{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 4. Spheres as manifolds\n",
"\n",
"This notebook is part of the [Introduction to manifolds in SageMath](https://sagemanifolds.obspm.fr/intro_to_manifolds.html) by Andrzej Chrzeszczyk (Jan Kochanowski University of Kielce, Poland)."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"'SageMath version 9.6, Release Date: 2022-05-15'"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"version()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## One dimensional sphere $\\mathbf{S}^1$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.1**\n",
"\n",
"In `manifolds.Sphere` in the one-dimensional case \n",
"$\\ \\ S^1=\\{(x,y)\\in R^2: x^2+y^2=1\\}\\ \\ $ the default coordinates are the spherical (polar) ones."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{1} & \\longrightarrow & \\mathbb{E}^{2} \\\\ \\mbox{on}\\ A : & \\phi & \\longmapsto & \\left(x, y\\right) = \\left(\\cos\\left(\\phi\\right), \\sin\\left(\\phi\\right)\\right) \\end{array}$"
],
"text/plain": [
"iota: S^1 → E^2\n",
"on A: phi ↦ (x, y) = (cos(phi), sin(phi))"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"S1=manifolds.Sphere(1) # 1-dimensional sphere\n",
"Phi=S1.embedding() # embedding S^1 -> E^2\n",
"Phi.disp() # show embedding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The spherical coordinates are predefined in `manifolds.Sphere` but we want to have a shorter name."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\phi :\\ \\left[ -\\pi , \\pi \\right] \\mbox{(periodic)}\\)"
],
"text/latex": [
"$\\displaystyle \\phi :\\ \\left[ -\\pi , \\pi \\right] \\mbox{(periodic)}$"
],
"text/plain": [
"phi: [-pi, pi] (periodic)"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"sph.=S1.spherical_coordinates() # spherical coordinates\n",
"sph.coord_range() # coordinate range"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"On $\\ [-\\pi,\\pi]\\ $ the map $\\varphi \\to (\\cos\\varphi, \\sin\\varphi) $ is not one-to one, so not homeomorphic."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To represent graphically the sphere $S^1$ we need the ambient space $R^2$ and the mapping\n",
"$S^1 \\to R^2$."
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"data": {
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\n",
"text/plain": [
"Graphics object consisting of 1 graphics primitive"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"E=S1.ambient() # ambient space E^2\n",
"c_cart. = E.cartesian_coordinates() # Cartesian coord in R^2\n",
"p=sph.plot(c_cart,mapping=Phi,number_values=15,\n",
" thickness=1,color='grey') # plot S^1\n",
"p.show(figsize=[3,2]) # show plot"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Restricting the coordinate to an open interval contained in $[-\\pi,\\pi]$ we obtain a homeomorphic map (but the image is a proper subset of the sphere)."
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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\n",
"text/plain": [
"Graphics object consisting of 1 graphics primitive"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"p=sph.plot(c_cart,mapping=Phi,ranges={phi:(-pi,pi/4)},\n",
" number_values={phi:50},color={phi:'grey'}) # plot subset of S^1\n",
"p.show(figsize=[3,2]) # show plot"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.2**\n",
"\n",
"$S^1$ can be defined as a manifold with **two-element atlas**. Note that in this new example we do not use the `manifolds.Sphere` command!"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\left(U,(u)\\right), \\left(V,(v)\\right)\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left(U,(u)\\right), \\left(V,(v)\\right)\\right]$"
],
"text/plain": [
"[Chart (U, (u,)), Chart (V, (v,))]"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# example from the Manifolds manual, sect 2.7.1 (p.546 in 9.5 version)\n",
"M = Manifold(1, 'S^1') # manifold S^1\n",
"U = M.open_subset('U') # the complement of one point (1,0)\n",
"c_u. = U.chart('u:(0,2*pi)') # the standard angle coordinate\n",
"V = M.open_subset('V') # the complement of the point (-1,0)\n",
"c_v. = V.chart('v:(0,2*pi)') # the angle u-pi\n",
"M.declare_union(U,V) # S^1 is the union of U and V\n",
"M.atlas() # atlas"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"On the intersection of $W=U\\cap V$ two \"new\" charts are defined, they are just restrictions of the \"old\" maps to $W$."
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\left(U,(u)\\right), \\left(V,(v)\\right), \\left(W,(u)\\right), \\left(W,(v)\\right)\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left(U,(u)\\right), \\left(V,(v)\\right), \\left(W,(u)\\right), \\left(W,(v)\\right)\\right]$"
],
"text/plain": [
"[Chart (U, (u,)), Chart (V, (v,)), Chart (W, (u,)), Chart (W, (v,))]"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"u_to_v = c_u.transition_map(c_v, (u-pi,), # transition map u->v\n",
" intersection_name='W',\n",
" restrictions1 = u!=0, \n",
" restrictions2 = v!=pi)\n",
"v_to_u = u_to_v.inverse() # inverse transition\n",
"M.atlas() # atlas"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"If we need to use $W=U\\cap V$ in calculations we can define it:"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Open subset W of the 1-dimensional differentiable manifold S^1\n"
]
}
],
"source": [
"W = U.intersection(V) # intersection of\n",
"print(W) # chart domains"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To make the plot in $R^2$ we need to define an ambient space and the embedding."
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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\n",
"text/plain": [
"Graphics object consisting of 1 graphics primitive"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"R2 = Manifold(2, 'R^2') # manifold R^2\n",
"X. = R2.chart() # Cartesian coordinates\n",
"# the embedding is not predefined this time, so it must be\n",
" # defined:\n",
"F = M.continuous_map(R2, {(c_u, X): [cos(u),sin(u)]}, name='F')\n",
"p=c_u.plot(X,mapping=F,color='grey') # plot image of S^1\n",
"p.show(figsize=(3,2)) # show plot"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"### Stereographic coordinates on $\\mathbf{S}^1$\n",
"\n",
"
\n",
"\n",
"**Example 4.3**\n",
"\n",
"In `manifold.Sphere` not only the Cartesian and polar but also the stereographic coordinates are predefined."
