# Derived metrics on surfaces of positive curvature

• I
Gold Member

## Summary:

Using the principal curvatures of an embedded surface of positive Gauss curvature, what is the curvature of the derived metric?

## Main Question or Discussion Point

Start with a closed surface of positive Gauss curvature embedded smoothly in $R^3$. At each point, choose two independent eigenvectors of the shape operator whose lengths are the corresponding principal curvatures. By declaring them to be orthonormal one gets - I think - a new metric on the surface.

- How does one compute the Gauss curvature of this new metric?
- Can the surface also be embedded in $R^3$?
- Does this metric also have positive Gauss curvature? If so repeat the process and calculate the Gauss curvature again. In case one gets an infinite sequence of surfaces of positive Gauss curvature what is the limiting metric?
- Does one always get a limiting metric?

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Infrared
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I'm a little confused by the question. You can rescale eigenvectors to have unit length, so you can always pick an orthonormal basis of eigenvectors for the shape operator.

Do you have a way of picking 'favored' eigenvectors that you want to declare as having unit length for your new metric?

Gold Member
I'm a little confused by the question. You can rescale eigenvectors to have unit length, so you can always pick an orthonormal basis of eigenvectors for the shape operator.

Do you have a way of picking 'favored' eigenvectors that you want to declare as having unit length for your new metric?
Take the vectors that are of length the principal curvatures and redefine the metric so that these vectors are length 1.

"Start with a closed surface of positive Gauss curvature embedded smoothly in R3. The eigenvectors of the shape operator are an orthogonal basis for the tangent plane at every point but they may not be of unit length. By declaring them to be orthonormal one gets - I think - a new metric on the surface."

How does this work? Eigenvectors of a linear operator don't usually come with a natural length, do they? And of course at an umbilic point — where the principal curvatures are equal — the choice of orthogonal directions would be arbitrary (not that this is important — it isn't).

(
But the idea of a closed Coo surface in R3 of positive Gaussian curvature everywhere is a beautiful thing. If we also consider its interior, this will be a convex body in R3 with a smooth boundary. The Minkowski sum of two smoth convex bodies is also a smooth convex body: By definition, the Minkowski sum is

B1 + B 2 = {v + w ∈ R3 | v ∈ B1 and w ∈ B2}.

So the set of such convex bodies forms a commutative monoid https://en.wikipedia.org/wiki/Monoid#Commutative_monoid.

Each such convex body B has a unique center of gravity or centroid. And at least 4 double normals. (A double normal is a chord, a line segment whose endpoints lie on the boundary ∂B of B, such that it is perpendicular to the tangent planes of ∂B at each of its endpoints.)
)

Gold Member
"Start with a closed surface of positive Gauss curvature embedded smoothly in R3. The eigenvectors of the shape operator are an orthogonal basis for the tangent plane at every point but they may not be of unit length. By declaring them to be orthonormal one gets - I think - a new metric on the surface."

How does this work? Eigenvectors of a linear operator don't usually come with a natural length, do they? And of course at an umbilic point — where the principal curvatures are equal — the choice of orthogonal directions would be arbitrary (not that this is important — it isn't).

(
But the idea of a closed Coo surface in R3 of positive Gaussian curvature everywhere is a beautiful thing. If we also consider its interior, this will be a convex body in R3 with a smooth boundary. The Minkowski sum of two smoth convex bodies is also a smooth convex body: By definition, the Minkowski sum is

B1 + B 2 = {v + w ∈ R3 | v ∈ B1 and w ∈ B2}.

So the set of such convex bodies forms a commutative monoid https://en.wikipedia.org/wiki/Monoid#Commutative_monoid.

Each such convex body B has a unique center of gravity or centroid. And at least 4 double normals. (A double normal is a chord, a line segment whose endpoints lie on the boundary ∂B of B, such that it is perpendicular to the tangent planes of ∂B at each of its endpoints.)
)
The principal curvatures are the eigenvalues and the principal directions are the eigen vectors. One gets two orthogonal vectors by multiplying orthonormal eigenvectors by their corresponding principal curvatures. Declare these two vectors to be orthonormal and this gives a new metric at that point. In the case of an umbilic there is no change of metric.

For instance for the sphere there is no change of metric anywhere. For an ellipsoid there is no change of metric at its four umbilics. Across the whole surface one gets a new Riemannian metric. It is clear intuitively I think that if the surface starts out to be close enough to a sphere then the new metric is still of positive Gauss curvature.

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