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How does one show that there exists no neighborhood of a point on a sphere that may be isometrically mapped into a plane? I understand that I can find the first fundamental form of the sphere $(u, v, \sqrt{r^2 - u^2 - v^2})$, for fixed $a>0$, which is given by: $E = 1 + \frac{u^{2}}{r^2 - u^2 - v^2}$, $F = \frac{u v}{r^2 - u^2 - v^2}$, and $G = 1 + \frac{v^2}{r^2 - u^2 - v^2}$. Meanwhile, for the plane $(u,v,0)$, the first fundamental form is given by: $E = 1$, $F = 0$, $G = 1$. Since these first fundamental forms are not equal, the isometry is impossible.

But is this enough? How do I account for the "neighborhood" condition?

In some sense, this is a follow up question to ones like that which is given here: There is no isometry between a sphere and a plane., but it is not truly derived therefrom. The actual problem statement reads, in its entirety as follows: "Show that no neighborhood of a point on a sphere may be isometrically mapped into a plane"; this claim is made in utter isolation. In particular, no mention of a metric is made.

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kevin
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  • did you learn about curvature already, and what isometries do to it? – Thomas Apr 10 '14 at 22:25
  • That is not enough. I can (for example) change to polar coordinate on $\mathbb R^2$ and say that the metric under polar coordinate is different from that under Euclidean coordinate. –  Apr 10 '14 at 22:26
  • @Thomas I know of $K$ (using the second fundamental form too, in ratio with the first). – kevin Apr 10 '14 at 22:29
  • @John, I suspected not, but how do I more fully account for these possibilities in order to derive the result? – kevin Apr 10 '14 at 22:29
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    if you know that $K$ is an intrinsic quantity which is preserved under isomtries, you are done. – Thomas Apr 10 '14 at 22:35
  • @kevin: I do not think it is possible to say only at the level of metric to show that they are not isometric (unless using the beautiful argument given by hardmath in his answer). The reason is that the local expression of the metric can be as different as possible. In general you need to consider the second derivative of the metric (which is the curvature) to say that the two are not isometric. –  Apr 10 '14 at 22:44
  • But, using the second derivatives/second fundamental form too, it would be possible (just by showing that their curvatures are different)? Hardmath's comment/response seemed (to me) to indicate that it is possible if the metrics used are left vague; on the other hand, the curvatures should be the invariant under a change of metric (right?) and therefore the isometry is impossible, regardless of the metrics (unspecified or otherwise). I thought that the question left it vague for a reason: namely, the metric is unnecessary information. – kevin Apr 10 '14 at 22:58
  • @Thomas In another problem, I had two surfaces with the same Gaussian curvature $K$ everywhere but mutually distinct first and second fundamental forms. No metric (to my knowledge) was induced. I was supposed to show that the map of one parametrization composed with the inverse of the other was not an isometry. I used the fact that the first fundamental forms were different in order to justify the result/claim. Is that wrong? Do I need to also use the second fundamental form, or more work? I can provide details if you wish.

    (Do the parametrizations induce a metric, by the way?)

     – kevin Apr 10 '14 at 23:05

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If you mean the great circle distances on the sphere, it's fairly simple to brute force it.

Consider a small region on the sphere bounded by a circle. If this were mapped isometrically to the plane, since the center of that spherical region is equidistant from the bounding circle, the same would be true in the plane: a simply connected region bounded by a circle, ie. a disk.

But now check the distance around the circle. If the mapping were isometric, the distance around the circle would be preserved, but for the given radius (great circle distance to boundary from center of region), the flat disk will have a greater circumference.

hardmath
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  • I am not sure about whether or not I mean the great circle distances on the sphere. I probably do, but we have not explicitly said so much. What other (common) options are there? – kevin Apr 10 '14 at 22:44
  • If you are not specific about the metric on a region of the sphere, then of course we can have an isometry with (the usual metric on) the plane. Take any continuous 1-1 map $f$ from spherical region $\mathscr{O}$ on the sphere into the plane. Define a metric on $\mathscr{O}$ as the (euclidean) distance between image points in the plane, and voila, an isometry. – hardmath Apr 10 '14 at 22:49
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    Can that be "translated" into mathematics rather than an English description? – kevin Apr 10 '14 at 22:49
  • The question says "Show that no neighborhood of a point on a sphere may be isometrically mapped into a plane." ^^Does that help at all? I am, at least otherwise, confused as to the apparent conflict between your statement and the assignment's statement. – kevin Apr 10 '14 at 22:53
  • Given $f:\mathscr{O} \to \mathbb{R}^2$, $f$ is 1-1 (monomorphism), then define a metric $d(x,y) = ||f(x) - f(y)||_2$ on $\mathscr{O}$. – hardmath Apr 10 '14 at 22:53
  • Obviously the question presumes some given metric on the sphere. Other than the great circle distance, the only "natural" metric I can think of is the metric inherited from the sphere's embedding in $\mathbb{R}^3$, ie. the straightline distance between points. – hardmath Apr 10 '14 at 22:58
  • Okay, so you think that there is some sort of implicit metric assumed but behind the scenes? I thought that the question (which I presented in its entirety) was vague about it because the metric being specified is an unnecessary condition that obscures the underlying truth. – kevin Apr 10 '14 at 23:01
  • You are my expert about what the Question is! You stated that this was a "follow-up" to a previous Math.SE question, so I assumed you were asking something you were curious about and would be in a position to correct the assumption about great circle distances on the sphere if appropriate. – hardmath Apr 10 '14 at 23:05
  • Oh, sorry, I was perhaps not entirely clear. I will edit my question statement accordingly. I meant that my question here is related to that one, but only in subject matter. As far as my actual question goes, it is in isolation and is only about the quoted text that I gave previously. In other words, the question is a single sentence and does not mention metrics: it only mentions a neighborhood, a sphere (not even param'd), a plane, and the lack of an isometric mapping from the former two to the lattermost. – kevin Apr 10 '14 at 23:09
  • I will assume the great circle metric condition if you think that it is necessary, but I was under the impression that no information beyond that given in the quoted claim was necessary and that no metric was assumed. – kevin Apr 10 '14 at 23:15
  • A negative result (impossibility of an isometry) depends on the choice of metric for the sphere. While you strongly assert that the Question here "is made in utter isolation," it seems likely that you tagged it as "differential-geometry" because that describes, at least in part, the context where the Question arose. My best guess was (and is) that the metric assumed for the sphere is the great-circle-distance metric, i.e. a Riemannian metric for which great circles are geodesics. – hardmath Apr 11 '14 at 01:59