Diffeomorphisms, Differential Structure, ETC.

1. Dec 11, 2012

friend

I'm trying to understand the difference between diffeomorphisms, diffeomorphism invariance, reparameterization invariance, and differential structures and how each of these terms relate to physics. Perhaps there's a book out there that explains the differences between these constructs and the relationship between them, with examples and counter examples. Or perhaps someone could run through how all these terms are defined and related. Any help is appreciated. Thanks.

Last edited: Dec 11, 2012
2. Dec 11, 2012

quasar987

Here it is in a nutshell:
-On a topological manifold (see wiki) M, a differential structure is an open covering of M by sets homeomorphic to R^n and such that the transition function between the respective coordinates that they define on M is a diffeomorphism (i.e. bijective and smooth with a smooth inverse). The meaning of this definition is that it allows one to speak unambiguously about smoothness of a function f:M-->N between differential manifolds.

-A diffeomorphism f:M-->N between two differential manifolds is, as on R^n, a map that is bijective and smooth with a smooth inverse. From the point of view of differential manifold theory, two diffeomorphic manifolds are "essentially the same object". In other words, the diffeomorphisms are the isomorphisms in the category of differential manifolds.

"Diffeomorphism invariance" just means that "something that remains invariant under all diffeomorphisms of a manifold to itself". It is a term used by physicists much more than by mathematicians. A good example, explained by marcus here:

is that the Einstein equations together with their solutions are diffeomorphism invariant in the sense that if you have a universe (smooth 4 manifold) with some matter in it and a metric that is a solution of the Einstein equation for this matter distribution, then if you apply a diffeomorphism to the universe, then you move the matter around and also the metric, and what you get is still a solution to Einstein's equations.

From the mathematical point of view this is an utter triviality not worth mentioning, but I guess that for physicists it's a condition they needs to be satisfies by their theory for them to make sense.

-Reparametrization invariance: I suspect this has to do with invariance under change of coordinates. To define a global object on a manifold, one way to go is to define it first on each coordinate chart. Then all these objects glue together to a global object on the whole manifold if and only if the objects "agree on overlaps" of the coordinate charts. In other words, they have to be independant of the coordinates, or "reparametrization invariant".

3. Dec 12, 2012

friend

Thank you. It was nice of you to respond. Yes, I've seen these definitions before. But I lack experience in how to use them, and I'm not sure how these ideas are interconnected, or how they apply to physics. So perhaps you'd be interested if I ask for some clarification.

The first thing that confuses me is what exactly Rn is. Is Rn flat space with the euclidean metric? Or is it just a list of numbers (x1,x2,x3,...) with no metric implied?

I take it that a differential structure is not local since it "covers" all of M. And it sounds like it refers to only coordinate maps since it covers M - a unique number for each point on M for each chart, like coordinates.

It sounds like a differential structure is a collection of diffeomorphisms.

I've heard that there is an infinite number of differential structures for R4 and even for Minkowski space M4, but there is only a finite number of differential structures for other dimensions. This is curious as it suggests a physical reason for differential structures since there are only 4 dimensions in the world. I wonder what that's all about.

There's not a lot on the web about differential structures. Do you have a reference? I'm trying to get to the differential structures and why there are an infinite number of them for R4.

Does diffeomorphism invariance mean that when you change variables (or coordinates) that you are left with the same formula with no extra terms or factors introduced by the change of variable (or coordinate)? In other words you could have simply changed y for x as if it were a dummy variable not changing the "form" of the equations?

Is reparameterization invariance just a type of diffeomorphism invariance? Thanks.

4. Dec 12, 2012

quasar987

In the definition, R^n is seen in the usual way as a normed vector space (so that we may make sense of derivatives of maps R^n-->R^n) with its topology induced by the norm (so that we can make sense of the coordinate charts on the manifold as being homeomorphisms).

Recall that in the definition of a differential structure, we start with a topological manifold M, and then we select from all the possible coordinate charts on M a bunch of them that covers M and that are "smoothly compatible" in the sense described in post #2. And it is only once such a choice has been made that we can start talking about maps M-->R^n being smooth, or being diffeomorphisms. But you're right in that once a differential structure has been chosen, then the coordinate charts that make it up are indeed, almost tautologically, diffeomorphisms.

