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#21 Xerxes

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Posted 16 February 2017 - 04:19 PM

OK, so let's talk a bit about tensors in differential geometry. Recall that, in normal usage, differential geometry is the study of (possibly) non-Euclidean geometry without reference to any sort of surrounding - or embedding - space.

First it is useful to know what is a manifold. No. Even firster, we need to know what is a topological space.

Right.

Suppose S a point set - a set of abstract points. The powerset \mathcal{P}(S) is simply the set formed from all possible subsets of S. It is a set whose members (elements) are themselves sets.
Note by the definition of a subset, the empty set \O and S itself are included as elements in \mathcal{P}(S)

So a topology T is defined on S whenever S is associated to a subset (of subsets of S) of \mathcal{P}(S) and the following are true

1. Arbitrary (possibly infinite) union of elements in T are in T

2. Finite intersections of elements in T are in T

3. S \in T

4.  \O \in T

The indivisible pairing S,T is called a topological space. Note that T is not uniquely defined - there are many different subsets that can be found for the powerset.

Now often one doesn't care too much which particular topology id used for any particular set, and one simply says "X is a topological space". I shall do that here.

Finally, elements of T are called the open sets in the topological space, and the complements in S,T of elements in T are called closed.

Ouch, this already over-long, so briefly, a manifold is M is a topological space for which there exists a continuous mapping from any open set in M to an open subset of some R^n which has a continuous inverse. This mapping is called a homeomorphism (it's not a typo!), so that when U \subseteq M one writes h:U \to R^n for this, and n is taken as the dimension of the manifold

Since R^n \equiv R \times R \times R \times..... the homeomorphic image of m \in U \subseteq M is, say, h(m)= (u^1,u^2,,....,u^n), a Real n-tuple

And really finally, one defines projections on each n-tuple \pi_j:(u^1,u^2,....,u^n)\to u^j, a Real number.

So the composite function is defined to be \pi_j \circ h = x^j:U \to \mathbb{R}

Elements in the set \{x^k\} are called the coordinates of m. They are functions, by construction
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#22 wtf

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Posted 16 February 2017 - 10:18 PM

So far so good at my end. I haven't forgotten this thread, I've been slowly working my way through the universal property of the tensor product applied to multilinear forms on the reals. I can now visualize the fact that \mathbb R \otimes \dots \otimes \mathbb R = \mathbb R, because you can pull out all the coefficients of the pure tensors so that the tensor product is the 1-dimensional vector space with basis 1 \otimes \dots \otimes 1. This is pretty simple stuff but I had to work at it a while before it became obvious.

I'm still curious to understand the significance of the duals in differential geometry and physics so feel free to keep writing, you'll have at least one attentive reader.

Edited by wtf, 16 February 2017 - 10:29 PM.

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#23 wtf

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Posted 17 February 2017 - 03:04 AM

ps -- Let me just say all this back and, being a pedantic type, clarify a couple of fuzzy locutions.
 

The indivisible pairing S,T is called a topological space. Note that T is not uniquely defined - there are many different subsets that can be found for the powerset.


Minor expositional murkitude. I'd say this as: For a given set S, various topologies can be put on it. For example if T = \mathcal P(S) then every set is open. That's the discrete topology. The discrete topology is nice because every function on it to any space whatsoever is continuous. Or suppose T = \{\emptyset, S\}. This is called the indiscrete topology. No sets are open except the empty set and the entire space. And the everyday example is the real numbers with the open sets being countable unions of open intervals. [This is usually given as a theorem after the open sets have been defined as sets made up entirely of interior points. But this is a more visual and intuitive characterization of open sets in the reals].
 

Now often one doesn't care too much which particular topology id used for any particular set, and one simply says "X is a topological space". I shall do that here.


This is actually interesting to me. Do they use unsual topologies in differential geometry? I thought they generally consider the usual types of open sets. Now I'm trying to think about this. Hopefully this will become more clear. I guess I think of manifolds as basically Euclean spaces twisted around in various ways. Spheres and torii. But not weird spaces like they consider in general topology.
 

Ouch, this already over-long


Well if anything it's too short, since this is elementary material (defined as whatever I understand :-)) and I'm looking forward to getting to the good stuff. But I hope we're not going to have to go through the chain rule and implicit function theorem and all the other machinery of multivariable calculus, which I understand is generally the first thing you have to slog through in this subject.

If you can find a way to get to tensors without all that stuff it would be great. Or should I be going back and learning all the multivariable I managed to sleep through when I was supposed to be learning it? I can take a partial derivative ok but I'm pretty weak on multivariable integration, Stokes' theorem and all that.
 

