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$a_{ij}$ is a rotation matrix that satisfies $\hat{e}'_i=a_{ij}\hat{e}_j$. Show that $\epsilon_{lmn}a_{mi}a_{nj}=\epsilon_{ijk}a_{lk}.$

Using the result from above, how can I show that $\vec{x}\times \vec{y}$ transforms like a rank$-1$ tensor under the rotation $a_{ij}?$

I am struggling understanding tensors.

So, I know that right-handed Cartesian coordinate system is defined by unit vectors that satisfy $\hat{e}_i \times\ \hat{e}_j = \epsilon_{ijk}\hat{e}_k$ and that $\epsilon$ is the Levi-Civita symbol. But I don't know how to solve this problem.

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  • $\begingroup$ @Semiclassical Unfortunately nothing. I was thinking of using the tensor notation of a determinant to solve this problem but I failed miserably. $\endgroup$ – Robben Nov 7 '14 at 2:47
  • $\begingroup$ Ok. My immediate thought is to show that something like $\epsilon_{lmn}a_{mi}a_{nj}\hat{e}_l=\epsilon_{ijk}a_{lk}\hat{e}_l$ is true, and then take the dot product with $\hat{e}_l$ on both sides to get the identity. (That form seems more tractable since it starts to look like the provided relation for the rotation matrix.) $\endgroup$ – Semiclassical Nov 7 '14 at 2:49
  • $\begingroup$ @Semiclassical Hm, I am not sure how to show that $\epsilon_{lmn}a_{mi}a_{nj}\hat{e}_l=\epsilon_{ijk}a_{lk}\hat{e}_l$ is true. $\endgroup$ – Robben Nov 7 '14 at 3:02
  • $\begingroup$ Well, note that $\epsilon_{lmn}\hat{e}_l$ can be replaced by a cross product as you noted above. That gives you terms like $\hat{e}_m a_{mi}$, which is nearly of the form required for the rotation matrix transformation. Problem is, the indices are out of order. (There may be an easy resolution of that, I'm just forgetting.) $\endgroup$ – Semiclassical Nov 7 '14 at 3:06
  • $\begingroup$ To my eye, I don't think this relationship is true unless some of the indices refer to different sets of bases. $\endgroup$ – Muphrid Nov 7 '14 at 6:27
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You mention "the tensor notation of a determinant" in the comments. I don't know quite how your definitions have been given to you, but I imagine that you already know something like this: $$\epsilon_{pmn}a_{pk}a_{mi}a_{nj}=\epsilon_{kij}\operatorname{det}(a).$$ Then in the case $a$ is a rotation matrix, we know $\operatorname{det}(a)=1$, so we have $$\epsilon_{pmn}a_{pk}a_{mi}a_{nj}=\epsilon_{kij}.$$ We also know that $a$'s inverse is its transpose, so that $a_{pk}a_{lk}=\delta_{pl}$. So if we multiply both sides of our equation by $a_{lk}$ we get $$\epsilon_{pmn}\delta_{pl}a_{mi}a_{nj}=\epsilon_{kij}a_{lk}$$ and hence $$\epsilon_{lmn}a_{mi}a_{nj}=\epsilon_{ijk}a_{lk}.$$ For the second bit of the question, start from the fact that $(\vec{x}\times\vec{y})_i=\epsilon_{ijk}x_jy_k$.

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