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Motivated by this question and by my own calculations where I try to find what is the curvature of a rigid transformation to my curve I get to the following equation ($\alpha$ is a curve, $\phi$ a linear transformation):

$\phi(\alpha') \times \phi(\alpha'') = \phi(\alpha' \times \alpha'')$

I wonder if it is true and conjecture that it is true because $\phi$ is an orthogonal linear transformation. So that the more general formula:

$Ma \times Mb = \det(M)(M^T)^{-1} (a \times b)$

holds. Now, what I ask you for is to prove this last formula (fix it if it is not true) or at least give a reference where I can encounter it.

user1868607
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    Please take a look at the addendum my answer here, though it might not be wholly satisfactory because of its reliance on multilinear algebra: https://math.stackexchange.com/questions/752774/generalized-cross-product/752800#752800 – Branimir Ćaćić Apr 10 '18 at 18:23

1 Answers1

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The general formula is correct. Your original one holds only when $\phi$ is a proper (orientation-preserving) orthogonal linear map. What underlies all of this is (a) the way linear maps act on the exterior algebra: $$(\Lambda^2M)(a\wedge b) = Ma\wedge Mb$$ and (b) the Hodge star isomorphism $\star\colon \Lambda^2\Bbb R^3 \to\Bbb R^3$. So the underlying action on the cross product of two vectors $a\times b$ is found by chasing around $\star (\Lambda^2M)(a\wedge b) = \star(\Lambda^2M)(\star^{-1}(a\times b))$.

We don't need to be so fancy. The crucial thing is the "classical adjoint" formula $$M^{-1} = \frac1{\det M}(\text{cof }M)^\top,$$ where $\text{cof } M$ is the matrix of cofactors. Here is the typical computation: \begin{align*} Me_1 \times Me_2 &= \sum m_{k1}e_k \times \sum m_{\ell 2}e_\ell = \sum_{k<\ell} \left|\begin{matrix} m_{k1} & m_{k 2} \\ m_{\ell 1} & m_{\ell 2}\end{matrix} \right| e_k\times e_\ell \\ &= \sum (\text{cof }M)_{i3}e_i = (\text{cof }M)e_3 = (\det M)(M^{-1})^\top e_3\\ &= (\det M)(M^{-1})^\top(e_1\times e_2). \end{align*}

Ted Shifrin
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