I remember that my GR professor said (~6 years ago) that no one (i.e. physicists) had yet managed to do anything (i.e. physically) with this intriguing fact.

Differential structures are also called "smooth structures". Maybe this will yield more results? They are defined first thing in almost any differential geometry book. I recommend John Lee's Introduction to Smooth Manifolds.

I think this would be an instance of reparametrization invariance!

No I think they refer to different phenomenon.

5. Dec 14, 2012

friend

It seems a metric and a norm have similar definitions. A metric is defined on a couple of points of a manifold, where those points are given by a list of components. But a norm is defined as the length of a single vector, where a vector is again given as a list of components. When you break things down to components, they seem to be the same thing. The difference seems to be that a metric is defined between a couple of points of a manifold, whereas a norm is defined on a vector space which can be viewed as the tangent space of a point on a manifold. If the manifold is already Rn, then the metric and norm seem to be the same thing. So since the tangent space is already Rn is the norm and the metric the same thing in a tangent space?

There seems to be some concept I'm missing about the equivalence class of atlases that make up a differential structure. I think it has to do with being able to construct whole atlases with different smoothly compatible charts. And each such atlas that covers M is some how equivalent to a different atlas made up of a different smoothly compatible set of charts that also cover M. If one differential structure is made of the equivalence class of atlases, then I fail to understand how any manifold (R4) could have more than one differential structure. An example would probably clear things up quickly.

Here's a guess. Perhaps like there are no prefered coordinates of spacetime, there also is no prefered manifold of space. Points of a manifold really have no physical meaning; there are no unique points in space. And so there is no unique differential structure to cover that non-unique manifold. So maybe there being no unique manifold (thus no unique differential structure) really means there must be allowed an infinite choice of manifolds (thus infinite number of differential structures) available to describe space from various perspectives. So space needs to be R4 or M4 in order to have an infinite number of differential structures. Obviously, I'm not sure of what I'm talking about. So one should easily be able to point out my error. Thank you.

That's a \$100 book. I've looked through the table of contents. It looks like a very thorough reference. Perhaps you own this book. Does it cover why there are a finite number of differential structures for 1D, 2D, 3D,... manifolds but an infinite number of structures for 4D?

I guess I'm asking what it means to have the "form" of physical laws be diffeomorphism invariant. What does invariant "form" mean if not that the expression itself does not change wrt to some transformation?

Last edited: Dec 14, 2012
6. Dec 15, 2012

lavinia

7. Dec 15, 2012

lavinia

On manifolds an L^2 norm on the tangent spaces is called a metric.

Two things.

- Two sets of atlasses may be incompatible but the differentiable structures may still be diffeomorphic. Two structures on a manifold are considered to be different is they are not diffeomorphic.

The classic simple example ot this is two incompatible atlasses on the real line, the atlas determined by the coordinate chart f(x) = x and the atlas determined by the coordinate chart f(x) = x^3. Still these two differentiable structures are diffoemorphic.

- I do not know whether different differentiable structures have been described in term of atlasses. One proof for the 7 sphere involves showing that certain homology invariants of associated vector bundles are different. See Milnor's Characterisitc Classes

For 1 manifolds the proof is easy. It uses theorems about solutions of ODE's. Try to figure this out for yourself.

For two dimensions I think this is hard. There is a grand theorem which says that any simply connected Riemann surface is conformally equivalent to the standard 2 sphere, the entire complex plane or the open unit disk. I think this gives you a proof for 2D.

Last edited: Dec 15, 2012
8. Dec 15, 2012

atyy

There is not a standard definition of "diffeomorphism invariance". You always have to see the definitions given by each author to know what they are talking about.

A good discussion of the issues is given by http://arxiv.org/abs/gr-qc/0603087 .

