Let $(M,g)$ and $(\overline M,\overline g)$ be riemannian manifolds of dimension $n$
and let $u:M\rightarrow\overline M$ be an immersion (we will also assume here that $u$ is orientation preserving).
Let $SO_p(g,\overline g)$ denote the space of orientation preserving isometries $T_p M\rightarrow T_{u(p)}\overline M$.
We denote by $g_p$ the matrix representation of the metric $g$ in normal coordinates centered at $p\in M$.
Now I want to write $\mathrm{dist}(du_p,SO_p(g,\overline g))$
as $\mathrm{dist}(du_p A^{-1}),SO(n))$ for some explicit isometry $A\in SO_p(g,\overline g)$.
For simplicity I will only write $SO(g,\overline g)$, but we will always consider the distance pointwise.
I read that if $\overline M=\mathbb R^n$ with the euclidean metric, then
$\sqrt g$ is an orientation preserving isometry.
Then the distance would become
$$\mathrm{dist}(du,SO(g,\overline g))
=\mathrm{dist}(du\sqrt g^{-1},SO(n))$$
However, I do not see why $\sqrt g$ is an isometry.
To generalize this to an arbitrary manifold $\overline M$, I thought about
looking at $u^* \overline g$, the pullback metric.
Is it true that then $\sqrt{u^*\overline g}\in SO(g,\overline g)$, so that
$\mathrm{dist}(du,SO(g,\overline g))$ becomes
$\mathrm{dist}(du,SO(g,\overline g))=\mathrm{dist}(du\sqrt{u^*\overline g}^{-1},SO(n))$?
I know that the Taylor expansion of the metric $g$ in normal coordinates is given by
$$g_{ij}=\delta_{ij}+\frac{1}{3}R_{kijl}x^k x^l+O(|x|^3)$$
hence
$$\sqrt g_{ij}=\delta_{ij}-\frac{1}{6}R_{kijl}x^k x^l+O(|x|^3)$$
but I do not see why this is an isometry…
Any help would be very much appreciated!
I could not find this in any book (it is possible that I overlooked it somewhere…), so references are also appreciated.
Best Answer
$\sqrt g$ is an isometry when considered as a linear map $(\mathbb{R}^n,g) \to (\mathbb{R}^n,e)$, where $e$ is the standard Euclidean inner product, and $g$ is the inner product defined by $g(v,w):=v^T g w=e(v , gw )$. (Here we think of $g$ as a symmetric positive-definite matrix).
Thus, one can observe that $$ g(v,w)=e(v , gw )=e(v , \sqrt{g} \sqrt{g}w )=e(\sqrt{g}^Tv , \sqrt{g} w )=e(\sqrt{g}v , \sqrt{g} w ),$$
so $\sqrt{g}:(\mathbb{R}^n,g) \to (\mathbb{R}^n,e)$ is an isometry, as required.
Note that this has nothing to do with the specific Taylor expansion of the metric.
Regarding the generalization you have suggested, I need to think a bit about this, but note that in general $u^*\bar g$ is a bilinear form, and not a linear transformation (so to get a transformation you should probably use some musical isomorphism).