# Isometry between Minkowski space and Tangent space

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?

• 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). Commented May 18, 2022 at 1:01

## 1 Answer

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.

• ...and of course you can identify Minkowski spacetime with its tangent space. Commented May 17, 2022 at 19:26
• "what he means is that it is isomorphic to the tangent space to Minkowski spacetime at any arbitrarily chosen point" that was my confusion Commented May 17, 2022 at 19:28
• @ kricheli what do you mean by identify Minkowski spacetime with its tangent space? Commented May 17, 2022 at 19:29
• @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. Commented May 17, 2022 at 19:35