9. Dec 15, 2012

friend

OK, so I'm trying to think of the prime example of diffeomorphism invariance, such that the mathematical "form" does not change with coordinate changes. Correct me if I'm wrong, but it seems the Dirac delta function serves as a prime example of diffeomorphism invariance. Let me try to explain:

Consider the defining property of the Dirac delta function,
$$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = 1$$
If we change coordinates to $y = y(x)$ so that $x = x(y)$, then $dx = \frac{{dx}}{{dy}}dy$.
Then we can use the notation
$$\frac{{dx}}{{dy}} = \sqrt {g(y)} = \sqrt {\frac{{dx}}{{dy}} \cdot \frac{{dx}}{{dy}}}$$
in order to be consistent with higher dimensional versions. We also have
$$x - {x_0} = \int_{{x_0}}^x {dx}$$
And by using the transformation, $y = y(x)$, so that ${y_0} = y({x_0})$, this integral becomes,
$$x - {x_0} = \int_{{x_0}}^x {dx} = \int_{{y_0}}^y {\sqrt {g(y)} dy}$$
And the original integral can be transformed to,
$$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (\int_{{x_0}}^x {dx} )dx} = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (\int_{{y_0}}^y {\sqrt {g(y)} \,dy} )\,\,\sqrt {g(y)} \,\,dy} = 1$$
where ${y^{ + \infty }} = y(x = + \infty )$ and ${y^{ - \infty }} = y(x = - \infty )$.

Then using the composition rule for the Dirac delta, we have
$$\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (\int_{{y_0}}^y {\sqrt {g(y)} \,dy} )\,\,\sqrt {g(y)} \,\,dy} = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\left( {\frac{{\delta (y - {y_0})}}{{|\sqrt {g({y_0})} \,|}}} \right)\,\,\sqrt {g(y)} \,\,dy = } \frac{1}{{|\sqrt {g({y_0})} \,|}}\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,\sqrt {g(y)} \,\,dy}$$
where ${{y_0}}$ is where $\int_{{y_0}}^y {\sqrt {g(y)} \,dy} = 0$.

And then using the sifting property of the Dira delta we have
$$\frac{1}{{|\sqrt {g({y_0})} \,|}}\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,\sqrt {g(y)} \,\,dy} = \frac{1}{{|\sqrt {g({y_0})} \,|}}\left( {\,\sqrt {g({y_0})} \,\int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,dy} } \right) = \int_{{y^{ - \infty }}}^{{y^{ + \infty }}} {\delta (y - {y_0})\,\,dy} = 1$$
So that finally,
$$\int_{ - \infty }^{ + \infty } {\delta (x - {x_0})dx} = \int_{ - \infty }^{ + \infty } {\delta (y - {y_0})\,dy} = 1$$
where I assume that using an integration interval of $- \infty \le y \le + \infty$ changes nothing.

So it appears here that any invertible coordinate transformation (diffeomorphism) leaves the integration of the Dirac delta unchanged in form, meaning it is diffeomorphism invariant. Does this look right?

If so, then I have a question. Does that mean any formulation built solely on the Dirac delta is automatically diffeomorphism invariant? If the Dirac delta is required on the basis of first principles, does that specify the necessary existence of an equivalence class of diffeomorphic coordinate transformations? What does it tell us about the underlying space in which the Dirac delta resides? Thanks.

Last edited: Dec 15, 2012
10. Dec 16, 2012

quasar987

No, it does not adress that subject at all, except maybe the 1-dimensional case in an exercice or something. The book "4-Manifolds and Kirby Calculus" appears to be talking about it and giving references. Donaldson-Kronheimer (The Geometry of Four-Manifolds) possibly also talks about it but it's not obvious from the table of content.

11. Dec 19, 2012

friend

And to continue this kind of thinking, I wonder if the dirac delta doesn't also specify a differential structure. Let me explain. The sifting property 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,
$$\int_{ - \infty }^\infty {\delta (x - {x_1})\delta ({x_1} - {x_0})d{x_1}} = \delta (x - {x_0})$$
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,
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {\delta (x - {x_n})\delta ({x_n} - {x_{n - 1}}) \cdot \cdot \cdot \delta ({x_1} - {x_0})} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1} = \delta (x - {x_0})$$
And since this results in a dirac delta function, integrating one more time gives us the constant 1.

So if each dirac delta function specifies an eqivalence class of diffeomorphism on a coordinate patch (i.e. the coordinate transformations), then does each dirac delta function in the last integral specify the existence of its own coordinate patch with diffeomorphic coordinate transformations? And does the last integral specify a differential structure consisting of an equivalence class of atlases constructed from the overlapping coordinate patches, one patch for each dirac delta function in the last integral?

