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sagemanifolds
GitHub Repository: sagemanifolds/IntroToManifolds
Path: blob/main/04Manifold_Spheres.ipynb
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Kernel: SageMath 9.6

4. Spheres as manifolds

This notebook is part of the Introduction to manifolds in SageMath by Andrzej Chrzeszczyk (Jan Kochanowski University of Kielce, Poland).

version()
'SageMath version 9.6, Release Date: 2022-05-15'

One dimensional sphere S1\mathbf{S}^1


Example 4.1

In manifolds.Sphere in the one-dimensional case   S1={(x,y)R2:x2+y2=1}  \ \ S^1=\{(x,y)\in R^2: x^2+y^2=1\}\ \ the default coordinates are the spherical (polar) ones.

%display latex S1=manifolds.Sphere(1) # 1-dimensional sphere Phi=S1.embedding() # embedding S^1 -> E^2 Phi.disp() # show embedding

ι:S1E2on A:ϕ(x,y)=(cos(ϕ),sin(ϕ))\displaystyle \begin{array}{llcl} \iota:& \mathbb{S}^{1} & \longrightarrow & \mathbb{E}^{2} \\ \mbox{on}\ A : & \phi & \longmapsto & \left(x, y\right) = \left(\cos\left(\phi\right), \sin\left(\phi\right)\right) \end{array}

The spherical coordinates are predefined in manifolds.Sphere but we want to have a shorter name.

sph.<phi>=S1.spherical_coordinates() # spherical coordinates sph.coord_range() # coordinate range

ϕ: [π,π](periodic)\displaystyle \phi :\ \left[ -\pi , \pi \right] \mbox{(periodic)}

On  [π,π] \ [-\pi,\pi]\ the map φ(cosφ,sinφ)\varphi \to (\cos\varphi, \sin\varphi) is not one-to one, so not homeomorphic.

To represent graphically the sphere S1S^1 we need the ambient space R2R^2 and the mapping S1R2S^1 \to R^2.

E=S1.ambient() # ambient space E^2 c_cart.<x,y> = E.cartesian_coordinates() # Cartesian coord in R^2 p=sph.plot(c_cart,mapping=Phi,number_values=15, thickness=1,color='grey') # plot S^1 p.show(figsize=[3,2]) # show plot
Image in a Jupyter notebook

Restricting the coordinate to an open interval contained in [π,π][-\pi,\pi] we obtain a homeomorphic map (but the image is a proper subset of the sphere).

p=sph.plot(c_cart,mapping=Phi,ranges={phi:(-pi,pi/4)}, number_values={phi:50},color={phi:'grey'}) # plot subset of S^1 p.show(figsize=[3,2]) # show plot
Image in a Jupyter notebook

Example 4.2

S1S^1 can be defined as a manifold with two-element atlas. Note that in this new example we do not use the manifolds.Sphere command!

# example from the Manifolds manual, sect 2.7.1 (p.546 in 9.5 version) M = Manifold(1, 'S^1') # manifold S^1 U = M.open_subset('U') # the complement of one point (1,0) c_u.<u> = U.chart('u:(0,2*pi)') # the standard angle coordinate V = M.open_subset('V') # the complement of the point (-1,0) c_v.<v> = V.chart('v:(0,2*pi)') # the angle u-pi M.declare_union(U,V) # S^1 is the union of U and V M.atlas() # atlas

[(U,(u)),(V,(v))]\displaystyle \left[\left(U,(u)\right), \left(V,(v)\right)\right]

On the intersection of W=UVW=U\cap V two "new" charts are defined, they are just restrictions of the "old" maps to WW.

u_to_v = c_u.transition_map(c_v, (u-pi,), # transition map u->v intersection_name='W', restrictions1 = u!=0, restrictions2 = v!=pi) v_to_u = u_to_v.inverse() # inverse transition M.atlas() # atlas

[(U,(u)),(V,(v)),(W,(u)),(W,(v))]\displaystyle \left[\left(U,(u)\right), \left(V,(v)\right), \left(W,(u)\right), \left(W,(v)\right)\right]

If we need to use W=UVW=U\cap V in calculations we can define it:

W = U.intersection(V) # intersection of print(W) # chart domains
Open subset W of the 1-dimensional differentiable manifold S^1

To make the plot in R2R^2 we need to define an ambient space and the embedding.

