The second fundamental form and derived metrics

[lavinia]

I have asked this question before but badly and just caused confusion. So I thought to ask it again but without muddled presentation.

If $M$ is a closed surface embedded in $R^3$ that has strictly positive Gauss curvature then its second fundamental form is positive definite and so is a Riemannian metric on $M$.

Can one give an example where this derived metric can not be embedded in $R^3$?

 

[zinq]

This suggests to me the general question of whether a closed surface with everywhere positive Gaussian curvature must embed in 3-space.

What if a hole is drilled through the north and south poles of a standard sphere centered at the origin of 3-space, and then the remaining portion having x-coordinate < 0 is deleted. Now identify the north-south edges (arcs of great circles) by a 180º rotation about the z-axis. There should be a C^∞ way to fill in the two circular holes with small round disks, keeping curvature > 0. Can this surface embed in 3-space?

PS This paper states that Hans Lewy proved in 1938 that all C^∞ compact surfaces of everywhere positive Gaussian curvature do embed isometrically in 3-space:

https://arxiv.org/pdf/1503.07556.pdf .

 

[lavinia]

This suggests to me the general question of whether a closed surface with everywhere positive Gaussian curvature must embed in 3-space.

What if a hole is drilled through the north and south poles of a standard sphere centered at the origin of 3-space, and then the remaining portion having x-coordinate < 0 is deleted. Now identify the north-south edges (arcs of great circles) by a 180º rotation about the z-axis. There should be a C^∞ way to fill in the two circular holes with small round disks, keeping curvature > 0. Can this surface embed in 3-space?

PS This paper states that Hans Lewy proved in 1938 that all C^∞ compact surfaces of everywhere positive Gaussian curvature do embed isometrically in 3-space:

https://arxiv.org/pdf/1503.07556.pdf .

I wonder how the surface differs from an ellipsoid of the form (1/2cos(u)sin(v),1/2sin(u)sin(v),cos(v))

Identifying the opposite moth eaten half great circles with a half twist gives a Mobius band of constant positive curvature. I wonder if that can be embedded in $R^3$

 

[zinq]

As the hole through the poles approaches radius = 0, your half-twist Möbius band approaches the standard projective plane S² / Z₂ minus a point.

In fact, any surface in R³ with everywhere positive curvature must have a trivial normal bundle. (Because, the positive curvature defines a nowhere-zero section of the normal bundle.)

So, any Möbius band with everywhere positive curvature cannot embed isometrically in R³.

 

[lavinia]

In fact, any surface in R³ with everywhere positive curvature must have a trivial normal bundle. (Because, the positive curvature defines a nowhere-zero section of the normal bundle.)

Since the normal curvatures are never zero in any direction at any point and vary continuously from point to point. Very cool.

If one tries to paste a strip of positive curvature to itself to make a Mobius band the two ends can't be lined up without flattening. I wonder if this can be done by allowing the strip to move in $R^4$.

 

[zinq]

Good question. Robert Bryant has stated that it's not currently know whether P² can be embedded in R⁴ with everywhere positive curvature. If it does, then of course just removing a point would be such a Möbius band. (I just learned that every smooth surface does have an isometric embedding into R⁵.)

 

[zinq]

PS The Veronese embedding P² → R⁶ can be adjusted so it is an isometry and so that its image is a minimal surface in the unit sphere S⁴ of R⁵.

See Remark 2 on page 3 of this paper: https://arxiv.org/pdf/1812.10173.pdf.

 

[zinq]

In case people are still interested in whether the Möbius band can be smoothly embedded in R^4 with constant positive curvature, a differential geometer has answered my query:

"Regarding your specific question about whether there is a Möbius strip of constant Gaussian curvature in Euclidean 4-space, the answer is 'yes, such an example does exist' but the 'natural' proof I know uses the Cartan-Kähler Theorem and thus is not constructive. I don't know of an explicit example."

In fact he pointed out that the constant curvature of the Möbius band could be any real number.

 

[lavinia]

In case people are still interested in whether the Möbius band can be smoothly embedded in R^4 with constant positive curvature, a differential geometer has answered my query:

"Regarding your specific question about whether there is a Möbius strip of constant Gaussian curvature in Euclidean 4-space, the answer is 'yes, such an example does exist' but the 'natural' proof I know uses the Cartan-Kähler Theorem and thus is not constructive. I don't know of an explicit example."

