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I'm not clear on how to interpret infinitesimal coordinate transformations of the form $x_\mu \mapsto x_\mu+\xi_\mu$. The first thing that comes to my mind is that the point having coordinates $x_\mu$ gets the coordinates $x_\mu+\xi_\mu$ after the transformation, but I feel that this is a bit awkward because for this to happen the coordinate chart needs to undergo an "inverse transformation" in the sense that the point which initially has coordinates $x_\mu+\xi_\mu$, the coordinates will shift from there in the opposite direction i.e. by $-\xi_\mu$ and get mapped to the point having coordinates $x_\mu$. Another way I thought of it was to think of it as a map on the manifold which maps the point at $x_\mu$ to the point $x_\mu +\xi_\mu$ which looks more natural but I don't know how to see this as a coordinate transformation.

Is there a neat way to look at these infinitesimal coordinate transformations?

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  • $\begingroup$ The post (v2) seems not really about infinitesimal transformations but rather about passive versus active transformations. $\endgroup$
    – Qmechanic
    Commented Feb 20, 2022 at 18:19

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What you should look up is flows of vector fields; this is essentially about taking "basic" ODE existence and uniqueness results and transporting them to the manifold setting.

So, what's going on is that you have a smooth manifold $M$, and a smooth vector field $\xi$ on $M$. The existence theorem tells you that associated to the vector field, there is a smooth mapping $\Phi:U\subset \Bbb{R}\times M\to M$ (called the flow/ integral flow of $\xi$) having the following properties:

  1. The domain $U$ of is an open set in $\Bbb{R}\times M$, containing $\{0\}\times M$
  2. $\Phi$ is smooth
  3. for all $x\in M$, $\Phi(0,x)=x$
  4. for all $s,t\in\Bbb{R}$ and $x\in M$, if $(t,\Phi(s,x))$ and $(t+s,x)$ belong to $U$, then $\Phi(t,\Phi(s,x))=\Phi(t+s,x)$.
  5. for all $x\in M$, the tangent vector to the curve $t\mapsto \Phi(t,x)$ (which is defined for sufficiently small $t$ due to point (1)) at $t=0$ is the vector $\xi(x)$.

Now, it is tradition to write $\Phi_t(x)$ instead of $\Phi(t,x)$. With this notation, (3) says that $\Phi_0=\text{id}_M$, and (4) roughly speaking says that $\Phi_s\circ \Phi_t=\Phi_{s+t}$, i.e roughly speaking, the mapping $t\mapsto \Phi_t$ is a group homomorphism from a subset of $\Bbb{R}$ (considered as an additive group) into the group of diffeomorphisms on $M$. The reason I said "roughly speaking" is that the map $\Phi$ is not defined on the whole of $\Bbb{R}\times M$, so one has to be slightly careful with domain issues. However, on a first reading, you may wish to ignore (even though it's an important matter) these domain issues, and for simplicity assume the domain is all of $\Bbb{R}\times M$ (these are called 'complete flows').

We often say that the vector field $\xi$ is the "infinitesimal generator" of the flow $\Phi$. The adjective "infinitesimal" refers to the fact that it lives at the level of tangent spaces, not at the manifold level (i.e $\xi(x)\in T_xM$ for all $x\in M$). The "generator" refers to the fact that from $\xi$, we managed to create a mapping $\Phi$ which has the group-like property $\Phi_0=\text{id}_M$ and $\Phi_s\circ \Phi_t=\Phi_{s+t}$. Finally, the term "flow" is indeed appropriate because if you imagine the vector field $\xi$ as describing the movement (flow) of water in a river, then the interpretation of the mapping $\Phi$ is that

For each $(t,x)\in U$, the output $\Phi_t(x)\in M$ is regarded as "the place in the manifold a small particle will end up if it initially started at point $x$, and it is left alone under the influence of the vector field $\xi$ for a 'parameter time' of $t$ units".

