What is a manifold on a Euclidean space?

Question

In this semester I study differential geometry and in this chapter we want to define what is a surface. In order to do that we first define what a manifold is on a Euclidean space, not generally what is a manifold, and Euclidean space I mean $(\mathbb{R}^{n},\left \| \cdot \right \|)$.

That been said, if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

Can we just say it's a homomorphic function from open sets to open sets ?

Older posts had been made on this topic (but I don't think they answer well my question, in my option) and have already been answered if there isn't anything new to add I will delete this.

in simple terms, a manifold is something that looks locally like $\mathbb R^m$ — J. W. Tanner, Nov 25 '20 at 13:20
Also, a manifold is not a function, neither a homomorphic or homeomorphic one. A manifold is the mathematical object giving rigorous sense to the sentance "looking locally to the euclidean space", or, in terms of sensible notions, "a surface is an object that looks locally like a plane, and a 3-manifold is something that looks locally like de three-dimensionnal space". Then, you can illustrate this saying "the earth is round, but locally, it looks like it is flat : the earth is a 2-manifold". — Didier, Nov 25 '20 at 14:01
@Ramanujan since the OP mentions Feynman, I'm reminded of a line in Feynman's story A Different Box of Tools: "We forgot to tell you that it's Class 2 Hausdorff homomorphic." — KCd, Nov 26 '20 at 04:14

score 10 · Accepted Answer · answered Nov 25 '20 at 14:10

So the OP asked this:

if someone who doesn't study math ask you "what this a manifold?" how would you answer, in simple terms (as Feynman says).

My emphasis. So clearly the interest is about informing people who have no practice with math and requires analogies and explanations that do not include jargon. Which is great, because I am not a mathematician so I will provide the answer that was provided to me when I had this same questions hovering in my head.

You can start by saying that, the simplest example of a manifold is surface that is, in general, globally curved. As an example you can give a sphere. The surface of a sphere. Then explain that despite the fact that the sphere is curved, you can divide it in several little squares or triangles and if your division is fine enough and each little triangle is small enough, you can tile the entire surface of the sphere very tightly even though its surface is curved and the small polygonal tiles are flat. So it is locally flat, despite being globally curved. This, incidentally, is why the surface of the Earth seems so flat when seen from our vantage point.

Here you make your point that if you can tile a surface with such minuscule flat polygons, then it is a manifold because even though it is generally curved, it is locally flat. As a general example you can give a waving flag, relief terrain maps etc. 2D manifolds for which global curvature isn't constant, the most general examples you can think of.

The next point in the instruction is harder because then you need to show your interlocutor that the notion is general. That is, there is no reason why manifolds should be 2D, tiled by flat polygons. They may very well be 3D, tiled by flat polygons or 4D, tiled by flat polytopes.

Once you do that you can say that an entire manifold can, in fact, be flat. In which case it will be both globally and locally flat. When you study a manifold embedded inside a flat Euclidean space it is just a fancy way of saying that you, as the observer, is watching your curved study manifold from an external flat manifold. For instance, when you study the curvature of the surface of a sphere, a 2D manifold, inserted in a 3D flat space like the room where you are studying it, which is a 3D flat manifold. You can contrast this type of study with a study of the surface of the sphere where the observer isn't in an external room, but constrained to walk on the surface of the sphere itself.

Again, the notion of observing a sphere inside a room can be extended to the general case of $n-1$ dimensional manifolds embedded in $n$ dimensional flat manifolds.

I will leave it as homework how to explain what you mean by "Euclidean space".

I hope this helps, good luck.

One slight catch with this answer is that it may leave people with the mistaken impression that manifolds are always embedded in higher-dimensional (Euclidean) spaces and that this is why they can be curved. — Harry Johnston, Nov 25 '20 at 22:30
I agree with you @HarryJohnson. But do realize that the OP specifically asked about how to approach a lay interlocutor about the study of embedded manifolds. In one of the paragraphs I say that the OP could contrast this study of a sphere from the vantage point of an observer in an external room with the study of a sphere from the vantage point of an observer constrained to the sphere's surface. This could illustrate that there is a different math to be found in each approach and that they are complimentary. — urquiza, Nov 26 '20 at 02:25

score 3 · Answer 2 · answered Nov 25 '20 at 14:05

We are all familiar with a line or a plane in $\mathbb R^n$.

Perhaps we are also familiar with the generalization to arbitrary dimension $k$ with $1 \le k \le n-1$ (the cases $k=0$ and $k=n$ could also be included, but for intuition's sake I'll leave them out of this discussion). Let's refer to this as a "flat $k$-dimensional subspace" in $\mathbb R^n$ (a more mathematical term is an "affine $k$-space").

