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I define a vector as any object $(a_i,a_j, a_k)$ such that it transforms the same way as the coordinates themselves. That is if $x'_i = R_{ij}x_j$, then $a'_i = R_{ij}a_j$ (using the Einstein convention of summing over repeated indices). Please correct me if this is not a proper definition.

I now want to take the cross product of two vectors, $\vec{c} = \vec{a}\times \vec{b}$ and prove (by this definition) that the output is also a vector. Using the Levi-Civita symbol, we have

$c'_i = R_{ij}c_j = det(R) R_{ij}\epsilon_{jlm}a_l b_m$

But this should somehow be the same as the cross product of the transformed $\vec{a}$ and $\vec{b}$. That is $c'_i = \epsilon_{ijk}a'_jb'_k = \epsilon_{ijk}R_{jl} a_l R_{km}b_m$

How do I reconcile these two? I know I should get $\rm{det(R)}$ in the second expression somehow but I can't see how to express it in this notation.

user1936752
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    Actually, from the point of view of transformation properties, the cross product is not really a vector, but rather a “pseudovector”: https://en.wikipedia.org/wiki/Pseudovector. Or better, if you want to avoid that mess: use the bivector $a \wedge b$ instead. – Hans Lundmark Apr 24 '18 at 07:51
  • Doesn't the det(R) take care of the pseudovector-ness of the output? In your link, it also says that $Rv_1 \times Rv_2 = det(R) R(v_1 \times v_2)$. I just can't see how to prove that statement using index notation. – user1936752 Apr 24 '18 at 08:18
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    Well, if you include the determinant, then yes. But you need to assume that $R$ is an orthogonal matrix too, maybe that's what's causing you trouble? (And then $\det R= \pm 1$.) – Hans Lundmark Apr 24 '18 at 08:33
  • I see your point. I've edited the question now. The trouble I'm having is to express det(R) in index notation and show the equivalence. – user1936752 Apr 24 '18 at 09:23
  • "Please correct me if this is not a proper definition." - It can be extended to a proper definition. – Filippo Jun 25 '23 at 22:48

2 Answers2

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Suppose $\mathbf c = \mathbf a \times \mathbf b$. Then, since $\mathbf a$ and $\mathbf b$ are vectors and $\varepsilon_{ijk}$ is a rank $3$ tensor, \begin{align*} a_i' &= R_{ij} a_j \\ b_i' &= R_{ij} b_j \\ \varepsilon_{ijk}' &= R_{ip}R_{jq}R_{kr} \varepsilon_{pqr}. \end{align*} Now under a coordinate transformation, \begin{align*} c_i& = \varepsilon_{ijk} a_j b_k \\ \implies c_i' &= \varepsilon_{ijk}' a_j' b_k' \\ &= R_{ip} R_{jq} R_{kr} \varepsilon_{pqr }R_{jm}a_m R_{kn}b_n \\ &= R_{ip} \delta_{qm} \delta_{rn} \varepsilon_{pqr} a_m b_n \tag{$*$} \\ &= R_{ip} \varepsilon_{pqr}a_q b_r \\ &= R_{ip} c_p, \end{align*} where $(*)$ follows since $R_{jq} R_{jm} = [\mathbf R^T]_{qj} \left[\mathbf R\right]_{jm} = \left[\mathbf R^T \mathbf T\right]_{qm} = \delta_{qm}$.

kipf
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How do I reconcile these two?

First of all, I would like to emphasize that in general $Ma\times Mb\neq M(a\times b)$ for an invertible matrix $M$. If $M$ is orthogonal (i.e. $MM{}^t=1$), then$$M(a\times b)=(\det{M})(Ma\times Mb)$$ (see the proof below). Note that $\det M=\pm 1$ (here is a proof), so we can equivalently write the determinant on the LHS of the equation. Hence we obtain $Ma\times Mb=M(a\times b)$ if $M$ is orthogonal and $\det M=1$.

Proof:

Set $$\forall i\in\{1,2,3\}:v_i:=\begin{pmatrix}M_{1i}\\M_{2i}\\M_{3i}\end{pmatrix}$$ then $$\epsilon_{ijk}\det{M}=\epsilon_{ijk}\det(v_1|v_2|v_3)=\det(v_i|v_j|v_k)=\epsilon_{pqr}M_{ip}M_{jq}M_{kr}$$ where in the last step we used the Leibniz formula and hence $$\forall k:(\det M)(Ma\times Mb)_k=(\det M)\epsilon_{ijk}(Ma)_i(Mb)_j=\epsilon_{pqr}M_{ip}M_{jq}M_{kr}M_{i\beta}a_\beta M_{j\alpha}b_\alpha\\ =\epsilon_{pqr}\underbrace{M_{ip}M_{i\beta}}_{=\delta_{p\beta}}\underbrace{M_{jq}M_{j\alpha}}_{=\delta_{q\alpha}}M_{kr}a_\beta b_\alpha=M_{kr}\epsilon_{pqr}a_pb_q=M_{kr}\epsilon_{pqr}a_pb_q=M_{kr}(a\times b)_r=(M(a\times b))_k$$

Filippo
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