We consider 2D metric: $ds^2={d\theta}^2+\sin^2(\theta)d{\phi}^2$ (1)
Is it possible to transform the above metric to the form: $ds^2=dx^2+dy^2$ (2)
Let's check: Initially we write equation (2) in the form: ds'^2=dx^2+dy^2 (3)
For relations (1) and (3) we use the following transformations:
$\theta=f_1(p,q)$
$\phi=f_2(p,q)$
$x=f_3(p,q)$
$y=f_4(p,q)$
$d(\theta)=[ \partial{f_1} / \partial{p} ]dp + [ \partial{f_1}/ \partial{q}] dq$
$d(\phi)=[\partial{f_2}/\partial{p} ]dp + [\partial{f_2}/\partial{q}] dq$
$dx=[\partial{f_3}/\partial{p} ]dp + [\partial{f_3}/\partial{q}] dq$
$dy=[\partial{f_4}/\partial{p} ]dp + [\partial{f_4}/\partial{q}] dq$
Using the above transformations in (1) and (3) we have: ds'^2=[(\partial{f_1}/\partial{p})^2+\sin^2(f_1)(\partial{f_2}/\partial{p})^2]dp^2+[(\partial{f_1}/ \partial{q})^2+\sin^2(f_1)(\partial{f_2}/\partial{q})^2]dq^2 $+2[(\partial{f_1}/\partial{p} )( \partial{f_1}/\partial{q})+\sin^2(f_1) (\partial{f_2}/\partial{p} )( \partial{f_2}/\partial{q})]dpdq \qquad$ (4)
And
$ds^2=[(\partial{f_3}/\partial{p})^2+(\partial{f_4}/\partial {p})^2]dp^2+[(\partial{f_3}/\partial{q})^2+ (\partial{f_4}/\partial{q})^2]dq^2$ $+2[(\partial{f_3}/\partial{p} ) (\partial{f_3}/\partial{q})+ (\partial{f_4}/\partial{p}) (\partial{f_4}/\partial{q})]dpdq \qquad$ (5)
To make ds^2=ds'^2 we may consider the following equations: (denoted by SET A): $(\partial{f_1}/\partial{p})^2+\sin^2(f_1)(\partial{f_2}/\partial{p})^2= (\partial{f_3}/\partial{p})^2+(\partial{f_4}/\partial{p})^2$ (A1)
$(\partial{f_1}/\partial{q})^2+\sin^2(f_1)(\partial{f_2}/\partial{q})^2= (\partial{f_3}/\partial{q})^2+(\partial{f_4}/\partial{q})^2$ (A2)
$\frac{\partial{f_1}}{\partial{p}}\frac{\partial{f_1}}{\partial{q}}+ \sin^2(f_1) \frac{\partial{f_2}}{\partial{p}} \frac{\partial{f_2}}{\partial{q}} =\frac{ \partial{f_3}}{\partial{p}}\frac{\partial{f_3}}{\partial{q}}+\frac{ \partial{f_4}}{\partial{p} } \frac{\partial{f_4}}{\partial{q}}$ (A3)
If SET A has solutions for the functions $f_1, f_2, f_3$ and $f_4$ we are passing from relation (1) to relation (2) by coordinate transformation ($ds^2$ is preserved and it is being carried from one manifold to another). Is it really possible, that is, can a sphere be flattened?
What are the conditions[from the theory of differential equations] for the existence/non-existence of solutions for SET A?
[You may consider the following calculations to assess the nature of the problem Separate numbering starts from here.
$ds^2=d\theta^2+Sin^2(\theta)d\phi^2$ ------------------- (1)
$ds’^2=dx^2+dy^2$ --------------------------- (2)
Transformations:
$\theta=\theta(x,y)$
$\phi=\phi(x,y)$
$d\theta=\frac{\partial \theta}{\partial x}{dx}+\frac{\partial \theta}{\partial y}{dy}$
$d\phi=\frac{\partial \phi}{\partial x}{dx}+\frac{\partial \phi}{\partial y}{dy}$
Using the above differentials in (1) we have:
$ds^2= [(\frac{\partial \theta}{\partial x})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial x})^2]dx^2$ $+[(\frac{\partial \theta}{\partial y})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial y})^2]dy^2$ $+[\frac{\partial \theta}{\partial x} \frac{\partial \theta}{\partial y}+ Sin^2(\theta)\frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial y}] dx dy$
-------------------- (3)
Relations (1) and (2) would become identical $[ds^2=ds’^2]$ if the following equations [SET A] are satisfied if:
$(\frac{\partial \theta}{\partial x})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial x})^2=1$ -------- A1
$(\frac{\partial \theta}{\partial y})^2+Sin^2(\theta)(\frac{\partial \phi}{\partial y})^2=1$ ------- A2
$[\frac{\partial \theta}{\partial x} \frac{\partial \theta}{\partial y}+ Sin^2(\theta)\frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial y}]=0$ ---------------- A3
We will look for a solution for which the following hold true:
SET B
$\frac{\partial \theta}{\partial x}=\psi$ ------------ B1
$\frac{\partial \theta}{\partial y}=\chi$ ------------ B2
$\frac{\partial \phi}{\partial x}=\frac{1}{Sin \theta}\chi$ -------------- B3
$\frac{\partial \phi}{\partial y}= - \frac{1}{Sin \theta}\psi$ ------------- B4
$\psi^2+\chi^2=1$ ----------- B5
Exact differential conditions:
[SET C]
$\frac{\partial \psi}{\partial y}=\frac {\partial \chi }{\partial x}$ ----------- C1
$\frac{\partial}{\partial y}\frac{1}{Sin \theta}{\chi}=-\frac{\partial}{\partial x}\frac{1}{Sin \theta}{\psi}$ -------------------- C2
Differentiating C2 we have:
$-\frac{Cos^2 \theta}{Sin \theta}\frac{\partial \theta}{\partial y}\chi+\frac{1}{Sin\theta}\frac{\partial \chi}{\partial y}= \frac{Cos^2 \theta}{Sin \theta}\frac{\partial \theta}{\partial x}\psi+\frac{1}{Sin\theta}\frac{\partial \psi}{\partial x}$
Using B1 and B2 in the above step we have,
$\frac{Cos^2 \theta}{Sin \theta}(\chi^2+\psi^2)=\frac{1}{Sin \theta}(\frac{\partial \psi}{\partial x}+\frac{\partial \chi}{\partial y})$
Since
$(\psi^2 + \chi^2)=1$
We may write:
$Cot \theta =\frac{\partial \psi}{\partial x}+\frac{\partial \chi}{\partial y}$
Therefore our final equation is:
$\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 \theta}{\partial y^2}=Cot \theta$ -------------- D
If the above equation is solvable we can flatten a sphere.
Theorema Egregium:The Gaussian curvature of a surface is invariant under local isometry.
Incidentally an isometry comes under diffeomorphisms which involve invertible functions which are differentiable 'r' times [or infinite times]. This is suggestive of linear functions[or bijections]. But the functions representing $\theta $and $\phi$ could be of non-linear nature]
[It is important to note that tensor equations are preserved in coordinate transformations where the value of the line element,ds^2, does not change. But in General Relativity the same tensor equations[ex: the Geodesic equation,Maxwell's Rquations in covariant form] are considered applicable in all manifolds--flat spacetime and curved spacetime. The stated equations are considered invariant in all manifolds]