Since you are going into GR I believe some elaboration might be good. In General Relativity spacetime is a smooth four dimensional Lorentzian manifold $(M,g)$.

Being a manifold means that $M$ is one Hausdorff topological space (don't mind too much about this by now) and that there is a collection of pairs $\mathcal{A}=\{(x_\lambda,U_\lambda)\}$ where $U_\lambda\subset M$ is open, $M$ is the union of all the $U_\lambda$ and $x_\lambda : U_\lambda \to \mathbb{R}^4$ is a coordinate system: i.e., it assigns to each point $p\in U_\lambda\subset M$ the coordinates

$$x_\lambda(p)=(x_\lambda^0(p),x_\lambda^1(p),x_\lambda^2(p),x_\lambda^3(p)).$$

There is the additional condition that given two coordinate systems $(x,U)$ and $(y,V)$ on $M$ if their domains overlap $U\cap V \neq \emptyset$ then $T_{y,x}=x\circ y^{-1}$ and $T_{x,y}=y\circ x^{-1}$ are both $C^\infty$ functions on $\mathbb{R}^4$. These are respectively the map that transforms $y$ coordinates in $x$ coordinates and vice versa.

In summary: being a topological space gives the notion of continuity and open sets, and the existence of these functions with the overlap condition, ensures you have local coordinates which are smoothly put together to make up the whole spacetime.

So in GR we have a clear separation between the points and their coordinates. The points lie in $M$, the coordinates lie in $\mathbb{R}^4$ and what brings them together are the coordinate systems as defined above.

Now, a real scalar field is a $C^\infty$ function $\phi : M\to \mathbb{R}$. Nothing more to say about it.

How can it be? Where the transformation law is gone? Well, I haven't introduced coordinates above. I've defined the object *intrinsicaly* to spacetime $M$.

Now let $p\in M$ and a coordinate system $(x,U)$ around $p$ be given, namely $p\in U$. Then, we can localy, on $U$, resolve $\phi$ in coordinates. Since $x : U\to \mathbb{R}^4$ we have that $x^{-1} : \mathbb{R}^4\to U$ takes coordinates to the corresponding point.

The composite function $\phi_{(x)}=\phi \circ x^{-1} : \mathbb{R}^4\to \mathbb{R}$ is the *coordinate expression of $\phi$*. Notice that it takes some coordinate values and gives $\phi$ on the point corresponding to them.

Now the issue with coordinate systems is that they are not unique nor there is any prefered one. Imagine I pick another coordinate system $(y,V)$ with $p\in U\cap V$. Then we can also get the coordinate expression $\phi_{(y)}=\phi\circ y^{-1}$.

Now focus on the overlap. You shall notice that the following happens:

$$\phi_{(y)}=\phi\circ y^{-1}=(\phi\circ x^{-1})\circ x \circ y^{-1}=\phi_{(x)}\circ (x\circ y^{-1})=\phi_{(y)}\circ T_{y,x}$$

Obviously $T_{x,y} = (T_{y,x})^{-1}$ Thus you finaly conclude that

$$\phi_{(y)}\circ T_{x,y} = \phi_{(x)}$$

This is precisely what you wrote, take a time to understand that.

When $\phi : M\to \mathbb{R}$ is already defined as a function on $M$ then this condition is satisfied as a result.

But generaly you want to define it from coordinates, because it is easier and more intuitive. Then this becomes a compatibility condition to ensure that the object you are defining is actually a well defined function on $M$, i.e., it doesn't depend on the coordinates themselves.

## Best Answer

A general diffeomorhpism is not an isometry. Or rather, it can be made into an isometry. Consider smooth manifolds $M$ and $N$, with metrics $g$ and $h$. Let $\phi:M\rightarrow N$ be a diffeo. We say that $\phi$ is an isometry if $g=\phi^*h$.

But now, let's forget about $h$. We define it instead as $$ h=(\phi^{-1})^*g. $$

Then $(M,g)$ and $(N,h)$ are automatically isometric as (semi-)Riemannian spaces.

With this said, consider $(M,g)$ to be Minkowski spacetime. Let $\phi:M\rightarrow M$ be a diffeo. Let $X,Y$ be vector fields.

Obviously, itistrue, that $$ (\phi^{-1})^*g(\phi_*X,\phi_*Y)=g(X,Y), $$ so applying a diffeo toeveryobject on the manifold will preserve relations. But is it true that $$ g(\phi_*X,\phi_*Y)=g(X,Y)? $$ Or alternatively, $$ (\phi^{-1})^*g(X,Y)=g(X,Y)? $$No. In general it is not true. Those transformations for which $ \phi^*g=g $ in Minkowski spacetime are PoincarĂ© transformations. The (homogenous) linear ones are Lorentz-transformations. This concludes my answer, but here is a (hopefully) illuminating aside.

Although this is in a slightly different context, here is an example, where the difference between isometries or general invertible & structure-preserving transformations make a difference:

Consider local Lorentzian geometry using local (possibly anholonomic) frames. What is the minimum information needed to give the local geometry exactly?

For coordinate frames:The metric components $g_{\mu\nu}$.For completely general frames:The metric components $g_{ab}$, and the relationship between any one coordinate frame and the general frame, which is $e^\mu_a$ or $\theta^a_\mu$ ($\theta^a=\theta^a_\mu dx^\mu$, $e_a=e^\mu_a\partial_\mu$).For orthonormal frames:The relationship between any one coordinate frame and the orthonormal frame. Why? Because if $\theta^a_\mu$ is given, then $g_{\mu\nu}=\eta_{ab}\theta^a_\mu\theta^b_\nu$.So you can see, that despite the fact that all frames are just tools, and they don't have physical/geometric relevance, and thus all frames are equally good, specifying a frame

and demanding it to be orthonormal actually gives a metric!There is valuable information content in the fact that a frame is orthonormal, and this info is lost if we use a general frame.We can of course cast this notion in the language of transformation by noting that given an initial frame, a system of orthonormal frames can be constructed by demaning that two valid frames differ by a Lorentz transformation: $e_{a'}=\Lambda^a_{\ a'}e_a.$ So, despite the fact that any $GL(n,\mathbb{R})$-valued transformation is a

goodframe transformation, the Lorentz-valued transformations are special. A system of frames for which Lorentz-transforms allowed only specifies a metric uniquely. The associated statement in modern, invariant differential geometry would be that any reduction of the frame bundle $F(M)$'s $GL(n,\mathbb{R})$ into a generalized orthogonal group uniquely yields a semi-Riemannian metric.