I'm going through the book "Tensors, Differential Forms, and Variational Principles" by David Lovelock and Hanno Rund (on my own, for fun) and got stuck on problem 2.9 page 51.
The referenced euqations are
2.11 $$\bar{A}_j=a_{jh}A_h$$ 2.25 $$\bar{C}_{jk}=a_{jh}a_{kl}C_{hl}$$
Where it is understood that an index appearing twice should be summed over (summation convention)
My first question is, what does "proper transformation" mean?
I think it means
$$\det(a_{jk})=+1$$
$$\delta_{jk}=a_{jh}a_{kh}$$
Is this correct?
My second questions is, how do I use that?
If I calculate $\bar{C}_j$ and $\bar{C}_{kl}$ component wise using 2.11 and 2.25 and compare them I get 9 relations that needs to be fullfilled, one for each entry in $a_{jh}$. For instance, for $C_1$
$$a_{11}=a_{22}a_{33}-a_{23}a_{32}$$ $$a_{12}=a_{23}a_{31}-a_{33}a_{21}$$ $$a_{13}=a_{32}a_{21}-a_{31}a_{22}$$
And sure enough, if I check these relations for a couple of orthogonal transformations they seem valid. But how do I prove this using the relations of a proper transformation. Is this even what I am suppoed to show?
Just in case someone else is interested or can provide feedback this is my analysis. I assume I can close this question/thread now.
$C_j=\epsilon_{jhk}A_hB_k$
The Levi-Civita symbol, $\epsilon_{jhk}$ is a tensor under proper orthogonal transformations.
$$\overline{\epsilon_{jhk}}=a_{ju}a_{hv}a_{kw}\epsilon_{uvw}=\det(a)\epsilon_{jhk}$$
Since $\det(a)=+1$ (proper transformation) $\overline{\epsilon_{jhk}}=\epsilon_{jhk}$ we have
$$\overline{C_j}=\epsilon_{jhk}a_{hu}A_ua_{kv}B_v$$
Using the the transformation property of Levi-Civitia once again (this time as an identity to make it esier to handle A and B):
$$\overline{C_j}=a_{ju}a_{hv}a_{kw}\epsilon_{uvw}a_{hu}a_{kv}A_uB_v$$
Now we remember that the transformation is orthogonal, hence $a_{ju}a_{hu}=\delta_{jh}$ and so on:
$$\overline{C_j}=\delta_{jh}\delta_{hk}a_{kw}\epsilon_{uvw}A_uB_v$$
$$\overline{C_j}=a_{jw}\epsilon_{uvw}A_uB_v$$
Expanding the possible values ($\epsilon_{uvw}\neq 0$) of u,v,w
$$\overline{C_j}=a_{j1}(A_2B_3-A_3B_2)+a_{j2}(A_3B_1-A_1B_3)+a_{j3}(A_1B_2-A_2B_1)$$
Finally, since $C_1=C_{23}=A_2B_3-A_3B_2$ and likewise for $C_2$, $C_3$
$$\overline{C_j}=a_{jh}C_h$$