I'm reading about the differential geometry of surfaces in $\mathbb{R}^3$. I keep seeing statements about certain surface properties being either "intrinsic" or "extrinsic".

Sometimes people say that the intrinsic properties are those that depend only on the coefficients of the first fundamental form. I don't see why you would ever make a definition like this. Why does it matter?

People say that the "intrinsic" properties are related only to the surface itself, whereas extrinsic ones depend on how the surface is "embedded in $\mathbb{R}^3$". I don't understand this at all. What does "embedded in $\mathbb{R}^3$" mean? If the surface isn't "embedded in $\mathbb{R}^3$", then where is it? Is there more than one way to embed a given surface in $\mathbb{R}^3$?

I see that Gauss was very happy when he managed to prove that Gaussian curvature is an intrinsic property. Gauss was a fairly practical fellow, so I assume that this result has some practical significance. In other words, being "intrinsic" must be a helpful property, somehow. But how?

I work with surfaces in engineering (ship hulls, turbine blades, airfoils, car bodies, etc.) so I like concrete physical explanations better than abstractions. I have read numerous texts on basic differential geometry, so repetition of the standard material probably won't help me much, unless you can throw some new light on it.


The terms "intrinsic" and "extrinsic" are confusing when trying to be defined in introductions to differential geometry, and for good reason: the standard definition of surfaces at this level float between things like "a subset of $\mathbb{R}^3$ such that etc etc".

A surface is a topological space. A topological space with a particular set of properties, and its definition generalizes to manifolds, which are "surfaces" of higher dimensions. A Riemannian manifold is a manifold together with a Riemannian metric, which is a kind of object defined on a space associated to the manifold. It is common to denote a Riemannian manifold by $(M,g)$, where $M$ is the manifold (and its underlying topology, which we usually omit from notation) and $g$ is the metric. I did not expose precisely what means to be a Riemannian manifold, but nowhere in the definition I'm alluding to is supposed that a Riemannian manifold is inside some $\mathbb{R}^n$.

Before proceeding, an analogy may go well (although, as all analogies, it is not perfect). Consider your nickname: "bubba", as defined by a concatenation of characters. Do you need a paper in order to conceive your name? Or a blackboard? Your nickname has an abstract existence on itself. If I were to ask, say: "How big is the 'u' on your name?", this question would make little sense. It depends on how you write it on paper. The length of the letters is an extrinsic property. However, having five letters is an intrinsic property: it doesn't matter how/where it is written, it is a result of how your name is defined.

Now, moving on. We then usually say that a property of a Riemannian manifold $(M,g)$ is intrinsic if it is a byproduct only of the topology on $M$, and $g$. One example of an intrisic property is the fact that any smooth function on the torus $T^2$ has at least two critical points (in fact, the lower bound is a little bigger). This is a consequence of the fact that $T^2$ is compact. You may say that we know that $T^2$ is compact since it is a subset of $\mathbb{R}^{3}$, but what if I told you that my $T^2$ is $S^1 \times S^1$? This lives inside $\mathbb{R}^4$ instead, and is completely different setwise than what you imagine as a standard "doughnut torus". If I said that my $T^2$ is the square with convenient identifications, then this $T^2$ isn't in any $\mathbb{R}^n$ setwise-ly speaking. Intrinsic properties receive this name because they do not depend on how you envision them, only on the structure the spaces have.

This has a lot of theoretical and practical applications. But I think there is a reason why this terminology is not so abundant in all mathematics, and it is due to the practical applications of geometry. For example, Gauss's result that the curvature is an intrinsic information is marvelous: it says that something that you can define using the way that a normal vector field varies (and a normal vector field clearly depends on how you put your surface in space) can be computed directly through measures which are related to the tangent space and the Riemannian metric (namely, the first fundamental form - it may not be clear how the tangent space is something intrinsic if you think about it geometrically, so I suggest you look up for one of the abstract definitions of tangent space), and therefore are intrinsic - it doesn't matter how you are "inside" space. In fact, it doesn't matter that you are "inside" space.