12. Dec 25, 2012

Bacle2

The number of differential structures (as well as many other aspects) of simply-connected 4-manifolds have to see with the properties of their intersection forms. There are aspects of the form , such as the parity, signature, etc., that dictate whether the corresponding manifold admits a differential structure or not. Of course, there is a correspondence, using Poincare-duality between the algebraic and topological aspects of a 4-manifold. There is also the fact that there are infinitely-many inequivalent quadratic forms, and every quadratic form can be made to correspond to a 4-manifold (you have to consider other invariants for more precision/ narrowing down).

13. Dec 25, 2012

Bacle2

Are you sure you can multiply distributions the way you're doing it in a meaningful way? Most of the time, distributions cannot be multiplied.

14. Dec 25, 2012

micromass

It's ok in this case. Something that you can't define is $\delta(x)^2$. But you can define $\delta(x)\delta(y)$ because the variables are "independent". You just set

$$\iint f(x,y)\delta(x)\delta(y)dxdy = f(0,0)$$

Something similar works in the post.

15. Dec 25, 2012

TrickyDicky

friend, there seem to be so many misconceptions in the way you mix Dirac delta distributions with terms from differential geometry that it's no wonder you are not getting any answers to your questions both here and in your physics thread on this.
Your hypothesis is so vague and confusing that a response is hard to give, but I would say that a "formulation built solely on the Dirac delta" whatever that means, is certainly not "automatically diffeomorphism invariant" unless you take the view that many hold here that any physical theory can be formulated in a diffeomorphism invariant way, wich makes your connection with dirac deltas totally superfluous. And you still would have to explain what it means to build a physical theory solely on the Dirac delta function. Certainly QM is not built that way and much less relativity.Besides a Dirac delta is not even strictly a function.

16. Dec 25, 2012

Bacle2

Ah, yea, duh. I should read more carefully.

17. Dec 25, 2012

friend

Perhaps you are right. That's why I put these concepts into the form of a question.
What's so vague about asking if some formula or integral is invariant wrt coordinate changes?
The integral,
$$\int_{ - \infty }^\infty {\delta (x - {x_1})\delta ({x_1} - {x_0})d{x_1}} = \delta (x - {x_0})$$
is shown on the wikipedia.org site here, just before the composition section.

The integral,
$$\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - }}{{\rm{x}}_n}){\rm{\delta (}}{{\rm{x}}_n}{\rm{ - }}{{\rm{x}}_{n - 1}}) \cdot \cdot \cdot {\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})} } } d{x_n}d{x_{n - 1}} \cdot \cdot \cdot d{x_1} = {\rm{\delta (x - }}{{\rm{x}}_0})$$
This is explicitly written out in Prof. Hagen Kleinert's book, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, page 91.

The Dirac delta function is usually given in applications as some kind of regular function that integrates to 1, but whose absolute value tends to infinity as some parameter approaches zero. They do the integration first as if the dirac delta was a regular function and then make the parameter approach zero after the integration.

If the first 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.

So naturally, it's tempting to consider whether there is a connection between the dirac delta function and relativity as well. If the dirac delta function is diffeomorphism invariant, then it appears that the path integral is diff. morph. inv. as well.

Last edited: Dec 25, 2012
18. Feb 8, 2013

friend

But I haven't justified the complex nature in $${\Delta ^2} = \frac{{2i\hbar }}{m}({t_1} - {t_0})$$
And it seems to me that if I could find a good reason for complex numbers here, then I could justify the use of complex numbers in quantum mechanics, since the Dirac delta function can be used to derive the Feynman Path Integral.

I suppose the only way to justify the use of complex numbers within any formulism is to show that this formulism has the same algebraic properties as the complex numbers. As I understand it, the complex numbers lose the ability to determine magnitude (is i > -i ?), just as the quaternions also lose commutativity, and the octonions then also lose associativity.

Now since it is not possible to say which Dirac delta function is less than or greater than any other Dirac delta function, the Dirac delta function loses the property of magnitude (when compared to other Dirac deltas) but retains commutativity and asscociativity, just like the complex numbers do. So the complex numbers are a fair representation of the Dirac delta function since they share the same algebra. And we see here how the complex numbers enter quantum mechanics.

This brings me to wonder whether the properties of diffeomorphisms and differential structures change if we are talking about structures on a complex manifold instead of a real valued manifold.

Last edited: Feb 8, 2013