transformations are more than rotations or Lorentz transformations (otherwise they would be quite boring!), in fact the conformal algebra in $d>2$ with signature $(p,q)$ is isomorphic to $\mathfrak{so}(p+1,q+1)$. Aug 24, 2015 at 13:03
• @ACuriousMind (Let's move to chat if you can't answer simply) OK: I'll delete if the above is wrong, but just a question first. Are you speaking globally? Because here I'm speaking about the pushforward of the transformation: so that (in 2D), yes, the algebra of vector fields that exponentiate to e.g. level curves of ${\rm Re}$ and ${\rm Im}$ of meromorphic functions is indeed infinite dimensional, but locally a conformal transformation is an isometry and dilation. Aug 24, 2015 at 22:26
• @ACuriousMind I'm well aware that the group of globally conformal transformations of $\mathbb{R}^{p+q}$ is bigger than the group of global isometries composed with dilations (but, astoundingly, not much bigger when $p+q>2$, as you are aware). Aug 25, 2015 at 6:59
• Ah, I misunderstood what you meant by "locally" - reading it again, it is clear and correct. Aug 25, 2015 at 15:58