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\mathbb{S}^{1}\\setminus\\{\\mathrm{NP}\\},(u)\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\mathbb{S}^{1}\\setminus\\{\\mathrm{NP}\\},(u)\\right)$"
],
"text/plain": [
"Chart (S^1-{NP}, (u,))"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"S1=manifolds.Sphere(1) # stereographic projection from the\n",
"S1.stereographic_coordinates(pole='north',names='u') # North pole"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\mathbb{S}^{1}\\setminus\\{\\mathrm{SP}\\},({u'})\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\mathbb{S}^{1}\\setminus\\{\\mathrm{SP}\\},({u'})\\right)$"
],
"text/plain": [
"Chart (S^1-{SP}, (up,))"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
" # stereographic projection from the\n",
"S1.stereographic_coordinates(pole='south',names='u') # South pole"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"### Stereographic projection of $S^1$ from the North pole $N = (0, 1)$"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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4cJuLl3zhwgXq6+vx9/dvdtzf35+S66ONXyUwMJDPP/+cDRs2sHHjRsLDw7n99ttJSUlptZy33noLjUZjSkFBQWZ9DnulT58+xMfHc+bMGY4fP95iHmlA9oNNGU9GRgaXLl1ixowZaktplesnMAohWp3UGB4ezqJFixg1ahRxcXF88sknzJo1i/fff7/V+y9fvhytVmtKRUVFZtVvz4SEhHDTTTexc+dO6uvrW80nDcj2sRnjqa2tJSUlhZiYGJsaUG7Ex8cHBweHG1o3paWlN7SC2iI2NpacnJxWzzs7O+Pp6dksSa4xffp0Ll68yMGDB9vNKw3IdrEZ40lPT8dgMDC56YZINoSTkxMxMTHs2LGj2fEdO3Z0av1YZmYmgdcHjZF0GH9/f6Kjo0lOTqaurq5D10gDsj1swnj0ej3p6emMGjXKpneI+Nvf/saXX37JV199xfHjx3n++ecpLCxk8eLFgLGb9FiTEHgfffQRP/30Ezk5ORw7dozly5ezYcMGljWG2JN0iSlTplBdXd2hVk9TpAHZDn3VFgCwb98+9Hq9zYe7eOihhygvL+eNN96guLiYyMhItmzZwtCrGz8VFxc3m9NTV1fHiy++yNmzZ3F1deWWW24hISGBu+66S61H6BH079+fESNGsHfvXsaMGdPpt5+NBrRs2bVwHF980fm94SVdR/VFovX19Xz44YeEh4cze/ZsS2mxW3rrItH2OH/+PKtWreL+++8nMjKyW/fqzt7wvYSet0g0KyuLqqoqYmNj1ZYisSP8/f0JCQkhPT292/eSXTDro7rx7Nu3j+DgYJt8k6UmvTEeT2cZN24cZ86c4ezZs2a5nzQg66Gq8RQXF1NUVMTYsWPVlGGTLF26lKysLPbt26e2FJtl+PDheHl5maXV0xRpQJZHVePZv38/Hh4evSZwu8S89OnTh5iYGI4fP45erzf7/aUBWQ7VjMdgMJCVlUV0dLRNhb2Q2BdRUVGm75KlkAZkflT7jT916hS1tbVERUWpJUHSA/D09CQkJITDhw9boSxpQOZCNeM5evQo/v7++Pn5qSVB0kOIjo6moKCAS5cuWaU8aUDdRxXjqaurIzs7W7Z2JGYhIiICBwcHq0d2lAbUdVQxntzcXAwGAxEREWoUbxfI1+kdx9nZmZCQELKzs1UpXxpQ51HFeHJycvD29mbAgAFqFG8XyNfpnSMsLIyCggJqa2tV0yANqONY3XiEEOTk5Mi4wxKzEhYWRkNDA6dOnVJbijSgDmB14zl//jyVlZWEhYVZu2hJD0aj0eDv78/p06fVlmJCGlDrWN148vPzcXBwkBv0SczO0KFDKSgoUFvGDUgDuhGrG09RURGDBg2ib1+biMgh6UEMGTKEiooKKisr1ZbSItKArmFV4xFCUFhYKFs7EovQ+L1qb58ztbGGAZWXg58f5Oeb534toShEKQpnFAX3zl5rVeOpqKjg8uXL0ngkFsHDwwMvLy+b7G61hCUN6K23YPZsCA7u+j0qKmDBAlAUtFfTGkWhf+N5ITgKZADPd/beVjWexkDpAwcOtGaxdomcx9M1Bg8e3Op2Q7aKuQ2opgZWr4annuqervnz4dAhAGZeTbcCa67L9jWwRFHoVBhIqxpPaWkp7u7uuLt3umXW65DzeLqGn58fpaWlN2z6Zw+Yy4B+/RX69oW4OOPnb76B/v2b5/npJ2hlVyYAjh+HrVvhyy9BCFKFIBVYBNytKDQNJ7ENGADc1nGFKhiPXJslsST+/v7o9Xq0Wq3aUrpMdw0oJQVGj+6ehtRU0Ghg3Lhrx4QgDdAC45scqwMOA5M6c39pPJIeReMeZ+fPn1dZSffpqgHl50N3RzNKSoyD0y1QCgRcd+wsENyZ+1vNeAwGAxcvXpTGI7EoHh4eODk5UV5errYUs9FZA6qpAReX7pfbSldM4caNHGoAt87c22rGo9PpEELg5eVlrSIlvRBFUUy7cvQ0OmpAPj7GN1KN9OkD1w95XbnSdlkBAdBKo9EXuP6MN1DW0ecAKxsPILdokVicnmo8jbRnQCNHQtOAjL6+UFkJVVXXjl19W9UqcXGg1UJGxrVjisI4QAPsvS57JJDZmWewmvE0fhGk8UgsTU83nkZaM6ADB+DYsWutnnHjwM3NmPfUKfjXv4xvutri5pth5kzjHmOKQqyiEAt8AfwiBCcb8ykKwcAg4LfOaLdqi8fV1RVHR0drFWnXyHk8XUej0VgtGqEtcL0Bbd8O9fUwZ46xBeTtDd9/D1u2QFSU0aBWrGh+j/x845hOUtK1Yz/8YMwPbL+ajgALrit+HrBdCDo1a9NqxlNZWWnT+6LbGnIeT9dxd3entraWhoYGtaVYlaYGtGAB7N4Nw4YZu2CjR0NOjnHgefNmY0um6bhPfr5xrk909LVjjYYlBJ5X06NCcKnxvKLgDCwB/t5ZrVYzHr1ej4s5htolknZwczO+YFEzKJiaeHrCt9/C//wPPPdcx17Db91qNK1OvvsZCrwpBHs6q9GqxuPs7Gyt4iS9mMY/cDU1NSorUZeXXoJ33+3Ya/i334b/+q/O3V8IsoXgs65ok8Yj6XE0hlypr69XWYltYIvhOKxmPFeuXJEDyxKr4OBgXK9oMBhUVmJb2JIBWc14hBAoba1Kk0jMRKPxXGlvllwvxRYMyCJhABVFUa6fR1FTU0N1dbVpIqGkZfR6fbN9wBuj6cl66zjFxcXU1tZy+PBhOVO+HZYtg8ceg88/h48/Nv6blgY33XQtj0aj8QQqhRmX/CuWCB+gKIonxlWsEomkZ6ARQpjtr5+ljEfRarXNJlGsXr2abdu28eWXX7Y6e3nMmDGtzltp61x3z+t0OoKCgigqKlJd2/UtnuLiYsaOHUtWVhaDBg1SVdv1tFdvliy7rfNlZWV8+eWXfPfdd6Snp9uUtu5+19TQptFoNJi5xWORrlZLAj08PHBzc8PT07PVCndwcOjSOXOcB2xam4eHh81qa63eLF12a+erqqpwcXFBCGFz2hrp6ndNDW3mbOk0YrXBZScnJ5ycnNrMs3Tp0i6dM8f59pDaOo+ly27tfOOgcluDy2pp6wi2rM1cWKSrdZVmN167di1bt27ln//8p80tFNXpdKaFhbam7cyZM6bm7+DBg9WW0wxbrbeTJ0/y448/8t5773Hu3Dmb0mardQZtajP762irtXjc3NwICQmxyUmEzs7OvPbaazarrem/toSt1lvjUokXX3zR5rTZap2BdbVZrcWTmJjIwYMHeeGFFyxVXo/Elv9C2iqpqakkJibyyiuvqC2lp2C/LR4PDw+qqqp63YrhriLDYnQdnU4nTdrGsarxCCGorq62VpF2jQyL0XWk8dg+VjUeoFdEhpOoi1arxTj1RGKrWM14BgwYAMDFixetVaSkl3Lp0iVpPDaORY3nzTffZPz48bi5ueHv70+/fv3a3XZECMGKFSsYOHAgrq6uTJkyhWPHjplVV0VFBQsWLECj0aDRaFiwYEG7oTIff/xxFEVplmJjY82qy9b45JNPCAkJwcXFhZiYGHbv3t1q3qSkpBvqR1EUTpw4YXZdKSkpzJ49m4EDB6IoCj/99JPpXFVVFVVVVfj6+ja7Jjk5mZiYGFxcXBg2bBirVq0yu672tLWEtertrbfeYsyYMXh4eODn58d9993HyZMn270uOTkZRVEOKIpSqyhKrqIoi82hx6LGU1dXxwMPPMCSJUsAY6unPeN59913+eCDD/j444/Zt28fAQEBzJgxw7RY0hzMnz+fQ4cOsXXrVrZu3cqhQ4dYsOD6ULI3MnPmTIqLi01py5YtZtNka6xbt47nnnuOV199lczMTCZNmsSdd95JYWFhm9edPHmyWR0NHz7c7NqqqqqIjo7m448/vuFcWZlxl5Wm+7fl5eVx1113MWnSJDIzM3nllVf4y1/+woYNG6yqrS0sXW/JycksXbqUtLQ0duzYgcFgID4+nqqmW09cR2O9AbuBkcD/AP+rKMr93RYkhLBUMvH1118LjUYjEhISxMcffyxao6GhQQQEBIi3337bdKy2tlZoNBqxatWqVq/rDFlZWQIQaWlppmOpqakCECdOnGj1uoULF4p7773XLBo6g1arFYDQarVWLXfs2LFi8eLFzY5FRESIl19+ucX8iYmJAhAVFRVWUHcNQGzatMn0OT09Xfz9738XBoPBdOy///u/RURERLPrnnnmGREbG2tVbS2hVr2VlpYKQCQnJ7eap0m9mX6vgVVAquimP1h1C+OBAwdy4cKFZosgm5KXl0dJSQnx8fGmY87Oztx2223s3Xv9Vj5dIzU1FY1Gw7gmm0LHxsai0WjaLSMpKQk/Pz/CwsJYtGgRpaWlZtFka9TV1XHgwIFmPweA+Pj4duto5MiRBAYGcvvtt5OYmGhJmS1SWlqKj4+PKSYPGH/m1z/LHXfcwf79+20mZo+1663xJY+3t3ereVqqN2AbMFpRlG5F9bO68YBxxXVLlJSUANf2v27E39/fdK67lJSUtLiNsp+fX5tl3Hnnnfzwww/s2rWLf/zjH+zbt49p06a1aqLdRc15PBcuXKC+vr5TP4fAwEA+//xzNmzYwMaNGwkPD+f2228nJSXFGpJNtPTzLSkpafFZDAYDFy5csKa8G1Cj3oQQ/O1vf2PixIlERka2mq+lesO4i2hfwKc7Gjq9Ol1RlBXAa+1kGyNamBHt4+ODo6MjZ8+eJTg4uK0ymn0WHYheuGLFCl5//fU28zTOiWnpXu2V8dBDD5n+HxkZyejRoxk6dCgJCQnMnTu3zXK7wtKlS1m6dKlp5rIadObnEB4eTnh4uOlzXFwcRUVFvP/++0yePNmiOhsxGAwUFxcTdXUzqKa09CwtHbc2atTbsmXLOHLkCL///nu7eVuon8YD3Vry0JWwGB8DP7aTJ7+lg3369CEwMJCzZ8+2eFFAQABgdNrAwEDT8dLS0pactxnLli3j4YcfbjNPcHAwR44c4XwLm0KXlZW1W0ZTAgMDGTp0KDk5OR2+xl5o7Kpc37rpyM+hKbGxsXz//ffmltcq586do6GhgaCgoGbHAwICWnyWvn37mqZ52BKWrLdnn32Wn3/+mZSUlHYXHbdUb4AfYADafkvUDp02HiHEBaDL7dMhQ4Zw8ODBFv96hoSEEBAQwI4dOxg5ciRgHG9ITk7mnXfeafO+Pj4++Pi03/qLi4tDq9WSkZHB2LFjAUhPT0er1TJ+/PgOP0d5eTlFRUXNDLKn4OTkRExMDDt27GDOnDmm4zt27ODee+/t8H0yMzOtWj95eXk4Ozub/oA1EhcXx+bNm5sd2759O6NHj7bJDQgsUW9CCJ599lk2bdpEUlISISEh7V7TUr0B8cB+IUT3Bse6OzrdRhIFBQUiMzNTvP7666Jfv34iMzNTbN++XaxYsUKUlJQIIYQIDw8XGzduNI2kv/3220Kj0YiNGzeKo0ePinnz5onAwECh0+k6MWbfNjNnzhQjRowQqampIjU1VURFRYm77767WZ6muiorK8ULL7wg9u7dK/Ly8kRiYqKIi4sTgwYNMquullDrrdaPP/4oHB0dxerVq0VWVpZ47rnnhLu7u8jPzxdCCPHyyy+LBQsWmPJ/+OGHYtOmTSI7O1v88ccf4uWXXxaA2LBhg9m1VVZWiszMTJGZmSkA8cEHH4jMzEyxatUqsXbt2hu05ebmCjc3N/H888+LrKwssXr1auHo6CjWr19vNW0FBQVCCPXqbcmSJUKj0YikpCRRXFxsStXV1aY8rdUb8AFwM/AEUAfcL7rpDxY1noULFwqMfUFT6tu3r3jjjTfEnj17hBDGV45ff/216WEbGhrEa6+9JgICAoSzs7OYPHmyOHr0aLcrvinl5eXikUceER4eHsLDw0M88sgjN7zObKqrurpaxMfHC19fX+Ho6CiGDBkiFi5cKAoLC82qqyXUMh4hhFi5cqUYOnSocHJyEqNGjWr26nXhwoXitttuM31+5513RGhoqHBxcRFeXl5i4sSJIiEhwSK6Gl9BN02Ojo7itddeE2lpaTdoE0KIpKQkMXLkSOHk5CSCg4PFp59+ajVtgFi4cKEQQr16a0nT9b97rdUbcBDQA3nAYmEGf7BaWIymrF27ltraWv785z9bquwegwyL0TFOnDjBunXrWLp0aYe63JJOYb9hMZoSHh5OUVGRXKkuMRvHjx/H19dXmo6doIrxhIWFIYQgOztbjeLtAhmPp+PU19eTnZ3NzTffrLYUSQdRxXj69evH4MGDLbKAsKcg4/F0nPz8fGpra6Xx2BGqGA/An/70J06dOmWKjyuRdJWsrCy8vLw6NcdIoi6qGU9kZCQNDQ0cPXpULQmSHkBdXR1//PEHkZGRqs9ClnQc1YzHw8ODsLAwDhw4gAXfrEl6OMeOHaOuro5Ro0apLUXSCVQzHoDRo0dz/vz5VpdQSCTtcfDgQUJDQ+nfv7/aUiSdQFXjafzC7N+/X00ZEjvl/PnznDlzhpiYGLWlSDqJqsajKAoxMTEcO3aMmpoaNaVI7JADBw7g7u5OWFiY2lIknURV4wFjAKSGhgYOHjyothSbQs7jaZvq6moOHTpETExMs6BfEvtAdeNxd3cnOjqatLQ0m4kGZwvIeTxtk5GRgRCiWSRJif2guvEATJw4kaqqKjIzM9WWIrED6urqyMjIYNSoUbi5uaktR9IFbMJ4vL29iYyMZM+ePdTX16stR2LjZGRkoNfriYuLU1uKpIvYhPEATJo0CZ1OZ7NjPXIvLtugpqaGPXv2MGrUKPkK3Y6xGePx9fXl1ltvJSkpySaXUci9uGyDxlbxbbfdprYUSTfoSsxlizFt2jSOHTvG7t27mTFjhtpyTBw/fpytW7eSlpZmGsz84osviIuL4+TJk82CdV9PS6E4JV2jsrKS9PR04uLi6Nevn9pyJN3AZlo8YFxGMWHCBNLT06moqFBbjglr7sWl1+vR6XTNksTI9u3bcXJy6lRsbIltYlPGA8YA025ubvz2229qSzFhzb243nrrLdM4kkajuWHHhN5Kbm4uf/zxBzNmzMDFxUVtOZJuYnPG4+TkxPTp08nKyuLUqVMWLWvFihU3DP5enxqXc3R1L65Zs2YRGRnJ7Nmz+fXXX8nOziYhIaHVa5YvX45WqzWloqKi7j+onWMwGNiyZQtDhgwhOjpabTkSM2BTYzyNREVFcejQIRISEliyZAlOTk4WKccW9+JydnbG2dm5w/fsDezdu5eKigoefPBBGfqih2CTxqMoCnfffTeffvop27dv5+6777ZIOXIvLtuntLSUlJQUYmNjW+zuSuwTm+tqNeLt7c3MmTM5cOAAx48fV1XLzTffzMyZM1m0aBFpaWmkpaWxaNEi7r777mZvtCIiIti0aRMAly9f5sUXXyQ1NZX8/HySkpKYPXs2Pj4+zTbJk7SOwWBg48aNeHt7M3XqVLXlSMyIzRoPwKhRo4iIiGDz5s2qv9354YcfiIqKIj4+nvj4eEaMGMGaNWua5Tl58iRarRYABwcHjh49yr333ktYWBgLFy4kLCyM1NRUPDw81HgEuyMxMZGysjLmzp1L37422TiXdBFV9tXqDNXV1axatQofHx8WLFjQ6/r4vXVfrby8PL777jumT5/OhAkT1JbT2+kZ+2p1Bjc3N+bMmUNeXl67c2Z6Er05LIZOp2PDhg0EBwfL9Vg9FJtv8TTy22+/kZqaymOPPcbQoUPNeWubpre1eAwGA99++y06nY6nn34ad3d3tSVJemOLp5GpU6cyZMgQ1q1bx8WLF9WWI7EQ27Zto7i4mAcffFCaTg/GbozHwcGBBx98EFdXV9Pe65Kexf79+9m/fz933nkngwYNUluOxILYjfEAuLq6Mn/+fC5fvsz69etpaGhQW5LETGRnZ7NlyxbGjRsng7f3AuzKeAAGDBjAAw88QG5uLgkJCXJPrh7A2bNnWb9+PeHh4cTHx6stR2IF7M54AIYNG8bs2bM5ePAgu3btUluOpBuUlJTw/fff4+/vz9y5c+nTxy6/kpJOYrezskaOHElNTQ07duygb9++MjCUHVJWVsaaNWvw8vLikUcewdHRUW1JEitht8YDMH78eAwGA4mJidTX1zN16tQeM8Fw5cqVrFy5ssfGoC4vL+e7776jX79+PProozLURS/DbubxtMXevXvZsWMH48ePZ/r06T3GfKBnzuNp7F65urqycOFCGU3Q9jH7L5Rdt3gaGT9+PA4ODmzduhWDwcDMmTN7lPn0JAoKCli7di0DBgxg/vz5cq5OL6VHGA/AuHHjcHBwYMuWLeh0OubMmWOxOD6SrnHy5EnWr19PUFAQDz30kIw71IvpEV2tpmRnZ7N+/Xp8fHyYN2+e3a8E7wldLSEE6enpbN++nYiICLna3P4we/ehxxkPQHFxMWvXrkVRFObNm2fXuzzYu/EYDAZ++eUXDh8+TGxsLDNmzJCvzO0PaTwdRafTsXbtWsrLy5k9ezZRUVFqyuky9mw8Op2OdevWUVpayuzZsxkxYoTakiRdQxpPZ6irqyMhIYEjR44wevRo7rjjDrtr4tur8eTm5rJx40YcHBx46KGHGDhwoNqSJF1HGk9nEUJw4MABtm7dire3N3PnzrWLrlfTeTzZ2dl2YzwGg4GkpCT27NnDsGHDmDt3rnxzZf9I4+kqpaWlbNy4kQsXLjB16lTi4uLsYqzBnlo8xcXF/PTTT6Y6njBhgpzW0DOQxtMdDAYDu3btIjU1lUGDBjF79uxObU+jBvZgPFeuXOH333/n999/x9fXlzlz5th8vUo6hTQec1BYWMjmzZu5ePEi48ePZ/LkyTa7TsjWjefkyZNs3boVnU7HhAkTuO2223BwcFBblsS8SOMxFwaDgT179rB7927c3NyYNm0aI0aMsLnul60az8WLF9m6dSs5OTmEhoYyc+bMDu1RJrFLpPGYm4qKCnbu3MmxY8fw9/dnxowZhIaGqi3LhK0ZT3V1NXv27CE9PZ1+/fpxxx13EBERIcdyejbSeCzFmTNn2LFjB4WFhYSGhjJlyhQGDx6stiybMZ7a2lpSU1NJS0sDjLurTpgwwWa7qBKzIo3HkgghOHnyJDt37uTChQsEBwczadIkQkJCVPuLrrbx6HQ6MjIy2L9/P/X19YwZM4YJEybIV+S9C2k81kAIwYkTJ9i9ezfFxcUEBAQQExNDVFSU1RY2qj2Pp6SkhLS0NI4ePYqjoyOjRo0iLi7O7te+SbqENB5rIoQgNzeXjIwMcnJy6Nu3L1FRUcTExFhtJq41Wzy1tbVkZWVx6NAhioqK8PT0JDY2llGjRsmV5L0baTxqodPpOHjwIJmZmeh0OgIDA4mKiiIiIgIvLy+LlmtJ4zEYDJw6dYojR46QnZ1NfX09w4YNIyYmhvDwcPlqXALSeNSnoaGBU6dOcfDgQU6dOkV9fT3+/v5EREQQERGBv7+/WceDLGE8NTU15OTkmJJerycgIIARI0YQGRkpu1OS65HGoxZvvvkmCQkJHDp0CCcnJy5duoRer+f06dOcOHGC7Oxs9Ho97u7uDB061JQ+/fRTPv/8cyoqKhg3bhwrV67klltu6XC55jCe2tpaCgsLyc/Pp6CggOLiYoQQBAYGMnz4cG655Rb8/Py6dG9Jr0Aaj1q89tpr9O/fnzNnzrB69WouXbrU7Hx9fT0FBQXk5eVRUFDA2bNnaWhooKamBl9fXwYNGsSuXbvIyMhg+/btHf5F74zxCCGorKzk/PnzlJSUmFLjls8eHh4EBwcTHBzMTTfdZBPzgiR2gTQetfnmm2947rnnbjCe69Hr9UyYMIE5c+YwbNgwiouLqaysNJ13cXHB29sbb29v3N3dcXNzw9XV1ZQcHBxQFIXKykqioqI4duwYzs7O6PV6U6qpqUGn06HVarl06RJarda0u6qzszMBAQGmNGTIELy8vOREP0lXkMHe7YWzZ89y4MABvvjiC0aOHAkYzWjhwoV4e3szb948Ll68SEVFBSUlJdTU1FBTU3PDtsyNe8SvWbOm2RYwjo6OuLi44OnpSf/+/QkMDKR///5oNBp8fX3p37+/NBmJzSKNx0KUlJQANFul7ezsjKenJ6dPn2bSpEk3XCOE4PXXX+ejjz4ytXgaW6SLFi3C29sbZ2dnnJ2dbW5NmUTSGXr1t3fFihUoitJm2r9/f7fKuL7VIYRotSWiKArLly+nsLCQvLw8cnNzOXz4MAA+Pj70798fV1dXaToSu6dXt3iWLVvGww8/3Gae4ODgLt27McphSUkJgYGBpuOlpaVtxqppbNFIJD0ZSw4u90gURXkc+EgI0b+dfApwDvhQCPHu1WNOQCnwkhDisw6W5wloAY0QQtcN6RKJzSDb7B1EUZQhiqLcCgwBHBRFufVq6tckzwlFUeYACKOjfwS8oijKHEVRIoFvgGrgX50ouhLQXP1XIukR9OquVid5A1jY5HPm1X+nAklX/x+O0SQaeRdwBT4BvIB0IF4I0WETuWpgsqUj6VHIrpZEIrE6sqslkUisjjQeiURidaTxSCQSqyONRyKRWB1pPBKJxOpI45FIJFZHGo9EIrE60ngkEonVkcYjkUisjjQeiURidaTxSCQSq/P/Er2bPDwSlJIAAAAASUVORK5CYII=\n",
"text/plain": [
"Graphics object consisting of 4 graphics primitives"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"var('t') # symb. variable\n",
"p1=parametric_plot((cos(t),sin(t)),(t,0,2*pi),color='grey') \n",
" # plot the circle\n",
"p2=plot(1-x/2,(x,0,2)) # half line through (0,1) and (x,y)\n",
"p3= text(\"(x,y)\",(0.88,0.72)) # point (x,y)\n",
"p4= text(\"(u,0)\",(2.0,0.15)) # point (u,0)\n",
"(p1+p2+p3+p4).show(figsize=[3,2]) # combine plots"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Consider the line which passes through the North pole $(0, 1)$, the point $(x, y)$ on the circle and intersects\n",
"the axis Ox at the point $(u, 0)$. The function $u = u(x, y)$ defines the\n",
"stereographic projection. Three points: North pole, the\n",
"point $(x, y)$ on the circle and the point $(u, 0)$ are on the same line.\n",
"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 $\\frac{u}{x}=\\frac{1}{1-y}$ and consequently $u=\\frac{x}{1-y}$.\n",
"Solving the system of equations\n",
"$$ x^2+y^2=1,\\quad u=\\frac{x}{1-y},$$ with respect to $(x,y)$: "
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left[x = \\frac{2 \\, u}{u^{2} + 1}, y = \\frac{u^{2} - 1}{u^{2} + 1}\\right], \\left[x = 0, y = 1\\right]\\right]$"
],
"text/plain": [
"[[x == 2*u/(u^2 + 1), y == (u^2 - 1)/(u^2 + 1)], [x == 0, y == 1]]"
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"forget() # forget previous assumptions\n",
"var('x, y, u') # symbolic variables\n",
"assume(y>0) # assume y>0\n",
"s=solve([x^2+y^2==1,u==x/(1-y)],[x,y]);s # solve eq. above"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"we obtain the relations\n",
"$$ x=\\frac{2u}{u^2+1},\\quad y=\\frac{u^2-1}{u^2+1}, \\quad u=\\frac{x}{1-y}.$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"Now consider the South pole case.\n",
"\n",
"### Stereographic projection of $S^1$ from the South pole $S = (0, -1)$"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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\n",
"text/plain": [
"Graphics object consisting of 4 graphics primitives"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"var('t') # symbolic variable\n",
"p1=parametric_plot((cos(t),sin(t)),(t,0,2*pi),color='grey') \n",
" # plot the circle\n",
"p2=plot(x/2-1,(x,0,2)) # halfline through (0,-1) and (x,y)\n",
"p3= text(\"(x,y)\",(0.88,-0.72)) # point (x,y)\n",
"p4= text(\"(u',0)\",(2.0,0.15)) # point (u',0)\n",
"(p1+p2+p3+p4).show(figsize=[3,2]) # combine plots"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"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 $\\frac{u'}{x}=\\frac{1}{1-(-y)}$ ($y$ is negative now) and consequently $u'=\\frac{x}{1+y}$.\n",
"Solving the system of equations\n",
"$$ x^2+y^2=1,\\quad u'=\\frac{x}{1+y},$$ with respect to $(x,y)$: "
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left[x = \\frac{2 \\, {u'}}{{u'}^{2} + 1}, y = -\\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\\right], \\left[x = 0, y = \\left(-1\\right)\\right]\\right]$"
],
"text/plain": [
"[[x == 2*up/(up^2 + 1), y == -(up^2 - 1)/(up^2 + 1)], [x == 0, y == -1]]"
]
},
"execution_count": 15,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"forget() # forget previous assumptions\n",
"var('x, y, up') # symbolic variables\n",
"assume(y<0) # assume y<0\n",
"s=solve([x^2+y^2==1,up==x/(1+y)],[x,y]);s # solve eq. above"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"we obtain the relations\n",
"$$ x=\\frac{2u'}{u'^2+1},\\quad y=\\frac{1-u'^2}{u'^2+1}, \\quad u'=\\frac{x}{1+y}.$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.4**\n",
"\n",
"As we mentioned, both the stereographic projections from North and South poles are predefined in `manifolds.Sphere`."