R2 = Manifold(2, 'R^2') # manifold R^2 X.<x,y> = R2.chart() # Cartesian coordinates # the embedding is not predefined this time, so it must be # defined: F = M.continuous_map(R2, {(c_u, X): [cos(u),sin(u)]}, name='F') p=c_u.plot(X,mapping=F,color='grey') # plot image of S^1 p.show(figsize=(3,2)) # show plot
Image in a Jupyter notebook

Stereographic coordinates on S1\mathbf{S}^1


Example 4.3

In manifold.Sphere not only the Cartesian and polar but also the stereographic coordinates are predefined.

S1=manifolds.Sphere(1) # stereographic projection from the S1.stereographic_coordinates(pole='north',names='u') # North pole

(S1{NP},(u))\displaystyle \left(\mathbb{S}^{1}\setminus\{\mathrm{NP}\},(u)\right)

# stereographic projection from the S1.stereographic_coordinates(pole='south',names='u') # South pole

(S1{SP},(u))\displaystyle \left(\mathbb{S}^{1}\setminus\{\mathrm{SP}\},({u'})\right)


Stereographic projection of S1S^1 from the North pole N=(0,1)N = (0, 1)

var('t') # symb. variable p1=parametric_plot((cos(t),sin(t)),(t,0,2*pi),color='grey') # plot the circle p2=plot(1-x/2,(x,0,2)) # half line through (0,1) and (x,y) p3= text("(x,y)",(0.88,0.72)) # point (x,y) p4= text("(u,0)",(2.0,0.15)) # point (u,0) (p1+p2+p3+p4).show(figsize=[3,2]) # combine plots
Image in a Jupyter notebook

Consider the line which passes through the North pole (0,1)(0, 1), the point (x,y)(x, y) on the circle and intersects the axis Ox at the point (u,0)(u, 0). The function u=u(x,y)u = u(x, y) defines the stereographic projection. Three points: North pole, the point (x,y)(x, y) on the circle and the point (u,0)(u, 0) are on the same line. Since the right triangle with the hypotenuse joining (0,1), (x,y) and the right triangle with the hypotenuse joining (0,1),(u,0) are similar, we have ux=11y\frac{u}{x}=\frac{1}{1-y} and consequently u=x1yu=\frac{x}{1-y}. Solving the system of equations x2+y2=1,u=x1y, x^2+y^2=1,\quad u=\frac{x}{1-y}, with respect to (x,y)(x,y):

%display latex forget() # forget previous assumptions var('x, y, u') # symbolic variables assume(y>0) # assume y>0 s=solve([x^2+y^2==1,u==x/(1-y)],[x,y]);s # solve eq. above

[[x=2uu2+1,y=u21u2+1],[x=0,y=1]]\displaystyle \left[\left[x = \frac{2 \, u}{u^{2} + 1}, y = \frac{u^{2} - 1}{u^{2} + 1}\right], \left[x = 0, y = 1\right]\right]

we obtain the relations x=2uu2+1,y=u21u2+1,u=x1y. x=\frac{2u}{u^2+1},\quad y=\frac{u^2-1}{u^2+1}, \quad u=\frac{x}{1-y}.


Now consider the South pole case.

Stereographic projection of S1S^1 from the South pole S=(0,1)S = (0, -1)

var('t') # symbolic variable p1=parametric_plot((cos(t),sin(t)),(t,0,2*pi),color='grey') # plot the circle p2=plot(x/2-1,(x,0,2)) # halfline through (0,-1) and (x,y) p3= text("(x,y)",(0.88,-0.72)) # point (x,y) p4= text("(u',0)",(2.0,0.15)) # point (u',0) (p1+p2+p3+p4).show(figsize=[3,2]) # combine plots
Image in a Jupyter notebook

Since the right triangle with the hypotenuse joining (0,-1), (x,y) and the right triangle with the hypotenuse joining (0,-1),(u',0) are similar we have ux=11(y)\frac{u'}{x}=\frac{1}{1-(-y)} (yy is negative now) and consequently u=x1+yu'=\frac{x}{1+y}. Solving the system of equations x2+y2=1,u=x1+y, x^2+y^2=1,\quad u'=\frac{x}{1+y}, with respect to (x,y)(x,y):

forget() # forget previous assumptions var('x, y, up') # symbolic variables assume(y<0) # assume y<0 s=solve([x^2+y^2==1,up==x/(1+y)],[x,y]);s # solve eq. above

