Manifold Questions

1. Mar 9, 2013

friend

I have a few questions about manifolds. According to Wikipedia.org, A topological manifold is a locally Euclidean Hausdorff space.

First question, does locally Euclidean mean that there are a continuous set of points in order that they can be mapped to an infinite set of coordinates in the Euclidean space? Or can a manifold consist of separate points with no point in between?

Last edited: Mar 9, 2013
2. Mar 9, 2013

WannabeNewton

A topological space $X$ is locally euclidean of dimension n if $\forall p\in X,$ there exists a neighborhood $U\subseteq X$ of $p$ and a homeomorphism $\varphi :U\rightarrow V$ where $V\subseteq \mathbb{R}^{n}$ is open.

3. Mar 9, 2013

Bacle2

Aren't the change-of-coordinate functions required to be homeomorphic, or at least continuous?

4. Mar 9, 2013

micromass

Doesn't that follow immediately from the charts being homeomorphisms?

If we're talking about smooth atlases, then I agree that we must demand separately that the change-of-coordinate functions are smooth. But here it follows from the definition.

5. Mar 9, 2013

WannabeNewton

Are you asking if a manifold can be disconnected? If so then yes it can. For example, we know $\mathbb{R}$ is a 1 - manifold. Any open subset of an n - manifold is also an n - manifold thus $(-\infty ,-1)\cup (1,\infty )\subset \mathbb{R}$ is a 1 - manifold but it is clearly disconnected. What IS true; however, is that any topological manifold is locally connected (and locally path connected as well). The way you stated it, it is hard to decipher if this is exactly what you are talking about. It seems you could also be talking about separation axioms and as you already stated, topological manifolds are usually required to be Hausdorff but not always.

6. Mar 9, 2013

friend

Just to clarify, does connected mean that every two points in the set can be connected by a continuous line of points contained in that set? Is this what is meant by path-connected - a continous line between them? Thanks.

7. Mar 9, 2013

Bacle2

Well, path-connected and connected(meaning there is a path in the set joining any two points in the set)are equivalent in R^n , but not in general. The topologist's sine curve is a standard (counter) example:

http://en.wikipedia.org/wiki/Topologist's_sine_curve

It is connected as the continuous image of a connected space --unit interval.

8. Mar 9, 2013

micromass

Path connected and connected are equivalent in topological manifolds as well. Manifolds tend to be really well behaved.

9. Mar 9, 2013

WannabeNewton

Path connectedness is probably more intuitive to visualize than connectedness.
What you described is path connectedness. We say a topological space $X$ is path connected if for any two points $p,q\in X$, there exists a continuous map $f:[0,1]\rightarrow X$ such that $f(0) = p$ and $f(1) = q$. We call $f$ a path from $p$ to $q$.

A topological space $X$ is connected if it CANNOT be expressed as the union of two disjoint non - empty open subsets of $X$.

It is rather easy to show that for any locally path connected space, path connectedness and connectedness are equivalent (i.e. if a topological space $X$ is locally path connected then $X$ is connected if and only if it is path connected).

10. Mar 9, 2013

Bacle2

Just to add that connectedness is a function of the choice of topology, so that, maybe being pedantic, one should say that a subset/subspace is or not connected in a given topology, tho the choice of topology is usually understood when not made explicit.

11. Mar 9, 2013

friend

Thanks, great stuff, guys.

I would find it extremely helpful if it turns out that the property of being Hausdorff is actually equivalent to path connectedness. I mean if a topology is connected, then I suppose you could always construct disjoint neighborhoods around separate points. Though maybe not visa versa.

12. Mar 9, 2013

WannabeNewton

It isn't sorry to say. Consider again the topologist's sine curve: it is Hausdorff in the subspace topology simply because any subspace of a Hausdorff space is Hausdorff and the topologist's sine curve is a subspace of euclidean 2 - space but it is NOT path connected (there are different ways to prove this but most methods are centered around the same idea).

13. Mar 9, 2013

friend

If you specify a connected path, then it's continuous, and so can the points of the path be seperably Hausdorff? I mean, it's hard to see how two points differentially separated could be gauranteed to also have differentially sized disjoint neighborhoods seperating differentially separate points, right? If the seperation is shrinking to zero, how could you be sure the neighborhoods were always shrinking even smaller to keep the neighborhoods separate?

Also, like analytic continuation, if a space is Hausdorff, then each point has a neighborhood that contains other points. And these other points also have neighborhoods that contain still other points. So it seem in this way that it must be possible to construct a path through points of adjacent neighborhoods. Although, maybe that does not gauranteed connectedness. It would if the neighborhoods could be allowed to shrink to near zero, right?

