Intrinsic and Extrinsic curvature

Question

I want to understand the basic conceptual idea about intrinsic and extrinsic curvature. If we consider a plane sheet of paper (whose intrinsic curvature is zero) rolled into a cylindrical shape, then we say that its extrinsic curvature is non-zero. So how can I visualize the extrinsic curvature? I read somewhere that the extrinsic curvature indicates how the 2D surface is embedded in 3D space. So what does it mean?

Answer 1

Explaining the idea of a 2 D surface embedded in 3 D Space.

You will get a (much) better answer than this, but a straightforward reason is that for using extrinsic curvature, we need an extra dimension to put the lower dimensional object "into".

Intrinsic curvature of a surface or manifold can be performed by using math techniques developed by Gauss and Riemann and allows us to do without the extra dimension in curvature calculations.

Obviously everything we see around us is actually in 3D, but we can pretend a sheet of paper is 2D, so looking at a sheet of paper rolled up into a cylinder is looking at a 2D surface in a 3 D space and we can measure directly how much it's curved.

Image Source: Descriptions of Curvature

From looking at these 3 shapes, it's quite obvious that in a 3 D world, we could physically measure all aspects of them, and determine which of them has, in any small region on it, positive curvature, (the sphere), negative curvature, (the "vase" shape) and which has no curvature, (the middle cylinder).

So 2D embedded objects in 3D space are easy to measure.

But we have to describe 4 D objects and events in a 4 D space, so we don't have the "elbow room" of an extra dimension.

So we give up on the visual attempt ( which is why most, if at all depictions of real 4D on TV and books are misleading) and we use expressions such as this:

$${\displaystyle R^{\rho }{}_{\sigma \mu \nu }=\partial _{\mu }\Gamma ^{\rho }{}_{\nu \sigma }-\partial _{\nu }\Gamma ^{\rho }{}_{\mu \sigma }+\Gamma ^{\rho }{}_{\mu \lambda }\Gamma ^{\lambda }{}_{\nu \sigma }-\Gamma ^{\rho }{}_{\nu \lambda }\Gamma ^{\lambda }{}_{\mu \sigma }}$$

to calculate how much spacetime is curved.

This above is an example using intrinsic curvature, we don't (and actually can't) "lift " ourselves above the space.

We are in 4 D space-time, so to perform calculations in this space, we must use intrinsic curvature math ( differential geometry) as we can't go to a higher dimension.

In slightly more specific terms,you might have read about the metric tensor, which is based around the idea of using basis vector and their components to calculate distances (in 4 dimensions) utilising differential geometry to measure deviations from flat space caused by massive objects.

The other benefit of the intrinsic based metric tensor is that it is applicable in all GR calculations, (invariance).

Answer 2

Here is a way of thinking about the difference that I find useful. It is not particularly rigorous but I think it can be made so.

Think about a manifold $M$ which is embedded in some higher-dimensional flat space: so the manifold locally looks like $\mathbb{R}^n$ and it's embedded in $\mathbb{R}^m$ with $m > n$. (This is being a bit casual because it really needs to be embedded in some affine space, not a vector space, but it will serve.)

So, now, the way you define curvature of $M$, both intrinsic and extrinsic is in terms of parallel transport. In particular what you need to do is to take a vector in the tangent space of $M$ at some point $P$ and parallel-transport it around a little loop, ending up back at $P$. Then you compare the initial and final vectors. if they are the same (for all loops, and all vectors) then the manifold is flat, and if they are not the same, then it's curved.

In order to actually quantify this notion, you then need to develop a lot of mechanism to describe what parallel transport is on the manifold and what geometrical object describes how vectors change when you parallel transport them around little loops: that object ends up being the Riemann tensor, of course. But the idea is easy to understand.

Intrinsic curvature is when you do this trick entirely in $M$: you take your vectors around little loops in $M$ and see what you get back: if you don't always get back the same vector then $M$ has intrinsic curvature, if you do, it doesn't.

Extrinsic curvature is when you do this partly in the space into which $M$ is embedded. It's easy to see that vectors in $M$'s tangent spaces correspond to vectors in the (tangent spaces of) the space into which $M$ is embedded, so at any point you can 'lift' a vector at a point in $M$ out into the larger space and start transporting it there instead, and then drop it back onto $M$ at some other point and continue transporting it in $M$ back to where it started. So, in particular you can construct loops which are partly in $M$ and partly in the larger space, and transport things around those. Note that there are two different notions of parallel transport in such a loop: while in $M$ you must transport using $M$'s rules, and while in the larger space you must transport using its rules (and the larger space is flat, of course).

$M$ has extrinsic curvature if you don't always get the same vector back when transporting a vector around such loops.

The best way to think about this is to consider an example. Think about a 2-dimensional cylinder embedded in ordinary flat 3-space in the obvious way.

Well, you can easily convince yourself that the cylinder has no intrinsic curvature.

But now think about the extrinsic case. Consider a loop which starts at a point on the cylinder, goes half-way around it, and then hops out into the embedding space and crosses back to the original point (by any path in the embedding space: it does not matter since it is flat). It is easy to see (draw pictures!) that moving a vector $\vec{v}$ around such a loop results in $-\vec{v}$: it is pointing in completely the opposite direction.

So the surface of a cylinder has no intrinsic curvature, but it does have extrinsic curvature when embedded in the obvious way into 3-space.

Answer 3

Exactly what you wrote. It means that a surface appears curved when looked from the space that contains it.

Take the circle, $S^1$, for example. The intrinsic curvature would be what people call Riemann tensor. Since you can cover this circle globally by a single coordinate, you would get Riemann=0. There aren't any surprises when you parallel transport a vector around any closed loop.

