# Expression with levi-civita symbol

1. Feb 17, 2010

### mnb96

Hello,
during a calculation I got the following term:

$$\varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n})$$

where $$\varepsilon$$ is the levi-civita symbol and $$a_i^j$$ are real numbers.

Is it possible to simplify that expression?

Last edited: Feb 17, 2010
2. Feb 17, 2010

### tiny-tim

Hi mnb96!

I think you can write it $${a_{[i_1 \ldots i_n]}}^{[j_1 \ldots j_n]}$$

(or maybe that's out by a factor of 1/n!2 ? )

3. Feb 17, 2010

### mnb96

?? is that the solution ??

...I was wondering if that expression might perhaps be equivalent to the determinant of something. Any ideas?

4. Feb 17, 2010

### tiny-tim

waa! I didn't notice that you changed the question a minute before I posted my answer.

5. Feb 17, 2010

### mnb96

...sorry :) :P
I realized of the mistake later, after re-reading my post :(

6. Feb 18, 2010

### mnb96

I thought about the following: since we have that,

$$\varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} = det(A)$$

then,

$$\varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) =$$

$$=\sum_{k=1}^{n!}sgn(\sigma_k(n))\cdot \varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} =$$

$$=\sum_{k=1}^{n!}sgn(\sigma_k(n))\cdot det(A)$$

where $$\sigma_k(n)$$ is the k-th permutation on the sequence $(1,\ldots,n)$.
Is that correct?
Isn't that supposed to be zero?!

7. Feb 18, 2010

### tiny-tim

Hi mnb9!

(just got up :zzz: …)

I think it's clearer (and better, because it's accurate ) to write it …

$$\varepsilon_{j_1 \ldots j_n} a_{1}^{j_1}\ldots a_{n}^{j_n} = det((a_1,\cdots a_n))$$

then you get

$$\varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) =$$

$$\varepsilon^{i_1 \ldots i_n}det((a_i_1,\cdots a_i_n))$$

which is … ?

8. Feb 18, 2010

### mnb96

if it is true that given (1,...,n) there are always as many odd-permutations as even-permutations, then we get...zero! :) right?

BTW, I have some troubles with your notation $$det((a_1,\cdots a_n))$$. What are those $$a_i$$ ?
I assume they are vectors.
I also assume those $$a_{i1}$$ and $$a_{in}$$ are instead the vectors $$\mathbf{a}_{i_1}$$ and $$\mathbf{a}_{i_n}$$
Did I get it right?

Finally, do you know a good source in which I can find useful exercises and tricks about manipulations with this notation?

Last edited: Feb 18, 2010
9. Feb 18, 2010

### tiny-tim

Hi mnb96!
I don't think so … doesn't interchanging two indices multiply both the epsilon and the determinant by -1 ?
Yup!

(I couldn't be bothered to write it properly )
Sorry, no idea.

10. Feb 18, 2010

### mnb96

Damn! that's true, whenever we have an odd permutation the determinant should change its sign, but also the epsilon will be -1, so we finally have (crossing fingers):

$$\varepsilon^{i_1 \ldots i_n}\varepsilon_{j_1 \ldots j_n} (a_{i_1}^{j_1}\ldots a_{i_n}^{j_n}) = n!\cdot det((\mathbf{a}_1, \ldots, \mathbf{a}_n))$$

...Oh, BTW is it true that "given (1,...,n) there are always as many odd-permutations as even-permutations"?

11. Feb 18, 2010

### tiny-tim

(btw, i think we usually use small letter for "btw" )

Only for n > 1 …

try starting the proof with "if there exists an odd permutation π, then … "

12. Feb 18, 2010

### mnb96

Ah...why?
if n=1 we have:

$$\varepsilon^{i_1}\varepsilon_{j_1}a_{i_1}^{j_1} = \varepsilon^{1}\varepsilon_{1}a_{1}^{1} = a_{1}^{1} = det(\mathbf{a}) = 1!det(\mathbf{a}) = n!\cdot det(\mathbf{a})$$

I assumed that permuting (1) zero times, means even permutation so that $$\varepsilon^{1}=+1$$

13. Feb 18, 2010

### tiny-tim

That's a different question.

(and there are no odd permuatations for n = 1 )

14. Feb 18, 2010

### mnb96

Uhm...I am just wondering if we can just define the levi-civita symbol being equal to 1 in case the sequence has just one element, but that's not terribly important in any case

This reinforces the fact that the levi-civita symbol cannot be -1 for n=1, but there is no repetition of indices, so it should not even be 0.