And really finally, one defines projections on each n-tuple \pi_j:(u^1,u^2,....,u^n)\to u^j, a Real number.

So the composite function is defined to be \pi_j \circ h = x^j:U \to \mathbb{R}

Elements in the set \{x^k\} are called the coordinates of m. They are functions, by construction


What you are doing with this symbology is simply putting a coordinate system on the manifold. We started out with some general topological space, and now we can coordinatize regions of it with familiar old \mathbb R^n. All seems simple conceptually. In fact my understanding is that "A coordinate system can flow across a homeomorphism."

I know these things are called charts, but where my knowledge ends is how you deal with the overlaps. If U, U' \subset S, what happens if the h's don't agree?

Ok well if you have the patience, this is pretty much what I know about this. Then at the other end, I do almost grok the universal construction of the tensor product and I am working through calculating it for multilinear forms on the reals. This is by the way a very special case compared to the algebraic viewpoint of looking at modules over a ring. In the latter case you don't even have a basis, let alone a nice finite one. So anything involving finite-dimensional vector spaces can't be too hard :-)

Edited by wtf, 17 February 2017 - 03:15 AM.

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#24 Xerxes

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Posted 17 February 2017 - 07:24 PM

For a given set S, various topologies can be put on it.

Yes, this is true.

For example if T = \mathcal P(S) then every set is open. That's the discrete topology.

Yes, but every set is also closed
 

Or suppose T = \{\emptyset, S\}. This is called the indiscrete topology. No sets are open except the empty set and the entire space.

Yes, but they are also closed, and they are the only elements in this topology.

I had rather hoped I wouldn't have to get into the finer points of topology, but I see now this is unavoidable - and not just as a consequence of the above.

If we want a "nice" manifold, we prefer that it be connected and have a sensible separation property, say the so-called Hausdorff property.

As to the first, I will assert - I am not alone in this! - that a topological space is connected if and only if the only sets that are both open and closed are the empty set and the space itself. The discrete topololgy cleary fails this test.

I will further assert that, if a topological space M has the Hausdorff property and there exist open sets U,\,\,V with, say x \in U,\,\,y \in V then if and only if U \cap V = \O then I may say that x \ne y
The indiscrete (or concrete) topology fails this test.

So these 2 topologies, while they undeniably exist, will be of no interest to us
 

I guess I think of manifolds as basically Euclean spaces twisted around in various ways. Spheres and torii.

Be careful. Euclidean space has a metric, so do spheres and tori. We do not have one so far - so we do not have a geometry i.e. a shape
 

But I hope we're not going to have to go through the chain rule and implicit function theorem and all the other machinery of multivariable calculus,

Point taken, I will try to be as intuitive as I can (though it's not really in my nature)

Later.....
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#25 Xerxes

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Posted 18 February 2017 - 02:58 PM

Let just, by way of clarification, that the Hausdorff property I just referred to is NOT transitive.

Specifically, if I have 3 points x,\,y,\,z with x \ne y and y \ne z by the Hausdorff property I gave, and writing U_x for some open set containing x etc, then by definition
U_x \cap U_y = \O and U_y \cap U_z = \O, but this does NOT imply that U_x \cap U_z = \O.

But of course if I want x \ne z then I must find new open sets, say V_x,\,V_z such thatV_x \cap V_z = \O.

Clearly x \in V_x,\, x \in U_x but then V_x \ne U_x.

As a consequence, for the point m \in M (our manifold) with coordinates x^1,x^2,....,x^n we may extend these coordinates to an open set U_m \subsetneq M.

Then U is called a ccordinate neighbourhood (of m). Or just a neighbourhood.

Edited by Xerxes, 18 February 2017 - 03:35 PM.

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#26 wtf

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Posted 18 February 2017 - 07:52 PM

I must say that this use of the term Hausdorff is quite different from what I've learned about the term. In my understanding, asking if that property is transitive is meaningless.

A topological space is Hausdorff if it separates points by open sets. That is, given any two points x, y, there are open sets U_x, U_y with x \in U_x, y \in U_y, and U_x \cap U_y = \emptyset.

For example the real numbers with the usual topology are Hausdorff; the reals with the discrete topology are Hausdorff; and the reals with the indiscreet topology are not Hausdorff.

I confess I have no idea what it means for the Hausdorff property to be transitive. It's not a binary relation. It's a predicate on topological spaces. Given a topological space, it's either Hausdorff or not. It would be like asking if the property of being a prime number is transitive. It's a meaningless to ask the question because being prime is a predicate (true or false about any individual) and not a binary relation.

Given a pair of points, they are either separated by open sets or not. Of course for each pair of points you have to find a new pair of open sets, which is what I think you are saying.