For instance, this has a lot of importance in general relativity (although the setup is not exactly Riemannian manifolds): you may have heard that spacetime is curved. This terminology can be quite confusing, and sometimes people try to explain the concept by analogy with how balls curve a rubber sheet etc. However, a big part of the success of the theory is precisely that we don't need that our space is curved inside anything: we don't need to ask "what is outside", and it doesn't make sense a priori (and it should not). It is "curved" in a way that we can define only by means of itself, making it measurable and not a pseudo-science concept.

Now, back to the beginning, it is perfectly understandable that the "intrinsic/extrinsic" duality (and its usefulness) is a little cloudy if you do not know the abstract definitions. If the above discussion does not clear some things up, I think it may be wise to wait for (or go for) the abstract definitions.

  • $\begingroup$ Thanks for the answer. My concept of "surface" is (roughly) a mapping $S$ from some subset $U$ of $\mathbb{R}^2$ into $\mathbb{R}^3$. In my business, we often just use $U =[0,1]\times[0,1]$. Or, sometimes we might say that $S(U)$ is the surface, and $S$ is a parameterization. But, anyway, surfaces are definitely $\mathbb{R}^3$ things, for me. So, your example of a torus in $\mathbb{R}^4$ is quite puzzling. If I restrict my attention to surfaces in $\mathbb{R}^3$, is the intrinsic/extrinsic discussion still valid? $\endgroup$
    – bubba
    Mar 28 '17 at 6:13
  • 1
    $\begingroup$ It is valid, but may feel a little artificial. The whole structure for you is derived from the fact that you are in $\mathbb{R}^3$, so there really is no good reason to call something which depends on the first fundamental form "intrinsic" and something which depends on a normal vector "extrinsic", for example, since both the first fundamental form and the normal vector are coming explicitly from the structure of $\mathbb{R}^3$, and even your definition of tangent space may be as a subset of $\mathbb{R}^3$. $\endgroup$
    – Aloizio Macedo
    Mar 28 '17 at 14:23

The other answer is great, but I think it may add something to be a little more informal and computationally minded. In fact, due to the extensive use of differential geometry in computer graphics and animation, the intrinsic vs extrinsic distinction can be given very real and visualizable meaning.

Intuitively, you would expect intrinsic properties to be some kind of "abstraction" of the shape. Consider a mesh model of a human (a 2D manifold in $\mathbb{R}^3$). If we rotate or translate the mesh, all of the coordinates of the vertices are altered and so too are the values of the surface normals. However, clearly, the shape itself is not altered in some abstract, intuitive sense. No matter what rigid transform we apply its human-ness is still present. We will see that intrinsic properties are actually a little more general than merely those that are invariant to rigid transforms, however.

Let's talk about embedding for a moment. The ancient Greeks doing geometry did not assign coordinates to the points of their shapes. Instead, they only considered lengths and angles between points. These properties are intrinsic to the geometry; if we want to put those triangles and circles into some concrete position (e.g. pick one point to be the origin) in some space, we can, but we do not have to; we can instead simply work with the intrinsic properties of the geometry, without embedding it into any space.

Consider drawing the outline of some shape (like the border of the silhouette of a cat, for instance, on a sheet of paper). This is some 1D manifold. Now, we can embed this in 2D by drawing it on paper. However, we can just as easily embed it in 3D because the paper itself is in 3D. In other words, we can easily imagine 1D and 2D objects in 3D with us. In the same way, manifolds of any dimension can be embedded into higher dimensional manifolds. In terms of the meaning of embedding, one can think of this like assigning coordinates to points on the manifold. Intuitively, a 1D curve only needs 1 coordinate per point to describe it. This is its intrinsic dimensionality. However, we can lift (embed) the 1D curve into $n$D; this gives each point $n$ coordinates. In this sense, we can take our human model and embed it into some higher dimension; in general relativity, this is a (curved) space of dimension 4. (As you mentioned the view of a 2D parametric surface in 3D as a map $f:U\subseteq\mathbb{R}^2 \rightarrow\mathbb{R}^3$. But one can for example just "add coordinates" to have a map $\tilde{f}:U\subseteq\mathbb{R}^2 \rightarrow\mathbb{R}^\ell$).

