A Connection between Reciprocal Space and Cotangent Space

[Phinrich]

TL;DR Summary

Is there a mathematical connection between the Real Space and the Tangent Space PLUS the relation between the Reciprocal Space and the Cotangent Space if any ?

[Moderator's note: Link to reference protected by copyright removed.]

 

[mathwonk]

You are not getting replies because we, or at least I, don't know what you are talking about. A google search I did turns up the term "reciprocal space" in connection with crystallography and lattices, but it is not, to me at least, a common term in English discussions of differential geometry. My guess is that it is some translation of a foreign language discussion of "dual space" and thus is essentially the same as cotangent space, but that is just a(n educated) guess.

 

[Phinrich]

Good day Mathwonk.

Thank you for your comment. You are correct that the term "Reciprocal Space " is not a common term in differential geometry. I am playing with ideas and concepts here. When you do this you may often try to match two things which seem very different. If we accept that the concepts of Tangent Space and Cotangent Space are differential geometry issues (they may not be and I admit that I am not an expert in differential geometry and that is why I come before the experts in this Forum). Now the Professor in question said in his paper that there was a link between all these concepts. Unfortunately he did not explain the link explicitly. That is why I am asking anyone on this Forum if they can enlighten me. In the meantime I have e-mailed the Professor directly and hope to receive a response from him soon, So as I say it may end up that there is NO connection between them, Then at least I will have learned something.

Thank you again.

 

[Phinrich]

In trying to answer my own question, and bearing in mind that I am no expert,

We recognise that for each vector (in the tangent space) there is an associated 1-Form (in the cotangent space). One possible 1-Form is the Gradient function (I hope). Now Gradient means "rate of change over a unit distance". Additionally, anyone familiar with the concepts of real space and reciprocal space will know that the units of a real space vector is "length". While the unit of a reciprocal space vector is "inverse length" or "linear density". Does this not give us a connection in that the tangent unit vector is the same as the unit real space vector while the unit reciprocal vector (being a linear density over a unit reciprocal vector) may be compared to the gradient 1-Form ? This is not very rigorous and I am REALLY displaying my ignorance here.

 

[mathwonk]

Unfortunately that is not helping much. For one thing there is some confusion in the use of the term "one - form". Some people use it as a synonym for "covector", or linear function on the tangent space, as you seem to be doing here, whereas I myself usually reserve it for a "field" or family of such covectors, one chosen from each tangent space of the given manifold.

Furthermore, even if I accept the word to mean a single covector, i.e. a single element of the dual tangent space, or cotangent space, it is still not true a priori that a single tangent vector gives rise to an associated covector, rather that only happens in the presence of an "inner product" on the tangent space. So it helps us to know what is the starting point of your discussion.

Now it is true in all cases (when a smooth manifold is given) that a smooth function does give rise to a one form, namely the gradient of the given function. This again leads one to believe that the term "one form" is being used in the sense I prefer, namely as a field of convectors. I.e. a smooth function normally is defined over some open set, or even over the whole manifold, and then it has a derivative, i.e. its gradient, which is also defined where the function is smooth. Thus a smooth function defined on a manifold has an associated gradient one form which is a field of covectors, one at each point of the manifold. Of course can then consider the action of the one foprm only at one point, obtaining a covector.

This gradient construction does not require any inner product for its definition. In general, given a vector space V, one can consider linear maps R-->V from the real numbers into V, and linear maps V-->R from V into the real numbers. The first sort of map can be identified with a vector in V (the image of the number 1), and the second sort of map can be iddentified with a covector.

In the same spirit, given a smooth manifold M, one can consider smooth maps s:R-->M from the reals R into M, and smooth maps M-->R from M into the reals. The first sort of map gives rise to a parametrized curve, and hence a tangent vector, or velocity vector, at the points of the image curve, or just at the image point s(0). Dually, given both such maps, i.e. a smooth curve s:R-->M, and a smooth function f:M-->R, one can compose them, and get a real valued function (fos):R-->R, of which one can take the derivative at 0, obtaining a number. This number is the resuilt of evaluating the gradient of f at the tangent vector representing the velocity vector of s, at the point s(0).

So "maps in" , or vectors, are dual, or reciprocal, to "maps out", or functions. This is true for linear maps on individual vector spaces, and by taking derivatives extends also to smooth maps of manifolds.

If one defines an inner product (v,w), or v.w, between any two vectors, v and w, then given a vector v, one can think of it as just the vector v, or as the operator, (v, ), or v.( ), on other vectors w. I.e. v operates on w yielding the number (v,w) or v.w. In this way, vector fields can also be regarded as covector fields, or one forms, but only if the inner product is given.

