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Exercise 11.9 Help
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Joined: 29 May 2010, 23:31
Posts: 5
So I know this is supposed to be a simple problem, but I keep working it and not getting anywhere. so by n=2 he means find the wedge product of two "vectors" whos "basis" has dimension 2. I set up the following.
a=xi+yk
b=wi+zk

where w,x,y,z run through the reals, and i,k represent my "basis" for the algebra.

so a^b=(xi+yk)(wi+zk)
which gives us:

a^b = xiwi+xizk+ykwi+ykzk

Now, by the definition of the grassman product xiwi = 0 and ykzk = 0
(This part is a leap for me i assumed i could move the terms communitavely since they will just go to zero)

Now i have

a^b = xizk+ykwi (Now I used the anti-communative laws to make the next line, which I think is wrong)

a^b = ik(xz-yw)

this is as far as I get. Any pointers?
Side note: Anyone know how to make subscripts. that would make this post alot less clustered.

30 May 2010, 00:02 Joined: 26 Mar 2010, 04:39
Posts: 109
I'm not sure what you were trying to do here but I think you are a bit confused about Grassmann algebras.

In a nutshell, if you take any n-dimensional vector space V you can form a Grassmann algebra by defining a special bilinear and associative operation called the wedge product which is defined on basis vectors according to e_i \wedge e_j = -e_j \wedge e_i (unfortunately Latex rendering is broken on the forum at the moment).

This problem was intended to show that the wedge product is well-defined (i.e. that it doesn't matter which basis you chose to define it on).

This allows us to define 1-forms, 2-forms, ... , n-forms, where a p-form is the wedge of a 1-form with a (p-1)-form and a 1-form is just an element of V and we also append the scalars as 0-forms. You'll notice that if you try to create anything more than an n-form you will get 0 since there are only n basis elements in V. You will also notice that for each p from 0 to n, the space of p-forms is also a vector space and has dimension p!(n-p)!/n! and the whole Grassmann algebra is a direct sum of these vector spaces and thus has dimension 2^n.

This problem was solved correctly here: http://www.roadtoreality.info/viewtopic.php?f=19&t=1393

30 May 2010, 05:23 Joined: 26 Mar 2010, 04:39
Posts: 109
Jay,

I had another look at what you were trying to do and I think I can make things a bit clearer for you.

Jayroberts wrote:
So I know this is supposed to be a simple problem, but I keep working it and not getting anywhere. so by n=2 he means find the wedge product of two "vectors" whos "basis" has dimension 2.

This question (in my edition at least) does not ask for the special case n=2. It looks like the numbering of the questions may have changed. In my book the special case for n=2 is asked for in exercise 11.10

Jayroberts wrote:
I set up the following.
a=xi+yk
b=wi+zk

where w,x,y,z run through the reals, and i,k represent my "basis" for the algebra.

so a^b=(xi+yk)(wi+zk)
which gives us:

a^b = xiwi+xizk+ykwi+ykzk

You are on track but it looks you are trying to interpret the elements that have been wedged together as scalars. In fact there is a natural identification between them in the case n=2, but I think it is much clearer if you leave all the wedges in.

With your notation, you should have:
a^b=(xi+yk)^(wi+zk)
= xi^wi+xi^zk+yk^wi+yk^zk (associativity)
= 0 + xz(i^k) + yw(k^i) + 0 (bilinearity)
= (xz-yw)(i^k)

Similarly
b^a=(wi+zk)^(xi+yk)
= (zx-wy)(k^i)
= -(zx-wy)(i^k)
= -a^b

Let me know if this helps.

31 May 2010, 02:16 Joined: 29 May 2010, 23:31
Posts: 5
Firstly sorry about the mix up I meant 11.10

And I see what I did wrong. Well not exactly, but how you wrote it makes sense. And I can see the reasoning. Ya I'm fairly new to advanced mathematics, so this book goes over my head at times, but I'm slowly trying to work through it.  