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 Excersise [12.08] 
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Joined: 07 May 2009, 16:45
Posts: 62
Post Excersise [12.08]
Try this again.

Justify the fact that if

\varphi = \alpha \wedge . . . \wedge \gamma
\chi = \lambda \wedge . . . \wedge \nu


\varphi \wedge \chi =  (\alpha \wedge . . . \wedge \gamma) \wedge (\lambda \wedge . . . \wedge \nu)

is equal to

\psi = \alpha \wedge . . . \wedge \gamma \wedge \lambda \wedge . . . \wedge \nu

I thought we already knew the wedge product was associative and

(\alpha \wedge . . . \wedge \gamma) \wedge (\lambda \wedge . . . \wedge \nu)  = \alpha \wedge . . . \wedge \gamma \wedge \lambda \wedge . . . \wedge \nu

I guess that is not the answer being sought here.

I think we are meant to show that the components of \varphi \wedge \chi,


when all summed up are equal to the components of \psi = \alpha \wedge . . . \wedge \gamma \wedge \lambda \wedge . . . \wedge \nu,


when all summed up.

What I mean by that is, if you first compute \varphi, that p-form has components

\varphi_{i,...,k} = \alpha_{_{[i}}...\gamma_{_{k]}}

Likewise, the q-form \chi has components

\chi_{l,...,n} = \lambda_{_{[l}}...\nu_{_{n]}}

Then, if you take the wedge product of the p-form and the q-form, you get a (p+q)-form with components

\varphi_{i,...,k}\chi_{l,...,n} =  \alpha_{_{[i}}...\gamma_{_{k]}}  \lambda_{_{[l}}...\nu_{_{n]}}

which is (generally) not equal to the component of \psi which are

\psi_{i,...,k,l,...,n} =  \alpha_{ [i  } ... \gamma_k \lambda_l ... \nu_{ n]}

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:

\psi_{i,...,k,l,...,n} = \alpha_{[i}...\gamma_k\lambda_l...\delta_{n]}

\varphi_{i,...,k}  =  \alpha_{[i}...\gamma_{k]}
\chi_{l,...,n}  =  \lambda_{[i}...\nu_{k]}

For this exercise, these expressions for are associated with n-forms (or p-forms, or what have you), but this situation is analogous to n-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\boldsymbol\varphi = \boldsymbol\alpha \wedge \boldsymbol\beta \wedge \boldsymbol\gamma and a 2-vector \boldsymbol\chi = \boldsymbol\delta \wedge \boldsymbol\epsilon

Each vector, \boldsymbol\alpha, \boldsymbol\beta, \boldsymbol\gamma, \boldsymbol\delta, and \boldsymbol\epsilon, are the sum of scalar multiples of five independent basis vectors. For example,

\boldsymbol\alpha = \sum_{i=1}^{5}\alpha_i\boldsymbol\eta_i

Here \boldsymbol\eta_i is the i-th basis vector. \alpha_i 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\boldsymbol\varphiwould likewise be

\boldsymbol\beta = \sum_{j=1}^5 \beta_i\boldsymbol\eta_j

\boldsymbol\gamma = \sum_{k=1}^5 \gamma_i \boldsymbol\eta_k

I am using different letters to represent the indices because we wedge the three vectors together to get the 3-vector \boldsymbol\varphi, which is a sum of simple 3-vectors of the form \left(\alpha_i\beta_j\gamma_k\right) \boldsymbol\eta_i\wedge\boldsymbol\eta_j\wedge\boldsymbol\eta_k. We can find the "average coefficient" is expressed in terms of i,j, and k, that is \varphi_{i,j,k}.

\varphi_{i,j,k} = \alpha_{_{[i}}\beta_{_{j}}\gamma_{_{k]}}

Since\{i,j,k\} 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 \frac{(p+q)!}{p!q!} such combinations. In our case p = 3 andq = 2, 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\boldsymbol\eta_i\wedge\boldsymbol\eta_j\wedge\boldsymbol\eta_k and multiplied by different product of scalar \alpha_i\beta_j\gamma_k . 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\varphi_{1,2,3} = \alpha_{_{[1}}\beta_{_{2}}\gamma_{_{3]}} 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 \{i,j,k\} (123,124, 125, etc.) each producing a different \varphi_{i,j,k} that will be applied to six dependent permutations of base vectors. Ten values of \varphi_{i,j,k} applied to 10 groups of dependent 3-vectors, for a total of 60 3-vectors in our sum. While I am being explicit,

\varphi_{1,2,5} (\boldsymbol\eta_1\wedge\boldsymbol\eta_2\wedge\boldsymbol\eta_5-\boldsymbol\eta_1\wedge\boldsymbol\eta_5\wedge\boldsymbol\eta_2+\boldsymbol\eta_2\wedge\boldsymbol\eta_5\wedge\boldsymbol\eta_1-\boldsymbol\eta_2\wedge\boldsymbol\eta_1\wedge\boldsymbol\eta_5

We can do the same thing for \boldsymbol\chi which in our case is a sum of 2-vecotrs. We have two indices \{l,m\}. Again these indices can be any value from 1 to 5 (general case 1 to p+q). 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,\frac{(p+q)!}{p!q!}.

