From: mjb
Newsgroups: sci.math
Subject: clifford / geometric algebra from 4D vector
Date: Tue, 13 Dec 2005 10:45:21 +0000
I have derived a table of products using 4 basis vectors, e1, e2, e3 and e4.
I have picked certain permutations to be bivectors:
e1^e2, e3^e1, e2^e3, e1^e4, e4^e2 and e3^e4,
and trvectors:
e1^e2^e3, e2^e1^e4, e1^e4^e3 and e2^e3^e4
But are these the best basis to use?
When I draw up a multiplication table of all permutations there are at least
4 apparent anomalies (values which commute when all the others of the same
type anticommute) and things like that. I could not find a way to show the
table in text here, so I have put it on a web page here:
https://www.euclideanspace.com/maths/algebra/clifford/d4/
I have colour coded the values that seem to be anomalies. I would appreciate
your comments about this,
Have I made any errors? (I don't claim to be a mathematician)
Are there any permutations of bivectors and trvectors that I could have
chosen that would not have apparent anomalies?
What bivectors and trvectors do people usually use?
I would like to go on to work out how to use this to represent solid body
mechanics (isometry transforms) and I would appreciate any help/clues about
how to do this but the main purpose of this message is to try to find out
if what I have done so far is correct so that I can build on it?
I would appreciate any thoughts/ideas/help,
Thanks,
Martin
From: "Ross A. Finlayson"
Newsgroups: sci.math
Subject: Re: clifford / geometric algebra from 4D vector
Date: 13 Dec 2005 03:59:55 0800
mjb wrote:
> I have derived a table of products using 4 basis vectors, e1, e2, e3 and e4.
>
> I have picked certain permutations to be bivectors:
> e1^e2, e3^e1, e2^e3, e1^e4, e4^e2 and e3^e4,
> and trvectors:
> e1^e2^e3, e2^e1^e4, e1^e4^e3 and e2^e3^e4
> But are these the best basis to use?
>
> When I draw up a multiplication table of all permutations there are at least
> 4 apparent anomalies (values which commute when all the others of the same
> type anticommute) and things like that. I could not find a way to show the
> table in text here, so I have put it on a web page here:
> https://www.euclideanspace.com/maths/algebra/clifford/d4/
> I have colour coded the values that seem to be anomalies. I would appreciate
> your comments about this,
>
> Have I made any errors? (I don't claim to be a mathematician)
> Are there any permutations of bivectors and trvectors that I could have
> chosen that would not have apparent anomalies?
> What bivectors and trvectors do people usually use?
>
> I would like to go on to work out how to use this to represent solid body
> mechanics (isometry transforms) and I would appreciate any help/clues about
> how to do this but the main purpose of this message is to try to find out
> if what I have done so far is correct so that I can build on it?
>
> I would appreciate any thoughts/ideas/help,
>
> Thanks,
>
> Martin
I don't know.
Hey, how about this, combinatorially enumerate vector bases and product
rules and then winnow those for various properties. Some of them would
be, for example, Clifford algebras and other groups, or Study numbers,
or some Mat(2,C) or something.
Excuse me, others here are more qualified to answer your questions, I
just wanted to comment. What are the products of your bases? Consider
a variety of possible products of the bases, eg 1, 1, e+2 mod 2,
etcetera. Each of them form various structures, some more or less
useless.
On your web page you say "cross multiplication does not apply in 3D" I
think you mean "4D", where the cross product appears in 3D and 7D, and
generally vector multiplication preserves the norm, which for a
1vector or vector of grade 1, odd, is the length of the vector or
distance and for the bivector similar to an area and so on for an
nplane.
I would suggest Pertti Lounesto's book "Clifford algebras and spinors"
and before the chapter on spinors E. Cartan's "Theory of Spinors". I'm
on about chapter four. Those books you reference, Hestenes', are also
very good I have borrowed them before.
Ross
Date: Tue, 13 Dec 2005 21:23:11 +0100
From: JEMebius
Newsgroups: sci.math
Subject: Re: clifford / geometric algebra from 4D vector
Ross A. Finlayson wrote:
>mjb wrote:
>
>>I have derived a table of products using 4 basis vectors, e1, e2, e3
and e4.
>>
>>I have picked certain permutations to be bivectors:
>>e1^e2, e3^e1, e2^e3, e1^e4, e4^e2 and e3^e4,
>>and trvectors:
>>e1^e2^e3, e2^e1^e4, e1^e4^e3 and e2^e3^e4
>>But are these the best basis to use?
>>
>>When I draw up a multiplication table of all permutations there are
at least
>>4 apparent anomalies (values which commute when all the others of the
same
>>type anticommute) and things like that. I could not find a way to
show the
>>table in text here, so I have put it on a web page here:
>>https://www.euclideanspace.com/maths/algebra/clifford/d4/
>>I have colour coded the values that seem to be anomalies. I would
appreciate
>>your comments about this,
>>
>>Have I made any errors? (I don't claim to be a mathematician)
>>Are there any permutations of bivectors and trvectors that I could have
>>chosen that would not have apparent anomalies?
>>What bivectors and trvectors do people usually use?
