Basic tensor equation

1. Sep 19, 2015

S. Moger

I'm new to working with tensors, and feel a bit uneasy about the nomenclature. I picture words like antisymmetry in terms of average random matrices where no symmetry can be found at all. However, if I understand it correctly, antisymmetry is a type of symmetry, but where signs are inverted. So with this nomenclature Reagan, in that famous joke, is right to label anticommunists as a type of communists. But, anyway that's just linguistic mishmash.

1. The problem statement, all variables and given/known data

$A_{ij} = k \epsilon_{ijk} a_k$

k is a constant. $\vec{a}$ is a vector.

What must k be to fulfill $A_{ij} A_{ij} = |\vec{a}|^2$ ?

3. The attempt at a solution

The dual use of k also confuses me somewhat but from what I understand the only constant is the non-index k.

Is it correct to assume that $\vec{a}$ is taken to exist in three dimensions, from the use of the 3-index levi cevita?

$A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijk} a_k = k^2 \epsilon_{ijk} \epsilon_{ijk} a_k a_k = 6 k^2 a_k a_k$

So if k is $\pm \frac{1}{\sqrt{6}}$ it appears to fulfill the requirements, but this isn't the correct answer.

2. Sep 19, 2015

CAF123

Hey, the first equality on the last equation you wrote violates the Einstein summation convention.

3. Sep 19, 2015

S. Moger

Ok,

Would this be a correction?

$A_{ij} A_{ij} = k \epsilon_{ijk} a_k \cdot k \epsilon_{ijn} a_n$

Meaning I need to find $\epsilon_{ijk} \epsilon_{ijn}$? It seems like $n \neq k$ terms amount to zero, because it implies that there's a duplicate index number in one of the levi-civitas, leaving only ones involving $a_k^2$'s.

With an interpretation of k like this:
$A_{ij} = k \sum_k \epsilon_{ijk} a_k$

I see I have double ij on both sides though, which may violate the summation convention still.

I know there's an identity that relates $\epsilon_{ijk} \epsilon_{lmn}$ to Kroneckers. And I can show that both sides are equal, but I don't understand how one would arrive at that equality by "brute force", i.e. what's the natural thought behind its derivation? I see that the positive contributions come from the permutations that return 1*1 (but not -1*-1 mysteriously enough - do they get redundant? edit: yes they seem to contribute a double) and that the negatives come from 1*(-1).

$\epsilon_{ijk} \epsilon_{lmn}= \delta_{il} \delta_{jm} \delta_{kn} + \delta_{im} \delta_{jn} \delta_{kl} + \delta_{in} \delta_{jl} \delta_{km} - \delta_{im} \delta_{jl} \delta_{kn} - \delta_{il} \delta_{jn} \delta_{km}- \delta_{in} \delta_{jm} \delta_{kl}$

In the books they never explain how they construct the above in the first place. It's a kind of a jump. I mean it's possible to memorize it by using mnemonics (rotation of indices etc), but I would like to understand how to derive it from a blank piece of paper, or do I have to accept that the "wheel has been invented" on this matter?

One attempt I made at that was to list all levi civitas that return 1, i.e. : {(1,2,3), {2,3,1}, {3,1,2}), then to pair them with (i,j,k) and (l,m,n). So, with i=1, j=2, k=3 I looked up what a selection of l, m and n would have to be, for example 1=m, 2=n, 3=l. Another way would be to make some kind of truth table to make the levi civita - kronecker translation.

Last edited: Sep 19, 2015
4. Sep 19, 2015

Ray Vickson

Let's say $A_{ij} = c \epsilon_{ijk} a_k$. Now $\epsilon \neq 0$ requires that $i,j,k$ all be different. Your quantity
$$A_{ij} A_{ij} = c^2 \sum_{i,j: i \neq j} \sum_{k,n} \epsilon_{ijk} \epsilon_{ijn} a_k a_n$$
Nonzero terms in the sum need $k \neq i,j$ and $n \neq i,j$, hence $k = n$. That is,
$$A_{ij}A_{ij} = c^2 \sum_{i,j: i \neq j} \sum_{k: k \neq i,j} (\epsilon_{ijk})^2 a_k^2,$$
and $\epsilon^2 = 1$ for the non-zero terms. For each pair $(i,j)$ there is only one $k$, and as we vary the pair $(i,j)$ we sweep over all $k = 1,2,3$. Furthermore, there will be an equal number of $k=1$, $k=2$ and $k=3$ terms. Thus, we have $A_{ij}A_{ij} = K c^2 \sum_{k=1}^3 a_k^2$, with a constant $K$ that is not too difficult to figure out.

5. Sep 19, 2015

S. Moger

Thanks, it's a bit tricky, but I think I get this part now. It's probably a good idea to write it out in sums like you do, at least at the moment. I get K=2, which will lead the the correct result.