At the most intuitive level one can now say:

A $k$-dimensional manifold in $\mathbb R^n$ is a subset of $\mathbb R^n$ which looks, locally, like a flat $k$-dimensional subspace in $\mathbb R^n$.

And I might go on to give examples to shore up the intuition, for example a sphere in $\mathbb R^3$.

To formalize this one further step but still keep some intuition, one can use the idea of graphs. A 1-manifold in $\mathbb R^2$ is a subset which looks, locally, like a line. But, we might also want to capture the intuition that the line can be curved. To do this, we add one more layer of formality, which might not be understandable to someone who studies no math, but would at least be understandable to an undergraduate who has had some multivariable calculus. Here's one special case:

A 1-manifold in $\mathbb R^2$ is a subset which looks, locally, like the graph of a smooth function, having one of two forms: $y=f(x)$; or $x=g(y)$.

And, more generally,

A $k$-manifold in $\mathbb R^n$ is a subset which looks, locally, like the graph of a smooth function, having one of several possible forms: for example, $(x_{k+1},...,x_n) = f(x_1,...,x_k)$; or more generally of the form $(x_{i(k+1)},...,x_{i(n)}) = f(x_{i(1)},...,x_{i(k)})$, where $i : \{1,...,n\} \to \{1,...,n\}$ is a permutation.

Again, to shore up the intuition I might write down some formulas for the sphere in $\mathbb R^3$, such as $z = \sqrt{1-x^2-y^2}$ for the upper half-hemisphere and five other similar formulas to cover the lower, left, right, front, and back hemispheres.

Only in the next step of formality would I bring in open subsets in order to formalize the meaning of the phrase "looks, locally". I wouldn't expect someone who studies no math to follow at this point, nor even an undergrad calc student, but at least this could make sense to an undergraduate math major. And at this stage the definition would no longer be just intuitive, instead it becomes the complete definition:

A subset $M \subset \mathbb R^n$ is a $k$-dimensional manifold if for each $p \in M$ there exists smooth function of the form $(x_{i(k+1)},...,x_{i(n)}) = f(x_{i(1)},...,x_{i(k)})$, where $i : \{1,...,n\} \to \{1,...,n\}$ is a permutation, and there exist open subsets $U \subset \mathbb R^n$ and $V$ of $\{(x_{i(1)},...,x_{i(k)})\}$, such that $U \cap M$ is equal to the graph of $f$.

Thank you for your answer this was also very helpful to me. – 領域展開 Nov 25 '20 at 16:52 — 領域展開, Nov 25 '20 at 16:52

Leonhard Newton · Answer 3 · 2020-11-25T13:53:16.313

In layman's terms, it's a space that at least for a point and it's immediate neighbors behaves like a Euclidean space (which is a set of n-tuples $\mathbb{R}^n$ equipped with a dot product, or if you're more comfortable with Physics, at least for a 3 dimensional space, is like the 3d space you use for Newtonian mechanics).

More specifically, a Manifold is a Topological space (a set of points and their neighbors satisfying some axioms), such that each point has a neighborhood of points that can be mapped with a continuous and invertible function (Homeomorphism) to a Euclidean space.

An example of what this last paragraph looks like graphically: Manifold mapping to Euclidean space example

To answer your question: no, it's not an Homeomorphic function. As I've written above, it is a space satisfying some particular properties.

Leonhard, thank you for clarifying that and provide me with an image to understand this better. — 領域展開, Nov 25 '20 at 16:35

score 1 · Answer 4 · answered Nov 26 '20 at 00:13

A two dimensional manifold in Euclidean space can be bent, stretched, and/or cut to make a flat surface (i.e., a subset of a plane).

There are, however, caveats as to what cuts are allowed, and it's hard to cover them while remaining "simple". When you cut the object, the region around the cut has to be bendable/stretchable to a flat surface. So if you take two pieces of paper and make a T-shape out of them, that's not a manifold, because where the two pieces of paper meet can't be bent or stretched to make a single piece of paper, and we're not allowed to fix this with cutting. Cutting is allowed in situations such a cylinder, where any small region can be bent into a flat surface, but the overall surface can't be bent flat; the pieces can't be pressed flat simultaneously.

A higher dimensional manifold is harder to describe in "simple" terms; it's the same basic idea of bending, stretching, and cutting into an n-dimensional flat surface, but it's harder to visualize what those mean in higher dimensions. A 1-dimensional manifold is quite simpler, as it's simply a curve, and a 0-dimensional manifold is trivial.

What is a manifold on a Euclidean space?

4 Answers4