Historical note. Felix Hausdorff was German mathematician in the first half of the twentieth century. In 1942 he and his family were ordered by Hitler to report to a camp. Rather than comply, Hausdorff and his wife and sister-in-law committed suicide. https://en.wikipedia...Felix_Hausdorff

Edited by wtf, 18 February 2017 - 08:03 PM.

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#27 Xerxes

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Posted 20 February 2017 - 06:23 PM

So you don't like my use of the term "transitive". I can live with being wrong aboutthat.

Let's move on to the really interesting stuff, closer to the spirit of he OP (remember it?)

The connectedness property mandates that, every m \in M there exist at least 2 overlapping coordinate neighbourhoods containing m. I write  m \in U \cap U'.

So suppose the coordinates (functions) in U are \{x^1,x^2,....,x^n\} and those in U' are \{x'^1,x'^2,....,x'^n\} and since these are equally valid coordinates for our point, we must assume functional relation between these 2 sets of coordinates.

For full generality I write

f^1(x^1,x^2,....,x^n)= x'^1
f^2(x^1,x^2,....,x^n) = x'^2
..............................
f^n(x^1,x^2,....,x^n)= x'^n

Or compactly f^j(x^k)=x'^j.

But since the numerical value of each x'^j is completely determined by the f^j, it is customary to write rhis as x'^j= x'^j(x^k), as ugly as it seems at fist sight*.

This is the coordinate transformation U \to U'. And assuming an inverse, we will have quite simply x^k=x^k(x'^h) for U' \to U

Notice I have been careful up to this point to talk in the most general terms (with the 2 exceptions above). Later I will restrict my comments to a particular class of manifolds


* Ugly it may be, but it simplifies notation in the calculus.
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#28 wtf

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Posted 20 February 2017 - 11:30 PM

So you don't like my use of the term "transitive".


You are using the term in a highly nonstandard way and your exposition is unclear on that point.

 

Let's move on to the really interesting stuff, closer to the spirit of he OP (remember it?)


Very much so. I'm interested in why differential geometers and physicists are so interested in using dual spaces in tensor products when the algebraic definition says nothing about them. The current exposition of differential geometry is very interesting to me but not particularly relevant (yet) to tensor products.
 

The connectedness property mandates that, every m \in M there exist at least 2 overlapping coordinate neighbourhoods containing m. I write  m \in U \cap U'.


I hope I may be permitted to post corrections to imprecise statements, in the spirit of trying to understand what you're saying. The indiscreet topology is connected but each point is in exactly one open set. Perhaps you need the Hausdorff property. Again not being picky for the sake of being picky, but for my own understanding. And frankly to be of assistance with your exposition. If you're murky you're murky, I gotta call it out because others will be confused too.

I'm still digesting the rest of your post.

Edited by wtf, 20 February 2017 - 11:32 PM.

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#29 wtf

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Posted 21 February 2017 - 05:30 AM

Are you talking about the transition maps? I'm working through that now. The Wiki page is helpful. https://en.wikipedia.org/wiki/Manifold
 
ps ... Quibbles aside I'm perfectly willing to stipulate that the topological spaces aren't too weird. Wiki says they should be second countable and Hausdorff. Second countable simply means there's a countable base. For example in the reals with the usual topology, every open set is a union of intervals with rational centers and radii. There are only countably many of those so the reals are second countable.
 
Interestingly Wiki allows manifolds to be disconnected. I don't think it makes a huge difference at the moment. I can imagine that the two branches of the graph of 1/x are a reasonable disconnected manifold. 

Edited by wtf, 21 February 2017 - 06:20 AM.

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#30 Xerxes

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Posted 21 February 2017 - 06:13 PM

I'm interested in why differential geometers and physicists are so interested in using dual spaces in tensor products when the algebraic definition says nothing about them.

We will get to that in due course (and soon). It has to do with the difference between Euclidean geometry (algebra) and non-Euclidean geometry (diff. geom.)

The current exposition of differential geometry is very interesting to me but not particularly relevant (yet) to tensor products.

I will make it so, I hope. Again in due course
 
 

I hope I may be permitted to post corrections to imprecise statements,

You are not only permitted, you are encouraged to do so.

The indiscreet topology is connected but each point is in exactly one open set. Perhaps you need the Hausdorff property.

You do. In my defence, I explicitly said in an earlier post that manifolds with the discrete and indiscrete topologies, being respectively not connected and not Hausdorff, were of no interest to us. But yes, I should have reiterated it. Sorry.
 

Again not being picky for the sake of being picky, but for my own understanding. And frankly to be of assistance with your exposition.

