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
<sup>3</sup> 1-vectors of 3-space, and there are also
<sup>3</sup> 2-vectors of 3-space...
p. 22:
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
<sup>3</sup> 1-vectors of 3-space...?
Then, combining this with the above quote we see that there are also
<sup>3</sup> 2-vectors of 3-space.
Further, on p. 33, Ludwig Eckhart states:
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
<sup>4</sup> 1-vectors of 4-space...?
Then, combining this with the last quote we see that there are also
<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 <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?