Now, intuitively the circle appears curved. This is because it has extrinsic curvature. But be mindful, this definition makes reference to the ambient space. You only see the circle curved because you are looking at it from outside. Again, if you were constrained to live on the circle, it would pretty much look flat to you.

Look here Wikipedia extrinsic curvature for more details.

Answer 4

Intrinsic curvature, which we normally describe in terms of the Riemann tensor $R^a_{bcd}$ and quantities related to it, fundamentally has to do with the effect of parallel transportation on the manifold, and the curvature is responsible for manifolds having non-trivial holonomy.

Now if we have a sub-manifold embedded in some other manifold, the extrinsic curvature, as you more or less say, can be used to describe the way $\Sigma \subset M$ curves in $M$. The best way I have had it put to me is that, extrinsic curvature corresponds to everyone's layman understanding of curvature before we were ever introduced to differential geometry.

If the difference in dimension (or co-dimension) is greater than one, we can define multiple normal vectors to the manifold $\Sigma$, and there is a third notion. In the case of two, $\{n^a,m^a\}$ we can define a normal fundamental form, $\beta_a = m_b \nabla_a n^b = -n_b \nabla_a m^b$ which can be used to describe the curvature as one moves around $\Sigma$ of the normals in orthogonal planes.

Answer 5

My understanding comes from Milnor’s Morse Theory Ch. 2. He identifies a small open set of a manifold M with the Euclidean space U. This gives him two things: he can now define points on the manifold M as well as tangent vectors originating from these points, all expressed in common coordinate and vector bases of the Euclidean space U, respectively. All local operations such as inner product or covariant differentiation are expressed in these common bases. (This is like using a positive definite matrix expressed in standard bases to compute norms. The inner products of the standard Euclidean basis vectors will likely be something other than 0 or 1.) The covariant derivative operations are happening in the invisible M, but the end result is that we can express it all in terms of vectors originating from points using common Euclidean coordinates and vector bases. The relationships among the weights on these bases in these expressions lead to the two Christoffel formulas. (The weights are all zero for Euclidean spaces. So everything is trivially symmetric.)

The chain rule and bilinearity basically mean that the covariant derivative is defined as long as we have a rule for any two basis vectors. He just asserts that the answer is a weighted sum of the basis vectors with these weights magically emerging from the invisible M for each point in U. So now he can directionally differentiate a vector field in U at any point against any vector emanating from that point, all in a common coordinate and vector Euclidean basis.

Then in Section 10, he doesn’t have a curve at hand, only a tangent vector. So he begins with the simplest geodesic, one that just sits at point p and never moves. Yet, being a geodesic, it parallely moves a vector v emanating from p to … itself, so its Jacobian is invertible. He thus gains a diffeomorphism mapping a point and a tangent emanating from it to the endpoints of a local geodesic path emanating from the same point. He then exploits the consequence that locally any path can be expressed as exp$(r(t)\overrightarrow{v(t)})$, so he has a characterization of arbitrary paths that he can use to show that the geodesic is locally the shortest. Globally of course he has to assume that local geodesics can be indefinitely extended, which by the Archimedean principle implies some minimal small extension locally. But the small extension he needs (a limit point, actually) happens locally in a chart, which he exploits to prove the existence of global minimal geodesics using a) local perturbation arguments and b) the global properties of the segment of the global geodesic he has built so far. He needs a global distance metric, which he shows is compatible with the usual topology using local arguments. (A sufficiently small $\epsilon-$ open ball in the tangent space at a point is diffeomorphic to an $\epsilon-$ open ball on the manifold using the geodesic distance definition.)

I found his covariant second derivative intuition footnote on p.48 to be very illuminating. Suppose you want to compute the ‘cross partial’ of a function f w.r.t. two vector fields X and Y. So you first compute the directional derivative Xf and then apply Y to it. But this ‘overdifferentiates’ in the sense of the chain rule. Y first ‘differentiates’ X and then the directional derivative of f. The first term (Y |- X)f needs to be removed.

The curvature ‘tensor’, he says, then measures how symmetric the covariant second derivative of a vector field is w.r.t the pairs of basis vector fields. It’s a purely algebraic trilinear map characterization because of all the weights and the invisibility of M. He uses the word ‘tensor’ in quotes presumably because this tensor has a lot of relations in the universal property. For one, the range is a vector space of dim U not dim $U^3$.

I would recommend reading Frank Morgan’s short book Riemannian Geometry for Beginners (the first edition has typos) along with Milnor. I found switching between them very useful. I also kept a running example of a tangent cone on a sphere touching at a high latitude. This was a useful check on my understanding of Milnor (although the intuition is not generalizable cuz the weights on the cone are all zero). E.g., the centripetal acceleration vector of a particle traversing the latitude at uniform speed has a nonzero projection on the cone, and the two become more equal as you move the latitude higher and less equal as you move to the equator. You then get parallel transport on the equator because the tangent cone is now a cylinder and the projection of the centripetal acceleration vector is thus zero on the cylinder. Of course, in Milnor’s calculations, you never see the centripetal acceleration vector, only its projection.

Finally, all this assumes the existence of a global Riemann metric but one can build one from local charts using partitions of unity. A simple intuition for the metric is that you cannot integrate a function on a manifold using charts. It doesn’t obey change of variables. By contrast, a density, e.g., the square root of the determinant of the Riemann metric matrix, does, so you can integrate it (even though its value at a point depends on the chart). For example, a Riemann surface has holomorphic transition maps that amplify all tangent vectors at a point in the chart by the same proportion and rotate them by the same angle. The Riemann metric compatible with this structure is a positive scalar multiple of the identity matrix. See Terry Tao https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/30231#30231