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While reading the book Modern Geometry — Methods and Applications Part I, I have a problem reconciling the definition of isometries with the usual version (an isometry preserves the distance between two points). Quoted from the book, an isometry (or a motion of the given metric) is a transformation preserving the Riemannian metric (Def. 4.1.1., page 24). That is,

$x^i = x^i(z^1, \dots, z^n)$ is called an isometry if $g'_{ij}(z^1, \dots, z^n) = g_{ij}(x^1(z), \dots, x^n(z))$,

where the $g_{ij}$ and $g'_{ij}$ are Riemannian metrics of corresponding coordinates systems.

My problem is, how is the concept of distance connected with that of $g_{ij}$? Assuming that the distance between $x_1 = (x_1^1, \dots, x_1^n)$ and $x_2 = (x_2^1, \dots, x_2^n)$ is defined as

$d(x_1, x_2) = \sqrt{\left< \xi, \xi \right>} = \sqrt{\sum_{i, j} g_{ij}(x_1) \xi^i \xi^j}$,

where $\xi$ is the vector originating at $x_1$ and terminating at $x_2$. Let $z_1, z_2$ be points corresponding to $x_1, x_2$ under an arbitrarily chosen transformation, respectively, and let $\eta = \vec{z_1z_2}$. Then, by the definitions of vector and inner product in this book,

$d(z_1, z_2) = \sqrt{\left< \eta, \eta \right>} = \sqrt{g'_{ij}(z_1) \eta^i \eta^j} = \sqrt{ \left[\frac{\partial x^k}{\partial z^i} g_{kl}\frac{\partial x^l}{\partial z^j}\right]_{z = z_1} \eta^i \eta^j} = \sqrt{g_{ij}(x_1) \xi^i \xi^j} = d(x_1, x_2)$.

This suggests that every transformation is an isometry, which is obviously absurd. Thus, either my assumed definition of distance is false or I have misunderstood the definitions of vectors and inner products. Can anyone point out my fallacy?

*This question is coming out when I am tackling with an exercise stating that every isometry of Euclidean n-space is affine, for which I have known a proof like in page 7~8 of this document. Unfortunately, I am unable to "translate" it into the context of Riemannian metric due to my shortage in concepts of distance.

Kelvin Hsu
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1 Answers1

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The distance you have written down ($\sqrt{g_{ij}(x)\xi^i \xi^j}$) is the distance in the tangent space at $x$, which has a Euclidean structure defined by the metric tensor $g_{ij}$.

The distance in the ambient Riemannian manifold is defined in a more complicated way as the infimum of length of curves joining to given points $x,y$:

$$d(x,y) = \inf \{\ell(\gamma): \gamma \text{ a piecewise $C^1$ curve joining } x,y\}$$

The metric enters via the length functional:

$$ \ell(\gamma) = \int_0^1 \sqrt{g_{ij}\circ \gamma (\gamma^\prime)^i(\gamma^\prime)^j}$$

So the question is which (differentiable) maps $\phi$ preserves this distance function, and a necessary condition is, obviously, that the differential of $\phi$ preserves the metric tensor, which means it is an isometry between tangent spaces.

Thomas
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  • Great appreciation! I now understand the difference between the two kinds of distance. Another small question: am I safe to say that, in the case of Euclidean space, $g_{ij}=\delta_{ij}$? (In fact I know it is probably a wrong assertion, since the Riemannian metric under spherical coordinates apparently doesn't resemble $\delta_{ij}$. However, I found myself not really sure about what "Euclidean" exactly means.) – Kelvin Hsu Sep 18 '16 at 14:11
  • @KelvinHsu In Euclidean space (which is $\mathbb{R}^n$ with the standard scalar product), with the standard coordinate system, the represesentation of the metric tensor is, in fact, $\delta_{ij}$. While the metric tensor itself does not depend on the choice of coordinate system, the representation as a tensor does. So if you choose spherical coordinates the metric remains the same, but the representation changes. – Thomas Sep 18 '16 at 14:48
  • I see. Big thanks!!!! – Kelvin Hsu Sep 18 '16 at 15:05