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{1} & \\longrightarrow & \\mathbb{E}^{2} \\\\ \\mbox{on}\\ A : & \\phi & \\longmapsto & \\left(x, y\\right) = \\left(\\cos\\left(\\phi\\right), \\sin\\left(\\phi\\right)\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{1}\\setminus\\{\\mathrm{NP}\\} : & u & \\longmapsto & \\left(x, y\\right) = \\left(\\frac{2 \\, u}{u^{2} + 1}, \\frac{u^{2} - 1}{u^{2} + 1}\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{1}\\setminus\\{\\mathrm{SP}\\} : & {u'} & \\longmapsto & \\left(x, y\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + 1}, -\\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"iota: S^1 → E^2\n",
"on A: phi ↦ (x, y) = (cos(phi), sin(phi))\n",
"on S^1-{NP}: u ↦ (x, y) = (2*u/(u^2 + 1), (u^2 - 1)/(u^2 + 1))\n",
"on S^1-{SP}: up ↦ (x, y) = (2*up/(up^2 + 1), -(up^2 - 1)/(up^2 + 1))"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"reset()\n",
"S1=manifolds.Sphere(1) # S^1 from manifolds.Sphere\n",
" # stereographic projection from the North pole:\n",
"stereoN. = S1.stereographic_coordinates(pole='north')\n",
" # stereographic projection from the South pole:\n",
"stereoS. = S1.stereographic_coordinates(pole='south')\n",
"\n",
"Phi=S1.embedding() # embedding S^1 -> E^2\n",
"Phi.disp() # show embedding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.5**\n",
"\n",
"To show graphically how the projection from the North pole acts, let us extract \n",
"an appropriate part of the definition from the previous cell."
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\Phi_N:& \\mathbb{S}^{1} & \\longrightarrow & \\mathbb{E}^{2} \\\\ \\mbox{on}\\ \\mathbb{S}^{1}\\setminus\\{\\mathrm{NP}\\} : & u & \\longmapsto & \\left(x, y\\right) = \\left(\\frac{2 \\, u}{u^{2} + 1}, \\frac{u^{2} - 1}{u^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"PhiN: S^1 → E^2\n",
"on S^1-{NP}: u ↦ (x, y) = (2*u/(u^2 + 1), (u^2 - 1)/(u^2 + 1))"
]
},
"execution_count": 17,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# continuation\n",
"E=S1.ambient() # ambient space E^2\n",
"c_cart. = E.cartesian_coordinates() # Cartesian coord.\n",
"fun=list(Phi.coord_functions(stereoN).expr()) # embedd. functions\n",
"PhiN = S1.continuous_map(E,{(stereoN, c_cart): fun},\n",
" name='PhiN',latex_name=r'\\Phi_N') # define embedding\n",
"PhiN.display() # show embedding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let us plot the set of points $(x,y)$ corresponding to $u\\in (-10,10)$."
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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\n",
"text/plain": [
"Graphics object consisting of 1 graphics primitive"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"p=stereoN.plot(c_cart,mapping=PhiN,ranges={u:(-10,10)},\n",
" number_values={u:50},color={u:'grey'},\n",
" plot_points=500) # image of (-10,10) under PhiN\n",
"p.show(figsize=[3,2]) # show the image"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Thus the points of the grey arc in the figure are projected onto the interval (-10,10)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.6**\n",
"\n",
"Now let us extract the definition of the projection from the South pole,"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\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}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\Phi_S:& \\mathbb{S}^{1} & \\longrightarrow & \\mathbb{E}^{2} \\\\ \\mbox{on}\\ \\mathbb{S}^{1}\\setminus\\{\\mathrm{SP}\\} : & {u'} & \\longmapsto & \\left(x, y\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + 1}, -\\frac{{u'}^{2} - 1}{{u'}^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"PhiS: S^1 → E^2\n",
"on S^1-{SP}: up ↦ (x, y) = (2*up/(up^2 + 1), -(up^2 - 1)/(up^2 + 1))"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# continuation\n",
"fun=list(Phi.coord_functions(stereoS).expr()) # embedd. functions\n",
"PhiS = S1.continuous_map(E,{(stereoS, c_cart): fun},\n",
" name='PhiS',latex_name=r'\\Phi_S') # define embedding\n",
"PhiS.display() # show embedding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"and check which points of the circle are projected onto (-10,10):"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {},
"outputs": [
{
"data": {
"image/png": 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\n",
"text/plain": [
"Graphics object consisting of 1 graphics primitive"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"# continuation\n",
"p=stereoS.plot(c_cart,mapping=PhiS,ranges={up:(-10,10)},\n",
" number_values={up:50},color={up:'grey'}, # plot image of (-10,10)\n",
" plot_points=500)\n",
"p.show(figsize=[3,2]) # show image"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"\n",
"### Transition map from u coordinate to u' coordinate"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Using the relations $\\ u=\\frac{x}{1-y},\\ \\ u'=\\frac{x}{1+y}$ and $y=\\frac{u^2-1}{u^2+1},\\ $ we obtain\n",
"$u'=\\frac{x}{1+y}=\\frac{u(1-y)}{1+y}=\\frac{1-y}{1+y}u.\\ \\ $ Replacing $y$ by $\\frac{u^2-1}{u^2+1}\\ \\ $ we get $\\ \\frac{1-y}{1+y}=\\frac{1-\\frac{u^2-1}{u^2+1}}{1+\\frac{u^2-1}{u^2+1}}=\n",
"\\frac{u^2+1-u^2+1}{u^2+1+u^2-1}=\\frac{2}{2u^2}=\\frac{1}{u^2}.\\ $ Accordingly $u'=\\frac{1-y}{1+y}u=\\frac{1}{u},\\ $ so the transition map from $u$ coordinate to $u'$ coordinate has the form $\\ u'=\\frac{1}{u}.\\ $ It is smooth if $\\ u\\not=0, u'\\not=0.$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here is SageMath solution (eliminate x,y variables from equations of the first line in the previous computations):"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[2 \\, {\\left(u {u'} - 1\\right)} u\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[2 \\, {\\left(u {u'} - 1\\right)} u\\right]$"
],
"text/plain": [
"[2*(u*up - 1)*u]"
]
},
"execution_count": 21,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"var('x,y,u,up')\n",
"maxima.eliminate([u==x/(1-y),up==x/(1+y), # eliminate var. x,y\n",
" y==(u^2-1)/(u^2+1)],[x,y]).sage() # from eqs of projections"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The answer means that the bracket vanishes.\n",
"Since $u\\not=0$, the expression in the bracket vanishes if $\\ \\ u'=\\frac{1}{u}$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.7**\n",
"\n",
"In `SageMath Manifolds` the transition from one coordinate system to the other can be defined as follows:"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}$"
],
"text/plain": [
"up = 1/u"
]
},
"execution_count": 22,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"S1=manifolds.Sphere(1) # S^1 from manifolds.Sphere\n",
" # stereographic projection from the North pole:\n",
"stereoN. = S1.stereographic_coordinates(pole='north')\n",
" # stereographic projection from the South pole:\n",
"stereoS. = S1.stereographic_coordinates(pole='south')\n",
"\n",
"\n",
"trans = stereoN.transition_map(stereoS, 1/u, # define transition\n",
" intersection_name='W', # stereoN -> StereoS\n",
" restrictions1= u!=0, # specify\n",
" restrictions2 = up!=0) # restrictions\n",
"trans.display() # show transition"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In some cases, the definition of the inverse transition can be left to the `SageMath Manifolds`."
