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Let $(M,g)$ and $(\overline M,\overline g)$ be riemannian manifolds of equal dimension $n$ and let $u:M\rightarrow\overline M$ be an immersion.
Let $p\in M$. Then, by choosing positively oriented orthonormal bases for the tangent spaces $T_p M$ and $T_{u(p)}\overline M$, there exist matrix representations for $g$ and $\overline g$ that we can view as linear maps $\mathbb R^n\rightarrow\mathbb R^n$.

I want to find an isometry between inner product spaces $$f:(\mathbb R^n,g_p)\rightarrow(\mathbb R^n,\overline g_{u(p)})$$ This is motivated by a problem that I am trying to simplify:

I want to do some computations (pointwise) with $$\text{dist}(du,SO(g,\overline g))$$ where $SO(g,\overline g)_p$ is the set of orientation preserving isometries $T_p M\rightarrow T_{u(p)}\overline M$, so that this reduces to $$\text{dist}(du,SO(g,\overline g))=\text{dist}(du\circ f^{-1},SO(n))$$ I am seeking an explicit description for $f$.
For example if $\overline M=\mathbb R^n$ with the standard inner product, then such a map would be $\sqrt g$, also see my previous question. However, I am clueless how to find such a map for arbitrary $\overline M$.

As I already stated in my previous question, for the generalization I was looking at the matrix representation of the pullback $u^*\overline g$.

I have been jumping between the metrics and the euclidean metric on $\mathbb R^n$, yet I did not get anywhere.

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Since $M$ and $\bar M$ have equal dimension $n$, $u$ is in fact a local diffeomorphism, so it is a diffeomorphism in some open neighborhood of $p$ in $M$.

Now, it is important to note that $g$ and $\bar g$ are in general not isometric at $p$, i.e. in general $$\left( u^*\bar g \right)_p \neq g_p \, .$$ One can, however, find a linear map $f_p$ that relates two given orthonormal bases in the tangent spaces. $g$ and $\bar g$ are then isometric at $p$ if and only if $f_p \in O_n$.

In the following, I will give you a construction that works not just at $p$, but in some neighborhood of it. You were referring to normal coordinates in the other post, so I suppose you actually want something that works not just at $p$ itself. Normal coordinates at $p$ are coordinates on the manifold itself, not in its tangent space at $p$. I suggest you take a look at the proof, where they are constructed, in your favorite differential geometry textbook. Of course, you can evaluate everything at $p$ and get the expressions in the two tangent spaces.

Now, by Silvester's law of inertia, we can find `orthonormal' coframe fields $\theta$ on $U$ and $\bar \theta$ on $u (U)$ (which is open in $\bar M$) such that $$g = \theta^T \cdot \delta \cdot \theta \quad \text{and} \quad \bar g = \bar{\theta}^T \cdot \delta \cdot \bar{\theta}$$ on $U$ and $u (U)$, respectively. $\delta$ is just $$\delta = \sum_{i,j} \delta_{ij} \, \tilde{e}^i \otimes \tilde{e}^j$$ with dual basis $(\tilde{e}^i)_i$ to the standard basis $({e}_i)_i$ in $\mathbb{R} ^n$.

To explicitly construct this decomposition for say $g$ in practice, you would usually look at a coordinate representation $\kappa^*g$ of $g$ in a neighborhood of $p$, then diagonalize $$\kappa^*g = Y^T \cdot \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \\ \end{pmatrix} \cdot Y $$ and rescale $Y$ by the square root of the above diagonal matrix $D$. This is laborious, but doable (especially with software). You then don't worry any more about the rest of the manifold and restrict yourself to coordinates, which means you identify $g$ with $\kappa^*g$ and thus $\theta$ with $\sqrt{D} \cdot Y$.

We have $$u^*\bar g = (u_*)^T \cdot \bar g \cdot u_* = \left(\bar{\theta} \cdot u_* \right)^T \cdot \delta \cdot \left( \bar{\theta} \cdot u_* \right) \, ,$$ which we compare with the above expression for $g$. We would have an isometry, if $\theta = \bar{\theta} \cdot u_*$. So if we want a map from $\mathbb{R}^n$ to $\mathbb{R}^n$ at each point on $U$, that "measures" how $g$ "differs from" $\bar g$, we set $$f = \bar{\theta} \cdot u_* \cdot \theta^{-1} \, .$$

So what does $f$ do? It takes the components of a tangent vector field on $U$ with respect to the given orthonormal frame field and spits out its components with respect to the given orthonormal frame field on $u(U)$. The respective lengths are then computed via the standard inner product $\delta$.

I leave it to you to translate everything to coordinate expressions.

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  • $\begingroup$ Thank you very much for the very detailed answer. A little question about your notation: When you write $\theta=\overline\theta\cdot u_*$, do you mean the pullback $\overline\theta\cdot u_*(v)=\overline\theta(du(v))$? Would we then have $f(v)=\overline\theta(du(\theta^{-1}(v)))$? $\endgroup$ – Pink Panther May 5 at 21:12
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    $\begingroup$ If you use $(d u)_p \colon T_p M \to T_{u(p)}\bar M$ to denote the differential at $p$, then, yes, you may write $f_p(v) = (\bar \theta)_{u(p)}((d u)_p (\theta^{-1}_p(v)))$, provided by $v$ you mean the components of your tangent vector at $p$ with respect to an orthonormal basis. $\endgroup$ – user510186 May 5 at 22:09

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