@TedShifrin either will do. Physicists often write line element as $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$ for example is line element of a photon in flat sapcetime
But proceeding with my question. lets I manage to find the transformations that enable me to take $ds^2 \to \lambda ds^2$. For example $ds^2 = dx^2 + dy^2$ becomes $\lambda^2 ds^2 $ when $x \to \lambda x $ and $y \to \lambda y $
Now all those coordinates which participate in the transformation are replaced with their integrated variable $dx \to x$ and $dy \to y$ and I get $K = x^2 + y^2$
Honestly, I am fairly close to applying for a physics masters. That would let me teach physics here (there is no one within 90 miles of me who is qualified to teach even introductory college physics).
The nearest physicist is 90+ miles to the west, at Northern Arizona University. The next nearest physicist is likely in Tempe, which is 175 miles south of here.
@TedShifrin So the output of $ds^2 = dx^2 + dy^2$ becomes $K = x^2 + y^2$ while that of $ds^2 = d r^2 + r^2 d \theta^2 $ becomes $K = r^2$. Since $K$ is coordinate invariant (as conjectured) $r^2 = x^2 + y^2$
the quantity $K$ which is the output of the algorithm is coordinate invariant and does not depend on the coordinates of the metric used
@Thorgott The algorithm says if a coordinate does not participate in taking $ds^2 \to \lambda^2 ds^2$ then we put $0$. For polar coordinates $r \to \lambda r$ but $\theta \to \theta $ it undergoes the identity so we put $0$
@MoreAnonymous Mechanically replacing the $2$-tensor $du\otimes dv$ with $uv$ makes no mathematical sense. All right, try the metric $cos^2 y\,dx^2 + e^x\, dy^2$.
I can understand integrating the exact 1-form $dx$, but other things you're doing are mathematically not meaningful.
Anyone know what $\mathbb{H}$ identified with $\mathbb{R}^4$ means? Does it mean to just replace quaternions with a vector representation in $\mathbb{R}^4$
Ok, that's just notation, but what kind of things are $i,j$? What space do they live in? My point is that you may as well define $i=(0,1,0,0)$ and $j=(0,0,1,0)$ (and then define the multiplication, which is the important additional structure). Anyway, whether there truly is a need to do something or not is a philosophical question I can't entertain. But $\mathbb{H}$ and $\mathbb{R}^4$ are so naturally identified that it won't make a genuine difference. It's just a matter of notation, I'm almost 100% certain even without even knowing what proof you're looking at.
You know the group homomorphism: $$\phi ; S^3 \rightarrow SO(3) ; q \mapsto \phi_q$$ Where $$\phi_q; \mathbb{H}^* \rightarrow \mathbb{H}^*; x \mapsto qxq^{-1}$$ for a non-zero pure quaternion $q$. How does it actually work, I read somewhere it's to do with these loops which collapse?
