Re: Differential Forms

I'm not sure but I think that your A and B should be sums over all 1<=i_1<i_2< .... i<p<=n=dimV - I think the summation is often omitted in textbooks so its easier to write down. Then I think when you take the exterior product you should get what you want (a lot of terms will cancel since w^i wedge w^i=0) Hope this is helpful.
g.s.

David Norton wrote:
Hi. I've got a quick question regarding differential forms that I'd
appreciate some help with. Suppose that I have a vector space V and a
dual vector space V*. We can take {e_a} as a basis for V and {w^a} as a
basis for V* such that

<w^a|e_b> = delta{ab}.

As a result, I take
{ p i_m } i_1 i_2 i_p
{ /\ w } = w /\ w /\ ... /\ w
{ m=1 }

as a basis for the vector space of p-forms L^p(V). Suppose, therefore,
that I have a p-form A and a q-form B. I can write these in terms of
their respective bases as

i_1 i_2 i_p
A = A w /\ w /\ ... /\ w
i_1 i_2 ... i_p

i_1 i_2 i_q
B = B w /\ w /\ ... /\ w
i_1 i_2 ... i_q

Now, since these are differential forms I can also write them in terms
of antisymmetrized indices:

1 i_1 i_2 i_p
A = --- A w /\ w /\ ... /\ w
p! [i_1 i_2 ... i_p]

1 i_1 i_2 i_q
B = ---B w /\ w /\ ... /\ w
q! [i_1 i_2 ... i_q]

Now suppose that I take the exterior product of A and B. As far as I
can see, what I get is

1 i_1 i_(p+q)
(A/\B) = ---- A B w /\ .. /\ w
p!q! [i_1 .. i_p] [i_(p+1) .. i_(p+q)]

However, I've also seen this written as follows

(p+q)! i_1 i_(p+q)
(A/\B) = ------A B w /\ ... /\ w
p!q! [i_1 .. i_p i_(p+1) .. i_(p+q)]

The problem I have is that I can't see how to go from the second last
line to this last one. In particular, in the second-last equation, I
antisymmetrize the indices of A and B *separately*, and get an overall
factor of 1/p!q! as a result. However, in the last line the
antisymmetrization is over all the indices and this seems to give an
extra factor of (p+q)!. Can anyone give me a hint how I can get this
last equation from the second-last?

Cheers,

dvn

.

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