Dimension confusion with k-forms...!

Higher-dimensional geometry (previously "Polyshapes").

Dimension confusion with k-forms...!

Postby Paul » Tue Dec 07, 2004 2:37 pm

Hello Pat and all,

I hope someone can help clarify what I'm reading in this new book I got... Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach. It's a good book, although there's allot of errata. However, I checked... there's nothing in the errata for these two pages.

It has to do with dimension stuff again... perhaps it has something to do with the concepts of dimension of a vector space and 'grade' of a form,... 'grade' as one might use in Clifford Algebra...?

Also... when the authors say such and such is a vector space of dimension 1, 2, etc., does that mean that one can interpret the vector space geometrically as being a 1D line, a 2D plane, etc. ...?

I know they're applying the "n choose k" formula, but I don't understand what this is all telling us about the geometry of these spaces...?

The yellow highlightings connect a statement concerning 'dimension' and the example it references. The purplish highlightings connect a statement concerning 'dimension' and an example it... might... be referencing... I'm not sure about the connection between the example and statement connected by the purplish highlighting...

Image

Image
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Postby Paul » Tue Dec 07, 2004 10:12 pm

Hello again Pat and all,

I think I'll post this in this thread since this thread concerns dimensionality issues.

Concerning Hodge duality, grades, and dimension of pure k-vectors... all associated with application of Hodge duality.

If I understand correctly these passages from Ludwig Eckhart's Four-Dimensional Space it appears the Hodge dualities that I'm familiar with, those in 3 and 4 space have an equal(?) infinity of objects paired for duality. That is, for instance, there's Image<sup>3</sup> 1-vectors of 3-space, and there are also Image<sup>3</sup> 2-vectors of 3-space...

p. 22:
Image


I included the part after the yellow highlighting because I'm not certain... but, I believe the number of 1-vectors in 3-space corresponds to the number of points of 3-space... all these 1-vectors can be assumed to have their tails anchored at the origin... then, extending rays to all the points of 3-space, including the origin (the null vector), will give you all the 1-vectors of 3-space.

Although there are other 1-vectors which can be formed which would not have their tails anchored at the origin, these are just translations of the 1-vectors whose tails are anchored at the origin. (prop. 1)

I included the rest of the quote beyond the yellow highlighting in case my logic here is incorrect. So first, I need to ask if the logic of prop. 1 is correct, and in fact, we can state that there are Image<sup>3</sup> 1-vectors of 3-space...?

Then, combining this with the above quote we see that there are also Image<sup>3</sup> 2-vectors of 3-space.

Further, on p. 33, Ludwig Eckhart states:

Image


Again, assuming prop. 1 is true, I think it easily follows that prop. 1 will be true for any R<sup>n</sup>, not just R<sup>3</sup>... so, I wish to similarly assert that there are Image<sup>4</sup> 1-vectors of 4-space...?

Then, combining this with the last quote we see that there are also Image<sup>4</sup> 3-vectors of 4-space.

And those are the two 1-vector operand cross products I've derived. I'm also aware that there's something of a trivial cross product in 4-space that maps 2-vectors to 2-vectors, but I've not derived that one.


Now, observations, questions...
    (ques. 1) Is it true, as it appears from the above, that the number(?) of distinct k-forms of R<sup>n</sup> is Image<sup>n choose k</sup>, that is, the order of infinity is simply the binomial coefficient?
    (ques. 2) Is it necessary that in order to form the Hodge dual between two spaces of j- and k-forms in R<sup>n</sup> that there must be an equivalence in the number(?) of distinct j- and k-forms of R<sup>n</sup>?
    (ques. 3) If ques. 2 is true, does this imply that the Hodge dual sets up a one-to-one correspondence between distinct j- and k-forms of R<sup>n</sup>?


PS Is there some easier way to get the infinity symbol other than pasting an image of it into the post?
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Re: Dimension confusion with k-forms...!

Postby pat » Wed Dec 08, 2004 7:50 pm

Paul wrote:It has to do with dimension stuff again... perhaps it has something to do with the concepts of dimension of a vector space and 'grade' of a form,... 'grade' as one might use in Clifford Algebra...?


Yes, the 'grade' of a form is just like 'grade' in Clifford Algebras.

Also... when the authors say such and such is a vector space of dimension 1, 2, etc., does that mean that one can interpret the vector space geometrically as being a 1D line, a 2D plane, etc. ...?


