Try this again.

Justify the fact that if

then

is equal to

I thought we already knew the wedge product was associative and

I guess that is not the answer being sought here.

I think we are meant to show that the components of

,

when all summed up are equal to the components of

,

when all summed up.

What I mean by that is, if you first compute

, that

p-form has components

Likewise, the

q-form

has components

Then, if you take the wedge product of the

p-form and the

q-form, you get a

(p+q)-form with components

which is (generally) not equal to the component of

which are

Note the difference in the placement of the brackets in the two expressions. The two expressions are not equal, but when you sum the components all up, you end up with the same result. I think trying to show this is what is what this exercise is about. I think I have it figured out. I have peered into that Platonic reality and seen the solution. It is a bit hard for me to explain, however.

Keep in mind what these expressions are:

For this exercise, these expressions for are associated with

n-forms (or

p-forms, or what have you), but this situation is analogous to

-vectors. These expressions are average coefficients of the dependent n-forms (n-vectors) in the wedge product, as explained on pages 213 and 214. This average coefficient is calculated and then applied to each of the dependent n-form (n-vector) equally. On pages 213 and 214 the two and three dimensional vector cases were shown explicitly, and my post for Exercise [11.11] spelled it out for the four dimensional vector case even more explicitly. In general, for each combination of values of the indices can take there are a bunch of dependent n-forms (n-vectors), one fore each permutation of those index values. You sum up all the coefficients of all those dependent n-forms (n-vectors), and divide by the number of permutations to arrive at the average coefficient, which is what those expressions above represent.

When I say dependent here, I mean not non-independent, that is one n-form (n-vector) can be expressed in terms of another n-form (n-vector).

And when I say that the case of n-forms is analogous to the case of n-vectors, I can not prove that. I am just going on what I gather from what Penrose says, like "A similar notation applies for the wedge product of a p-vector with a q-vector."

As illustrative of the general case, I am going to show explicitly how the thing works for a 3-vector wedged with a 2-vector, which is analogous to a 3-form wedged with a 2-form.

So let us have a 3-vector

and a 2-vector

Each vector,

, and

, are the sum of scalar multiples of five independent basis vectors. For example,

Here

is the i-th basis vector.

is the i-the scalar multiple of the i-th basis vector.

Here we are we are summing over 5 dimensions. In the general case, where we are wedging a p-vector with a q-vector, we would sum over (p+q) dimensions, each vector being the sum of scalar multiples of (p+q) basis vectors.

So the other two vectors that make up the 3-vector

would likewise be

I am using different letters to represent the indices because we wedge the three vectors together to get the 3-vector

, which is a sum of simple 3-vectors of the form

. We can find the "average coefficient" is expressed in terms of i,j, and k, that is

.

Since

can each take a value from one to five, there are ten combinations of index values: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345. In general, there will be

such combinations. In our case

and

, so we have ten combinations.

For each of these combinations there is a group of dependent 3-vectors in the wedge product sum, with each dependent 3-vector made from a different permutation of

and multiplied by different product of scalar

. So, for instance, for combination 123, we have

In general there will be p! dependent vectors for each combniation. In our case 3! = 6, so there are six dependent vectors associated with combination 1,2, 3. We find the average coefficient for these 6 vectors

where we add up the six (p!) scalar coefficients, applying the

or

associated with of the coefficient, and divide the sum by six, (again p! in the general case), and apply this average coefficient to all the dependent 3-vectors:

We do the same thing for each of the ten possible combinations of

(123,124, 125, etc.) each producing a different

that will be applied to six dependent permutations of base vectors. Ten values of

applied to 10 groups of dependent 3-vectors, for a total of 60 3-vectors in our sum. While I am being explicit,

=

+

+

+

+

+

+

+

+

+

We can do the same thing for

which in our case is a sum of 2-vecotrs. We have two indices

. Again these indices can be any value from 1 to 5 (general case 1 to

). Again the same number of combinations, that is ten: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45. In the general case the formula is the same for the p-vector,

.

So we will have ten average coefficients

, each will be the average of the coefficients of 20 (in general

) dependent 2-vectors. Shall I? Well, just a bit:

=

+

+

+

So now we wedge

with

. We do this distributively, each component of

wedged with each component of

. But look, only when the

indices of a

component are all different from all of the

indices of a

component will the resulting wedge product be non zero. For instance

=

=

=

=

So in the sum of wedge products that is

, one group of dependent 3-vectors will only combine significantly with a corresponding group of 2-vectors. There are ten such groups in

and ten complementary groups in

, but again in the general case there will be \frac{p+q)!}{p!q!} such groups.

So the group belonging to

will only combine with the group belonging to

etc. So you will have ten groups, each will contain twelve 5-vectors, five basis vectors wedged together, made up from the six 3-vectors belonging to a

group and two 2-vectors belonging to a complementary

group. Since all of the ten groups contain 5-vectors, and we a dealing with five dimensional space, then all the vectors are dependent. That is, any of the 120 (ten times twelve) 5-vectors can be expressed in terms of any other of the 5-vectors. That means that the ten

coefficients can be summed together and divided by ten, and that will be the average coefficient for the whole 120 vectors.

Each of the

average coefficients was the sum of six scalars

put together in every permutation all divided by six. Each

was the sum of two scalars

put together both ways and divided by two. When I say summing the permutations here, I actually mean adding the even permutations and subtracting the odd permutations. So

is equal to the sum of twelve scalars

multiplied together in every

permutation, (that is, the values of

permute amongst themselves as do the values of

and

) all divided by twelve. Summing all ten

values together is like summing 120 scalars

in all possible permutations, all divided by 12. We just need to divide this sum by ten as explained in the previous paragraph to show this average coefficient for all 120 vectors is exactly \psi_{1,2,3,4,5}, the same you would get for

.