In fact he pointed out that the constant curvature of the Möbius band could be any real number.

I am still interested. I wonder why it is difficult to visualize. So does this mean that the projective plane of constant curvature also embeds in $R^4$?

 

[zinq]

Whether the projective plane of constant curvature embeds in R⁴ is apparently unknown at present, according to Robert Bryant.

 

[zinq]

But my favorite Möbius band is the so-called "Sudanese Möbius band", embedded minimally in S³ as follows. In

S³ = {(z,w) ∈ ℂ² | |z|²+|w|² = 1}

consider the family of all great hemispheres that have the same great circle C as their boundary. This family is parametrized by a circle. Each great hemisphere has one "pole" — its center. On each of these great hemispheres, draw a great semicircle passing through its pole, in such a way that these semicircles rotate at a constant rate, making a 180º turn as the circle parameter comes back to where it started from.

This defines a minimal Möbius band in S³ whose boundary is the great circle C.

It can be defined via the equation

M(s,θ) = (cos(s) e^iθ, sin(s) e^2iθ) ∈ S³, 0 ≤ s ≤ π, 0 ≤ θ ≤ π.

Of course for any point * ∈ S³ not on this Möbius band, a stereographic projection from S³ - {*} → R³ will take this Möbius band to one in R³ whose boundary is a round circle. Such a Möbius band is surprisingly hard to visualize!

 

[lavinia]

As the hole through the poles approaches radius = 0, your half-twist Möbius band approaches the standard projective plane S² / Z₂ minus a point.

In fact, any surface in R³ with everywhere positive curvature must have a trivial normal bundle. (Because, the positive curvature defines a nowhere-zero section of the normal bundle.)

So, any Möbius band with everywhere positive curvature cannot embed isometrically in R³.

Hmmm.. I am lost . The argument here is that one can not embed the Mobius band of positive curvature in $R^3$ because one could then add a point to get a projective plane. And also the a Mobius band can not have a trivial normal bundle. But in $R^4$ there is no problem and the Mobius band of positive curvature lives happily in $R^4$. So it should seem that the same geometric argument would allow that band to be closed off.

 

[zinq]

"... because one could then add a point to get a projective plane"

I don't see that argument. (Where would you add a point to a skinny Möbius band that stays close to its core circle?)

The argument against a Möbius band M ⊂ R³ with everywhere positive curvature is just that the positive curvature defines a normal direction globally, meaning that the normal bundle ν(M) in R³ would have a section, so it would be trivial: ν(M) = ε¹(M).

But this leads to a contradiction: Its Whitney sum ν(M) ⊕ T(M) = ε¹(M) ⊕ T(M) with the tangent bundle T(M) of M must be the restriction to M of the tangent bundle of R³, therefore trivial: ε¹(M) ⊕ T(M) = ε³(M). Since T(M) is not stably trivial, this is a contradiction.

 

[lavinia]

"... because one could then add a point to get a projective plane"

I don't see that argument. (Where would you add a point to a skinny Möbius band that stays close to its core circle?)

The argument against a Möbius band M ⊂ R³ with everywhere positive curvature is just that the positive curvature defines a normal direction globally, meaning that the normal bundle ν(M) in R³ would have a section, so it would be trivial: ν(M) = ε¹(M).

But this leads to a contradiction: Its Whitney sum ν(M) ⊕ T(M) = ε¹(M) ⊕ T(M) with the tangent bundle T(M) of M must be the restriction to M of the tangent bundle of R³, therefore trivial: ε¹(M) ⊕ T(M) = ε³(M). Since T(M) is not stably trivial, this is a contradiction.

Right. I understand the normal bundle argument. Nice argument.

But the idea of twisting the half sphere with poles removed did not restrict how much of it one would use so the intuition was that if a skinny strip should be made into a Mobius band then the whole thing could and you would then be able to top it off into a projective space. This was what I thought the first argument was. I wondered how this argument would work and thought that the positive curvature might somehow form a nearly complete strip without self intersection which is impossible.

In $R^4$ it seems that one can make a skinny strip but maybe not a projective plane with one point removed. So how skinny? And what can go wrong? An immersion with self-intersection that still has positive curvature? Regions that are not positively curved? A singularity at the missing point?