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:





$d(\theta)=[ \partial{f_1} / \partial{p} ]dp + [ \partial{f_1}/ \partial{q}] dq$

$d(\phi)=[\partial{f_2}/\partial{p} ]dp + [\partial{f_2}/\partial{q}] dq$

$dx=[\partial{f_3}/\partial{p} ]dp + [\partial{f_3}/\partial{q}] dq$

$dy=[\partial{f_4}/\partial{p} ]dp + [\partial{f_4}/\partial{q}] dq$

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)


$ds^2=[(\partial{f_3}/\partial{p})^2+(\partial{f_4}/\partial {p})^2]dp^2+[(\partial{f_3}/\partial{q})^2+ (\partial{f_4}/\partial{q})^2]dq^2$ $+2[(\partial{f_3}/\partial{p} ) (\partial{f_3}/\partial{q})+ (\partial{f_4}/\partial{p}) (\partial{f_4}/\partial{q})]dpdq \qquad$ (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)




$d\theta=\frac{\partial \theta}{\partial x}{dx}+\frac{\partial \theta}{\partial y}{dy}$

$d\phi=\frac{\partial \phi}{\partial x}{dx}+\frac{\partial \phi}{\partial y}{dy}$

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:


$\frac{\partial \theta}{\partial x}=\psi$ ------------ 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:


$\frac{\partial \psi}{\partial y}=\frac {\partial \chi }{\partial x}$ ----------- 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,

$\frac{Cos^2 \theta}{Sin \theta}(\chi^2+\psi^2)=\frac{1}{Sin \theta}(\frac{\partial \psi}{\partial x}+\frac{\partial \chi}{\partial y})$


$(\psi^2 + \chi^2)=1$

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]

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    $\begingroup$ Seeing as the sphere isn't homeomorphic to a disk, no. $\endgroup$ – simplicity Nov 14 '11 at 2:25
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    $\begingroup$ Any metric can locally be flattened, but it won't work globally. $\endgroup$ – Cheerful Parsnip Nov 14 '11 at 2:42
  • $\begingroup$ If you want to leave the world of smooth manifolds with smooth metics, you can flatten the sphere almost everywhere. Think of a cone on a standard disk. I imagine that this has a curvature tensor that is distribution valued. $\endgroup$ – Baby Dragon Jul 16 '13 at 21:13

This is not possible, not even locally.

The first metric you mention is that of a piece of a sphere, which has constant, positive curvature. The second one, that of the plane, has zero curvature. The two are not locally isometric, because of the Theorema Egregium.

  • $\begingroup$ Tensor equations preserve their form in coordinate transformations where the line element,where ds^2 is preserved.But in General Relativity tensor equations like the Geodesic equation,Maxwell's Equations[expressed in covariant form]are preserved in all manifolds--flat spacetime and curved spacetime.How does one explain this? $\endgroup$ – Anamitra Palit Nov 14 '11 at 4:51
  • $\begingroup$ @AnamitraPalit, I have no idea what you mean! $\endgroup$ – Mariano Suárez-Álvarez Nov 14 '11 at 4:53
  • $\begingroup$ @AnamitraPalit: The form is the same, but the values are different. $\endgroup$ – Zhen Lin Nov 14 '11 at 7:34
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    $\begingroup$ Just for clarification, in my comment I was referring to diffeomorphism, not isometry. There certainly is a Riemanninan metric on the sphere which is flat in a neighborhood of some point. Clearly it is not isometric to the standard one! $\endgroup$ – Cheerful Parsnip Nov 14 '11 at 12:02
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    $\begingroup$ @AnamitraPalit, as I wrote above, the reason is Gauss's Theorema Egregium. This is explained in essentially every textbook on differential geometry of surfaces. $\endgroup$ – Mariano Suárez-Álvarez Nov 15 '11 at 5:55

Let’s solve the PDE:$$\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}=\cot (u)$$ ------------(1) Subject to the constraint :$$(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2=1$$ ----------------- (2)

We write PDE (1) as: $$\frac{\partial^2 [(u+F(u))-F(u)]}{\partial x^2}+ \frac{\partial^2 [(u+F(u))-F(u)]}{\partial y^2}=\cot (u)$$ ----------- (3)