In particular it is not assumed in general that a notion of length of a tangent vector is given unless there is something like an inner product. I.e. in a smooth manifold, there are initially no "units".

 

[mathwonk]

" Additionally, anyone familiar with the concepts of real space and reciprocal space will know that the units of a real space vector is "length". While the unit of a reciprocal space vector is "inverse length" or "linear density"."

Clearly I am not familiar with these concepts, so it would help if you would define them so that we will know what structures you are assuming to be present. I.e. in mathematics a "real vector space" does not have a unit of length until you specify one. and as i said, I have never heard of the concept of "reciprocal space". so that would be a useful definition as well.

 

[Phinrich]

Thank you Mathwonk for your explanation. I should have mentioned that an inner product was defined. I will need some time to study your response. It clarifies many things for me. I guess when you say ""maps in" , or vectors, are dual, or reciprocal, to "maps out" " you are speaking in the context of a "reverse" (reciprocal) operation rather than in the sense that both are geometric objects and the magnitude of 1 is 2 pi divided by the magnitude of the other ?

 

[Phinrich]

OK I will find a definition and send it your way. Thank you.

 

[mathwonk]

I have no idea what your last sentence in #7 means. When I say maps out of M, I mean a map fom M to something else, such as R, i.e. a map M-->R. similarly a map into M is a map from some other space like R with values in M, i.e. R-->M. neither one has any magnitude.

 

[Phinrich]

Sorry - that`s the "amateur physicist" in me trying to talk to the mathematician in you. I have a degree in Physics (1985) but don't work in an academic environment so my language could be very rusty. my language is certainly not as rigorous and precise as yours. I am actually a radiation protection specialist by trade. I will study your responses and try to firm-up my language.

 

[mathwonk]

for one thing it would help to know whether you are working in a single vector space, or on a smooth manifold where there is a tangent space at each point. If there is only one vector space around, you are doing linear algebra, and do not need any diffrential geometry, which in some sense is a tool for dealing with non linear smooth spaces by approximating them using a different vector space at each point.

 

[Phinrich]

I would say that my intention is to work with a smooth manifold on which an inner product is defined. Also a metric but I guess that to define an inner product we must have a metric defined (or am I missing it again?).

 

[mathwonk]

a gradient also is to me a very precise concept only roughly approximated by the phrase "rate of change over a unit length". The pairing I have given of a function with a curve is a precise definition of thje gradient of that function, but only if you know the prcise definition of the derivative of the real valued function of a real variable. The gradient is a precise version of the idea of rate of change. I.e. one meausres the rate of change over smaller and smaller intervals, takes the limit, and then extrapolates that limiting rate of change as if it were taking place over unit length. I.e. the velocity vector of a moving particle, at a given point, is the arrow stretching from that initial point, to the point the particle would reach, if only it stopped accelerating at that instant and continued to move uniformly for one second, at the same rate it was moving at that instant.

E.g. if the particle were being steered and propelled by some engine, we could stop steering and turn off the engine at that instant, and wait one unit of time, and wherever it ends up would be the velocity vector it has at that instant. In that sense it is a rate of change over a unit interval, at least to me, a physics novice.

 

[mathwonk]

a metric is a length. an inner product defines a length on the tangent spaces, and hence identifies them with the dual spaces. then i believe one can integrate and obtain also a length on curves in the manifold, and by taking shortest curves, also a length on the manifold. but maybe now some more expert differential geometers can weigh in. (hopefully they will refrain from invoking more abstract versions of length such as "connections", unless those are also part of your toolkit.)

 

[Phinrich]

OK ! I understand that COMPLETELY ! Also, as you call yourself a "physics novice" I would call myself a "mathematics novice". I did a 4-year physics degree with 3 years applied mathematics and 2 years mathematics. However that was some 30 years ago. Today I dabble in physics for the fun of it and only use the mathematics I think I need. I am GREATLY interested in General Relativity. If you Google there's a Wikipaedia article on the "Reciprocal Lattice". This has equations and concepts but in the framework of a physical crystal lattice. I guess I am trying to take these concepts and see if I can write them down as relating to a smooth manifold rather than a discrete lattice. Dont know if that makes any sense?

 

[Phinrich]

Talking about "connections" - I understand a connection in the Framework of General Relativity where the Christoffel Symbols are the "connection" relating a vector defined at one point in a smooth manifold to its image defined at a second point in the manifold. Hope I got that right at least. So in this sense a connection would be a part of my toolkit. However I suspect that you have a better more mathematically precise definition of a connection. This idea of a connection was given to me by Professor G F R Ellis who taught me GR. Dont know if his name rings a bell with you?