No, you are quite right to correct me if I am unclear or wrong. I welcome it
 

I'm still digesting the rest of your post.

Take your time - the incline increases from here on!
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#31 wtf

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Posted 22 February 2017 - 04:59 AM

I'm replying to your post #27 which said ...
 

Let's move on to the really interesting stuff ...


I commented on the first half earlier. Now to the rest of it.

First there's a big picture, which is that if we have a manifold M and a point m \in M, then we may have two (or more) open sets U, U' \subset M with m \in U \cap U'. So m has two different coordinate representations, and we can go up one and down the other to map the coordinate representations to each other.

My notation in what follows is based on this excellent Wiki article, which I've found enlightening.

https://en.wikipedia...Transition_maps

The notation is based on this picture.

Screen_shot_2017_02_21_at_8_29_43_PM.png

We have two open sets U_\alpha, U_\beta \subset M with corresponding coordinate maps \varphi_\alpha : U_\alpha \rightarrow \mathbb R^n and \varphi_\beta : U_\beta \rightarrow \mathbb R^n. I prefer the alpha/beta notation so I'll work with that.

Also, as I understand it the coordinate maps in general are called charts; and the collection of all the charts for all the open sets in the manifold is called an atlas.

If m \in U_\alpha \cap U_\beta then we have two distinct coordinate representations for m, and we can define a transition map \tau_{\alpha, \beta} : \mathbb R^n \rightarrow \mathbb R^n by starting with the coordinate representation of m with respect to U_\alpha, pulling back (is that the correct use of the term?) along \varphi_\alpha^{-1}, then pushing forward (again, is this the correct usage or do pullbacks and pushforwards refer to something else?) along \varphi_\beta.

So we define \tau_{\alpha, \beta} = \varphi_\beta \varphi^{-1}_\alpha . Likewise we define the transition map going the other way, \tau_{\beta, \alpha} = \varphi_\alpha \varphi^{-1}_\beta .

I found it helpful to work through this before tackling your notation.

 

So suppose the coordinates (functions) in U are \{x^1,x^2,....,x^n\} and those in U' are \{x'^1,x'^2,....,x'^n\} and since these are equally valid coordinates for our point, we must assume functional relation between these 2 sets of coordinates.


Now I feel equipped to understand this.

We have m \in U_\alpha \cap U_\beta. Then I can write

\varphi_\alpha(m) = (\alpha^i) and \varphi_\beta(m) = (\beta^i), with the index in both cases is the n in \mathbb R^n. I don't think we talked about the fact that the dimension is the same all over but that seems to be part of the nature of manifolds.

Question: You notated your ordered n-tuple with set braces rather than tuple-parens. Is this an oversight or a feature? I can't tell. I'll assume you meant parens to indicate an ordered n-tuple.

Also you referred to the coordinates as functions, and you did that earlier as well. I'm a little unclear on what you mean. Certainly for example \alpha_i = \pi_i \varphi_\alpha(m), in other words the i-th coordinate with respect to \varphi_\alpha is the i-th projection map composed on \varphi_\alpha.

Are you identifying each coordinate with its respective projection map? That's perfectly sensible. You probably said that earlier.


 

For full generality I write

f^1(x^1,x^2,....,x^n)= x'^1
f^2(x^1,x^2,....,x^n) = x'^2
..............................
f^n(x^1,x^2,....,x^n)= x'^n


Aha. This took me a while to sort out. What is f^i? Putting all this in my notation, we have

f^i(\alpha^1, \alpha^2, \dots, \alpha^n) = \beta^i.

So we seem to be starting with the \alpha-coordinates of m, using the transfer map \tau_{\alpha,\beta} to get to the corresponding \beta-coordinates; then taking the i-th coordinate via the i-th projection map.

Therefore we must have f^i = \pi_i \tau_{\alpha,\beta} = \pi_i \varphi_\beta \varphi_\alpha^{-1}.

As far as I can tell this is the equation that relates your notation to mine. Have I got this right?



 

Or compactly f^j(x^k)=x'^j.


I undersand that. But note that it's ambiguous. Does f^j act on the real number x^k? No, actually it acts on the n-tuple (x^k)_{k=1}^n. So if we are pedants (and that's a good thing to be when we are first learning a subject!) it is proper to write f^j((x^k)_{k=1}^n). Whenever we see f^j(x^k) we have to remember that we are feeding an n-tuple into f^j, and not a real number.
 

But since the numerical value of each x'^j is completely determined by the f^j, it is customary to write rhis as x'^j= x'^j(x^k), as ugly as it seems at fist sight*.


