In the case of fields, it is clear what a local transformation in the internal space of the field is:

$$\phi(x) \to \phi'(x')= G(x) \phi(x),$$

as opposed to a global transformation, where $G$ would not depend on $x$.

But I don't understand the difference between local and global coordinate transformations.

It is said that general coordinate transformations, $$x^\mu \to x'^\mu =x^\mu+\epsilon^\mu(x)$$ are local, while rotations $$x^\mu \to x'^\mu =x^\mu+\sigma^\mu_\nu x^\nu$$ are global.

How can we distinguish between the two? In both cases, the variation depends on $x$.

  • $\begingroup$ Related: physics.stackexchange.com/q/48188/2451 , physics.stackexchange.com/q/179592/2451 $\endgroup$
    – Qmechanic
    Nov 9, 2016 at 22:19
  • $\begingroup$ A transformation is said to be global if it acts the same at every point; it is said to be local otherwise. In your example the former is local, as $\epsilon(x)$ is any function of the point, whereas the latter is global, since the $\sigma^{\mu}_{\nu}$ are constant and play the role of multiplicative coefficients. $\endgroup$
    – gented
    Nov 9, 2016 at 23:12
  • $\begingroup$ @GennaroTedesco But a rotation acts different at every point, since the variation is $\sigma^\mu_\nu x^\nu$. Points far from the origin are displaced more than those near the origin. I'm guessing that we restrict ourselves to affine transformations, so any terms that would involved higher powers of $x$ are absorbed into the coefficients. $\endgroup$
    – jinawee
    Nov 10, 2016 at 13:47
  • $\begingroup$ @jinawee In that example the rotation acts on the point (that reflects into the multiplication $\sigma\cdot x$) but the rotation itself doesn't depend on it (the $\sigma$). $\endgroup$
    – gented
    Nov 10, 2016 at 14:29
  • $\begingroup$ You are correct. These terms are not great. $\endgroup$ Jun 7 at 18:05

1 Answer 1


In the vector notation, $x \rightarrow x' = (1+\sigma)x$. The transformation acts identically on each point. So, the rotation is global, as long as $\sigma$ is independent of $x$. Each point is given same the angular shift.

Consider, in the general transformation, if $\epsilon(x) = kx$ and $k$ is a constant scaler, then this isotropic scaling transformation is global as well. Each line segment is stretched by same factor, wherever it is situated.

So, the distinction between global and local is not dependent on variation of coordinates, as such. Instead, its dependent on the variation of the transformation $T,c$ over the manifold $M$ its acting upon. $$x \mapsto Tx + c ; \ x,c \in M$$


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