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Exercise 11.9 Help
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Joined: 29 May 2010, 23:31
Posts: 5
Exercise 11.9 Help
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
Re: Exercise 11.9 Help
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
Re: Exercise 11.9 Help
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
Re: Exercise 11.9 Help
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.

01 Jun 2010, 02:58
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