5
$\begingroup$

I am struggling with the term manifold. It's the first time, I am studying this. I have tried some sites and videos on youtube but I didn't really get the idea. Roughly speaking, Manifold is something that looks flat when we zoom it a lot or that is locally flat. It seems very easy but I am still confused.

  1. Why actually do we need to define the manifolds while every shape can be seen as flat except the corners part? Is there any easy way to understand the concept of the manifold?
  2. Or in any case, how to check practically if any space is manifold or not.

here is a problem that I found somewhere.

For $\lambda \in \mathbb R$ Let $M_{\lambda}=\{ (x,y,x)\in\mathbb R^3 : x^2+y^2-z^2=\lambda\}$. Determine the paremters $\lambda$ for which $M_{\lambda}$ is a sub-manifold of $\mathbb R^3$.

for $\lambda = 0$, we get something like the cone shape, and I feel like there is a problem at the origin because no matter how much we zoom at the origin, it will not homeomorphic to a flat thing. So

  1. Is that true $M_{\lambda}$ is not a manifold for $ \lambda = 0$?
  2. How to find all such $\lambda$? (Is it possibly Only by looking at the shapes in $\mathbb R^3$ or there is some other way. Like some particular mathematical conditions that the set does not satisfy)

I am sorry if it seems a very easy question for you guys. Looking forward to hearing from you people. Many Thanks.

$\endgroup$
9
  • 1
    $\begingroup$ You didn't actually cite the definition of a manifold. Just look at the definition of a manifold that your course is using, and this will tell you what a manifold is. $\endgroup$ Nov 11, 2021 at 21:34
  • 2
    $\begingroup$ For the first question, it's not true that every shape is locally flat "except for the corners"! What locally flat really means is "locally Euclidean", which isn't satisfied by lots of reasonable topological spaces. For example, the figure 8 space (bouquet of 2 circles) is not a manifold because it's not locally homeomorphic to any euclidean space near the intersection of the two circles. To be locally Euclidean is a rather strong condition to impose. $\endgroup$
    – YiFan Tey
    Nov 11, 2021 at 22:56
  • $\begingroup$ Conider surfaces, i.e. $2$-dimensional submanifolds in the euclidian space. You could try to get familiar with surfaces, maybe that helps in building up the required intuition you're looking for. $\endgroup$
    – Zest
    Nov 11, 2021 at 23:01
  • 3
    $\begingroup$ Instead of watching videos, pick up a textbook and try to read. Make sure you have at least some background in analysis and topology. $\endgroup$ Nov 11, 2021 at 23:36
  • $\begingroup$ @MoisheKohan Thank you for your suggestion. Would you please kindly suggest any book also? How about the book "An Introduction to Manifolds" by "Loring W. Tu". Thanks $\endgroup$ Nov 12, 2021 at 0:06

1 Answer 1

5
$\begingroup$

$1.$ It is true that a lot of spaces (not all of them) that we deal with are locally flat. A sphere in three dimensions is for example. That doesn't mean you can do calculations on it like it were a flat space though, because the curvature is gonna come into play.

I'm not going to give the definition of a manifold cause you can find it anywhere, but the main idea of that definition is to use the fact that your space is locally flat in a neighbourhood around any point, and make sure those neighbourhoods are "consistent" with each other, ie, you can glue everything together nicely.

If you require that "gluing" to be smooth, then you produce a smooth manifold, and you exclude cases such as the cone that you described in $\mathbb{R}^3$.

$2, 3, 4.$ In general, working from the definition to prove that a space is a manifold can get a little messy, so there are a lot of various theorems. The one you're looking for here is the preimage theorem:

if $f: X \to Y$ is a smooth map between smooth manifolds, $y \in Y$ satisfying some nice enough conditions, then $f^{-1}(y)$ is a submanifold of $X$.

Then, you would have to rewrite $M_\lambda$ as the level set $f^{-1}(y)$ of some well-chosen function, and check for which values of $\lambda$ y satisfies the "nice enough conditions" I mentioned above (which you can find in more details in the wikipedia article). I hope that helps :)

$\endgroup$
1
  • $\begingroup$ OK, it is a pretty nice way, but let me try it first. Thank you very much for your explanation. $\endgroup$ Nov 12, 2021 at 0:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .