# Conformal transformation/ Weyl scaling are they two different things? Confused!

+ 7 like - 0 dislike
266 views

I see that the weyl transformation is $g_{ab} \to \Omega(x)g_{ab}$ under which Ricci scalar is not invariant. I am a bit puzzled when conformal transformation is defined as those coordinate transformations that effect the above metric transformation i.e $x \to x' \implies g_{\mu \nu}(x) \to g'_{\mu \nu}(x') = \Omega(x)g_{\mu \nu }(x)$. but any covariant action is clearly invariant under coordinate transformation? I see that what we mean by weyl transformation is just changing the metric by at a point by a scale factor $\Omega(x)$. So my question is why one needs to define these transformations via a coordinate transforms. Is it the case that these two transformations are different things. In flat space time I understand that conformal transformations contain lorentz transformations and lorentz invariant theory is not necessarily invariant under conformal transformations. But in a GR or in a covariant theory effecting weyl transformation via coordinate transformations is going to leave it invariant. Unless we restrict it to just rescaling the metric?

I am really confused pls help.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prag1

+ 6 like - 0 dislike

A conformal transformation is a space-time transformation which leaves the metric invariant up to scale and thus preserves angles. A Weyl transformation actively scales the metric.

More formally:

Let $M, N$ be two manifolds with inner products $g, h$ and coordinates $x=(x^i), y=(y^j)$ respectively.

A map $f:M\rightarrow N$ is called conformal if there exists a function $\Omega\in\mathcal C^\infty(M)$ so that the pullback meets $$f^*h=\Omega g$$ which reads in coordinates $$h_{ij}(y)\frac{\partial y^i}{\partial x^r}\!(x)\frac{\partial y^j}{\partial x^s}\!(x) = \Omega(x)g_{rs}(x)$$ where $y=f(x)$.

In case of conformal transformations, $M=N$ and $g=h$ and thus $$f^*g = \Omega g$$ which reads in coordinates $$g_{ij}(y)\frac{\partial y^i}{\partial x^r}\!(x)\frac{\partial y^j}{\partial x^s}\!(x) = \Omega(x)g_{rs}(x)$$ and looks like a coordinate transformation $x\mapsto y$, but is the coordinate expression of an actual one (see my other answer).

In case of Weyl transformations, we have again $M=N$. However, the map will be given by $f=\mathrm{id}_M$ yielding $$h=\Omega g$$ with the trivial coordinate expression $$h_{ij}(x)=\Omega(x)g_{ij}(x)$$ which cannot be regarded as a coordinate transformation as the coordinates do not change.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Christoph
answered Sep 24, 2012 by (210 points)
+ 5 like - 0 dislike

The Weyl transformation and the conformal transformation are completely different things (although they are often discussed in similar contexts).

A Weyl transformation isn't a coordinate transformation on the space or spacetime at all. It is a physical change of the metric, $g_{\mu\nu}(x)\to g_{\mu\nu}(x)\cdot \Omega(x)$. It is a transformation that changes the proper distances at each point by a factor and the factor may depend on the place – but not on the direction of the line whose proper distance we measure (because $\Omega$ is a scalar).

Note that a Weyl transformation isn't a symmetry of the usual laws we know, like atomic physics or the Standard Model, because particles are associated with preferred length scale so physics is not scale invariant.

On the other hand, conformal transformations are a subset of coordinate transformations. They include isometries – the genuine geometric "symmetries" – as a subset. Isometries are those coordinate transformations $x\to x'$ that have the property that the metric tensor expressed as functions of $x'$ is the same as the metric tensor expressed as functions of $x$. Conformal transformations are almost the same thing: but one only requires that these two tensors are equal functions up to a Weyl rescaling.

For example, if you have a metric on the complex plane, $ds^2=dz^* dz$, then any holomorphic function, such as $z\to 1/z$, is conformally invariant because the angles are preserved. If you pick two infinitesimal arrows $dz_1$ and $dz_2$ starting from the same point $z$ and if you transform all the endpoints of the arrows to another place via the transformation $z\to 1/z$, then the angle between the final arrows will be the same. Consequently, the metric in terms of $z'=1/z$ will be still given by $$ds^2 = dz^* dz = d(1/z^{\prime *}) d (1/z') = \frac{1}{(z^{\prime *}z')^2} dz^{\prime *} dz'$$ which is the same metric up to the Weyl scaling by the fraction at the beginning. That's why this holomorphic transformation is conformal, angle-preserving. But a conformal transformation is a coordinate transformation, a diffeomorphism. The Weyl transformation is something else. It keeps the coordinates fixed but directly changes the values of some fields, especially the metric tensor, at each point by a scalar multiplicative factor.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Luboš Motl
answered Sep 24, 2012 by (10,278 points)
Thanks a lot! it is clear when they are different. My confusion has to do with a paper I read. " Rev. Mod. Phys. 34, 442–457 (1962) Conformal Invariance in Physics" by L.Witten et al.. They talk about active point transformation and define a corresponding coordinate transformation to define a conformal transformation, which amounts to I think, rescaling the metric thus weyl transformation in our lingo. they call this $C_{g}$ however they say that special conformal transformation is a subgroup of these transformations with the usual lie-algebra in our lingo "conformal is a subgroup of weyl" ??