Is there a concept of neighborhoods shrinking to a single point... almost, maybe?

14. Mar 9, 2013

Bacle2

Or take a discrete space --take it, please!. It is metrizable, with the trivial metric, and so it is Hausdorff--(Haus Dorff-on- golf!). But not only is it not connected, it is totally-disconnected. So this is an example of a Hausdorff space that is totally-disconnected , and so "nowhere path-connected".

Another way: let p,q be points in your discrete space X. A path joining them is a map f: I=[0,1] -->X . By connectedness of I ( I is --I am--connected, as a subspace of the reals), the image of f must be connected. But the connected components of X are the singletons.

15. Mar 9, 2013

Bacle2

There are interesting , unexpected properties of manifolds: there can be a homeomorphism between manifolds that is nowhere-differentiable.

16. Mar 9, 2013

WannabeNewton

Bacle answered your second question so I'll answer this one. No it isn't true that a path connected space is necessarily Hausdorff. A rather standard example is taking the set of non - negative reals $[0,\infty )$ and adding a second copy of 0, calling it say 0', and defining a partial order by saying 0' < x for all positive x but you don't define a way to compare 0 and 0' (0 is compared to all positive x in the usual way). Giving this set the order topology then produces a topological space that is path connected but not Hausdorff.

17. Mar 10, 2013

wisvuze

This is when your manifold is not already embedded in another topological space ( for example, R^N ). If you notice, the way WBN defined "locally euclidean", he said gave a condition that the coordinate maps must be homeomorphisms with respect to a prescribed topology on X.
The most general definition of manifold does not require the manifold to be embedded in any topological space.
Then, the conditions on coordinate maps are that there has to be an injective map to a set around a point from an open neighbourhood of R^n, and the union of the images of the coordinate maps must cover your manifold. Then, you must also have the transition conditions that you are talking about.

18. Mar 10, 2013

friend

I read on wikipedia.org, "Being locally compact Hausdorff spaces, manifolds are necessarily Tychonoff spaces." This is very interesting to me since Tychonoff spaces have the elements of lines and sets and counts of 1 for set membership and 0 otherwise. How close is this concept to Dirac measures, since this also prescribes 1 for set membership and 0 otherwise?

19. Mar 10, 2013

micromass

What? Could you clarify?

20. Mar 10, 2013

friend

I'm only noticing similarities here. So any underlying agenda I might have would have to be speculative. I don't know what it all means yet, but if you'd indulge me, you might see that the math is correct and the motivation is clear. If it turns out that I am wrong about certain concepts, I would appreciate your correction on these issues...

I noticed that the Dirac measure returns a count of 1 if an element is a member of a specified set, and it returns 0 if the element is not a member of the set. However, if the set used in the Dirac measure consists of only one element, then the measure is 1 if the element that defines the set is equal to the element whose membership is in question, and 0 otherwise. When each element is assigned a coordinate, this one-element form of the Dirac measure equates to the Kronecker delta, which generalizes to the Dirac delta if the space is connected.

The Dirac delta function has the property that it can be iterated an infinite number of times that still gives another Dirac delta function. When the complex gaussian form of the Dirac delta is used, it can be manipulated into the Feynman path integral of quantum mechanics. Fascinating! I've posted the math of all this many times on PF, and can post again if you wish. The math of this paragraph is not in dispute. And we see here a connection between paths, sets, and things that count 1 for set membership.

Anyway... I don't have a justification for the underlying set on which this one-element Dirac measure is used. Nor do I have a justification for using the Dirac measure if I had an underlying set. Then I noticed that the definition of a manifold includes that it is also a Tychonoff space which seems to be defined in terms of functions that count between 0 and 1 depending on whether an element is included in a set or not. This seems very close to the concept of the Dirac measure. And I think, wow! I'd like to learn more about that. Is this just another way of specifying the Dirac measure? Could it be that simply specifying a manifold requires quantum mechanical processes? Is this a way to connect the manifolds of GR to QM? That would be a miracle.

I think one question that would help resolve the issues would be whether the set, F, involved in the definition of the Tychonoff space could consist of one element? Any comments would be greatly appreciated. Thank you.

Last edited: Mar 10, 2013
21. Mar 11, 2013

WannabeNewton

Uh.....what? I don't even...

