1

In page-38 of Visual Differential Geometry by Tristan Needham, the following equation for the metric-curvature formula is introduced:

$$ \kappa= - \frac{1}{AB} \left[ \partial_v \left[ \frac{\partial_v A}{B} \right] + \partial_u \left[ \frac{\partial_v B}{A} \right] \right] \tag{1}$$

For a metric: $$ ds^2 = A^2 du^2 + B dv^2$$

For example, ne can show through (1) that the metric $ds^2 = dr^2 + r^2 d \theta^2$ and $ ds^2 = dx^2 + dy^2$ correspond to zero curvature i.e: a flat space.

My question is, if we have two metrics who agree whose curvature evaluated by (1) agrees everywhere, will shortest path between two points in the manifold also match?

For example, in the two examples I gave, we can verify that whatever coordinates we put on the cartesian grid, we find that straight line is shortest distance between points. But is this generally true that just by metric's curvature agreeing, the straight line/ geodesic looks same?

For a person who found this question by search, I would suggest reading this wikipedia article of a related topic known as Einstein's hole arguement.

tryst with freedom
  • 11,538
  • If I understand the spirit of the question, "no." Think of the rectangle $(0, 2\pi) \times (-1, 1)$ wrapped around a flat cylinder. Both the Euclidean rectangle and the cylinder are flat, but the shortest path from the image of $(0.001, 0)$ to the image of $(2\pi - 0.001, 0)$ is not the image of the Euclidean segment in the rectangle. – Andrew D. Hwang Mar 22 '22 at 02:13
  • What is the pecularity of the two metrics that I gave which makes it so that straight lines agree? @AndrewD.Hwang – tryst with freedom Mar 22 '22 at 02:38
  • I think the question is asking, "Given two metrics $g,\tilde{g}$ on a manifold $M$, if both metrics give the same curvature at every point of $M$, then are the metrics equal?" Changing the manifold as in the rectangle-cylinder example is an important consideration, but I don't think it's what the question is about. – 1Rock Mar 22 '22 at 02:52
  • @Buraian: for your specific example, you aren't actually describing two different metrics. Those both represent Euclidean distance - given two fixed points, the distance between them is always the same no matter which metric you use. A different metric would be something like $ds^2=4dx^2+4dy^2$: this metric also has zero curvature, but now all the distances are doubled. Obviously, doubling the distances doesn't change any of the geodesics. – 1Rock Mar 22 '22 at 02:57
  • In what sense are you considering the metric to be same/ equal? @1Rock – tryst with freedom Mar 22 '22 at 03:03
  • If you mean equality in sense they give same shortest geodesic then I agree with you – tryst with freedom Mar 22 '22 at 03:03
  • They're equal in the sense they assign the same length to each curve. (Really, a "metric" is an inner product on the tangent spaces, and I mean that the metrics are equal in that they give the same inner product to each pair of tangent vectors, but it amounts to the same thing). – 1Rock Mar 22 '22 at 03:04
  • Right yes. Your interpretation is absolutely correct @1Rock – tryst with freedom Mar 22 '22 at 03:10

1 Answers1

3

There are manifolds which have two (actually, infinitely many) metrics with the same curvature, but different geodesics. For example, this answer describes a "short cylinder" $\mathbb{S}^1 \times [0,1]$ and a "long cylinder" $\mathbb{S}^1 \times [0,10^{10}]$, each with a spherical cap on their ends. Call the rounded short cylinder $M_1$ and the rounded long cylinder $M_2$. The cylindrical parts of $M_1$ and $M_2$ have curvature 0. As discussed in that answer, you can find a curvature-preserving diffeomorphism $\phi:M_1 \to M_2$, by keeping the ends the same and shrinking the distances parallel to the cylinder in the middle. You can define a pullback metric on $M_1$ by $dist(x,y):=dist(\phi(x),\phi(y))$, i.e. you can a metric on $M_1$ in terms of the distances between the images of points on $M_2$ under $\phi$. Then these two different metrics have the same curvature. But the geodesics between points at opposite ends and on opposite sides of the cylinders will be different: the geodesic under the $M_2$ metric will "rotate around the cylinder" almost entirely in the middle section, while the the geodesic under the $M_1$ metric will "rotate" in the normal way.

1Rock
  • 2,046
  • Could you tell me why the two euclidean metrics agreed on shortest path? Generally what hypothesis required for equality? – tryst with freedom Mar 22 '22 at 03:11
  • The metrics don't just "agree on the shortest paths", they are the same metric, described in two ways. Using a single coordinate system (your choice of $r$ and $\theta$), you can construct all sorts of different metrics $f_1(r,\theta)dr^2+f_2(r,\theta)d\theta^2+f_3(r,\theta)drd\theta$. Given a fixed metric $ds^2$, there is exactly one choice of $f_1,f_2,f_3$ such that $ds^2=f_1(r,\theta)dr^2+f_2(r,\theta)d\theta^2+f_3(r,\theta)drd\theta$ holds. In your case, your choice of $f_1(r,\theta)=1,f_2(r,\theta)=r^2,f_3(r,\theta)=0$ is what makes your polar metric equal your Cartesian metric. – 1Rock Mar 22 '22 at 03:24
  • Another illustrative examples are the non-isometric flat tori $ \Bbb C / \Lambda$ where $\Lambda = \Bbb Z \cdot 1 + \Bbb Z \cdot \tau$, with $0\leqslant Re(\tau) < \frac{1}{2}$ and $Im(\tau) \geqslant 1$. They are pairwise non-isometric flat tori while they are all flat. This gives infinitely many flat metrics on the torus. – Didier Mar 22 '22 at 10:43
  • I came back and read this answer again.. now I'm like "what?!". " : the geodesic under the M2 metric will "rotate around the cylinder" almost entirely in the middle section, while the the geodesic under the M1 metric will "rotate" in the normal way." Could you please explain this with a picutre? I can't seem to visualize what you are saying – tryst with freedom Apr 16 '22 at 09:22
  • Sorry, I don't know how to upload pictures. Imagine the rectangle $[0,10^{10}] \times [0,\pi]$, and imagine a diagonal line from $(0,0)$ to $(10^{10},\pi)$. This passes through the points $(\frac{1}{4},\frac{\pi}{4\times 10^{10}})$ and $(10^{10}-\frac{1}{4},\pi-\frac{\pi}{4 \times 10^{10}})$. Then if you diffeomorphically map the rectangle onto $[0,1] \times [0,\pi]$, preserving the first and last quarter-unit lengths of the rectangle, then those points map to $(\frac{1}{4},\frac{\pi}{4\times 10^{10}})$ and $(\frac{3}{4},\pi-\frac{\pi}{4 \times 10^{10}})$. – 1Rock Apr 20 '22 at 02:21