# Extrinsics vs Intrinsics geometry, intuitive explanation

I was wondering if anyone can give me an insight of what is meant with "Intrinsics Geometry" and "Extrinsics Geometry". At the beginning I thought this was like a distinction between differentiation and integration (in order to distiguish local analysis from global analysis), however I have the feeling I might have got this completely wrong.

So I looked this up on wikipedia and I quote what I've found:

From the beginning and through the middle of the 18th century, differential geometry was studied from the extrinsic point of view: curves and surfaces were considered as lying in a Euclidean space of higher dimension (for example a surface in an ambient space of three dimensions). The simplest results are those in the differential geometry of curves and differential geometry of surfaces. Starting with the work of Riemann, the intrinsic point of view was developed, in which one cannot speak of moving "outside" the geometric object because it is considered to be given in a free-standing way. The fundamental result here is Gauss's theorema egregium, to the effect that Gaussian curvature is an intrinsic invariant.

The intrinsic point of view is more flexible. For example, it is useful in relativity where space-time cannot naturally be taken as extrinsic (what would be "outside" of it?). However, there is a price to pay in technical complexity: the intrinsic definitions of curvature and connections become much less visually intuitive.

These two points of view can be reconciled, i.e. the extrinsic geometry can be considered as a structure additional to the intrinsic one. (See the Nash embedding theorem.) In the formalism of geometric calculus both extrinsic and intrinsic geometry of a manifold can be characterized by a single bivector-valued one-form called the shape operator.

I still don't think like I get it, can anyone clarify? From the quote above what I'm getting is that intrinsics geometry assumes that the geometry analyzed is embedded in a bigger space (for example in differential geometry of curve and surface when we define the first fundamental form we rely on the euclidean metric of $$\mathbb{R}^3$$ to provide a definition, while we don't do this in Riemannian geometry where we define the Riemann tensor).

I'm not sure again though I fully understand the difference.

• Are your questions solved by the question/answers here: math.stackexchange.com/questions/2206328/… ? – Aloizio Macedo Sep 5 '19 at 15:20
• They give examples but not a rigorous definition of what an intrinsics/extrinsic property is. – user8469759 Sep 5 '19 at 21:38
• Maybe my answer can shed some light on this: math.stackexchange.com/a/3123669/2002 – Yuri Vyatkin Sep 6 '19 at 11:46
• I'm not sure I've ever seen a rigorous definition of intrinsic/extrinsic, but probably you could formulate one. The key idea is that an intrinsic feature of an object depends on (in principle, can be computed using only) the data on the structure itself---that is, we can talk about an intrinsic feature without thinking about whether our object is embedded in another or not (and if it the feature happens to arise in a context where it is embedded, it doesn't matter what the embedding is). – Travis Willse Sep 6 '19 at 16:16
• By contrast, an extrinsic feature requires an embedding, or put another way, an extrinsic feature is a feature not just of the object itself but of the embedding (or, if you like, a feature of the relationship between the object and object in which it is embedded). Thus, while embedding an object in another in two different ways by definition does not change its intrinsic features, it may change its extrinsic features. – Travis Willse Sep 6 '19 at 16:20

The mean curvature of a surface at a point is an extrinsic quantity. The Gaussian curvature is an intrinsic quantity. The mean curvature is the average of the two principal curvatures; the Gaussian curvature is the product of the principal curvatures. The principle curvatures are extrinsic quantities.

It is not at all obvious that the Gaussian curvature is an intrinsic property, but it is.

You asked (in a comment) for a formal definition of extrinsic vs. intrinsic. Here goes, at least for scalar-valued functions on surfaces. Suppose we have a definition for such a function, so for any surface $$S$$ the definition gives us a function $$f_S$$ on $$S$$. For an intrinsic function (like Gaussian curvature), the following holds:

For any distance-preserving bijection $$\psi:S\rightarrow T$$ to another surface, and for any point $$p$$ on $$S$$, the value of $$f_S$$ at $$p$$ will equal the value of $$f_T$$ at $$\psi(p)$$: $$f_S(p)=f_T(\psi(p))$$.

For example, let $$S$$ be a (flat) rectangle, and let $$T$$ be half a cylinder, obtained by bending the rectangle along one axis, so that that axis becomes a semi-circle of radius $$r$$. So we have a bijection $$\psi:S\rightarrow T$$ (the "bending map"). Let $$p$$ be a point in the middle of the rectangle. The principal curvatures at $$p$$ are 0 and 0, since any two lines through $$p$$ are straight. The principal curvatures at $$\psi(p)$$ are 0 and $$1/r$$. The Gaussian curvature is 0 in both cases, but the mean curvature is 0 on the rectangle and $$1/2r$$ on the semi-cylinder. (Of course, this example doesn't prove in general that the Gaussian curvature is intrinsic, but it does show that mean curvature is not intrinsic---i.e., extrinsic.)

