Diffeomorphism invariance of GR

[haushofer]

I'm not sure about that. A diffeomorphism is a transformation on the manifold $M$. A coordinate transformation is a transformation in $R^n$. But a diffeomorphism does induce a coordinate transformation, so they are not independent of each other (that's why there is a 1-to-1 correspondence between diffeormorphisms and coordinate transformations, "active and passive transformations", in the first place! See again Wald). So the only thing you can do imo is to first perform a diffeomorphism (which induces a coordinate transformation), and only after that perform a coordinate transformation (or the other way around).

Maybe I'm pedantic here, but from experience I know that there is no overload of being pedantic in these discussions. ;) Maybe it helps for you to do all these steps explicitly (and in both orders, to see the hang of it).

 

[Frank Castle]

I'm not sure about that. A diffeomorphism is a transformation on the manifold MM. A coordinate transformation is a transformation in RnR^n. But a diffeomorphism does induce a coordinate transformation, so they are not independent of each other (that's why there is a 1-to-1 correspondence between diffeormorphisms and coordinate transformations, "active and passive transformations", in the first place! See again Wald). So the only thing you can do imo is to first perform a diffeomorphism (which induces a coordinate transformation), and only after that perform a coordinate transformation (or the other way around).

Maybe I'm pedantic here, but from experience I know that there is no overload of being pedantic in these discussions. ;) Maybe it helps for you to do all these steps explicitly (and in both orders, to see the hang of it).

Ok. So if one starts with a diffeomorphism $\phi:M\rightarrow M$ such that $p\mapsto\phi(p)=q$, then in a local coordinate chart this induces a coordinate transformation $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$. We can then choose a coordinate transformation such that $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$. Would this be correct at all?

 

[haushofer]

Yes, I think that 's exactly what is done in your notes. First the diffeo, and then the coordinatetranso (or, first the active and then the passive transformation).

 

[Frank Castle]

Yes, I think that 's exactly what is done in your notes. First the diffeo, and then the coordinatetranso (or, first the active and then the passive transformation).

Ok cool. So does one then transform a tensor $T\mapsto\phi^{\ast}T$ and subsequently evaluate it in the final coordinate chart such that $\left(\phi^{\ast}T\right)(x')=\left(\phi^{\ast}T\right)(x)$?

With a diffeomorphism is the whole philosophy that one can either view it in an active sense as actually moving points around on the manifold, such that for an infinitesimal diffeomorphism, points are mapped to different positions on the manifold, with the coordinates $x^{\mu}$ and $x'^{\mu}$ labelling two different points in the same coordinate chart. Or, one can view it passively as a simple coordinate transformation between two coordinate charts such that the coordinates $x^{\mu}$ and $x'^{\mu}$ label the same point in different coordinate charts?

In the former case, i.e. the active sense, the points on the manifold are actually moved around, with $x^{\mu}$ and $x'^{\mu}$ labelling two different points $p$ and $q$ in the same coordinate chart, however, we can always carry out a coordinate transformation, $x'\rightarrow\tilde{x}$, such that in this new coordinate chart the coordinate values of the point $q$ are the same as the coordinates of the point in the old coordinate chart, i.e. $\tilde{x}^{\mu}=x^{\mu}$? For an infinitesimal diffeomorphism, does this happen as follows: An infinitesimal diffeomorphism is generated by a vector field $X$ such that the orbit of a point $p$ under this diffeomorphism is along the integral curves of $X$. This induces a coordinate transformation $$x^{\mu}\rightarrow y^{\mu}=x^{\mu}+\epsilon X^{\mu}$$ where $\epsilon<<1$. Thus if we carry out the coordinate transformation $$y^{\mu}\rightarrow x'^{\mu}=y^{\mu}-\epsilon X^{\mu}$$ then we find that $x'^{\mu}=x^{\mu}$. Is this then why the author claims that $d^{4}x$ does not transform under a diffeomorphism?