PDE (3) may be represented as the following two equations: $$\frac{\partial^2 (u+F(u))}{\partial x^2}+ \frac{\partial^2 (u+F(u))}{\partial y^2}=0$$--------- (4)


$$\frac{\partial^2 F(u)}{\partial x^2}+ \frac{\partial^2 F(u)}{\partial y^2}=-\cot (u)$$--------------- (5)


$$\frac{\partial F(u)}{\partial x}=\frac{dF}{du} \frac{\partial u}{\partial x}$$ $$\frac{\partial^2 F(u)}{\partial x^2}=\frac {d^2 F}{du^2} (\frac {\partial u}{\partial x})^2+\frac{dF}{du}\frac{\partial^2 u}{\partial x^2}$$


$$\frac{\partial F(u)}{\partial y}=\frac{dF}{du} \frac{\partial u}{\partial y}$$ $$\frac{\partial^2 F(u)}{\partial y^2}=\frac {d^2 F}{du^2} (\frac {\partial u}{\partial y})^2+\frac{dF}{du}\frac{\partial^2 u}{\partial y^2}$$ PDE (5) may be written as:

$$\frac {d^2 F}{du^2} [(\frac {\partial u}{\partial x})^2+(\frac {\partial u}{\partial y})^2]+\frac{dF}{du}[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]=-\cot(u)$$ Or, $$\frac {d^2 F}{du^2}\times {1}+\frac{dF}{du}\times \cot(u) =-\cot(u)$$ Or, $$\frac {d^2 F}{du^2}+\cot(u)\frac{dF}{du} =-\cot(u)$$

The above ODE gives us the form for F to be used in PDE (4) which is Laplace’s equation in two dimensions

We obtain: $$ u+F(u)=Soln \;of \;Laplace’s\; eqn.\; in \; two \;dimensions$$

[Differentiating the above relation twice wrt to x and y and adding the results we have, $$F''(u)[(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2]+F'(u)[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]=-[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}]$$ Again F(u) satisfies relation (5):$$F''(u)+\cot(u)F'(u)=-\cot(u)$$ The simplest choice would be that the PDE's (1) and (2) are getting satisfied simultaneously.This a reflection of the fact that the constraint $\psi^2+\chi^2=1$ has been used in the construction of relation D in the Question]

In the final step the particular solutions for F and that for Laplace's equation are chosen in a manner such that equation (2) is satisfied.

[u has been used instead of $\theta$ ,which has been used in the previous postings]


A sphere may be flattened by a one to many mapping[strictly speaking these mappings are not functions because of their multiple valued nature]which preserves the value of $ds^2$ . But the metric coefficients are not preserved. The manifold itself changes. The transformed manifold contains multiple images of each point of the sphere.

In our example $\theta$ and $\phi$ [I am referring to the link in the question]are single-valued functions of the ordered pairs(x,y) but x and y may not be single-valued functions of ordered pairs $(\theta,\phi)$.

You may separate out different subsets from the original image set to get a bijective [one -one ,onto-mapping] for each subset . But for these subsets the curvature elements are not supposed to change[Theorema Ergregium]

So far as the differential equations in the question [or in the link provided in the question] are concerned $\theta$ and $\phi$ are single valued functions of the ordered pairs(x,y). So there is really nothing is wrong or unrigorous about the differential equations.

But we may start our calculations[referring to the ones in the link in the question] with

$x=x(\theta,\phi)$ And $y=y(\theta,\phi)$ The existence of solutions in either of the procedures will not indicate towards the violation of Theorema Egregium. But we can definitely flatten a sphere by suitable mappings for which the value of $ds^2$ is preserved.

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    $\begingroup$ To be honest, I do not understand what all this even means at all. $\endgroup$ – Mariano Suárez-Álvarez Nov 20 '11 at 15:50
  • $\begingroup$ We do have certain mappings by which a sphere may be flattened with the preservation of $ds^2$. This does not violate Theorema Egregeim.We can use mappings which do not belong to the category of one to one correspondence In fact your answer in incomplete [and incorrect] in the fact that it completely denies the possibility of flattening a sphere $\endgroup$ – Anamitra Palit Nov 21 '11 at 1:51
  • $\begingroup$ Well, I do not understand your comment at all either. I guess the simplest will be to simply drop the subject :) $\endgroup$ – Mariano Suárez-Álvarez Nov 21 '11 at 2:03
  • $\begingroup$ Did you see the calculations on the link[in the question].The equations are not unsolvable ones. $\endgroup$ – Anamitra Palit Nov 21 '11 at 3:29
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    $\begingroup$ -1: This answer is wrong. Gauss's theorema egregium asserts that curvature is a local isometry invariant, so there is no map which preserves the value of $ds^2$ and makes it flat. $\endgroup$ – Zhen Lin Nov 21 '11 at 8:52

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