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{lcl} u & = & \\frac{1}{{u'}} \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{lcl} u & = & \\frac{1}{{u'}} \\end{array}$"
],
"text/plain": [
"u = 1/up"
]
},
"execution_count": 23,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"trans.inverse().display()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"As we can see the transition map and its inverse are smooth."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Some transition maps are predefined in `manifolds.Sphere`.
\n",
"\n",
"
\n",
"\n",
"**Example 4.8**\n",
"\n",
"Let us check that the transitions from the previous example are predefined in `manifolds.Sphere`."
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}, \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}, \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}\\right)$"
],
"text/plain": [
"(up = 1/u, up = 1/u)"
]
},
"execution_count": 24,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"dim=1 # dim of the sphere\n",
"Sph=manifolds.Sphere(dim) # sphere S^1 \n",
"spher = Sph.spherical_coordinates() # spherical coord.\n",
"stereoN, stereoS = Sph.coordinate_charts('stereographic',\n",
" names=['u']) # sterogr. projections\n",
"A=stereoN.domain() # domain of stereoN\n",
" # intersection of domains of stereoN and stereo S:\n",
"W = Sph._stereoN_dom.intersection(Sph._stereoS_dom)\n",
" # transition from stereoN to stereoS:\n",
"FNS=Sph.coord_change(stereoN.restrict(W),stereoS.restrict(W))\n",
" # transition from stereoS to stereoN:\n",
"FSN=Sph.coord_change(stereoS.restrict(W),stereoN.restrict(W))\n",
"FNS.disp(),FSN.disp() # show transitions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can check that both transitions can be obtained as compositions of other transitions."
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\begin{array}{lcl} u & = & \\frac{1}{{u'}} \\end{array}, \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\begin{array}{lcl} u & = & \\frac{1}{{u'}} \\end{array}, \\begin{array}{lcl} {u'} & = & \\frac{1}{u} \\end{array}\\right)$"
],
"text/plain": [
"(u = 1/up, up = 1/u)"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# continuation\n",
"A = spher.domain() # domain of spher.coord\n",
" # intersection of domains of stereoN and stereo S:\n",
"W = Sph._stereoN_dom.intersection(Sph._stereoS_dom)\n",
"V=W.intersection(A) # intersection of domains of \n",
" # stereoN, stereoS and spher \n",
"F1=Sph.coord_change(spher.restrict(V), \n",
" stereoN.restrict(V)) # transition spher -> stereoN\n",
"\n",
"F2=Sph.coord_change( stereoN.restrict(V),\n",
" spher.restrict(V)) # transition stereoN -> spher\n",
"\n",
"F3=Sph.coord_change(spher.restrict(V), stereoS.restrict(V))\n",
" # transition spher -> stereoS\n",
"F4=Sph.coord_change( stereoS.restrict(V),spher.restrict(V))\n",
" # transition stereoS -> spher\n",
"(F1*F4).disp(),(F3*F2).disp() # show compositions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.9**\n",
"\n",
"Knowing the transition maps we are ready to define $S^1$ as a manifold with two maps. \n",
"\n",
"Note that this time, our calculations are independent of `manifolds.Sphere`. "
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\begin{array}{lcl} v & = & \\frac{1}{u} \\end{array}, \\begin{array}{lcl} u & = & \\frac{1}{v} \\end{array}\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\begin{array}{lcl} v & = & \\frac{1}{u} \\end{array}, \\begin{array}{lcl} u & = & \\frac{1}{v} \\end{array}\\right)$"
],
"text/plain": [
"(v = 1/u, u = 1/v)"
]
},
"execution_count": 26,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"M = Manifold(1, 'S^1') # sphere S^1 \n",
"U = M.open_subset('U') # complement of the North pole\n",
"cU. = U.chart() # sterogr. proj. from North pole\n",
"V = M.open_subset('V') # complement of the South pole\n",
"cV. = V.chart() # sterogr. proj. from South pole\n",
"M.declare_union(U,V) # M is the sum of U and V\n",
"trans = cU.transition_map(cV, 1/u, intersection_name='W',\n",
" restrictions1= u!=0, # transition cU -> cV\n",
" restrictions2 = v!=0)\n",
"trans_inv=trans.inverse() # inverse transition\n",
"trans.disp(),trans_inv.disp() # show transitions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"\n",
"
\n",
"\n",
"## Two dimensional sphere $S^2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Spherical coordinates in $\\mathbf{S}^2$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In `manifolds.Sphere` in two-dimensional case \n",
"$\\ \\ S^2=\\{(x,y,z)\\in R^3: x^2+y^2+z^2=1\\}\\ \\ $ the default coordinates are the spherical ones."
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {},
"outputs": [],
"source": [
"var('u v t') # symb.var.\n",
"\n",
"po1={'thickness':5,'color':'darkblue'} # parameters\n",
"po2={'fontsize':20,'color':'black'}\n",
"po3={'size':7,'color':'black'}\n",
"\n",
"ax =line3d([(0,0,0), (1+0.15,0,0)], **po1) # axes\n",
"ax+=line3d([(0,0,0), (0,1+0.15,0)], **po1)\n",
"ax+=line3d([(0,0,0), (0,0,1+0.15)], **po1)\n",
"ax+=text3d(\"x\",(1.25,0,0),**po2)\n",
"ax+=text3d(\"y\",(0,1.25,0),**po2)\n",
"ax+=text3d(\"z\",(0.,0.,1.25),**po2)"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
" # semisphere:\n",
"s=parametric_plot3d((cos(u)*cos(v), sin(u)*cos(v), sin(v)),\n",
"(u,0,2*pi), (v,0,pi/2),opacity=0.9,color='lightgrey')\n",
"\n",
"a=0.59 # triangle\n",
"tr=line3d([(0,0,0),(a,a,0),(a,a,a),(0,0,0)],**po1)\n",
"\n",
"dots=point3d([(0.5*cos(t),0.5*sin(t),0) # dots\n",
" for t in srange(0,pi/4,0.1)], **po3)\n",
"dots+=point3d([(0.5*cos(t),0.5*cos(t),0.5*sin(t))\n",
" for t in srange(pi/4,pi/2,0.1)],**po3)\n",
" \n",
"t=text3d(\"(x,y,z)\",(0.6,0.8,0.7),**po2) # variables\n",
"t+=text3d(\"φ\",(0.7,0.3,0.0),**po2) # names\n",
"t+=text3d(\"θ\",(0.,0.2,0.5),**po2)\n",
" # combine plots:\n",
"(ax+s+tr+dots+t).rotateZ(-pi/8).show(frame=False)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Example 4.10**\n",
"\n",
"Let us show how to use the spherical coordinates in $\\ S^2$."
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ A : & \\left(\\theta, \\phi\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\cos\\left(\\theta\\right)\\right) \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ A : & \\left(\\theta, \\phi\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\cos\\left(\\theta\\right)\\right) \\end{array}$"
],
"text/plain": [
"iota: S^2 → E^3\n",
"on A: (theta, phi) ↦ (x, y, z) = (cos(phi)*sin(theta), sin(phi)*sin(theta), cos(theta))"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"S2=manifolds.Sphere(2) # two-dimensional sphere\n",
"Phi=S2.embedding() # embedding into E^3\n",
"Phi.disp() # show embedding"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(A,(\\theta, \\phi)\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(A,(\\theta, \\phi)\\right)$"
],
"text/plain": [
"Chart (A, (theta, phi))"
]
},
"execution_count": 30,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"S2.default_chart() # default chart"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let us check the ranges of variables $\\theta, \\phi$."
