Diffeomorphisms, Differential Structure, ETC.by friend Tags: diffeomorphisms, differential, structure 

#1
Dec1112, 12:57 PM

P: 962

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.




#2
Dec1112, 08:05 PM

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PF Gold
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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: http://www.physicsforums.com/showthread.php?t=222112 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
Dec1212, 07:42 PM

P: 962

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.
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 R^{4} and even for Minkowski space M^{4}, 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 R^{4}. 



#4
Dec1212, 09:48 PM

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Diffeomorphisms, Differential Structure, ETC. 



#5
Dec1412, 05:04 PM

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#6
Dec1512, 06:36 AM

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#7
Dec1512, 08:24 AM

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On manifolds an L^2 norm on the tangent spaces is called a metric.  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 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. not sure about 3 D 



#8
Dec1512, 09:48 AM

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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/grqc/0603087 . 



#9
Dec1512, 07:30 PM

P: 962

Consider the defining property of the Dirac delta function, [tex]\int_{  \infty }^{ + \infty } {\delta (x  {x_0})dx} = 1[/tex] If we change coordinates to [itex]y = y(x)[/itex] so that [itex]x = x(y)[/itex], then [itex]dx = \frac{{dx}}{{dy}}dy[/itex]. Then we can use the notation [tex]\frac{{dx}}{{dy}} = \sqrt {g(y)} = \sqrt {\frac{{dx}}{{dy}} \cdot \frac{{dx}}{{dy}}} [/tex] in order to be consistent with higher dimensional versions. We also have [tex]x  {x_0} = \int_{{x_0}}^x {dx} [/tex] And by using the transformation, [itex]y = y(x)[/itex], so that [itex]{y_0} = y({x_0})[/itex], this integral becomes, [tex]x  {x_0} = \int_{{x_0}}^x {dx} = \int_{{y_0}}^y {\sqrt {g(y)} dy} [/tex] And the original integral can be transformed to, [tex]\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[/tex] where [itex]{y^{ + \infty }} = y(x = + \infty )[/itex] and [itex]{y^{  \infty }} = y(x =  \infty )[/itex]. Then using the composition rule for the Dirac delta, we have [tex]\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} [/tex] where [itex]{{y_0}}[/itex] is where [itex]\int_{{y_0}}^y {\sqrt {g(y)} \,dy} = 0[/itex]. And then using the sifting property of the Dira delta we have [tex]\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[/tex] So that finally, [tex]\int_{  \infty }^{ + \infty } {\delta (x  {x_0})dx} = \int_{  \infty }^{ + \infty } {\delta (y  {y_0})\,dy} = 1[/tex] where I assume that using an integration interval of [itex]  \infty \le y \le + \infty [/itex] 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. 



#10
Dec1612, 08:25 AM

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#11
Dec1912, 10:55 AM

P: 962

[tex]\int_{  \infty }^{ + \infty } {f({x_1})\delta ({x_1}  {x_0})d{x_1}} = f({x_0})[/tex] And if we let [itex]f({x_1}) = \delta (x  {x_1})[/itex] in the above we get, [tex]\int_{  \infty }^\infty {\delta (x  {x_1})\delta ({x_1}  {x_0})d{x_1}} = \delta (x  {x_0})[/tex] And when this process is iterated again, we get [tex]\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})[/tex] Iterating this process an infinite number of times gives us, [tex]\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})[/tex] 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
Dec2512, 03:57 AM

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The number of differential structures (as well as many other aspects) of simplyconnected 4manifolds 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 Poincareduality between the algebraic and topological aspects of a 4manifold. There is also the fact that there are infinitelymany inequivalent quadratic forms, and every quadratic form can be made to correspond to a 4manifold (you have to consider other invariants for more precision/ narrowing down).




#13
Dec2512, 03:59 AM

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#14
Dec2512, 04:02 AM

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[tex]\iint f(x,y)\delta(x)\delta(y)dxdy = f(0,0)[/tex] Something similar works in the post. 



#15
Dec2512, 07:55 AM

P: 2,891

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
Dec2512, 10:35 AM

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#17
Dec2512, 04:40 PM

P: 962

[tex]\int_{  \infty }^\infty {\delta (x  {x_1})\delta ({x_1}  {x_0})d{x_1}} = \delta (x  {x_0})[/tex] is shown on the wikipedia.org site here, just before the composition section. The integral, [tex]\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})[/tex] 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 ChapmanKolmogorov equation whose solution is a gaussian distribution as found here, for example. So if we use the gaussian form of the Dirac delta function, [tex]{\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}}}[/tex] with [tex]{\Delta ^2} = \frac{{2i\hbar }}{m}({t_1}  {t_0})[/tex] the dirac deltas become [tex]\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][/tex] which can be manipulated to [tex]\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][/tex] or, [tex]\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}}}}[/tex] When this is substituted for each of the dirac deltas in the above we get [tex]\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}[/tex] with the appropriate limits implied. Then since the exponents add up, the above becomes [tex]\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}[/tex] 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. 



#18
Feb813, 11:25 AM

P: 962

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. 


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