In other words, the flow literally moves points in the manifold around, just as much as particles in a river are moved due to the motion of the river. Now, how does this relate to coordinates? Well, start with a coordinate chart $(V,\alpha)$ (and for simplicity assume $V$ has compact closure, so that there is some fixed $\epsilon>0$ such that $[-\epsilon,\epsilon]\times \overline{V}$ is contained in the domain $U$ of the flow). Then, for any $t\in [-\epsilon,\epsilon]$, we can consider a different coordinate chart $((\Phi_t)[V], \alpha\circ (\Phi_t)^{-1})=(\Phi_t[V], \alpha\circ \Phi_{-t})$. Note that the inverse composition comes up because if we flow for time $t$, the set $V$ gets mapped to $\Phi_t[V]$, so the coordinate chart map $\alpha$ needs to get "pushed forward", and push forwards are achieved by composing with the inverse map (this should also make sense just set theoretically $\alpha\circ \Phi_{-t}$ is the composition $\Phi_t[V]\to V\to \alpha[V]$). In local coordinates, we usually never write the parameter $t$, and just say that (in a fixed coordinate system) a point with coordinates $(x^{\mu})$ is shifted/displaced/transformed to a point with coordinates $(x^{\mu}+\xi^{\mu})$ (this is just saying we take a point $x\in M$ and then the flow transforms it to $\Phi_t(x)$, for various values of $t$).

To summarize: vector fields $\xi$ on a manifold $M$ give rise (by solving many initial-value problem ODEs) to the integral flow map $\Phi$. This map $\Phi$ is interpreted as moving points in the manifold around. Ignoring domain issues, each $\Phi_t$ is a diffeomorphism, so you can compose with charts to get new charts charts (this final bit is a standard instance of the "transport of structure" phenomenon in all of math).

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  • $\begingroup$ Thanks a lot for your answer, this was really helpful. I had one last question, when a vector $V^\mu$ is pushed forward under this transformation to get $V^\mu+\xi^\mu_{,\nu} V^\nu$, is this in the old coordinate chart or the new pushed forward one? $\endgroup$ Commented Feb 21, 2022 at 14:31
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    $\begingroup$ @AkshatSharma the old coordinate chart. More explicitly, start with a vector $v\in T_pM$, and consider the flow map $\Phi_t$ for small $t$. This induces a linear mapping between tangent spaces $T(\Phi_t)_p:T_pM\to T_{\Phi_t(p)}M$. So, we can now consider the new vector $w:= T(\Phi_t)_p(v)\in T_{\Phi_t(p)}M$. For small $t$, the point $\Phi_t(p)$ will still lie in the domain of the original coordinate chart map $\alpha(\cdot)=(x^1(\cdot),\dots, x^n(\cdot))$. So, if $v=v^i\frac{\partial}{\partial x^i}(p)$, then we can write $w=w^i\frac{\partial}{\partial x^i}(\Phi_t(p))$. $\endgroup$
    – peek-a-boo
    Commented Feb 21, 2022 at 16:40
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    $\begingroup$ For some numbers $v^1,\dots, v^n,w^1,\dots, w^n\in\Bbb{R}$. You can verify by unwinding the definitions that $w^i=v^j\frac{\partial (x^i\circ \Phi_t)}{\partial x^j}(p)$. So, if we expand in powers of $t$, using that $\Phi_0$ is the identity map (and point (5)), $w^i=v^j\left(\delta^i_j+ t \frac{\partial \xi^i}{\partial x^j}(p)+ \mathcal{O}(t^2)\right)= v^i+t\cdot \xi^i_{,j}(p)v^j + \mathcal{O}(t^2)$. (In your notation you're thinking of $\xi^{\mu}$ as "infinitesimal" i.e the small parameter $t$ is already absorbed into your definition, whereas in my case I like to keep it explicitly written). $\endgroup$
    – peek-a-boo
    Commented Feb 21, 2022 at 16:48
  • $\begingroup$ Ah got it, thanks a lot! $\endgroup$ Commented Feb 21, 2022 at 17:57

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