This is very interesting. Let me say this back to you. m has \beta-coordinates (\beta^i). And now what I think you are saying is that we are going to identify the coordinate \beta^i with the map f^i = \pi_i \varphi_\beta \varphi_\alpha^{-1}. Is that right? We identify each \beta-coordinate with the process that led us to it! Very self-referential :-)

This is what I understand you to be saying, please confirm.

ADDENDUM: No I no longer understand this. f^i doesn't play favorites with some particular \beta^i. It makes sense to say that f^i maps \varphi_\alpha(m) to the i-th coordinate of \varphi_\beta(m). But it's a different f^i for each m.

I think I am confused. I should sort this out before I post but I'll just throw this out there.
 

This is the coordinate transformation U \to U'. And assuming an inverse, we will have quite simply x^k=x^k(x'^h) for U' \to U


Ok I had to think about this. Two points.

* Each f^i is a map from \mathbb R^n to the reals. It inputs an n-tuple that is the \alpha-representation of a point m; and outputs a single real number, the i-th coordinate of the \beta-representation of m.

So the only way to make sense of what you wrote is to that the the collection of all the f^i 's are the coordinate transformations.

Actually what I understood from the Wiki article is that the transfer maps were the coordinate transformations. So maybe I'm confused on this point. Can you clarify?

* There's actually a little swindle going on with \varphi_\alpha. At first it was a map from U to some open subset of \mathbb R^n. But in order to pull back along \varphi_\alpha^{-1} we have to restrict the domain to the image \varphi_\alpha(U_\alpha \cap U_\beta). So we don't really have a map from U to U' in your notation; but only from their intersection to itself.

Can you clarify?
 

Notice I have been careful up to this point to talk in the most general terms (with the 2 exceptions above). Later I will restrict my comments to a particular class of manifolds


It doesn't seem to matter at this point what the topological conditions are. It's all I can do to chase the symbols.

 

* Ugly it may be, but it simplifies notation in the calculus.


I think I'm with you so far. Just the questions as indicated. Two key questions:

* How the transition maps can be said to be from U to U' when in fact they're only defined from the \alpha and \beta images, respectively, of the intersection. I'm just a little puzzled on this.

* Your notation x'^j= x'^j(x^k). First I thought I understood it and now I've convinced myself x'^j depends on m.

* And now that I think about it, the transition maps are from Euclidean space to itself, they're not defined on the manifold.

I'm more confused now than when I started working all this out.

Edited by wtf, 22 February 2017 - 05:00 AM.

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#32 wtf

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Posted 23 February 2017 - 03:26 AM

I think I understand what you're saying. In my notation, you are using \beta^i as both the value of the i-th coordinate of the \beta-representation of some point m \in U_\alpha \cap U_\beta; and also as the function \pi_i \varphi_\beta \varphi_\alpha^{-1} that maps the \alpha-representation of some point m to the i-th coordinate of the \beta-representation of m.

That's how I'm understanding this. You're taking the i-th coordinate to be both the function and the specific value for a given m. It's a little bit subtle. The REAL NUMBER \beta^i changes as a function of m; but the FUNCTION \beta^i does not.

Is that right? I want to make sure I'm nailing down this formalism.

Secondly I believe that you are a little confusing or inaccurate when you say the transfer maps (without the extra projection at the end) go from U to U'. Rather the transition maps go from \varphi_\alpha(U_\alpha \cap U_\beta) to \varphi_\beta(U_\alpha \cap U_\beta) and back.

Since the charts are homeomorphisms so are the transfer maps in both directions. And I've read ahead on Wiki and a couple of DiffGeo texts I've found, and I see that if the transfer maps are differentiable or smooth then we call the manifold differentiable or smooth. That makes sense. We already know how to do calculus on Euclidean space.

So I'm a litle confused again ... the charts themselves don't have to be differentiable or smooth as long as the transfer maps (on the restricted domain) are. Is that correct? So for example the charts could have corners outside the areas of overlap? Perhaps you can help me understand that point.
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#33 Xerxes

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Posted 23 February 2017 - 10:22 PM

Wow wtf, so much to respond to. Sorry for the delay but we have been without power until now. The best I can do for now is to re-iterate my earlier post in a slightly different form

I am aware that, on forums such as this it is considered a hanging offence to disagree with the sacred Wiki, so let us say I have confused you. Specifically FORGET the term "transition function". But I am quite willing to use the Wiki notation, as you say you prefer it......

So.

We have 3 quite different mappings in operation here. The first is our homeomorphism: given some open set U \subsetneq M that \varphi:U \to R^n. Being a homeomorphism it is by definition invertible.

