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This may be just a terminology question (or not).

A 2-cylinder is intrinsically flat. Its curvature cannot be detected from inside (although its topology can be studied by making various round trips).

When we view the cylinder as isometrically embedded in $\Bbb{R}^3$, we see its cylindrical shape and its extrinsic curvature. In this case the intrinsic and extrinsic curvatures are separate and independent, so no confusion.

Now consider a 2-sphere. It's curvature can be precisely studied from inside by measuring angles and distances, so it's curvature is intrinsic.

When we view the sphere as isometrically embedded in $\Bbb{R}^3$, we see its spherical shape and its [...] curvature. What term fills the blank? On one hand, the embedded "shape" implies extrinsic curvature (at least in the cylinder example). On the other hand, extrinsic curvature is inconsequential inside, but the curvature of the sphere has clear consequences inside.

What is the accepted terminology here? Neither extrinsic nor intrinsic seems to fit. What term should be used to describe the curvature of an embedded sphere?

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  • $\begingroup$ Extrinsic curvature: How much do curves have to accelerate to stay on the surface? Intrinsic curvature: How much of that acceleration is detectable on the surface itself? $\endgroup$ – Neal Feb 23 '19 at 14:10
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Oh, that is a really nice question. Thank you for asking :)

Now, let start dissecting your claims and confusions.

First of all, topology is irrelevant here since curvature is a purely local phenomenon. Essentially, curvature is the second derivative of a geometric structure involved. More precisely, curvature comes from a specific expression that exhibits an invariant behavior. This expression, as I said, is built from the second derivatives of some quantity which is thought as a geometric structure on the given space. I deliberately use a very vague language in order to be able to refine my statements for the specific cases that will come shortly.

Secondly, there are two kinds of geometric structures that we can observe on surfaces embedded into $\mathbb{R}^3$ (or, in any other ambient Riemannian manifold: just keep in mind that implicitly we regard $\mathbb{R}^3$ as a Riemannian manifold endowed with the standard Euclidean metric). The first structure is the induced Riemannian metric on the embedded surface (or, more generally, on a submanifold). The second structure is the embedding itself (the way how the surface has been placed into the ambient manifold).

With this said, one can also notice that if we embed our surface (submanifold) into a more general ambient Riemannian manifold, with a Riemannian metric on it, then we actually have the third (or, the zeroth, if you like to count this way) geometric structure involved, the ambient Riemannian metric.

If we now forget about the ambient manifold and the embedding, the induced Riemannian metric will be still there, that is defined as a certain quantity, on the surface (submanifold), but now this surface (submanifold) looks to us as an abstract manifold with some Riemannian metric given on it. We can even move on and embed something else into our new ambient space, which just a minute ago used to be embedded! This Riemannian metric that remains defined on our space is intrinsic in the sense that it does not need to have any external structures anymore.

In this sense, the Riemannian metric on the ambient space is intrinsic to this ambient space, but we refer to this metric as the ambient metric when we look from the perspective of the embedded submanifold. As I said, in the case of $\mathbb{R}^3$, the ambient metric is the standard Euclidean metric.

The embedding itself is also a quantifiable information, which can be differentiated (provided, it is smooth), so that we can obtain expressions of the derivatives of this information. This information is very delicate and hard to reason about, if your view it from the traditional perspective, as a mapping from $\mathbb{R}^m$ into $\mathbb{R}^n$, because it quickly brings us to a coordinate hell, with a lot of technical difficulties blowing up. A smart trick which I learned from Prof. A.Rod Gover, is to use defining functions.

We call a function $\varphi \colon M \to \mathbb{R}$ a (local) defining function for a hypersurface $S$ in an ambient manifold $M$, if (in a neightborhood of some point on $S$) S is the zero locus of function $\varphi$, that is $S = \{ p \in M | \varphi(p) = 0 \}$, and the differential $\mathrm{d}\varphi$ never vanishes at all points on $S$ (within the given neighborhood). This definition can be extended to submanifolds of any co-dimension. Of course, this construction is known for ages.