Caveat: the definition above is clumsy and crude (though not wrong) in ways that would take too long to explain fully. Briefly, intrinsic vs. extrinsic still makes sense locally. (Curvature after all can be defined locally.) Also, dealing only with scalar-valued functions is too restrictive. However, we need a coordinate-independent definition of tensors for a "good" definition, which is another whole story.

The definition for general manifolds is pretty much the same: isometric invariants. In other words, just replace the word "surface" with "$$n$$-dimensional manifold".

You also asked for a reference. I looked in a few books, but they don't provide formal definitions of intrinsic vs. extrinsic. Here is a typical discussion from Tu's Differential Geometry: Connections, Curvature, and Characteristic Classes:

For a surface in $$\mathbb{R}^3$$ we defined its Gaussian curvature $$K$$ at a point $$p$$ by taking normal sections of the surface, finding the maximum $$\kappa_1$$ and the minimum $$\kappa_2$$ of the curvature of the normal sections, and setting $$K$$ to be the product of $$\kappa_1$$ and $$\kappa_2$$. So defined, the Gaussian curvature evidently depends on how the surface is isometrically embedded in $$\mathbb{R}^3$$.

On the other hand, an abstract Riemannian manifold has a unique Riemannian connection. The curvature tensor $$R(X,Y)$$ of the Riemannian connection is then completely determined by the Riemannian metric and so is an intrinsic invariant of the Riemannian manifold, independent of any embedding...

You'll also find good discussions in Gravitation (Misner, Thorne, and Wheeler, $$\S21.5$$, "Intrinsic and Extrinsic Curvature"), and a historical treatment in Ch.4 of Wells, Differential and Complex Geometry: Origins, Abstractions and Embeddings.

The basic idea is that anything defined using only the metric (and the differential manifold structure) must be an isometric invariant. That's why you'll find the phrase "intrinsically defined" often used.

Finally, let me address one possible source of your confusion. I've been talking about "intrinsic" vs. "extrinsic" in the context of differential geometry, and for local properties (like curvature). But the terms are generally used informally, to contrast properties that depend only on the "abstract manifold" vs. an imbedding of the manifold. The other answer to your question (by gandalf61) gives a couple of good topological illustrations. The knottedness property depends on the imbedding of the circle in $$\mathbb{R}^3$$. Orientability on the other hand is a homeomorphism invariant, depending only on the topology of the space.

• If you had a definition for general manifolds that would make me happier, also any reference for your definition would be useful as well. – user8469759 Sep 6 '19 at 9:23
• OK, I've added some stuff. – Michael Weiss Sep 7 '19 at 13:11
• I've just asked a question that is related to this one: computergraphics.stackexchange.com/questions/9129/… If you could give an insight please do. – user8469759 Sep 8 '19 at 15:17
• I see someone has already given the answer I'd give. I'll just add this: the Laplace-Beltrami operator can be defined using only the metric (and the associated volumetric form, and of course the diff-manifold structure). For the normal derivative $\nabla_n$, you need the embedding of the surface in 3-space. – Michael Weiss Sep 9 '19 at 12:56
• If the surface is only given as an abstract 2d manifold---a topological space with an atlas of charts---how do you define the normal derivative? OTOH, given an embedding of $S$ in $\mathbb{R}^3$ and a point on $S$, we can talk about a vector (in $\mathbb{R}^3$) that is orthogonal to the surface at that point, and use that to define the normal derivative. – Michael Weiss Sep 9 '19 at 13:35

Imagine a closed curve sitting in three dimensional Euclidean space $$\mathbb{R}^3$$. This closed curve could form a simple loop, which is easily transformed into a circle. Or it could be tangled up in some complicated way. And maybe it is so tangled that it cannot be transformed into a circle without breaking it and joining the ends - in this case the embedding of the curve forms a (non-trivial) knot.

The property of being knotted (or not) is a property of the extrinsic geometry of the curve - it depends on exactly how the curve is embedded in $$\mathbb{R}^3$$. As far as their intrinsic geometry goes, all closed curves are circles. An ant walking along the curve could not tell whether the curve was knotted or not.

Now consider a cylinder and a Mobius strip. An ant exploring the two surfaces could tell that they are different - the cylinder has two boundaries, whereas the Mobius strip has only one boundary. So the difference between the two surfaces is intrinsic - it does not just depend on how the surfaces are embedded in $$\mathbb{R}^3$$.

• Can you give an example of property in elementary differential geometry that is extrinsics and another that is intrinsics? So I can look these up and have some more thoughts about them. – user8469759 Sep 5 '19 at 20:47
• These are good examples, but it's perhaps worth mentioning that these are both examples of global properties---whether a curve is closed, and whether a space is homeomorphic to a cylinder or a Mobius strip---and that there are examples of local properties than are intrinsic and local properties that are extrinsic, like Gaussian and mean curvature of a surface. – Travis Willse Sep 6 '19 at 16:23