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What Constitutes something being coordinate free

by saminator910
Tags: constitutes, coordinate free
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saminator910
#1
Jan29-13, 07:22 PM
P: 94
People say that exterior calculus ie. differentiating and integrating differential forms, can be done without a metric, in without specifying a certain coordinate system. I don't really get what qualifies something to be 'coordinate free', I mean in the differential forms I do, one still references components ie. x1,x2, etc., yet I never specified a metric, so is this classified as 'coordinate free'. Also, how does one do differential geometry without a coordinate system, in my mind once you don't specify a coordinate system or a metric, and things become vague, it sort of turns into differential topology, is there a 'middle ground' I am missing, keep in mind I have never taken a coarse in differential geometry. Also, in differential geometry, it has always been pertinent to give specific parametrization in order to find tangent vectors, metrics, etc.
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CompuChip
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Jan30-13, 08:57 AM
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If you use xi, that (implicitly) means that you have chosen a coordinate system, so it is not coordinate free.

To give a very simple example, consider a linear transformation (e.g. rotation) T on a vector x. You could write this in coordinate-free form as
x' = T x
This does not depend on which basis you choose for the space that x lives in - it just means: apply this transformation.

When you calculate the result on a vector, you usually pick a coordinate system by choosing a set of basis vectors (x, y, z-axis) and write the action of T as a matrix M.
You then calculate
[tex]\mathbf{x'}_i = \sum_{j = 1}^n M_{ij} x_j [/tex]
This is not coordinate-free, because both the components of x and x' as well as the entries of M depend on the coordinate system.

The advantage of the coordinate-free form is that it looks the same in any choice of basis. If you and I both chose a different coordinate system and wrote down M, we would get two different bunches of numbers but it would not be immediately clear that we're talking about the same "physical" operation.
HallsofIvy
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Jan30-13, 09:08 AM
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Another example: Let X be the set of all quadratic polynomials from R to R and define T by T(f)= df/dx+ f. That is a "coordinate free" definition. If I had chosen [itex]\{1, x, x^2\}[/itex] as basis (essentially choosing a "coordinate system" by choosing a basis), say that T(1)= 1, T(x)= x+ 1, T(x2)= x2+ 2x, then say T is "extended by linearity", T(af+ bg)= aT(f)+ bT(g), that is not "coordinaate free" because I have used a coordinate system (a basis) to define it. Or course, those are exactly the same definition.

Stephen Tashi
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Jan30-13, 11:41 AM
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What Constitutes something being coordinate free

It would be interesting to sort out the distinction between "having a coordinate system" and "having a metric". The two definitions are not identical, but what common situations allow us to proceed from having one to having the other?
CompuChip
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Jan30-13, 12:27 PM
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Aren't they two completely different things?
Sure, for a lot of "common" metric spaces the metric is defined in terms of coordinates. But for Rn, for example, that's mostly because people usually learn Pythagoras before inner products, so
[tex]\sum_i (y_i - x_i)^2[/tex]
is a little more intuitive than
[tex]\langle \vec y - \vec x, \vec y - \vec x\rangle[/tex]
Stephen Tashi
#6
Jan30-13, 12:39 PM
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Quote Quote by CompuChip View Post
Aren't they two completely different things?
Of course they are, that's why sorting out their relation is complicated. The idea of a metric on a set of things is standardized, but I'm not sure whether there is an standard definition for a set of things to "have a coordinate system".

The pythagorean idea won't necessarily work for a coordinate system where the same thing can have two different coordinates (as is the case in the polar coordinate system).
saminator910
#7
Jan30-13, 09:07 PM
P: 94
Okay so, your standard differential form, written with coordinates say for example a standard 1 form, [itex]\alpha = \sum^{n}_{i=1}f_{i}du_{i}[/itex], you are still referencing a it's local coordinates, but you don't necessarily need to know what the metric is ie. euclidean space vs. it being embedded in some other manifold, so the coordinates don't necessarily need any intrinsic value. then if you know the [itex]u_{i}[/itex] coordinates in terms of euclidean or other coordinates, you pullback/pushforward the form. Does this count as 'coordinate free' since you don't specify a basis: the coordinates in the form aren't in terms of anything.
HallsofIvy
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Jan31-13, 09:42 AM
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Quote Quote by Stephen Tashi View Post
It would be interesting to sort out the distinction between "having a coordinate system" and "having a metric". The two definitions are not identical, but what common situations allow us to proceed from having one to having the other?
Once we have a coordinate system, so that every point, x, can be identified with [itex](x_1, x_2, ..., x_n)[/itex], there is a "natural" metric, [itex]d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2- y_2)^2+ ...+ (x_n- y_n)^2}[/itex].

Given a metric space, even finite dimensional [itex]R^n[/itex], we would need a choice of "origin", and n-1 "directions" for the coordinate axes (after choosing n-1 coordinate directions, the last is fixed) in order to have a coordinate system. So a "coordinate system" is much more restrictive, and stronger, than a "metric space".
Stephen Tashi
#9
Jan31-13, 10:44 AM
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Quote Quote by HallsofIvy View Post
Once we have a coordinate system, so that every point, x, can be identified with [itex](x_1, x_2, ..., x_n)[/itex], there is a "natural" metric, [itex]d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2- y_2)^2+ ...+ (x_n- y_n)^2}[/itex].
But does the definition of a "coordinate system" ( if there is such a standard definition) include the idea that each element of "the space" can be identified with a unique finite vector of coordinates? And must the vector consist of a vector of real numbers?
saminator910
#10
Jan31-13, 06:31 PM
P: 94
okay, so now can someone give an example of using a differential form without a metric?


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