Suppose there exist 2 such open sets, say U_\alpha,\,\,U_\beta with U_\alpha \cap U_\beta \ne \O. In fact suppose the point m \in U_\alpha \cap U_\beta, so that \varphi_\alpha:U_\alpha \to V \subseteq R^n and \varphi_\beta:U_\beta \to W \subseteq R^n.

So the composite function \varphi_\beta \circ \varphi_\alpha^{-1} \equiv \tau_{\alpha,\beta}:V \to W \in R^n. One calls this an "induced mapping" (but no, \varphi_\alpha^{-1} is not a pullback, it's a simple inverse)

Your Wiki calls this a transition, I do not. So let's forget the term.

But note that single points in V,\,\,W are Real n-tuples, say (\alpha^1,\alpha^2,....\alpha^n) and (\beta^1,\beta^2,....,\beta^n), so that image of \tau_{\alpha,\beta}((\alpha^1,\alpha^2,....,\alpha^n))= (\beta^1,\beta^2,....\beta^n)

So the second mapping I defined as: for the point m \in U_\alpha , say, the image under \varphi_\alpha is the n-tuple (\alpha^1,\alpha^2,....,\alpha^n) likewise for \varphi_\beta(U_\beta)= (\beta^1,\beta^2,....,\beta^n) then there always exist projections \pi_\alpha^1(\alpha^1,\alpha^2,....,\alpha^2)= \alpha^1 and so on, likewise for the images under \pi_\beta^j of the n-tuple (\beta^1,\beta^2,....,\beta^n).

Note that since the \alpha^j, say, are Real numbers this is a mapping R^n \to R.

So the composite mapping (function) \pi_\alpha^j \circ \varphi_\alpha \equiv x^j is a Real-valued mapping (function) U_\alpha \to R and the n images under this mapping of m \in U_\alpha is simply the set \{\alpha^1,\alpha^2,....,\alpha^n\} and the images under this mapping of m \in U_\beta is the set \{\beta^1,\beta^2,....,\beta^n\} so that x^j(m) = \alpha^j and x'^k(m) = \beta^k

The x^j,\,x'^k are coordinate functions, or simply coordinates

The coordinate transformations I referred to are simply mappings from \{x^1,x^2,....,x^2\} \to \{x'^1,x'^2,....,x'^n\}, they map (sets of) coordinates (functions) to (sets of) coordinates (functions) if and only if they refer to the same point in the intersection of 2 open sets. This mapping is multivariate - that is, it is NOT simply the case that say f^1(x^1)=x'^1 rather f^1(x^1,x^2,....,x^2)=x'^1.

Note that the argument of f^j is a set, not a tuple, appearances to the contrary

I hope this helps.

It also seems I may have confused you slightly with my index notation - but first see if the above clarifies anything at all.

P.S I am generally very careful with my notation. In particular I will always b careful to distinguish a tuple from a set

Edited by Xerxes, 23 February 2017 - 10:24 PM.

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#34 wtf

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Posted 24 February 2017 - 12:01 AM

Sorry for the delay but we have been without power until now. The best I can do for now is to re-iterate my earlier post in a slightly different form


Sorry about your power loss but the recent pace is fine for me. It might have been me who pulled the plug :-)

 

I am aware that, on forums such as this it is considered a hanging offence to disagree with the sacred Wiki, so let us say I have confused you. Specifically FORGET the term "transition function".

I'm just grasping at straws to follow your posts. FWIW here is a screen shot from Introduction to Differential Geometry by Robbin and Salamon. This is from page 59 of this pdf. https://people.math....NTS/diffgeo.pdf

Screen_shot_2017_02_23_at_3_19_08_PM.png

They use the term transition map exactly as I've used it. But no matter, we can call them something else. But it's clear what they are, you are in agreement even if you prefer to use a different name.

 

We have 3 quite different mappings in operation here. The first is our homeomorphism: given some open set U \subsetneq M that \varphi:U \to R^n. Being a homeomorphism it is by definition invertible.

Suppose there exist 2 such open sets, say U_\alpha,\,\,U_\beta with U_\alpha \cap U_\beta \ne \O. In fact suppose the point m \in U_\alpha \cap U_\beta, so that \varphi_\alpha:U_\alpha \to V \subseteq R^n and \varphi_\beta:U_\beta \to W \subseteq R^n.

So the composite function \varphi_\beta \circ \varphi_\alpha^{-1} \equiv \tau_{\alpha,\beta}:V \to W \in R^n. One calls this an "induced mapping" (but no, \varphi_\alpha^{-1} is not a pullback, it's a simple inverse)

Your Wiki calls this a transition, I do not. So let's forget the term.

