Show that triple product is invariant under rotations

Question

Let $a$, $b$, $c$ be vectors from $\mathbb R^3$. The triple product $a \cdot (b \times c)$ is the (oriented) volume in the parallel piped spanned by the three vectors.

Let $O$ be an skew-symmetric, unit-determinant rotation matrix with $O^{-1} = O^T$ (transpose). I know intuitively that the volume must not change if I rotate all the three vectors with the same rotation. Yet I fail to see how to show this.

Writing it in components, I can also write it as $$ \epsilon_{ijk} O_{il} O_{jm} O_{kn} a_l b_m c_n $$ with implicit summation over repeated indices. Somehow one needs to reduce all this to the original expression again.

What is the trick to do this?

Answer 1

This is easy if we use determinants. Let $A$ be the matrix with columns $a, b, c$, then the volume you are interested in is $V_A = \det A$.

On the other hand, multiplying $A$ from the left with the orthogonal rotation matrix $O$ results in a matrix $B$ with columns $Oa, Ob, Oc$. The rotated volume $V_B$ fulfils $V_B = \det B$.

Now using the fact that the determinant is multiplicative and $\det O = 1$ implies

$$V_B = \det B = \det (OA) = \det O \det A = \det A = V_A.$$

Answer 2

You can use the Leibniz formula in the form

$$\varepsilon_{i j k} O_{i l} O_{j m} O_{k n} = \mathrm{det}(O) \varepsilon_{l m n},$$

so that the expression you wrote down becomes

$$\varepsilon_{i j k} O_{i l} O_{j m} O_{k n} a_l b_m c_n = \varepsilon_{l m n} a_l b_m c_n $$

for $O \in \mathrm{SO}(3)$, i.e. $\mathrm{det}(O) = +1$.