Consider a function $f(a_i)$ of 6 Cartesian variables $a_i$. In the presence of rotational symmetry, we say that $f$ really just depends on the modulus $r=|a|$ and, particularly if we wish to look for critical points, it suffices to consider $f(|a|)$; there is only one true degree of freedom.
I happen to have a function which can be written in terms of $r=|{a}|$ as well as $c=|\vec{a1}\times \vec{a2}|$, where $\vec{a1}=(a_1,a_2,a_3)$ and $\vec{a2}=(a_4,a_5,a_6)$: $$f(r,c)=f(|a|,|\vec{a1}\times \vec{a2}|).$$ My question is, are there really just 2 degrees of freedom to my function, can I find all critical points by differentiating w.r.t. $r$ and $c$?
My attempt:
It seems to me to depend on whether I can make a coordinate transformation on 6d space $\{a_1,a_2,a_3,a_4,a_5,a_6\}$ to a new coordinate system where two of the coordinates are $r$ and $c$ (the remaining 4 combinations of the initial $a_i$ not appearing in my function). What makes a good coord transformation? (After all not everything is allowed.. I can't simply call $f(a_i)$ itself a new coordinate to end up with a 1d problem). I believe the coordinate mapping needs to be 1-1 (and onto), at least in a network of overlapping open subsets.
Consider the candidate mapping $$\{a_1,a_2,a_3,a_4,a_5,a_6\}\rightarrow \{a_1,a_2,a_3,a_4,r(a_i),c(a_i)\}$$ The mapping will be 1-1 if $$a_{1,2,3,4}=b_{1,2,3,4} \quad \quad r(a_i)=r(b_i), \quad \quad c(a_i)=c(b_i)$$ $$\implies a_i = b_i$$ The equalities on $r$ and $c$ boil down to $$a_5^2 + a_6^2 = b_5^2 + b_6^2, \quad \quad a_2 a_5 + a_3 a_6 = a_2 b_5 + a_3 b_6$$ and solving one has $$ \left\{ b_5=a_5,\; b_6=a_6 \right\}\quad \text{or} \quad \left\{b_5=\frac{a_2^2 a_5 - a_3^2 a_5 + 2 a_2 a_3 a_6}{a_2^2 + a_3^2}, \; b_6=\frac{2 a_2 a_3 a_6 - a_2^2 a_6 + a_3^2 a_6}{a_2^2 + a_3^2}\right\}$$ Thus it would seem that transformation is not globally 1-1. However, even it could still perhaps be a good coordinate transformation using a network of overlapping open set 'patches', where only one of the two alternatives was valid. I'm not sure how to go about proving this.. and feel that maybe I am overcomplicating things...and there might be easier ways to realise if $r$ and $c$ are good variables..