22. Mar 11, 2013

friend

I used the dirac delta function to derive the path integral of QM. And then I noticed that the Tychonoff spaces require constructions similar to the dirac measure (from which the dirac delta can be derived). This is all suggestive of some connection between manifolds and quantum mechanics.

23. Mar 11, 2013

dx

There's nothing quantum mechanical about the Dirac delta. Its just a piece of mathematics. It could have been introduced before we knew anything about quantum mechanics.

24. Mar 11, 2013

WannabeNewton

You do know that the dirac delta isn't something exclusive to QM? For example it is used when talking about impulse reactions for forced harmonic oscillators and when writing down solutions to Poisson's equation via a green's function. Your logic is so far off base it is probably not worth pursuing this particular line of thought. It makes no sense at all.

25. Mar 11, 2013

friend

You are absolutely right. The Dirac delta function simply reflects the way values are distributed in a perfect manufacturing process, at least when expressed as a gaussian dirac delta function. And as seen in the math below, one might think that quantum mechanics would have arisen out of simply wondering if anything in nature was a perfect process.

The sifting property of the Dirac delta is
$$\int_{ - \infty }^{ + \infty } {f({x_1})\delta ({x_1} - {x_0})d{x_1}} = f({x_0})$$
And if we let $f({x_1}) = \delta (x - {x_1})$ in the above we get,
$$\int_{ - \infty }^\infty {\delta (x - {x_1})\delta ({x_1} - {x_0})d{x_1}} = \delta (x - {x_0})$$This is shown on the wikipedia.org site here, just before the composition section.

And when this process is iterated again, we get
$$\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {\delta (x - {x_2})\delta ({x_2} - {x_1})\delta ({x_1} - {x_0})d{x_1}} d{x_2} = } \int_{ - \infty }^\infty {\delta (x - {x_2})\delta ({x_2} - {x_0})d{x_2}} = \delta (x - {x_0})$$
Iterating this process an infinite number of times gives us,
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {\delta (x - {x_n})\delta ({x_n} - {x_{n - 1}}) \cdot \cdot \cdot \delta ({x_1} - {x_0})} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1} = \delta (x - {x_0})$$This is explicitly written out in Prof. Hagen Kleinert's book, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, page 91, and other books as well.

If the second integral above is at least temporarily treated like a regular function, then it becomes a Chapman-Kolmogorov equation whose solution is a gaussian distribution as found here, for example.

So if we use the gaussian form of the Dirac delta function,
$${\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0}) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{1/2}}}}{e^{ - {{({x_1} - {x_0})}^2}/{\Delta ^2}}}$$
with
$${\Delta ^2} = \frac{{2i\hbar }}{m}({t_1} - {t_0})$$
the dirac deltas become
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im{{({x_1} - {x_0})}^2}}}{{2\hbar ({t_1} - {t_0})}}} \right]$$
which can be manipulated to
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{{t_1} \to {t_0}} {\left[ {\frac{m}{{2\pi i\hbar ({t_1} - {t_0})}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{{x_1} - {x_0}}}{{{t_1} - {t_0}}})}^2}({t_1} - {t_0})} \right] = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {\left[ {\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}}} \right]^{1/2}}\exp \left[ {\frac{{-im}}{{2\hbar }}{{(\frac{{\Delta {x_{1,0}}}}{{\Delta {t_{1,0}}}})}^2}\Delta {t_{1,0}}} \right]$$
or,
$$\delta ({x_1} - {x_0}) = \mathop {\lim }\limits_{\Delta {t_{1,0}} \to 0} {(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}$$
When this is substituted for each of the dirac deltas in the above we get
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta {t_{,n}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{,n}})}^2}\Delta {t_{,n}}}}{{(\frac{m}{{2\pi i\hbar \Delta {t_{n,n - 1}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{n,n - 1}})}^2}\Delta {t_{n,n - 1}}}} \cdot \cdot \cdot {{(\frac{m}{{2\pi i\hbar \Delta {t_{1,0}}}})}^{1/2}}{e^{\frac{{-im}}{{2\hbar }}{{({{\dot x}_{1,0}})}^2}\Delta {t_{1,0}}}}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}$$
with the appropriate limits implied.

Then since the exponents add up, the above becomes
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{{(\frac{m}{{2\pi i\hbar \Delta t}})}^{n/2}}{e^{\,\,{\textstyle{-i \over \hbar }}\int_0^t {\frac{m}{2}{{(\dot x)}^2}dt} }}} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1}$$
Which is Feynman's path integral for a free particle. So we see that there is a very easy connection between the dirac delta function and quantum mechanics. I'm not sure yet how to include potentials in the lagrangian.