>>
>>I would like to go on to work out how to use this to represent solid body
>>mechanics (isometry transforms) and I would appreciate any help/clues
about
>>how to do this but the main purpose of this message is to try to find out
>>if what I have done so far is correct so that I can build on it?
>>
>>I would appreciate any thoughts/ideas/help,
>>
>>Thanks,
>>
>>Martin
>
>
>
>I don't know.
>
>Hey, how about this, combinatorially enumerate vector bases and product
>rules and then winnow those for various properties. Some of them would
>be, for example, Clifford algebras and other groups, or Study numbers,
>or some Mat(2,C) or something.
>
>Excuse me, others here are more qualified to answer your questions, I
>just wanted to comment. What are the products of your bases? Consider
>a variety of possible products of the bases, eg 1, 1, e+2 mod 2,
>etcetera. Each of them form various structures, some more or less
>useless.
>
>On your web page you say "cross multiplication does not apply in 3D" I
>think you mean "4D", where the cross product appears in 3D and 7D, and
>generally vector multiplication preserves the norm, which for a
>1vector or vector of grade 1, odd, is the length of the vector or
>distance and for the bivector similar to an area and so on for an
>nplane.
>
>I would suggest Pertti Lounesto's book "Clifford algebras and spinors"
>and before the chapter on spinors E. Cartan's "Theory of Spinors". I'm
>on about chapter four. Those books you reference, Hestenes', are also
>very good I have borrowed them before.
>
>Ross
>
I am not very familiar with Clifford algras, but nevertheless I would
like to make some remarks that may be helpful.
Please start with recalling the general === Laplace expansion theorem
=== for determinants. Look for instance at
http://planetmath.org/encyclopedia/LaplaceExpansion2.html .
In 2D one has only one Laplace expansion, which is just the definition
of a 2ndorder determinant.
In 3D there is also only one Laplace expansion, but this one involves
elements and their associated 2x2 minors (a (1+3) Laplace expansion).
The geometrical interpretation of this expansion goes loosely speaking
as follows: the first two columns yield three 2x2 minors, which are the
components of the cross product of these column vectors. The Laplace
expansion is the dot product of the cross product and the last column,
i.e. the wellknown box product of the three column vectors.
In 4D one has the (1+3) and the (2+2) Laplace expansions. The (1+3)
expansion cosists of four terms, together making the dot product of the
first column and the vector consisting of the four 3rdorder
determinants that arise from the remaining columns of the original
4thorder determinant. Beware however that this dot product is actually
the wedge product of a 1form and a 3form and therefore is
anticommutative.
This expansion was explained as follows in an item that I posted in
sci.math on November 17th 2005 (*).
The (2+2) Laplace expansion consists of six terms, which together from
the (6D) dot product of two bivectors. It takes some patience to make a
drawing showing the 4D parallelepiped as an algebraic sum (disjoint
union with the appropriate signs for the volumes) of the six polytopes
that are generated from origin of coordinates and the (12)(34),
(23)(14) and (31)(24) pairs of 2D facets of the parallelepiped. I did
not yet make this drawing.
I guess that Hermann Grassmann Sr. had exactly this kind of geometrical
interpretations of the Laplace expansion in mind when he constructed his
Ausdehnungslehre (Theory of Extensions).
As regards the anomalies in your multiplication table: recall that in
exterior algebra Pforms and Qforms anticommute if both P and Q are
odd, and commute in all other cases. This is due to the parity of the
permutation which transforms the sequence (1a, 2a, ... Pa; 1b, 2b,
...Qb) into (1b, 2b, ...Qb; 1a, 2a, ... Pa).
So do not be surprised if it appears that some pairs of terms must
commute instead of anticommute in order to obtain a wellbehaved
algebraic system.
Ciao: Johan E. Mebius
(*) (Excerpt from my news:sci.math newsgroup post of November 17th 2005)
==================================================
(...)
To make {A, B, C} into a basis of R^4: just add a vector D that is
linearly independent of A, B, C.
One way to do this is as follows:
write vectors A, B, C down as three column vectors, making together a
3column 4row matrix.
Denote the rows as I, II, III and IV and take
D1 = det (II, III, IV), D2 = det (I, IV, III), D3 = det (IV, I, II) and
D4 = det (III, II, I).
This vector might be denoted as the 4D box product of A, B, C. It is
zero if and only if A, B, C are linearly dependent.
Note the even permutations (2, 1, 4, 3), (3, 4, 1, 2) and (4, 3, 2, 1)
of the initial sequence (1, 2, 3, 4).
Now you are almost ready to absorb the algebra accompanying Einstein's
Special Theory of Relativity.
(...)
==================================================
From: mjb
Newsgroups: sci.math
Subject: Re: clifford / geometric algebra from 4D vector
Date: Thu, 15 Dec 2005 09:40:34 +0000
Thank you very much for your replies, I think you have given me the clues I
need, but due to deficiencies in my maths, it will take me some time to
work through it.
I am very encouraged to do so by the last line: "Now you are almost ready to
absorb the algebra accompanying Einstein's Special Theory of Relativity"
Thanks for pointing out the error, which I have corrected, also (I hope you
don't mind) I have put I copy of your replies here in the hope that they
also help others who come across the webpage:
https://www.euclideanspace.com/maths/algebra/clifford/d4/news1.htm
Cheers,
Martin
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