Let $M$ and $N$ be Riemannian manifolds. My understanding is that strictly speaking, a diffeomorphism $\phi:M \to N$ only acts on the smooth manifold structure, not the metric tensor. But there is a natural action of $\phi$ on the metric tensor given by the pullback under the diffeomorphism, which in fact makes the diffeomorphism an isometry. If no diffeomorphism between $M$ and $N$ is an isometry, is it standard to still refer to the Riemannian manifolds as "diffeomorphic"?
Are Riemannian manifolds only referred to as diffeomorphic if the diffeomorphism is an isometry
diffeomorphismdifferential-geometryisometrymanifoldsterminology
Related Solutions
It seems hard to me to give an "answer" to this quesiton, but let me try to share some thoughts. On a Riemannian manifold, the Riemannian metric can be used to define a distance function (sometimes also referred to as a metric) which in turn defines a topology on the manifold. As you note, it is a nice feature, that this topology coincides with the manifold topology. But this also implies that the topology is completely independent of the Riemannian metric in question. For example, you can put a Riemannian metric on $\mathbb R^4$ which does not admit a single isometry. But still, this will induce the usual topology (which of course is locally homogeneous).
In the pseudo-Riemannian case, this goes wrong in several respects. If you try to define a notion of "distance" parallel to the Riemannian case, you can get positive and negative distances and (even worse) different points may have distance zero (if they lie on a lightlike geodesic). This certainly does not fit into the setting of metric spaces, and while one could try to use it to define a topology, this topology would certainly be very badly behaved. (Any point which can be reached from $x$ by a lightlike geodesic would be considered as being arbitrarily close to $x$.)
The basic point in my opinion is that in mathematics you usually look at the same object from different perspectives. This is usually expressed by looking at different classes of "(iso-)morphisms". Any Riemannian or pseudo-Riemannain manifold has an underlying smooth manifold, which in turns has an underlying topological space (and if you want to push things further you can look at the underlying measure spaces or even the underlying set). The difference between these pictures is whether you look at isometries or at diffeomorphisms (respectively smooth maps), homeomorphisms (respectively continous maps) or measurable maps and isomorphisms. Now any smooth manifold is homogeneous under its group of diffeomorphisms (and much more than that is true), whereas (pseudo-)Riemannian manifolds which are homogeneous under their isometry group are rather rare. This of course implies that the topologies of smooth manifolds are always homogeneous. Even worse, any two compact manifolds (regardless of their dimension) are isomorphic as measure spaces.
So the "answer" from my point of view would be that Minkowski space is the smooth manifold $\mathbb R^4$ endowed with a flat Lorentzian metric. There are many properties of this metric, which are by no means reflected in the topology of $\mathbb R^4$, but this is also true for Riemannian metrics on $\mathbb R^4$. There certainly are some structures resembling topologies, which can be used to encode interesting properties of Minkowski space (I don't know the Zeeman and Hawking topologies you refer to, but I would guess that they are such structures, and causal structures are another example). But I don't think that they should be used as a replacement for the vector space topology on $\mathbb R^4$ ...
Let me finally remark that Minkowski space is homogeneous as a pseudo-Riemannian manifold (since translations are isometries). They are not homogeneous on an infinitesimal level, since there are different directions emanating from a point. (This is another bit of confusion. Minkowski space should not be considered as a vector space endowed with an inner product. Otherwise there would be a distinguished point - the origin. Mathematically speaking it is an affine space togehter with the inner product on each tangent space.)
An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.
Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles: $$ T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}. $$ We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by $$ \pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n. $$ For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have $$ d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}). $$ where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by $$ J(x) = \begin{bmatrix} ie^{ix_1} & 0 & \cdots & 0 \\ 0 & ie^{i x_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & i e^{i x_n} \end{bmatrix}. $$
We can just define $$ \langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n $$ where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.
We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then
\begin{align*}
\langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\
& = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\
& = \sum_1^n u_j v_j \\
& = \langle u, v \rangle,
\end{align*}
which completes the proof.
Best Answer
The answer to your last question ("If no diffeomorphism between $M$ and $N$ is an isometry, is it standard to still refer to the Riemannian manifolds as "diffeomorphic"?) is Yes. (As @TedShifrin pointed out, this is at odds with the question in your title, to which the answer is No.)
To say that two smooth manifolds are diffeomorphic means there exists a diffeomorphism between them, irrespective of what that diffeomorphism does to any metric. If the manifolds are also endowed with Riemannian metrics, we say that they are isometric if there exists a diffeomorphism between them that pulls one metric back to the other.