The traditional way of representing the embedding can be viewed as explicit, and using a defining function is then implicit.

From the first derivative of the defining function we can construct the unit normal along the hypersurface, namely by setting $N := \frac{\mathrm{d}\varphi}{|\mathrm{d}\varphi|}$. It is easy to verify that (up to the sign) the value of $N$ does not depend on the choice of $\varphi$ along $S$. This $N$ is a vector field along $S$, and its ambient covariant derivative ( i.e. with respect to the Levi-Civita connection of the ambient metric) is a tensorial quantity, called the shape tensor $L$ of the embedding: $L := \nabla N$ (I am a little bit sloppy here for the sake of brevity). If you expand the definitions, you will get (a long and ugly) expression in terms of the second derivatives of $\varphi$ and the ambient Riemannian metric.

This shape tensor, up to certain index-juggling identifications, is also known as the second fundamental form. The induced metric on the submanifold in this tradition is referred to as the first fundamental form (of the embedded submanifold).

If you think about the definition of the shape tensor, you will be able to visualize it as a rate of change of the unit normal vector along the surface. This is exactly how you perceive the shape of the embedded surface!

The quantities that depend on the way the surface (submanifold) is embedded are referred to as extrinsic.

In this sense, the shape tensor is the extrinsic curvature of the embedding.

The curvatures of the Riemannian metric, on the other hand, are intrinsic with respect to the manifolds that they are defined on, but we adhere to the convention to refer to the metric induced on the submanifold as the intrinsic metric, and the metric on the ambient manifold as the ambient metric, but this is just a terminological convention.

As you can "feel" the shape by the unit normal's wobbling, the same way one can "feel" the intrinsic curvature from how the geodesics behave within the surface (manifold), regardless of the way the surface (manifold) is embedded, provided the embedding was isometric (that is the intrinsic metric was preserved, in other word the embedding induces the original metric).

To conclude, I will add some comments on your statements.

A 2-cylinder is not necessarily flat, it depends on the embedding.

The curvature can be detected from inside, because, for instance, we can detect the intrinsic metric. Alternatively, we can observe the diverging geodesics.

When you measure angles and distances, you essentially study the intrinsic metric. This can be done on a cylinder too.

When you see a spherical shape, you see the extrinsic curvature. When you study lengths and angles within the surface, you detect the intrinsic curvature.

I hope that my remarks can help to clarify this terminology a little bit.

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  • $\begingroup$ Your detailed answer is fantastic, as it helps with my broader enthusiast research that has motivated this question. I hope you don't mind if I ask for more help since you are an expert in this field. In the meantime, can you please give an example of a non-flat cylinder (if it exists) isometrically embedded in $\Bbb{R}^3$? Thanks! $\endgroup$ – safesphere Feb 23 '19 at 18:17
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    $\begingroup$ If you mean something homeomorphic (topologically equivalent) to a cylinder, take a hyperboloid of one sheet; it has (non-constant) negative curvature everywhere. If you mean a literal cylinder, the fact that the tangent plane is constant along the lines dictates that the curvature must be $0$. $\endgroup$ – Ted Shifrin Feb 23 '19 at 21:47
  • $\begingroup$ @safesphere I am glad that my answer was helpful. $\endgroup$ – Yuri Vyatkin Feb 23 '19 at 22:23
  • $\begingroup$ @TedShifrin Thank you, Professor. This is exactly what I mean. The standard cylinder is flat, but I thought about non-standard embeddings which are diffeomorphic to the standard cylinder. Thank you for the clarification! $\endgroup$ – Yuri Vyatkin Feb 23 '19 at 22:34
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    $\begingroup$ Of course the usual $2$-torus can be so written, locally. You really need to specify that you're talking about the flat $2$-torus, and this becomes a global question, not a local one. $\endgroup$ – Ted Shifrin Feb 26 '19 at 17:56

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