[[x=2uu2+1,y=u21u2+1],[x=0,y=(1)]]\displaystyle \left[\left[x = \frac{2 \, {u'}}{{u'}^{2} + 1}, y = -\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\right], \left[x = 0, y = \left(-1\right)\right]\right]

we obtain the relations x=2uu2+1,y=1u2u2+1,u=x1+y. x=\frac{2u'}{u'^2+1},\quad y=\frac{1-u'^2}{u'^2+1}, \quad u'=\frac{x}{1+y}.


Example 4.4

As we mentioned, both the stereographic projections from North and South poles are predefined in manifolds.Sphere.

reset() S1=manifolds.Sphere(1) # S^1 from manifolds.Sphere # stereographic projection from the North pole: stereoN.<u> = S1.stereographic_coordinates(pole='north') # stereographic projection from the South pole: stereoS.<up> = S1.stereographic_coordinates(pole='south') Phi=S1.embedding() # embedding S^1 -> E^2 Phi.disp() # show embedding

ι:S1E2on A:ϕ(x,y)=(cos(ϕ),sin(ϕ))on S1{NP}:u(x,y)=(2uu2+1,u21u2+1)on S1{SP}:u(x,y)=(2uu2+1,u21u2+1)\displaystyle \begin{array}{llcl} \iota:& \mathbb{S}^{1} & \longrightarrow & \mathbb{E}^{2} \\ \mbox{on}\ A : & \phi & \longmapsto & \left(x, y\right) = \left(\cos\left(\phi\right), \sin\left(\phi\right)\right) \\ \mbox{on}\ \mathbb{S}^{1}\setminus\{\mathrm{NP}\} : & u & \longmapsto & \left(x, y\right) = \left(\frac{2 \, u}{u^{2} + 1}, \frac{u^{2} - 1}{u^{2} + 1}\right) \\ \mbox{on}\ \mathbb{S}^{1}\setminus\{\mathrm{SP}\} : & {u'} & \longmapsto & \left(x, y\right) = \left(\frac{2 \, {u'}}{{u'}^{2} + 1}, -\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\right) \end{array}


Example 4.5

To show graphically how the projection from the North pole acts, let us extract an appropriate part of the definition from the previous cell.

# continuation E=S1.ambient() # ambient space E^2 c_cart.<x,y> = E.cartesian_coordinates() # Cartesian coord. fun=list(Phi.coord_functions(stereoN).expr()) # embedd. functions PhiN = S1.continuous_map(E,{(stereoN, c_cart): fun}, name='PhiN',latex_name=r'\Phi_N') # define embedding PhiN.display() # show embedding

ΦN:S1E2on S1{NP}:u(x,y)=(2uu2+1,u21u2+1)\displaystyle \begin{array}{llcl} \Phi_N:& \mathbb{S}^{1} & \longrightarrow & \mathbb{E}^{2} \\ \mbox{on}\ \mathbb{S}^{1}\setminus\{\mathrm{NP}\} : & u & \longmapsto & \left(x, y\right) = \left(\frac{2 \, u}{u^{2} + 1}, \frac{u^{2} - 1}{u^{2} + 1}\right) \end{array}

Let us plot the set of points (x,y)(x,y) corresponding to u(10,10)u\in (-10,10).

p=stereoN.plot(c_cart,mapping=PhiN,ranges={u:(-10,10)}, number_values={u:50},color={u:'grey'}, plot_points=500) # image of (-10,10) under PhiN p.show(figsize=[3,2]) # show the image
Image in a Jupyter notebook

Thus the points of the grey arc in the figure are projected onto the interval (-10,10).


