Re: Definition of a algebra generated by a set ?

On Sun, 10 Jun 2007 14:10:56 -0700, Randy Poe wrote:

On Jun 10, 3:21 pm, Marc Mertens <marc.mert...@xxxxxxxxxxx> wrote:
Hello,

I'm reading a book about mathematical physics and have a lot of
troubles understanding some examples in this book. First they define a
algebra as vector space A over a field K with a product . defined that
satisfies A.(bB+cC)=bA.B+cA.C and (bB+cC).A=bB.A +cC.A, the algebra is
commutative (associative) if the product is associative (commutative).
No problem here. However at one moment they start to give examples
like:
Let Q be the associative algebra over R generated by the four elements
{1,i,j,k} satisfying i^2=j^2=k^2=-1, ij=k,jk=i,ki=j,
1^2=1,1i=i,1j=j,1k=k which leads to the quaternions.

This multiplication table among these four elements defines the algebra,
and in the process defines what "multiplication" will mean for these
four objects. Nothing is implied about how you might represent them.
It's just a set of symbols with a "multiplication" table.

Let V be a real vector space with the inner product u.v and e1,...en a
orthonormal basis, the Clifford algebra associated with the inner
product space is the associative algebra generated by 1,e1,..en with
product rules eiej+ejei=2gij 1 (gij=ei.ej)
Now I understand that you can create the quaternions as a set of tuples
of real numbers with the correct multiplication, but I fail to see how
you can take just 4 elements {1,i,j,k} and then generate a vector space
over a field together with a product (unless you define 1=(1,0,0,0) ,
i= (0,1,0,0), j=(0,0,1,0) and k=(0,0,0,1) and define addition in the
usual way (product is of course a little bit more complex), but you can
hardly call this generating.

That happens to be a representation which, with "multiplication"
suitably defined, gives the same algebra. But the multiplication table
itself is the algebra, regardless of the representation.

I was aware of the fact that my example is one of many algebras that
fullfills the multiplaction table and are isomorphic with my algebra.
Maybe I'm incorrect but the book defined a algebra as a vector space with
a product defined on it (and some distribution properties), does that not
mean that the generated algebra must be a concrete algebra i.e a vector
space etc... (although I agree that the multiplication table implies that
you can have a lot of other algebras that fullfills the same
multiplication rules). I must confess that I'm not well versed in
category theory so maybe this is what I'm missing (or as I mentioned in a
previous post I miss the stuff about free algebras).


Also when I look up Clifford algebras on the internet, It is never
defined as the associative algebra generated by ....

There seems to be a pattern here where given a set of undefined
elements {1,a1,...an) you go on and generate a algebra out of this. So
how do you proceed to define a vector space (definition of the sum,
multiplication by a scalar (what is the field used) and the algebra
product. I looked really hard but could not find such a definition in
the book neither in other books I have on algebra neither on the
internet. Either I'm missing something obvious or my book is based on
some loosy mathematical principles.


PS. The previous sentence is not a critic on the book itself, but I
found that a lot of books about physics do not always use rigorous
mathematical definitions.

I'm not familiar with Clifford algebras, but this looks like group
theory, and you seem to be hung up on the distinction between an
abstract group and a representation of that group (there may be more
than one natural representation). I'd suggest looking under keywords
like "group theory" and "representation", such as here:
http://en.wikipedia.org/wiki/Group_representation

- Randy

Actually the book contains some stuff about representation theory (for
finite groups) but they do not refer to this when talking about Clifford
algebras. I will try to look deeper in this, thanks for the hint.




Thanks a lot for your help

Marc Mertens


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