Ok, recall for a minute that a Riemannian manifold $(M,g)$ has a metric tensor $g$ which allows measuring distances on $M$ at an infinitesimal scale. Our human mesh model can be viewed as a discretized Riemannian manifold (with the vertices sampled from the true underlying $M$). I'll denote $x$ and $y$ to be different points on $M$.

Let's go back to our example of different embeddings via rigid transforms. Notice that one property that is not affected by rotation or translation of the shape is the geodesic distance between points. It doesn't matter where in space the shape is. Thus, intuitively, the geodesic distances $d_g(x,y)$ are intrinsic to the shape. The geodesic distance is the length of the shortest curve connecting the points, where the curve does not leave the manifold. That is, how the shape is embedded into the "larger space" (called the ambient space) it has been "placed" does not affect this curve. The ambient positions of the curve will change, but its length will not.

Recall the definition of geodesic distance: $$ d_g(x,y) = \inf_\gamma \int_a^b \sqrt{g_{\gamma(t)}(\gamma'(t),\gamma'(t))}\,dt $$ In other words, the geodesic distance between points is essentially summing the local infinitesimal distances computed via the metric tensor $g$. In this sense, the metric is "more fundamental" than $d_g$ (conceptually anyway), which we have shown to be intuitively intrinsic already.

Thus, a reasonable definition of intrinsic should be related to properties that depend only on the metric tensor. In fact, notice that whenever people define Riemannian manifolds they say, "consider a Riemannian manifold $(M,g)$"; this is because $g$ encapsulates all of the intrinsic information about the manifold, which is really what mathematicians care about.

Finally, let me talk about the concept of isometries. In the context of Riemannian geometry, isometries are transformations that preserve the distance between points on the manifold (i.e. $g$ or $d_g$). Thus by definition isometric transformations preserve the intrinsic properties of the manifold. Clearly, rigid transformations are isometric. But there are others. For instance, consider our human mesh model bending their arm. This is almost isometric (in reality, few transformations of physical objects will be truly isometric). Obviously the Euclidean (ambient) distances between points will change drastically by the arm bending, but the geodesic lengths of curves between points involving the arm will be changed little.

Let me address your questions now:

Sometimes people say that the intrinsic properties are those that depend only on the coefficients of the first fundamental form. I don't see why you would ever make a definition like this. Why does it matter?

The first fundamental form is the metric tensor, which we have essentially defined to control the intrinsic properties of the object (with good intuitive reasons). Mathematically it matters because the notion of intrinsic properties allows succinct expression of the abstract, core properties of the manifold that truly characterize it, regardless of where/how you have embedded it. But in more practical terms, it matters because often you want to be able to transform your object and know which properties will be preserved. Also, in physics, extrinsic properties tend not to be useful; you really only care about the true physical reality, not worrying about how the underlying coordinates affect things, for instance.

People say that the "intrinsic" properties are related only to the surface itself, whereas extrinsic ones depend on how the surface is "embedded in R3". I don't understand this at all. What does "embedded in R3" mean?

Intuitively, I would say it means taking the "abstract notion" of the shape and placing it into some specific space, i.e. giving the points on it coordinates in the ambient space. Like giving Cartesian coordinates to the geometric drawings of the ancient Greeks. The formal notion of embedding uses topological methods to properly define the fuzzy idea of an "abstract notion" of the mesh.

If the surface isn't "embedded in R3", then where is it?

Can be anywhere you want :3 The choice of what space to put your object is arbitrary, and affects only the extrinsic properties of the object. The intrinsic ones will be preserved.

Is there more than one way to embed a given surface in $\mathbb{R}^3$?

Yes. Translate, rotate, or isometrically transform your mesh, and you will have another embedding in $\mathbb{R}^3$.