Ok. I agree with all your notation so far. As I say it took me the duration of your power outage for all this to become clear so feel free to pretend the power's out as I work to absorb subsequent posts.
 

But note that single points in V,\,\,W are Real n-tuples, say (\alpha^1,\alpha^2,....\alpha^n) and (\beta^1,\beta^2,....,\beta^n), so that image of \tau_{\alpha,\beta}((\alpha^1,\alpha^2,....,\alpha^n))= (\beta^1,\beta^2,....\beta^n)

Yes, entirely clear.
 

So the second mapping I defined as: for the point m \in U_\alpha , say, the image under \varphi_\alpha is the n-tuple (\alpha^1,\alpha^2,....,\alpha^n) likewise for \varphi_\beta(U_\beta)= (\beta^1,\beta^2,....,\beta^n) then there always exist projections \pi_\alpha^1(\alpha^1,\alpha^2,....,\alpha^2)= \alpha^1 and so on, likewise for the images under \pi_\beta^j of the n-tuple (\beta^1,\beta^2,....,\beta^n).

Perfectly clear.
 

Note that since the \alpha^j, say, are Real numbers this is a mapping R^n \to R.

So the composite mapping (function) \pi_\alpha^j \circ \varphi_\alpha \equiv x^j is a Real-valued mapping (function) U_\alpha \to R and the n images under this mapping of m \in U_\alpha is simply the set \{\alpha^1,\alpha^2,....,\alpha^n\} and the images under this mapping of m \in U_\beta is the set \{\beta^1,\beta^2,....,\beta^n\} so that x^j(m) = \alpha^j and x'^k(m) = \beta^k

Yes.
 

The x^j,\,x'^k are coordinate functions, or simply coordinates

Ok so we are identifying the coordinates with the projection mappings composed on the charts that produce them.
 

The coordinate transformations I referred to are simply mappings from \{x^1,x^2,....,x^2\} \to \{x'^1,x'^2,....,x'^n\}, they map (sets of) coordinates (functions) to (sets of) coordinates (functions) if and only if they refer to the same point in the intersection of 2 open sets. This mapping is multivariate - that is, it is NOT simply the case that say f^1(x^1)=x'^1 rather f^1(x^1,x^2,....,x^2)=x'^1.

Yes this is clear to me.
 

Note that the argument of f^j is a set, not a tuple, appearances to the contrary

I take this to mean that \{f^j\}_{i=1}^n is a set of maps where f^j = \pi_j \varphi_\beta \varphi_\alpha^{-1}, is that right?
 

I hope this helps.

Yes very much.
 

It also seems I may have confused you slightly with my index notation - but first see if the above clarifies anything at all.

Yes much better. Of course the couple of days I spent working through this in my own mind helped a lot too.
 

P.S I am generally very careful with my notation.


Maybe I should leave that remark alone :-) Let me just say that I sometimes find it productive to work through points of murkiness in your exposition. I'm ready for the next step and do feel free to take this as slowly as you like. Also if you have any particular text you find helpful feel free to recommend it. There are so many different books out there.
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#35 Xerxes

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Posted 24 February 2017 - 10:18 PM

OK, good. We have both worked hard to arrive at a very simple conclusion: if a point in our manifold "lives" jointly into 2 different "regions", then it is entitled to 2 different coordinate representations, and these must be related by a coordinate transformation.

I will say this to our nearly 1000 lurkers: you have seen an example of rigourous mathematics at work, far from the hand waving of my simple (but true) statement above.

wtf. I had planned to say more about the finer points of differentiable manifolds, but on reflection have decided to try and get back to the matter at hand - tensors in the context of differential geometry, since geodief stated his interest was started by an attempt to understand the General Theory.

I will say no more tonight as I collided with a bottle wine earlier, causing serious (but temporary) brain damage
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#36 wtf

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Posted 25 February 2017 - 01:47 AM

wtf. I had planned to say more about the finer points of differentiable manifolds, but on reflection have decided to try and get back to the matter at hand - tensors in the context of differential geometry, since geodief stated his interest was started by an attempt to understand the General Theory.


Thanks Xerxes for all your patience.

That is actually my interest too so this direction is perfect for me. My goal is to understand tensors in differential geometry and relativity at a very simple level, but sufficient to understand the connection between them and the tensor product as defined in abstract algebra.

In fact lately I've been finding DiffGeo texts online and flipping to their discussion of tensors. Sometimes it's similar to what I've seen and other times it's different. It's all vaguely related but I think it will all come together for me if I can see an actual tensor in action. And if it's the famous metric tensor of relativity, I'll learn some physics too. That's a great agenda.