Example 4.6

Now let us extract the definition of the projection from the South pole,

# continuation fun=list(Phi.coord_functions(stereoS).expr()) # embedd. functions PhiS = S1.continuous_map(E,{(stereoS, c_cart): fun}, name='PhiS',latex_name=r'\Phi_S') # define embedding PhiS.display() # show embedding

ΦS:S1E2on S1{SP}:u(x,y)=(2uu2+1,u21u2+1)\displaystyle \begin{array}{llcl} \Phi_S:& \mathbb{S}^{1} & \longrightarrow & \mathbb{E}^{2} \\ \mbox{on}\ \mathbb{S}^{1}\setminus\{\mathrm{SP}\} : & {u'} & \longmapsto & \left(x, y\right) = \left(\frac{2 \, {u'}}{{u'}^{2} + 1}, -\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\right) \end{array}

and check which points of the circle are projected onto (-10,10):

# continuation p=stereoS.plot(c_cart,mapping=PhiS,ranges={up:(-10,10)}, number_values={up:50},color={up:'grey'}, # plot image of (-10,10) plot_points=500) p.show(figsize=[3,2]) # show image
Image in a Jupyter notebook

Transition map from u coordinate to u' coordinate

Using the relations  u=x1y,  u=x1+y\ u=\frac{x}{1-y},\ \ u'=\frac{x}{1+y} and y=u21u2+1, y=\frac{u^2-1}{u^2+1},\ we obtain u=x1+y=u(1y)1+y=1y1+yu.  u'=\frac{x}{1+y}=\frac{u(1-y)}{1+y}=\frac{1-y}{1+y}u.\ \ Replacing yy by u21u2+1  \frac{u^2-1}{u^2+1}\ \ we get  1y1+y=1u21u2+11+u21u2+1=u2+1u2+1u2+1+u21=22u2=1u2. \ \frac{1-y}{1+y}=\frac{1-\frac{u^2-1}{u^2+1}}{1+\frac{u^2-1}{u^2+1}}= \frac{u^2+1-u^2+1}{u^2+1+u^2-1}=\frac{2}{2u^2}=\frac{1}{u^2}.\ Accordingly u=1y1+yu=1u, u'=\frac{1-y}{1+y}u=\frac{1}{u},\ so the transition map from uu coordinate to uu' coordinate has the form  u=1u. \ u'=\frac{1}{u}.\ It is smooth if  u0,u0.\ u\not=0, u'\not=0.

Here is SageMath solution (eliminate x,y variables from equations of the first line in the previous computations):

%display latex var('x,y,u,up') maxima.eliminate([u==x/(1-y),up==x/(1+y), # eliminate var. x,y y==(u^2-1)/(u^2+1)],[x,y]).sage() # from eqs of projections

[2(uu1)u]\displaystyle \left[2 \, {\left(u {u'} - 1\right)} u\right]

The answer means that the bracket vanishes. Since u0u\not=0, the expression in the bracket vanishes if   u=1u\ \ u'=\frac{1}{u}.


Example 4.7

In SageMath Manifolds the transition from one coordinate system to the other can be defined as follows:

S1=manifolds.Sphere(1) # S^1 from manifolds.Sphere # stereographic projection from the North pole: stereoN.<u> = S1.stereographic_coordinates(pole='north') # stereographic projection from the South pole: stereoS.<up> = S1.stereographic_coordinates(pole='south') trans = stereoN.transition_map(stereoS, 1/u, # define transition intersection_name='W', # stereoN -> StereoS restrictions1= u!=0, # specify restrictions2 = up!=0) # restrictions trans.display() # show transition

u=1u\displaystyle \begin{array}{lcl} {u'} & = & \frac{1}{u} \end{array}

In some cases, the definition of the inverse transition can be left to the SageMath Manifolds.

trans.inverse().display()

u=1u\displaystyle \begin{array}{lcl} u & = & \frac{1}{{u'}} \end{array}

As we can see the transition map and its inverse are smooth.

Some transition maps are predefined in manifolds.Sphere.


Example 4.8

Let us check that the transitions from the previous example are predefined in manifolds.Sphere.

dim=1 # dim of the sphere Sph=manifolds.Sphere(dim) # sphere S^1 spher = Sph.spherical_coordinates() # spherical coord. stereoN, stereoS = Sph.coordinate_charts('stereographic', names=['u']) # sterogr. projections A=stereoN.domain() # domain of stereoN # intersection of domains of stereoN and stereo S: W = Sph._stereoN_dom.intersection(Sph._stereoS_dom) # transition from stereoN to stereoS: FNS=Sph.coord_change(stereoN.restrict(W),stereoS.restrict(W)) # transition from stereoS to stereoN: FSN=Sph.coord_change(stereoS.restrict(W),stereoN.restrict(W)) FNS.disp(),FSN.disp() # show transitions

(u=1u,u=1u)\displaystyle \left(\begin{array}{lcl} {u'} & = & \frac{1}{u} \end{array}, \begin{array}{lcl} {u'} & = & \frac{1}{u} \end{array}\right)

We can check that both transitions can be obtained as compositions of other transitions.