I see that Gauss was very happy when he managed to prove that Gaussian curvature is an intrinsic property. Gauss was a fairly practical fellow, so I assume that this result has some practical significance. In other words, being "intrinsic" must be a helpful property, somehow. But how?

Well in computer graphics, intrinsic properties are preserved under operations that often we don't care about. If I want to, say, search for mesh models of people similar to a mesh model $S$ I already have, then I don't care if they are bending their arms that much; I care that they are not e.g. way fatter than $S$ or something.

Generally, in math or physics, calculating things using intrinsic properties just tends to make things way easier. For physics especially, one expects certain quantities to be invariant to coordinate transforms. (This is the motivation for the use of tensors in physics, because how they transform under coordinate change was understood, and they preserved certain properties under these changes.)

For example, consider the Einstein Field Equations: $$ \frac{c^4}{8\pi G}\left( R_{\alpha\beta} - \frac{1}{2} R g_{\alpha\beta} + \Lambda g_{\alpha\beta} \right) = T_{\alpha\beta} $$ where $R$ is Ricci curvature, $g$ is the metric tensor, $c$ is the speed of light, $G$ is the gravitational constant, $\Lambda$ is the cosmological constant, and $T$ is the stress-energy tensor. Recall that (in GR) the space itself in which we live has curvature (kind of like "living" on a curved part of a mesh model), caused by gravity. This curvature is intrinsic, because it can be entirely computed using only the metric tensor. (Note that on 2D surfaces the Ricci curvature is twice the Gaussian curvature). That is part of what makes this equation so pretty: everything is somehow intrinsic to the system. The LHS in particular describes the intrinsic properties of space itself in a sense (e.g. its curvature, the speed of light within it, etc...).

Final notes

  • One thing that is confusing is the difference between the intrinsic and ambient dimensions. Surface meshes are 2D objects (which we usually embed in 3D). People like to talk about imagining little people in the 2D manifold, measuring things, unaware of the ambient space. This is a little off-putting to imagine because people are fundamentally 3D, so we think of a little person standing on the manifold, sticking up into the ambient space like a surface normal. But this is wrong: instead, we should imagine the person being "part" of the manifold, i.e. stuck inside it and being of the same dimensionality (visually, one can imagine a projector shining an image of a person onto the manifold; then think of that projection itself as the person). People discussing relativity tend to do this and it drove me nuts forever.

  • Feel free to check out some CS-oriented literature on this: for instance, this paper or this paper.

Anyway, hopefully this didn't tell you too many things you didn't already know; apologies if so!

  • $\begingroup$ Thanks very much. What I got from your answer is that intrinsic properties are those that are unchanged by isometries. That is helpful. In particular, intrinsic properties are invariant under rigid motions, or changes of coordinate system. But, actually, I think that almost everything I care about is independent of coordinate system. $\endgroup$
    – bubba
    Oct 5 '20 at 1:44

One humble addition: I think the following is obvious but may be helpful if explicitly stated. Extrinsic representations tend to be those where the manifold can be represented in a higher dimension but using the simple intuitive Euclidean metric of sqrt(deltax^2+deltay^2+…) and constraints of the form f(x,y,…)=0. Intrinsic representations tend to have non-Euclidean metrics without resorting to higher dimensions. So a 2D sphere can have an extrinsic representation in 3D using the intuitive Euclidean metric of sqrt(deltax^2+deltay^2+deltaz^2) but with a constraint of (x-A)^2+(y-B)^2+(z-C)^2-R^2=0 (where the center of the sphere is at (A,B,C) and the radius is R).

An intrinsic representation of a 2D sphere uses a non-Euclidean metric without resorting to higher dimensions. Look at any 2D map of the world and you will see “distortion.”

So if you want the intuition of the Euclidean metric, imagine curvature in extra dimension(s). After all, the surface of the earth is actually curved in 3D! If you want to stay in the dimensionality of the manifold, imagine a non-Euclidean metric and distortion.


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