# Cofactor expression with Levi Civita symbol

From A. J. McConnell's Applications of Tensor Analysis, the cofactor $A^m_n$ of an element in a second order system $a^m_n$ in three dimensions is derived using the Levi-Civita symbol, $\epsilon$. The book uses the smbol $e$ actually, but $\epsilon$ seems much more common.

From what's been presented in the book, the determinant $|a^m_n|$ is given by $$|a^m_n| = \epsilon_{rst} a_1^r a_2^s a_3^t$$ Where Einstein notation is used. The cofactor of an element, then, for example $A^r_1$, is given by $$A^r_1 = \epsilon_{rst}a_2^s a_3^t .$$

So far so good, but what happens next is what is confusing me. The author changes the expression using the fact that $\epsilon^{123}\epsilon_{rst}$ does not change the expression at all. Then this is modified. It goes like this. $$\epsilon_{rst} a_2^s a_3^t$$ $$= \epsilon^{123}\epsilon_{rst} a_2^s a_3^t$$ $$=\frac{1}{2!} \epsilon^{1jk}\epsilon_{rst} a_j^s a_k^t$$ This last step has me quite confused. The $2!$ seems to come out of replacing $a_2^s a_3^t$ with $a_j^s a_k^t$, which makes sense to some degree in that the repeated $jk$ indices imply summation, and this summation will apparently result in the expression $2! [\epsilon_{rst} a_2^s a_3^t$]. The fact that the $2!$ is not simplified to $2$, I can imagine, is to explicitly show the fact that this factorial term comes from the two indices $jk$ being summed. But I cannot seem to get from the $\epsilon^{123}$ step to the $\epsilon^{1jk}$ step.

I have seen that $\epsilon_{rst}\epsilon_{rst} = 3!$ for a three dimensional system (that is, indices take values from 1-3 ), so it would stand to reason that $\epsilon^{ijk}\epsilon^{ijk} = 3!$, and $\epsilon^{1jk}\epsilon^{1jk} = 2!$. But I don't see where that product would arise in the given expressions. This is not covered in the book, however, so perhaps there is another way for this term to come up.

Could someone express this in a more clear way? More explicitly, so that I can make sense of this?

-Edit-

I've found that, as expected, they are equal. Expanding the second expression: $$\epsilon^{1jk}\epsilon_{rst} a_j^s a_k^t$$ $$= \epsilon^{123}\epsilon_{rst} a_2^s a_3^t - \epsilon^{132}\epsilon_{rst} a_3^s a_2^t$$ $$= \epsilon_{rst} a_2^s a_3^t - \epsilon_{rst} a_3^s a_2^t.$$ Because switching two indices will change the sign, we get $$= \epsilon_{rst} a_2^s a_3^t + \epsilon_{rts} a_3^t a_2^s$$ $$= 2 [\epsilon_{rst} a_2^s a_3^t] .$$ Where $2 = 2 \times 1 = 2!$. But I'm still puzzled why the factorial is there and made explicit, when the $2$ arises from addition?

• $\epsilon^{1jk}a_ja_k$ expands to two terms because there are $2=2!$ permutations of the indices $j,k$. If you were to consider e.g. the four-dimensional analogue, you'd find $\epsilon^{1jkl}a_ja_ka_l$ expands to $6=3!$ terms. – Anthony Carapetis Feb 16 '18 at 0:25

As you say there's an extra factor of $2$ resulting from the two terms $\epsilon_{rst} a_2^s a_3^t$ and $\epsilon_{rts} a_3^t a_2^s$. The reason the factorial symbol is given is that one term comes from each possible permutation of $2$ and $3$.
Similarly if we were talking about $4 \times 4$ matrices we would have $$\epsilon^{1jkl}\epsilon_{rstu}a^s_ja^t_ka^u_l$$ $$= \epsilon^{1234}\epsilon_{rstu}a^s_2a^t_3a^u_4 +\epsilon^{1243}\epsilon_{rstu}a^s_2a^t_4a^u_3 +\epsilon^{1342}\epsilon_{rstu}a^s_3a^t_4a^u_2 +\epsilon^{1324}\epsilon_{rstu}a^s_3a^t_2a^u_4 +\epsilon^{1423}\epsilon_{rstu}a^s_4a^t_2a^u_3 +\epsilon^{1432}\epsilon_{rstu}a^s_4a^t_3a^u_2$$ $$= \epsilon_{rstu}a^s_2a^t_3a^u_4 -\epsilon_{rstu}a^s_2a^t_4a^u_3 +\epsilon_{rstu}a^s_3a^t_4a^u_2 -\epsilon_{rstu}a^s_3a^t_2a^u_4 +\epsilon_{rstu}a^s_4a^t_2a^u_3 -\epsilon_{rstu}a^s_4a^t_3a^u_2$$ $$= \epsilon_{rstu}a^s_2a^t_3a^u_4 -\epsilon_{rstu}a^s_2a^u_3a^t_4 +\epsilon_{rstu}a^u_2a^s_3a^t_4 -\epsilon_{rstu}a^t_2a^s_3a^u_4 +\epsilon_{rstu}a^t_2a^u_3a^s_4 -\epsilon_{rstu}a^u_2a^t_3a^s_4$$ $$= \epsilon_{rstu}a^s_2a^t_3a^u_4 -\epsilon_{rsut}a^s_2a^t_3a^u_4 +\epsilon_{rtus}a^s_2a^t_3a^u_4 -\epsilon_{rtsu}a^s_2a^t_3a^u_4 +\epsilon_{rust}a^s_2a^t_3a^u_4 -\epsilon_{ruts}a^s_2a^t_3a^u_4$$ $$= 6\epsilon_{rstu}a^s_2a^t_3a^u_4$$ $$= 3!\epsilon_{rstu}a^s_2a^t_3a^u_4$$