# 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.