coordinate based vs noncoordinate based differential geometryby Mike706 Tags: based, coordinate, differential, geometry, noncoordinate 

#1
Feb1912, 08:57 AM

P: 52

Hello Everyone,
I am just wondering what the difference in these is. Could someone please give a brief example of noncoordinate based differential geometry vs the equivalent in coordinate based, or explain the difference (whichever is easier)? Also, what advantages does one have over the other? Thanks for your help, Mike Edit  Also, are there specific names for these two types of DG? Thanks again. 



#2
Feb1912, 09:23 AM

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Hi,
There is no difference btw coordinatefree DG and coordinate DG. They are the same thing (namely, DG!) but viewed from two different perspective. The situation is exactly analogous to coordinate free linear algebra vs coordinate linear algebra. For instance, given an abstract vector space V, the coordinatefree approach sees a linear operator T:V>V as a map s.t. T(ax+by)=aTx+bTy. The coordinate approach results from choosing a basis v1,...,vn of V, which sets up an isomorphism with R^n via v_j >e_j. Then a linear operator T:V>V is one such that in coordinates, if x=(x1,...,xn), then [tex] (Tx)_i=\sum_{j=1}^na_{ij}x_j. [/tex] for some set of nē numbers a_ij. Whereas the coordinate approach of linear algebra arises from the noncoordinate one from choosing a basis for V, likewise the coordinate approach of DG arises from the noncoordinate one from choosing a coordinate system on the manifold M. In DG as in linear algebra, sometimes it is preferable to work in the coordinate free approach either because it conveys the ideas of what's going on more transparently, or because the computations are easier/less messy to carry out. And some other times the same holds true of the coordinate approach over the coordinate free approach. 



#3
Feb1912, 09:35 AM

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P: 1,716

I can describe examples if you like. 



#4
Feb1912, 10:21 AM

P: 52

coordinate based vs noncoordinate based differential geometry
Thank you for the quick replies!
quazar987: That made a lot of sense, I think I understand it now. Could you just confirm my understanding please? I am using this book. It uses Clifford algebra instead of differential forms. In the beginning of the book, the basis vectors were described explicitly in terms of matrices. The main purpose this is done, it is stated, is to show concretely that we can actually use matrices to form a linearly independent basis of the required degree. So using this basis is then coordinatebased DG. In chapter 4 the author just said we will not consider any more explicit matrices, and also states without proof (beyond scope of the book) that it is possible to define Clifford numbers without matrices at all. So we will no longer consider matrix products, but Clifford products. This is then the coordinatefree approach, since we are not considering any specific basis anymore. Am I correct? Edit  Actually, I'm not so sure that was correct, now that I think about it further. Would it be possible for you to provide a simple concrete example? Lavinia: Thank you for the reply as well. I've heard that said before in the past and have been very curious about it. However, I only started studying DG within the last few weeks so I have not come across it yet. I would definitely be interested in an example. Thanks again Mike Edit  Formatting 



#5
Feb1912, 10:44 AM

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I am not familiar with this version of DG, but what you wrote makes sense to me.




#6
Feb1912, 11:02 AM

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#7
Feb1912, 11:16 AM

P: 52

Lavinia I think you just answered about 3 of my questions with what you said after recommending the book. Things clicked  it was good. I will definitely check out that book as well, thanks. 


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