That's what I meant the other day when I said I hoped we didn't have to slog through the calculus part. I don't want to have to do matrices of partial derivatives and the implicit function theorem and all that jazz, even if it's the heart of the subject. I just want to know what the metric tensor in relativity is and be able to relate it to the tensor product. Partial derivatives make my eyes glaze over even though I've taken multivariable calculus and could explain and compute them if I had to.

Along the way, maybe I'll figure out where the duals come from. Because with or without the duals you get the same tensor product; but the duals are regarded as important in relativity. That's the part I'm missing ... why we care about the duals when they're not needed in the definition of tensor product.


 

I will say no more tonight as I collided with a bottle wine earlier, causing serious (but temporary) brain damage


Was that collision between the glass container and your skull? Or of the wine molecules with your brain cells? Or did you use the latter to mitigate the effects of the former?

Edited by wtf, 25 February 2017 - 01:59 AM.

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#37 Xerxes

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Posted 3 March 2017 - 07:08 PM

Partial derivatives make my eyes glaze over


I am very sorry to hear that. I cannot at present see how to proceed without a lot of it. Differential geometry - yes even in the bastard version that physicists use - involves a lot of partial derivatives.

I have been quiet here recently as I have been working overseas. Home tomorrow, when I will try to work out a strategy
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#38 wtf

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Posted 3 March 2017 - 09:44 PM

I am very sorry to hear that. I cannot at present see how to proceed without a lot of it. Differential geometry - yes even in the bastard version that physicists use - involves a lot of partial derivatives.

I have been quiet here recently as I have been working overseas. Home tomorrow, when I will try to work out a strategy


I'm perfectly happy to have some "character building opportunities" as they say :-) Partial differentiate away. No hurry on anything.

ps -- In case I'm being too oblique ... just write whatever you want and I'll work through it.

Edited by wtf, 4 March 2017 - 01:34 AM.

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#39 Xerxes

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Posted 5 March 2017 - 10:40 PM

I'm perfectly happy to have some "character building opportunities" as they say :-) Partial differentiate away.

OK, time to "man up" all readers.

First the boring bit - notation. One says that a function is of class C^0 if it is continuous. One says it is of class C^1 if it is differentiable to order 1. One says it is of class C^{\infty} if it is differentiable to all imaginable orders, in which case one says it is a "smooth function". I denote the space of all Real C^{\infty} functions at the point m \in M by C^{\infty}_m

So recall from elementary calculus that, given a C^1 function f:\mathbb{R} \to \mathbb{R} with a \in \mathbb{R} then \frac{df}{da} is a Real number.

Recall also that this can be interpreted as the slope of the tangent to the curve f(a) vs a.

Using this I make the following definition:

For any point m \in U \subsetneq M with coordinates (functions) x^1,x^2,....,x^n then I say a tangent vector at the point m \in U \subsetneq M is an object that maps C^{\infty}_m \to \mathbb{R} so that, for any f \in C^{\infty}_m and since m = \{x^1,x^2,...,x^n\} we may write v=\frac{\partial}{\partial x^1}f + \frac{\partial}{\partial x^2}f+....+\frac{\partial}{\partial x^n}f.

Or more succinctly as v= \sum\nolimits^n_{j=1} \frac{\partial}{\partial x^j}f \in \mathbb{R}.

As an illustration, recall the mapping (homeomorphism) h:U \to R^n where h(m)=(u^1,u^2,....,u^n)\in R^n and the projections \pi_1((u^1,u^2,....,u^n))=u^1 \in \mathbb{R} and so on. Recall also I defined the coordinate functions in U \subsetneq M by x^j= \pi_j \circ h so the x^j really are functions.

So I may have that \frac{\partial}{\partial x^j}x^k= \delta^k_j where \delta^k_j = \begin{cases}1\quad j=k\\0\quad j \ne k\end{cases}.

So in fact, since this defines linear independence, we may take the \frac{\partial}{x^h} to be a basis for a tangent vector space. At the point m \in U \subsetneq M one calls this as T_mM

Good luck!
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#40 wtf

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Posted 6 March 2017 - 03:57 AM

Good luck!


<Star Trek computer voice> Working ...

Actually I read through it and it looks pretty straightforward. I'll work through it step by step but I didn't see anything I didn't understand. The tangent space is an n-dimensional vector space spanned by the partials. I understand that, I just need practice with the symbology.

I see at the end you bring in the Kronecker delta. This is something I'm familiar with as a notational shorthand in algebra. I've heard that it's a tensor but at the moment I don't understand why. I can see that by the time I work through your post I'll understand that. This seems like a fruitful direction for me at least.

Edited by wtf, 6 March 2017 - 04:03 AM.

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