# Effect of the inverse exponential map on the curvature of a given curve

Suppose you have a curve $$\alpha$$ in a manifold $$\mathcal{M}$$. You are at a point $$\alpha(t)$$ of that curve. The curvature of $$\alpha(t)$$ is the same as the curvature of the curve $$exp^{-1}_{\alpha(t)}(\alpha(s))$$ at $$s=t$$. The previous curve is what you see from the tangent space of $$\alpha(t)$$.

Now imagine you move to another point $$x$$, and want to determine how the curvature of $$\alpha(t)$$ has changed when viewed from the tangent space of $$x$$.

That is, which is the curvature of $$exp^{-1}_{x}(\alpha(s))\in T_x\mathcal{M}$$ at $$s=t$$? (when $$x$$ and $$\alpha(t)$$ are close enough so we have a local diffeomorphism, for example).

It is fine for me for now to know the answer when the manifold is a space of constant sectional curvature. I suppose that in such a case it will only depend on the distance between $$x$$ and $$\alpha(t)$$ and the relative position between the geodesic joining $$x$$ and $$\alpha(t)$$ with respect to the hyperplane (in the tangent space of $$\alpha(t)$$) tangent to $$\alpha$$ at $$t$$.

Edit:

Let me add an example in a $$2$$-dimensional sphere.

After some computations, one can see that the first order distortion of the curve is the following. If we go from $$\alpha(t)$$ to $$x$$ following the unique minimal geodesic between them (they are not far) and the angle between this geodesic and $$\alpha'(t)$$ is $$\beta$$, then we can represent $$\alpha'(t)$$ viewed from $$T_{\alpha(t)}$$ and viewed from $$T_{x}$$ as $$(\sin(\beta), \cos(\beta))$$ and $$(\sin(\beta)\cdot a/\sin(a), \cos(\beta))$$ where $$a = d(x, \alpha(t))$$. This makes sense, because due to Gauss' lemma, the second coordinate does not change and the first one is rescaled. This captures the first order information of the deformation caused by the inverse exponential map.

What I am interested in is about the second order information. How does the curvature change. I have done the computations for the curvature change in the case of a sphere and the result is a not very pleasant long formula.

Qualitative results would be interesting, for instance, if I fix the distance $$a$$, intuitively the maximum curvature change would happen when you move perpendicularly to $$\alpha'(t)$$, it's like the Gauss' lemma deformation concentrates completely into flattening the curve. But I have not being able to prove this intuitive result or any other. What is the best way to look at this?

• Could you clarify? The curvature of $\alpha$ does not depend on where you observe it from? – Deane Yang Aug 15 '19 at 13:30
• @DeaneYang, I am asking about the curvature of $exp_x^{-1}(\alpha(s))$ (at $s=t$), which is a curve on the tangent space of $x$ and depends on $x$ – Damaru Aug 15 '19 at 13:38
• @DeaneYang I have edited the question and stated now where the new curve lives in. I hope is it now more clear – Damaru Aug 15 '19 at 13:43
• Thanks. It's clear now. – Deane Yang Aug 15 '19 at 14:04