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The Cartesian product of ((0, pi), [-pi, pi))\n"
]
}
],
"source": [
"sph.=S2.spherical_coordinates() # spherical coordinates\n",
"print(sph.codomain())"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let us note that that the spherical coordinates in `manifolds.Sphere` are defined on an open subset of the sphere, they do not cover the whole sphere. To cover the entire sphere with charts, we will use the stereographic coordinates instead."
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Open subset A of the 2-sphere S^2 of radius 1 smoothly embedded in the Euclidean space E^3\n"
]
}
],
"source": [
"A=sph.domain() # domain of definition of spher. coordinates\n",
"print(A)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Check the Jacobian of the embedding into $E^3$,"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\begin{array}{rr}\n",
"\\cos\\left(\\phi\\right) \\cos\\left(\\theta\\right) & -\\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right) \\\\\n",
"\\cos\\left(\\theta\\right) \\sin\\left(\\phi\\right) & \\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right) \\\\\n",
"-\\sin\\left(\\theta\\right) & 0\n",
"\\end{array}\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\begin{array}{rr}\n",
"\\cos\\left(\\phi\\right) \\cos\\left(\\theta\\right) & -\\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right) \\\\\n",
"\\cos\\left(\\theta\\right) \\sin\\left(\\phi\\right) & \\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right) \\\\\n",
"-\\sin\\left(\\theta\\right) & 0\n",
"\\end{array}\\right)$"
],
"text/plain": [
"[ cos(phi)*cos(theta) -sin(phi)*sin(theta)]\n",
"[ cos(theta)*sin(phi) cos(phi)*sin(theta)]\n",
"[ -sin(theta) 0]"
]
},
"execution_count": 33,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Phi.jacobian_matrix() # Jacobian matrix of embedding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"and the 2x2 minors of this matrix."
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\cos\\left(\\phi\\right)^{2} \\cos\\left(\\theta\\right) \\sin\\left(\\theta\\right) + \\cos\\left(\\theta\\right) \\sin\\left(\\phi\\right)^{2} \\sin\\left(\\theta\\right), -\\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right)^{2}, \\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right)^{2}\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\cos\\left(\\phi\\right)^{2} \\cos\\left(\\theta\\right) \\sin\\left(\\theta\\right) + \\cos\\left(\\theta\\right) \\sin\\left(\\phi\\right)^{2} \\sin\\left(\\theta\\right), -\\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right)^{2}, \\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right)^{2}\\right]$"
],
"text/plain": [
"[cos(phi)^2*cos(theta)*sin(theta) + cos(theta)*sin(phi)^2*sin(theta),\n",
" -sin(phi)*sin(theta)^2,\n",
" cos(phi)*sin(theta)^2]"
]
},
"execution_count": 34,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Phi.jacobian_matrix().minors(2) # minors of Jacobian"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The sum of squares of 2x2 minors allows us to check the rank of the Jacobian matrix."
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\sin\\left(\\theta\\right)^{2}\\)"
],
"text/latex": [
"$\\displaystyle \\sin\\left(\\theta\\right)^{2}$"
],
"text/plain": [
"sin(theta)^2"
]
},
"execution_count": 35,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"ll=[Phi.jacobian_matrix().minors(2)[k].full_simplify() \n",
" for k in range(3)] # list of minors of Jacobian\n",
"sum([ll[k]^2 for k in range(3)]).full_simplify()\n",
" # sum of squares of minors"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"so the rank is 2 if $\\theta\\in (0,\\pi)$.\n",
"\n",
"
\n",
"\n",
"If we are not interested in all details, we can check the rank of Jacobian using the command:"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle 2\\)"
],
"text/latex": [
"$\\displaystyle 2$"
],
"text/plain": [
"2"
]
},
"execution_count": 36,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Phi.jacobian_matrix().rank()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"As we can see spherical coordinates define transformation with Jacobian of rank 2 on the subset A of the sphere defined by $(\\theta,\\varphi)\\in (0,\\pi)\\times [-\\pi,\\pi)$.\n",
"\n",
"
\n",
"\n",
"**Example 4.11**\n",
"\n",
"Let us plot the coordinate lines of the spherical coordinates."
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"%display latex\n",
"S2=manifolds.Sphere(2) # two-dimensional sphere\n",
"sph.=S2.spherical_coordinates() # spherical coordinates\n",
"Phi=S2.embedding() # embedding into E^3\n",
"E=S2.ambient() # ambient space E^3\n",
"c_cart. = E.cartesian_coordinates() # Cartesian coordinates\n",
"p=sph.plot(c_cart,mapping=Phi,number_values=15,\n",
" thickness=2,color='grey', # plot coordinate lines\n",
" label_axes=False,frame=False)\n",
"p.show(frame=False) # show plot"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.12**\n",
"\n",
"We can restrict the ranges of parameters (for example to $\\ \\theta\\in (0,\\pi/2),\\ \\ \\phi \\in (0,\\pi/4)$ ):"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {},
"outputs": [],
"source": [
"po1={'thickness':5,'color':'darkblue'} # param.\n",
"po2={'fontsize':20,'color':'black'}\n",
"ax =line3d([(0,0,0), (1+0.15,0,0)], **po1) # axes\n",
"ax+=line3d([(0,0,0), (0,1+0.15,0)], **po1)\n",
"ax+=line3d([(0,0,0), (0,0,1+0.15)], **po1)\n",
"ax+=text3d(\"x\",(1.25,0,0),**po2)\n",
"ax+=text3d(\"y\",(0,1.25,0),**po2)\n",
"ax+=text3d(\"z\",(0.,0.,1.25),**po2)"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"S2=manifolds.Sphere(2) # sphere S^2\n",
"Phi=S2.embedding() # embedding S^2->E^3\n",
"sph.=S2.spherical_coordinates() # spherical coord.\n",
"A=sph.domain() # domain of sph\n",
"U=A.open_subset('U',coord_def={sph:[th>0,th0,ph = E.cartesian_coordinates() # Cartesian coord.\n",
"p=sphU.plot(c_cart,mapping=Phi,number_values=10,thickness=2,\n",
" color='grey',label_axes=False) # plot sphU coord lines \n",
"(p+ax).show(frame=False,label_axes=False) # plot coord lines and ax"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"**Example 4.13**\n",
"\n",
"We can investigate $S^2$ and its parts without `manifolds.Sphere`,\n",
"but the transition maps and the corresponding restrictions must be defined by the user in that case.\n",
"\n",
"We have to define the manifolds, charts and the embedding $S^2\\to R^3$."
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {},
"outputs": [],
"source": [
"# spherical coordinates in S^2 without manifolds.Sphere setup\n",
"M = Manifold(2, 'S^2') # sphere S^2\n",
"U = M.open_subset('U') # the complement of a meridian, \n",
" # domain of spherical coordinates\n",
"c_spher. = U.chart(r'th:(0,pi):\\theta ph:(0,2*pi):\\phi') \n",
" # spherical coordinates on U\n",
"N = Manifold(3, 'R^3', r'\\RR^3') # manifold R^3\n",
"c_cart. = N.chart() # Cartesian coord. on R^3\n",
"Phi=U.continuous_map(N,(sin(th)*cos(ph),sin(th)*sin(ph), cos(th)),\n",
" name='Phi',latex_name=r'\\Phi') # define embedding S^2 -> R^3"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can make previous plots without `manifolds.Sphere` setup."