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prag1
Apologies, I don't have access to the paper. But the conformal group is simply not a subgroup of Weyl – the former mixes points, the latter only changes magnitude of fields locally. A conformal symmetry is a coordinate transformation, whether active or passive, doesn't matter, it's always a convention, whose effect on the metric of the background may be undone by a Weyl transformation. But the previous sentence doesn't say that "conformal transformation is a Weyl transformation". You must read and understand the words properly, not sloppily.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Luboš Motl
Thanks I am clear about this point and I understand "undoing" by multiplying by the inverse of the function still gives a different space time. Where as conformal transformation does not change space-time just different coordinate label. I would try and attach a link to that paper then you might be able to tell where I am going wrong but not on this point. Thanks

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Prag1
+ 2 like - 0 dislike

There is an appendix (appendix D) on conformal transformations in Wald's book on general relativity. First paragraph of that is relevant to your question. However his terminology is different from yours and by term conformal transformations he simply means Weyl transformations.

Metric tensor (or any other tensor) will of course not change in any way under coordinate transformations. However if $M$ is a manifold, and $f:M\rightarrow M$ if a differentiable function then for any covariant tensor field $T$ on $M$ we can define a corresponding pulled back tensor field $f^*T$ on $M$ (again see Wald's appendix C for definitions). In particular if $g$ is metric tensor then we can define pulled back metric tensor $f^*g$ on $M$. If $f$ is a diffeomorphism and $g$ is nondegenerate then $f^*g$ too will be nondegenerate, and will have same signature as that of $g$. Also $f^*g$ will in general be different from $g$, i.e. pull back under a map is a different from coordinate transformation. (difference is something similar to "passive" and "active" coordinate tranformations, but this terminology could be very confusing here).

If $f$ is a diffeomorphism such that pulled back metric $f^*g$ is equal to $\Omega g$ for some positive function $\Omega$, then it is called a conformal isometry. Conformal field theory is by definition that whose symmetry group contains group of (possibly only "local") conformal isometries as a subgroup.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001
answered Sep 24, 2012 by (635 points)
+ 1 like - 0 dislike

This is my original answer, which, sadly, missed the main point of the question. However, as I did invest some time to write it and it actually does answer at least part of the question, I'll leave it as-is.

This is a fallacy of the hands-on approach to differential geometry using coordinate expressions only and one of the reasons why I prefer the abstract-geometric approach.

Let's assume for simplicity that out abstract manifold $M$ permits global coordinate systems \begin{align} \varphi:M&\rightarrow\mathbb R^n \\ p&\mapsto x^\mu \end{align} and \begin{align} \varphi':M&\rightarrow\mathbb R^n \\ p&\mapsto x'^\mu \end{align} The coordinate transformation from unprimed to primed coordinates is given by \begin{align} \varphi'\circ\varphi^{-1}:\mathbb R^n&\rightarrow\mathbb R^n \\ x^\mu&\mapsto x'^\mu \end{align} Now, a real transformation would be a diffeomorphism \begin{align} f:M&\rightarrow M \\ p&\mapsto q \end{align} which comes with a coordinate expression $f^\varphi=\varphi\circ f\circ\varphi^{-1}$ \begin{align} f^\varphi:\mathbb R^n&\rightarrow\mathbb R^n \\ x^\mu&\mapsto y^\mu \end{align} where $x^\mu=\varphi(p)$ and $y^\mu=\varphi(q)$. Even though $f^\varphi$ looks like any other coordinate transformation, we remain in the same unprimed coordinate system.

A coordinate transformations won't change the value of scalar expressions - eg contraction of the metric tensor with two vectors to compute their inner product - by definition of the transformation laws for tensors.

This is not the case for real transformations: As we do not change coordinate systems, the components of the metric tensor won't transform and thus can't balance the change in coordinates of the vectors.

After a coordinate transformation, we're still computing the same quantity similar to using a different set of units, whereas after a real transformation, we'll actually compute a different quantity as we evaluate at different points on the manifold, ie move around in spacetime.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user Christoph
answered Sep 24, 2012 by (210 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.