# finding isometries between tangent spaces

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.

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.

• 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 $f(v)=\overline\theta(du(\theta^{-1}(v)))$? – Pink Panther May 5 at 21:12
• 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. – user510186 May 5 at 22:09