Cross (?) product in 5-space...?

[Paul]

Hello Pat and all,

I've been trying to find some cross(?) product in 5-space. I found something... I haven't worked out an example yet,... it might be laborious.

In Cl_5,0 of R⁵ (It is proper to say the first space here is 'of' the second?), we have:

1 5 10 10 5 1

for the dimension (in one of the other senses of dimension...) of the various multivector spaces... scalar, vector, bivector,... etc.

I'm trying to find a cross(?) product of some nature between the bivectors and trivectors. I think these two spaces may be isomorphic, and if so a Hodge dual should be able to be set up...? (I think...)

One reason I'm not really sure if the two spaces are isomorphic,... and also perhaps share a 1-to-1 correspondence between elements is because there are non-simple bivectors and trivectors in 5-space. I'm not sure if it's safe to say that there are only going to be ¹⁰ cardinality of elements in each of these spaces. (I think this probably means a direct product 10 's. I'm still not sure about this cardinality stuff... See first reply to original post here) I'm concerned that even if we can expect this cardinality of elements (whatever precisely it might mean, or tell us...) that the non-simple elements might throw this cardinality out of whack... I'm just not sure...?

Anyway... so far this is what I'm come up with. It isn't really a relation between the bivectors or trivectors of 5-space per se, but rather between 3 1-vectors and bivectors... although one question I have is whether this is essentially still between bivectors or trivectors since wedging 3 1-vectors produces a trivector...? However, even if that's essentially accurate, I think it's always going to be a simple trivector, so we'll not get all 5-space trivectors this way.



I think I also found a simpler formula for resolving permutations like these which so often come up in various product calculations... See last reply by me in post Minimum transpositions of permutations...?

I'll summarize my questions here:

Are the spaces of bivectors and trivectors in 5-space isomorphic?

Do these two spaces have the same cardinality of elements? The non-simple elements don't alter the cardinality? What cardinality do the simple elements, non-simple, elements comprise?

Can a Hodge duality be set up between these two spaces?

Does what I've derived here qualify as a 'cross' product?

If yes to the last question, will the resulting bivector here be perpendicular to each of the 3 1-vector operands? Can these 3 1-vector operands also be seen as a trivector?

[Paul]

Hello again Pat and all,

Now I've done one example by hand, and run several with a program I wrote... I believe this is indeed some manner of cross product.

First of all, these are the completed formulas... we the final negations absorbed after reflecting through the pseudoscalar to get the dual bivectors...



I did one example by hand, and things did work out... once I caught an arithmetic error...



I also calculated the inner product for the first of these 1-vectors, vector a, and also derived the formula for the inner product (Hestenes semi-commutative inner product) for a 5-space 1-vector and 2-vector. It's not a general formula... I just derived it to test this cross product derivation.

I haven't transferred my handwritten inner product derivation to MathType yet, so I hope you can still read it...



I wrote a short Perl program to test the cross product. If you're interested, I posted the program to this webpage... http://paul-mccarthy.us/Misc/cross5D.htm

Some confusions might result from reading the code... First of all, this inner product is not, in general, commutative. In this specific case, for a 1-vector and 2-vector in 5-space it's anti-commutative, or anti-symmetric.

What might confuse you with the code is... Well, suffice it to say that sub Sp_inner is a real hack job just to get this product to commute properly... in this specific case only, however. It's not anything for general use... it'll likely just to cause confusion.

Here's some output of the Perl program... The first example is the example I worked by hand, and the second example is just a random one I made up...



For some reason, the Perl interpreter on the server sometimes likes to put a negative sign in front of zeros... hence, the -0's.

In this case, the inner product is 1-vector-valued... that's why there's a chain of 5 zeros.

I think the way to think about the duality part of this is... The 3 1-vectors wedge together to make a trivector, and then we reflect it through the pseudoscalar to come up with the dual of this trivector, which is a bivector... in 5-space.

So, the duality in 5-space that I'm utilizing is between bivectors and trivectors. I could reformulate all this easily enough to take a general trivector and find it's dual bivector... or, vice-versa, take a general bivector and find it's dual trivector.

The reason I didn't do that... at least to start... is because it's easier to work with the 1-vectors. All 1-vectors, as well as (n-1)-vectors, are simple. But, bivectors and trivectors in 5-space are not all simple... that makes the duality formulation much more extensive.

So, in essence, wedging 1-vectors together makes the computations easier, but there can't be a one-to-one correspondence between 1-vectors and trivectors (or bivectors either) in 5-space. So, in this formulation, the duality relationship between bivectors and trivectors is not illuminated. So, reformulating all this to illuminate this duality relationship seems worthwhile.

In 3-space, the 2 1-vectors wedged together to make a bivector... and the one-to-one duality relationship in 3-space is also between 1-vectors and bivectors. But, with this 5D cross product (there are other 5D cross products... one of which, the 1-vector and (n-1)-vector one, has this similar property that I'm saying that the 3D cross product has), the multivector grades having the duality relation and the 1-vectors wedging together to form the cross product... are different. This can't happen in our 3-space. Further, I don't believe it can happen in 4-space either. (In 4-space you have a duality relation, and some cross product, between 1-vectors and trivectors... and bivectors and bivectors. Perhaps wedging two 1-vectors together in 4-space would form a bivector, but than it can only dual with... bivectors. Although... perhaps the bivectors would be orthogonal to the 1-vector operands of the wedge... I'm not sure. But still there can't be a one-to-one duality between bivectors and 1-vectors in 4-space. 1 4 6 4 1...)

Anyway... if anyone has any thoughts, or insights that might correct false impressions I've garnered, or illuminate what's going on here... any input would be appreciated.

[quickfur]

Hello Pat and all,

I've been trying to find some cross(?) product in 5-space. I found something... I haven't worked out an example yet,... it might be laborious.

Have you ever looked at Prof. Andrew J. Hanson's formula for the N-dimensional cross product? You can find it here: http://ftp.cs.indiana.edu/pub/hanson/Siggraph01QuatCourse/ggndgeom.pdf. (Note: acrobat reader is needed.)

I wrote an N-dimensional vector calculator that computes cross products using that formula. The nice thing about his formula is that it generalizes nicely to N dimensions, and always gives you the last basis vector of N-space if you cross the first (N-1) in respective order. (E.g., <1,0,0,0> X <0,1,0,0> X <0,0,1,0> = <0,0,0,1>. Other formulas sometimes give you vectors pointing in the wrong direction for odd/even dimensions.)

[Paul]

Hello quickfur,

That looks like an interesting article... I'll need some time to read and absorb it.

Thanks for the link to the article.

[wendy]

i used to use the name 'cross-product' for what is now in the PG called the tegum-product. The cross-product is one of vectors, while the tegum-product is one of polytopes.

W

[houserichichi]