Diffeomorphisms and pullbacks

Question

First of all, I understand that this will be mostly a mathematics questions. However, I'm asking this in the context of General Relativity, which comes with its own language, conventions and applications. Furthermore, understanding diffeomorphisms and coordinate changes is one of the most important theoretical aspects of General Relativity so I think this is the best place to ask.

Setup

Let there be a manifold M with coordinates $x^\mu$ with $\mu=1,\dots,d$ and a diffeomorphism $\phi:M\rightarrow N$ that takes points in M to points in N. In the context of general relativity, it is most likely that N is physically the same manifold as M and $\phi$ is only moving points around. Specifically, if $x^\mu$ are the coordinates on M then there are coordinates $y^\mu$ on N such that

$$\phi^\mu (x)=y^\mu$$

How to act with a diffeomorphism on a function

I always thought that if we are given a function $f(x)$ on M, acting with the diffeomorphism $\phi$ is just

$$\phi^* f(x)=f\big[\phi(x)\big]$$

That is: we are moving every point $x^\mu$ to $f^\mu(x)$. This is a physically relevant transformation since the function used to be evaluated at some point in the manifold, and now it's evaluatedon a different one.

Let me clarify and say that I understand that this definition doesn't make sense mathematically since f is a function that takes points on M, not N. However, since $\phi$ is a diffeomorphism and M and N are the same manifold, I think it makes sense physically. For example, doing a simple traslation

$$f(x)\rightarrow f(x+\epsilon)$$

would fit with this definition, we send every point $x$ to a new point $\phi(x)=x+\epsilon$.

However, when looking at the definition of the pullback of a function, it works in a different way. Given a function $g(y)$ on N, the pullback by $\phi$ is given by

$$\big(\phi^* g\big)(x)=g\big[\phi(x)\big]$$

I understand that with the pullback we changed the domain of the function: $g(y)$ is a function on N, $g\big[\phi(x)\big]$ is a function on M. However, fundamentally, the two objects are physically the same, this transformation is just a change of coordinates but the coordinates $y^\mu$ and $\phi^\mu(x)$ represent the same point on the manifold. I would be even tempted to simply remove the pullback symbol and write $g(y)=g\big[\phi(x)\big]$ since both functions will give numerically the same result, even if their domains of evaluation are technically different manifolds (N for the one on the left and M for the one on the right).

Question

Is there a sense in which both of these are correct? Are there different names for them? Does this relate to active and passive coordinate transformations?

Answer 1

Diffeomorphisms in math is what active transformations are about in GR. A physical point is mapped to another physical point. If it is a diffeomorphism on the same space then the points of that space are moved around. A passive transformation in GR is what in math is essentially a chart. Two different charts will give two different coordinates to the same physical point. So although the two charts report something has changed - the coordinates - physically, nothing has changed.

The definition of a pullback of a function is as you assert in the later part of the post. The notion you explain in the earlier part isn't correct for the reason you gave, the mismatch of domains. It's worth stating by the way that math uses the notion of pullback in several different ways. It's the name of a specific construction in category theory and it's also used for any functor that acts contravariantly. The functor maybe left implicit, as it is here, or made explicit.

The thing that was confusing me was that people usually say things like "Under a diffeomorphism, the metric transforms as ..."

This is because they are - as mathematicians say - working locally. This means they are using charts, that is coordinate systems. In fact, the metric globally - and so without charts - tautologically remains the same, but locally, and so in each individual chart, it will look different. When we move to a different manifold, the global metric will - pace the renaming of points - will be exactly the same. This renaming is defined by the pullback of the metric which we can do since we can pullback not only functions but also tensors and a metric is a tensor as is a function - it is a tensor of rank 0.

However, locally, we have to say how the metric in all the charts of the first manifold must change - that is transform - when we map them to the charts of the second diffeomorphic manifold.