You have to be careful here. The answer is basically 'yes' if your scalars are real numbers (as opposed to integers or complex numbers or what-have-you). But, it may not mean geometrically the kind of thing you want it to mean. Particularly, take the Clifford Algebra Cl<sub>3,0</sub>. It forms an eight-dimensional vector space, but its 1-vectors are three-dimensional.

In other areas like topology, you end up being able to do things like form a seven-dimensional vector space (technically, a module I suppose, but they're almost the same thing... just that the scalars don't have to be commutative or have inverses... and these topological ones have integer scalars) out of an oriented triangle. So, the vectors are the three vertexes, the three sides, and the whole triangle. Now, you could form a vector subspace out of two vertexes and one side. Technically, this is a three-dimensional subspace. You can try to picture it "geometrically". But, the "geometry" of this vector space doesn't look anything like the Euclidean geometry of a triangle in a plane.

In the yellow-highlighted section of your first post, they start off saying that V is a plane embedded in R<sup>3</sup>. So, you'll be thinking at the same time in terms of the three-dimensional geometry of this plane sitting in three-space and the two-dimensional geometry of the plane.

The line between your yellow-highlighted line and the magenta-highlighted line is very key. "We saw that the three restricted elementary 2-forms ... are all multiples of each other." What this is saying to me is that there's really only one 2-form... really only two dimensions left if we restrict ourself to the plane. This is as it should be. Now, if you think of a 2-form's coefficient as sort of telling you how big some parallelogram of planar area is, then what this is telling you is that if you restrict yourself to the plane V, then the only real choice you have is 'how big' (with a negative being the flipped over version of the corresponding positive).

If you hadn't restricted yourself to this plane, then you'd have more freedom than that. You would have control over the size of its projections on each of the fundamental planes (based on how you oriented it in three-space).

The dimension of a vector space is the number of degrees of freedom that you have in that vector space. A typical way that it's described is 'the number of elements in a basis (a minimal spanning set).' If A<sup>2</sup>(V) is one-dimensional, then you know that all of the possible 2-forms are just scalar multiples of each other. If A<sup>2</sup>(V) is two-dimensional, then you know that all of the possible 2-forms are expressible as a combination ax + by of two fixed 2-forms x and y and some scalars a and b. Now, you can think of that as a plane if you like. But, since x and y are already 2-forms, then it's not clear exactly what the "geometry" of this plane is. It's a plane but the axises are measured in 2-forms instead of distance units.

Also, I said 'two fixed 2-forms x and y'. You can actually pick any two-forms... so long as one isn't just a scalar multiple of the other... and they will span the space of 2-forms.

People will refer to this space of 2-forms as a 2-dimensional vector space (or k-dimensional or whatever) depending on how many of the 2-forms like 'x' and 'y' one needs to cover every allowed 2-form. In this vector space, 2-forms are the vectors! So, the nomenclature gets pretty incestuous.

A vector space is a very abstract thing. Euclidean space is one example. But, things get pretty wiggy if you try to think of all vector spaces as analogous to Euclidean space.
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Postby pat » Wed Dec 08, 2004 7:54 pm

I should have also mentioned that the fact that you end up with a one-dimensional A<sup>2</sup>(V) when V is a plane (even though the plane is in R<sup>3</sup>) is exactly what we'd hope. The plane is basically equivalent to R<sup>2</sup>. And, there's only one basis 2-vector in R<sup>2</sup>: e1 wedge e2. So, even though we put this plane in R<sup>3</sup>, there's still effectively only one 2-vector.
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Postby Paul » Thu Dec 09, 2004 1:02 am

Hello Pat,

Thanks for your responses.

I'll probably have questions on your responses once I have a little more time to digest them,... but, there's another question about differential forms that's be troubling me...

I've kinda been assuming that with differential forms when they use the notation dx ^ dy that you can consider this to essentially be equivalent to the Clifford Algebra notation e<sub>1</sub> ^ e<sub>2</sub>, assuming that the axes associated with the basis 1-vectors e<sub>1</sub>, e<sub>2</sub> do correspond with the x,y axes in the vector space that both are in...?

I guess I've started to think that dx, dy are the differential from Calculus... and that the change in the unit basis 1-vectors essentially correspond to the first partial derivatives of x and y... dx and dy.

Are these views accurate? They're not likely to lead to some confusion, are they?
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