# continuation A = spher.domain() # domain of spher.coord # intersection of domains of stereoN and stereo S: W = Sph._stereoN_dom.intersection(Sph._stereoS_dom) V=W.intersection(A) # intersection of domains of # stereoN, stereoS and spher F1=Sph.coord_change(spher.restrict(V), stereoN.restrict(V)) # transition spher -> stereoN F2=Sph.coord_change( stereoN.restrict(V), spher.restrict(V)) # transition stereoN -> spher F3=Sph.coord_change(spher.restrict(V), stereoS.restrict(V)) # transition spher -> stereoS F4=Sph.coord_change( stereoS.restrict(V),spher.restrict(V)) # transition stereoS -> spher (F1*F4).disp(),(F3*F2).disp() # show compositions

(u=1u,u=1u)\displaystyle \left(\begin{array}{lcl} u & = & \frac{1}{{u'}} \end{array}, \begin{array}{lcl} {u'} & = & \frac{1}{u} \end{array}\right)


Example 4.9

Knowing the transition maps we are ready to define S1S^1 as a manifold with two maps.

Note that this time, our calculations are independent of manifolds.Sphere.

M = Manifold(1, 'S^1') # sphere S^1 U = M.open_subset('U') # complement of the North pole cU.<u> = U.chart() # sterogr. proj. from North pole V = M.open_subset('V') # complement of the South pole cV.<v> = V.chart() # sterogr. proj. from South pole M.declare_union(U,V) # M is the sum of U and V trans = cU.transition_map(cV, 1/u, intersection_name='W', restrictions1= u!=0, # transition cU -> cV restrictions2 = v!=0) trans_inv=trans.inverse() # inverse transition trans.disp(),trans_inv.disp() # show transitions

(v=1u,u=1v)\displaystyle \left(\begin{array}{lcl} v & = & \frac{1}{u} \end{array}, \begin{array}{lcl} u & = & \frac{1}{v} \end{array}\right)


Two dimensional sphere S2S^2

Spherical coordinates in S2\mathbf{S}^2

In manifolds.Sphere in two-dimensional case   S2={(x,y,z)R3:x2+y2+z2=1}  \ \ S^2=\{(x,y,z)\in R^3: x^2+y^2+z^2=1\}\ \ the default coordinates are the spherical ones.

var('u v t') # symb.var. po1={'thickness':5,'color':'darkblue'} # parameters po2={'fontsize':20,'color':'black'} po3={'size':7,'color':'black'} ax =line3d([(0,0,0), (1+0.15,0,0)], **po1) # axes ax+=line3d([(0,0,0), (0,1+0.15,0)], **po1) ax+=line3d([(0,0,0), (0,0,1+0.15)], **po1) ax+=text3d("x",(1.25,0,0),**po2) ax+=text3d("y",(0,1.25,0),**po2) ax+=text3d("z",(0.,0.,1.25),**po2)
# semisphere: s=parametric_plot3d((cos(u)*cos(v), sin(u)*cos(v), sin(v)), (u,0,2*pi), (v,0,pi/2),opacity=0.9,color='lightgrey') a=0.59 # triangle tr=line3d([(0,0,0),(a,a,0),(a,a,a),(0,0,0)],**po1) dots=point3d([(0.5*cos(t),0.5*sin(t),0) # dots for t in srange(0,pi/4,0.1)], **po3) dots+=point3d([(0.5*cos(t),0.5*cos(t),0.5*sin(t)) for t in srange(pi/4,pi/2,0.1)],**po3) t=text3d("(x,y,z)",(0.6,0.8,0.7),**po2) # variables t+=text3d("φ",(0.7,0.3,0.0),**po2) # names t+=text3d("θ",(0.,0.2,0.5),**po2) # combine plots: (ax+s+tr+dots+t).rotateZ(-pi/8).show(frame=False)