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In this notes Geometric Wave Equations by Stefan Waldmann at page 70 they have

Having a fixed Lorentz metric $g$ on a spacetime manifold $M$ we can now transfer the notions of special relativity, see e.g. 50 , to $(M, g)$. In fact, each tangent space $\left(T_{p} M, g_{p}\right)$ is isometrically isomorphic to Minkowski spacetime $\left(\mathbb{R}^{n}, \eta\right)$ with $\eta=\operatorname{diag}(+1,-1, \ldots,-1)$, by choosing a Lorentz frame: there exist tangent vectors $e_{i} \in T_{p} M$ with $i=1, \ldots, n$ such that $$ g_{p}\left(e_{i}, e_{j}\right)=\eta_{i j}=\pm \delta_{i j} . $$

We say that two manifolds $M$ and $N$ are isometric if we have vectors $v \in T_pM$ ,$u \in T_{\phi(p)}N$ and a map $\phi:M\rightarrow N$ such that

$g(v,v)=g'(\phi^*v,\phi^*v)$ where $g$ is a metric in $M$ , $g'$ is a metric in $N$ and $\phi^*$ denotes a pushfoward.

Now the definition of isometry refers to two manifolds, but in the notes they are claiming an isometry between a manifold and a tangent space.

How is this isometry constructed?

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  • $\begingroup$ Your definition of isometry is incorrect. Your upper-stars should be lower-stars (otherwise the equation makes no sense), and it needs to hold for all $v$, not just one (otherwise you could take $v=0$ and every map would be an isometry). $\endgroup$
    – WillO
    Commented May 18, 2022 at 1:01

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Two (metric) manifolds $(M,g_M)$ and $(N,g_N)$ are isometric if there exists a diffeomorphism $\varphi:M\rightarrow N$ such that $g_M = \varphi^*g_N$.

On the other hand, two pre-Hilbert spaces $(V, \langle \cdot,\cdot\rangle_V)$ and $(S,\langle\cdot,\cdot\rangle_S)$ (that is, vector spaces equipped with inner products) are isometric if there exists an invertible linear map $A:V\rightarrow S$ such that $\langle A(X),A(Y)\rangle_X = \langle X,Y\rangle_V$.

What Waldmann is saying is that at each point $p\in M$, the vector spaces $(T_pM,g_p)$ and $(\mathbb R^n, \eta)$ are isometric to one another because we can choose a basis $\{\hat e_i\}$ for $T_p M$ such that $g_p(\hat e_i,\hat e_j) =\eta_{ij}$ (such a basis is called an orthonormal frame). From there, we can construct a linear isometry $A$ via $$A: X\in T_pM \mapsto \pmatrix{-g_p(\hat e_1,X)\\g_p(\hat e_2,X)\\\vdots\\g_p(\hat e_n,X)}\in \mathbb R^n$$

Waldmann's wording is slightly confusing because he says that $(T_pM,g_p)$ is isometric to Minkowski spacetime $(\mathbb R^n,\eta)$; what he means is that it is isomorphic to the tangent space to Minkowski spacetime at any arbitrarily chosen point.

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    $\begingroup$ ...and of course you can identify Minkowski spacetime with its tangent space. $\endgroup$
    – kricheli
    Commented May 17, 2022 at 19:26
  • $\begingroup$ "what he means is that it is isomorphic to the tangent space to Minkowski spacetime at any arbitrarily chosen point" that was my confusion $\endgroup$ Commented May 17, 2022 at 19:28
  • $\begingroup$ @ kricheli what do you mean by identify Minkowski spacetime with its tangent space? $\endgroup$ Commented May 17, 2022 at 19:29
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    $\begingroup$ @kricheli If you wish, but I tend to regard that as being conceptually muddled. For instance, there is no such thing as the tangent space to a manifold - only the tangent space to a manifold at a point. The fact that the manifold $\mathbb R^n$ and the vector space $(\mathbb R^n,+,\cdot,\mathbb R)$ have the same carrier set can be the source of great confusion. $\endgroup$
    – J. Murray
    Commented May 17, 2022 at 19:35

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