So we will have ten average coefficients \chi_{l,m}, each will be the average of the coefficients of 20 (in general \frac{(p+q)!}{p!}) dependent 2-vectors. Shall I? Well, just a bit:

\chi_{1,2}(\boldsymbol\eta_1\wedge\boldsymbol\eta_2 - \boldsymbol\eta_2\wedge\boldsymbol\eta_2)
\chi_{1,3}(\boldsymbol\eta_1\wedge\boldsymbol\eta_3 - \boldsymbol\eta_3\wedge\boldsymbol\eta_1)
\chi_{4,5}(\boldsymbol\eta_4\wedge\boldsymbol\eta_5 - \boldsymbol\eta_5\wedge\boldsymbol\eta_4)

So now we wedge \boldsymbol\varphi with \boldsymbol\chi. We do this distributively, each component of \boldsymbol\varphi wedged with each component of \boldsymbol\chi. But look, only when the i,j,k indices of a \boldsymbol\varphi component are all different from all of the l,m indices of a \boldsymbol\chi component will the resulting wedge product be non zero. For instance

\varphi_{1,3,5}\chi_{4,5}(\boldsymbol\eta_3\wedge\boldsymbol\eta_1\wedge\boldsymbol\eta_4\wedge 0)

So in the sum of wedge products that is \boldsymbol\varphi\wedge\boldsymbol\chi, one group of dependent 3-vectors will only combine significantly with a corresponding group of 2-vectors. There are ten such groups in\boldsymbol\varphi and ten complementary groups in \boldsymbol\chi, but again in the general case there will be \frac{p+q)!}{p!q!} such groups.

So the group belonging to \varphi_{1,3,5} will only combine with the group belonging to \chi_{2,5} 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\varphi group and two 2-vectors belonging to a complementary\chi 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 \varphi_{i,j,k}\chi_{l,m} coefficients can be summed together and divided by ten, and that will be the average coefficient for the whole 120 vectors.

Each of the \varphi_{i,j,k} average coefficients was the sum of six scalars\alpha_i\beta_j\gamma_k put together in every permutation all divided by six. Each \chi_{l,m} was the sum of two scalars \delta_l\epsilon_n 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 \varphi_{i,j,k}\chi_{l,m} is equal to the sum of twelve scalars \alpha_i\beta_j\gamma_k\delta_l\epsilon_n multiplied together in every \{i,j,k\}\{l,n\} permutation, (that is, the values of i,j,k permute amongst themselves as do the values ofl and n) all divided by twelve. Summing all ten \varphi_{i,j,k}\chi_{l,m} values together is like summing 120 scalars \alpha_i\beta_j\gamma_k\delta_l\epsilon_n 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 \boldsymbol\alpha\wedge\boldsymbol\beta\wedge\boldsymbol\gamma\wedge\boldsymbol\delta\wedge\boldsymbol\epsilon.

Last edited by DimBulb on 29 Jul 2009, 08:36, edited 21 times in total.

23 Jul 2009, 14:37

Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Excersise [12.08]

(Those brackets go around the indices. They don't render too well.)

Try it like this:


which is rendered thus:


26 Jul 2009, 05:58

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Excersise [12.08]
Thanks. I am having a heck of a time trying to figure out what I am trying to say in this post.

27 Jul 2009, 02:17

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Excersise [12.08]
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.
Really? Their whole import is magnItude? ThEta's called the "INSIDE" or scalar product? The two expressions lie on the line of non-cummunative algebras! The Einstein rule of matrix indexing , here, at this point,,, is poorly used!

If one wishes to model the world then the concept of ROTATION with it's need to establish the direction of the ROTATION CONCEPT. Now if one establishes that a constant of nature (DOES NOT DEPEND ON THE SOURCE OR OBSERVER) is the speed of light!

And "ONE" might further investigate the X,Y---->R, Theta BS!!

25 Nov 2009, 21:52
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