But to go back to your earlier remark about the definition of the pullback, philosophically speaking - in the sense that math & physics use it - we should use all equivalent points in all diffeomorphic manifolds at once, rather than just a single point in a single avatar of a manifold. One could argue that this is a further broadening of the equivalence principle. Making this intuition is full of subtleties - for example, the hole argument by Einstein - and also in your argument of the definition of the pullback - but this is one of the motivations behind category theory, the ordinary and higher kind.

Physically, the idea of a diffeomorphism on a space being physical is also questionable. What does it mean to physically move around a point of space? Generally, this is resolved by thinking it some fluid or gas filling space and it is this that is moving around. The other way to resolve it is to think about it is to consider it as a relabelling of points. But spacetime is not a fluid occupying spacetime, it's the very stage of place itself - both of time and place. Philosophically, it leads to difficult questions of the ontology of spacetime points as well as the whole manifold of spacetime.

Answer 2

Diffeomorphisms do not involve moving each point to a new point. Instead, what they do is give a new label to the very same point. For example, consider a very non-symmetric manifold, such as the spacetime for the solar system. You don't have any kind of equivalence between different points in the spacetime, so you can't physically move them around. Hence, diffeomorphisms are passive coordinate transformations. When you write the expression for the pullback of a scalar function, notice it simply says that the function keeps its value on a point after relabeling the coordinates. That is because the point is still the same, only the coordinates changed.

Answer 3

I think I have an answer for my own question. I also want to clarify what was confusing me.

Diffeomorphisms

A diffeomorphism is an isomorphism $\phi:M\rightarrow N$ that is invertible, continuously differentiable and all those nice properties. Let's define coordinates $x^\mu$ on M and $y^\mu$ on N, with $\mu=1,\dots,d$ such that the diffeomorphism maps both coordinate charts

$$\phi(x)=y$$

Now, if we have a scalar function $f:N\rightarrow\mathbb{R}$ that takes values on $N$, we can do a pullback $\phi^* f$ as

$$\big(\phi^* f\big)(x)=f\big[\phi(x)\big]$$

This is it, this is the correct definition. However, moving points around also has physical meaning, although it is not a diffeomorphism.

Diffeomorphisms vs. Physical transformations

Let's use a specific example. I want to perform a rigid traslation on the manifold M and see if the fields have symmetry under that transformation. First, I'll define the map $\phi$ as

$$\phi^\mu(x)=x^\mu-\epsilon^\mu$$

Note that this is simply a change in the labelling of the points, we are not traslating anything, simply changing coordinates. In this case, the new coordinates are $$y^\mu=x^\mu-\epsilon^\mu$$

Now, if we have a scalar field $f$ that takes values on $N$, we can call the pullback $\phi^*f=f'$ and get

\begin{equation} \begin{aligned} \big(\phi^*f)(x)&=f\big[\phi(x)\big]\\ f'(x)=f(x-\epsilon) \end{aligned} \end{equation}

which we can write as

$$f(x)=f'(x+\epsilon)$$

This doesn't mean anything, physically. It is just the very trivial statement that the transformed field in the transformed coordinates is equal to the old field in the old coordinates. This can be made even more obvious if we write it as

$$f(x)=f\circ \phi\big[\phi^{-1}(x)\big]$$ The physical thing to do is to evaluate both the transformed field $f'$ and the original field $f$ at the same point and compare them

$$f(x)-f'(x)=f(x)-f(x-\epsilon)=\epsilon^\mu \partial_\mu f(x)$$

so if the function $f$ doesn't change along the direction $\epsilon^\mu$ then we get a symmetry.

The thing that was confusing me

The thing that was confusing me was that people usually say things like "Under a diffeomorphism, the metric transforms as

$$g_{\mu\nu}\rightarrow g_{\mu\nu}+2\nabla_{(\mu}\xi_{\nu)}$$

However, what they mean is that this is what you get if you do a diffeomorphism and then evaluate the components of the metric back in the original point. which is equivalent to moving points around, instead of simply doing a coordinate change.