 

[mathwonk]

I googled reciprocal lattice and saw it is a Fourier transform of another lattice. Now Fourier transforms, which i am also a novice at, are to me something requiring even more structure, namely that of a group. One translates calculus on one group over to calculus on the dual group, but in simple cases, maybe like the real numbers, the group is self dual. no wait even then maybe it is dual to the circle? i ghuess i need the definition of dual group, but it will be in terms of some maps "out" of the group probably, but into what?

no wait, maybe a Fourier series is a representation of a periodic function, hence represents a function on the circle as a series, i.e. a function on the integers. so it must be the circle is dual to the integers, both groups, and maybe the reals are indeed self dual? i'll check further, but this is not my field, at least not any more.

but in general transforms try to transform calculus from one space where it seems difficult, to another space where it seems easier. There are Fourier transforms, Paley - Weiner transforms, etc... in the cases I knew of, there is a family of functions satisfying a given diffrential equation, and the transform changes them into functions satisfying an (hopefully) easier Diff Eq. (My career as a harmonic analyst was brief and less than flourishing.)

 

[Phinrich]

Yes I am familiar with the concepts of the Fourier Transform used here. Perhaps I am mistaken and there is no connection andI should treat the Real and Reciprocal spaces as separate from the tangent and cotangent spaces.

 

[mathwonk]

here is an article, maybe a bit abstract, on general duality and Fourier transforms.

https://en.wikipedia.org/wiki/Pontryagin_duality

 

[Phinrich]

Thanks. I am aware of this article and have read it but as you correctly say it is a bit abstract for me to understand. Perhaps I should tr to understand as you say it discusses general duality and Fourier transforms. Perhaps if I understood it I may have a clue !

 

[Phinrich]

OK tried reading it again and its WAY over my head !

 

[mathwonk]

you no doiubt know more than this, but i will go out on a limb anyway and just say that the exponential under the integral sign in a Fourier transform, is looked at a map to the circle, this generalizes to the notion of dual group as maps to the circle. sorry for rushing.

 

[Phinrich]

Please explain the "is looked at a map to the circle " ?

 

[Phinrich]

Is this anything to do with the Argand diagram representation of a complex number ?

 

[Phinrich]

Or maybe its because the exponential may be written as a sum of sines and cosines

 

[Phinrich]

Try this link;

http://bilimneguzellan.net/fuyye-serisi/

 

[mathwonk]

yes, sorry for more math speak. we take it for granted that e^(it) = cos(t) + i sin(t), which is a point of the unit circle in the complex plane. so the "circle group" means the group of unit length complex numbers, i.e. those of form cos(t) + i sin(t) = e^(it), for real t.

thus t-->e^(it) is a map from the additive group of reals to the multiplicative group U of unit complex numbers. hence by definition of the dual group, it is an element of the dual group of the reals. for any real s, t-->e^(its) is another one, and apparently all such occur this way, so this would identify the dual group of maps R-->U with the original group R. but i am really not at all proficient in this topic.

 

[Phinrich]

Thanks. Dont worry about the "maths speak". The rigour of your arguments reminds me to try to be more careful in how I phrase my arguments. Yes I enjoyed the theory of complex numbers when I was at University so I understand what you say.

 

[Phinrich]

Summary:: Is there a mathematical connection between the Real Space and the Tangent Space PLUS the relation between the Reciprocal Space and the Cotangent Space if any ?

[Moderator's note: Link to reference protected by copyright removed.]

Hi all.

I asked this this question last year September. I have just found an interesting paper at the link

https://www.iucr.org/__data/assets/pdf_file/0017/13193/4.pdf

Published by the International Union of Crytallography

Where in Section 3 the author states that the basis vectors of the Dual Space are identical to the basis vectors of the Reciprocal Space. So that clears it up for me.

Thank you to all for your valued contributions to the discussion.

Paul

 

[Phinrich]

Good day, all.

Further to this old discussion (in case anyone is interested), If you Google "Dual Basis" you will find a Wikipedia Article which says the following;

In Linear Algebra, given a vector space, V with a basis B of vectors indexed by an index set I, the dual set of B is a set B∗ of vectors in the dual space V∗ with the same index set I such that B and B∗ form a biorthogonal system. The dual set is always linearly independent but does not necessarily span V∗. If it does span V∗, then B∗ is called the dual basis or reciprocal basis for the basis B.

In the case of finite-dimensional vector spaces, the dual set is always a dual basis and it is unique. These bases are denoted by B = { e1, …, en } and B∗ = { e1, …, en }. If one denotes the evaluation of a covector on a vector as a pairing, the biorthogonality condition becomes:

I may be wrong but I believe this answers my original question concerning whether there is a link between the dual space and reciprocal space.

Any comments would be welcome as I do not consider myself a Mathematician.

Thanks

Paul