]
},
{
"cell_type": "code",
"execution_count": 41,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"# Previous graph without manifolds.Sphere setup\n",
"forget()\n",
"M = Manifold(2, 'S2') # sphere S^2\n",
"c_pol. = M.chart() # spherical coordinates\n",
"D = M.open_subset('D', coord_def={c_pol: [0 = N.chart() # Cartesian coordinates on R^3\n",
"Phi = M.continuous_map(N,{(c_pol,c_cart): \n",
" (cos(ph)*sin(th),sin(ph)*sin(th), cos(th))},\n",
" name='Phi',latex_name=r'\\Phi') # define embedding S^2->R^3 \n",
"Phi1 = Phi.restrict(D) # restrict embedding to D\n",
"p=c_pol_D.plot(c_cart,mapping=Phi1, # plot the coordinate lines\n",
" ranges={ph:(0,pi/4),th:(0,pi/2)}, # contained in D\n",
" number_values={ph:11,th:10},thickness=2,\n",
" color='grey',label_axes=False)\n",
"(p+ax).show(frame=False) # use ax defined in the previous plot!"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Stereographic coordinates in $\\mathbf{S^2}$\n",
"\n",
"### Stereographic projection of $S^2$ from the North pole"
]
},
{
"cell_type": "code",
"execution_count": 42,
"metadata": {},
"outputs": [],
"source": [
"var('t u v')\n",
"\n",
"po1={'thickness':5,'color':'darkblue'} # parameters\n",
"po2={'fontsize':20,'color':'black'}\n",
"po3={'fontsize':20,'color':'white'}\n",
" # half line from North pole:\n",
"p=parametric_plot3d((t,t,1-t/0.85),(t,0,0.85),**po1)\n",
"p+=point3d([(1.2*cos(t),1.2*sin(t),0) # dotted arc\n",
" for t in srange(0,pi/2,0.087)], size=7,frame=False)"
]
},
{
"cell_type": "code",
"execution_count": 43,
"metadata": {},
"outputs": [],
"source": [
"ax =line3d([(0,0,0), (1+0.15,0,0)], **po1) # axes\n",
"ax+=line3d([(0,0,0), (0,1.1+0.15,0)], **po1)\n",
"ax+=line3d([(0,0,0), (0,0,1+0.15)], **po1)\n",
"ax+=text3d(\"x\",(1.25,0,0),**po2)\n",
"ax+=text3d(\"y\",(0,1.3,0),**po2)\n",
"ax+=text3d(\"z\",(0.,0.,1.25),**po2) "
]
},
{
"cell_type": "code",
"execution_count": 44,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
" # hemisphere:\n",
"s=parametric_plot3d((cos(u)*cos(v), sin(u)*cos(v), sin(v)),\n",
"(u,0,2*pi), (v,0,pi/2),opacity=0.9,color='lightgrey')\n",
"\n",
"t=text3d(\"(x,y,z)\",(0.8,0.67,0.3),**po3) # distinguished\n",
"t+=text3d(\"(u,v,0)\",(1.,1.,0.0),fontsize=20) # points\n",
"t+=text3d(\"(0,0,1)\",(0.6,0.2,0.98),**po3)\n",
"\n",
"(p+ax+s+t).rotateZ(-pi/6).show(frame=False) # combine plots"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Consider the line which passes through the North pole $(0,0,1)$, the point $(x,y,z)$ on the sphere and intersects\n",
"the plane Oxy at the point $(u,v,0)$. The functions $\\ u = u(x,y,z),\\ v=v(x,y,z)\\ $ define the stereographic projection from the North pole. Using the rotations around z-axis and appropriate similar triangles in $Oxz$ and $Oyz$ planes we can see that (as in the case of $S^1$) $\\quad\\frac{u}{x}=\\frac{1}{1-z},\\quad \\frac{v}{y}=\\frac{1}{1-z},\\quad$ so the stereographic projection can be described by $$u=\\frac{x}{1-z},\\quad v=\\frac{y}{1-z}.$$\n",
"Solving this system combined with the equation of the unit sphere: $x^2+y^2+z^2=1$ with respect to $x,y,z$:"
]
},
{
"cell_type": "code",
"execution_count": 45,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\left[x = 0, y = 0, z = 1\\right], \\left[x = \\frac{2 \\, u}{u^{2} + v^{2} + 1}, y = \\frac{2 \\, v}{u^{2} + v^{2} + 1}, z = \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right]\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left[x = 0, y = 0, z = 1\\right], \\left[x = \\frac{2 \\, u}{u^{2} + v^{2} + 1}, y = \\frac{2 \\, v}{u^{2} + v^{2} + 1}, z = \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right]\\right]$"
],
"text/plain": [
"[[x == 0, y == 0, z == 1], [x == 2*u/(u^2 + v^2 + 1), y == 2*v/(u^2 + v^2 + 1), z == (u^2 + v^2 - 1)/(u^2 + v^2 + 1)]]"
]
},
"execution_count": 45,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"var('u v x y z') # symbolic variables\n",
"assume(z>0) # assume z>0\n",
" # solve the system above, \n",
" # with respect to x,y,z\n",
"solve([u==x/(1-z),v==y/(1-z),x^2+y^2+z^2==1],[x,y,z])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"we obtain the transformation\n",
"$$x=\\frac{2u}{u^2+v^2+1},\\quad y=\\frac{2v}{u^2+v^2+1},\\quad z=\\frac{u^2+v^2-1}{u^2+v^2+1}.$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"### Stereographic projection of $S^2$ from the South pole"
]
},
{
"cell_type": "code",
"execution_count": 46,
"metadata": {},
"outputs": [],
"source": [
"var('t u v')\n",
"po1={'thickness':4,'color':'darkblue'} # parameters\n",
"po2={'fontsize':20,'color':'black'}\n",
"po3={'fontsize':20,'color':'black'}\n",
"\n",
"p=parametric_plot3d((t,t,-1+t/0.85),(t,0,0.85), # half line\n",
" thickness=3,color='green') # from South pole\n",
"p+=point3d([(1.2*cos(t),1.2*sin(t),0) # dotted arc\n",
" for t in srange(0,pi/2,0.075)],\n",
" color='darkblue',size=7,frame=False)"
]
},
{
"cell_type": "code",
"execution_count": 47,
"metadata": {},
"outputs": [],
"source": [
"ax =line3d([(0,0,0), (1.3,0,0)], **po1) # axes\n",
"ax+=line3d([(0,0,0), (0,1.3,0)], **po1)\n",
"ax+=line3d([(0,0,-1.), (0,0,0.55)], **po1)\n",
"ax+=text3d(\"x\",(1.4,0,0),**po2)\n",
"ax+=text3d(\"y\",(0,1.5,0),**po2)\n",
"ax+=text3d(\"z\",(0.,0.,0.7),**po2)"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
" # hemi-sphere:\n",
"s=parametric_plot3d((cos(u)*cos(v), sin(u)*cos(v), sin(v)), \n",
" (u,0,2*pi), (v,-pi/2,0),opacity=0.4,color='lightgrey')\n",
"\n",
"t=text3d(\"(x,y,z)\",(0.4,0.67,-0.2),**po3) # distinguished \n",
"t+=text3d(\"(u,v,0)\",(1.,1.2,0.0),**po3) # points\n",
"t+=text3d(\"(0,0,-1)\",(-0.5,0.4,-2.2),**po3)\n",
" # intersection\n",
"P=point3d([(0.6,0.67,-0.29)],size=20) # point\n",
"P+=point3d([(0,0,-1)],size=25,color='red') # South pole\n",
"\n",
"(p+ax+s+t+P).rotateZ(-pi/1.9).show(frame=False)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Consider the line which passes through the South pole $(0,0,-1)$, the point $(x,y,z)$ on the sphere and intersects\n",
"the plane Oxy at the point $(u,v,0)$. The functions $u = u(x,y,z), v=v(x,y,z)$ define a stereographic projection from the South pole. Using the rotations around z-axis and appropriate similar triangles in $Oxz$ and $Oyz$ planes we can see that (as in the case of $S^1$) $\\quad\\frac{u}{x}=\\frac{1}{1+z},\\quad \\frac{v}{y}=\\frac{1}{1+z},\\quad$ so the stereographic projection from the South pole can be described by $$u=\\frac{x}{1+z},\\quad v=\\frac{y}{1+z}.$$\n",
"Solving this system combined with the equation of the unit sphere: $x^2+y^2+z^2=1$ with respect to $x,y,z$:"
]
},
{
"cell_type": "code",
"execution_count": 49,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\left[x = 0, y = 0, z = \\left(-1\\right)\\right], \\left[x = \\frac{2 \\, u}{u^{2} + v^{2} + 1}, y = \\frac{2 \\, v}{u^{2} + v^{2} + 1}, z = -\\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right]\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left[x = 0, y = 0, z = \\left(-1\\right)\\right], \\left[x = \\frac{2 \\, u}{u^{2} + v^{2} + 1}, y = \\frac{2 \\, v}{u^{2} + v^{2} + 1}, z = -\\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right]\\right]$"
],
"text/plain": [
"[[x == 0, y == 0, z == -1], [x == 2*u/(u^2 + v^2 + 1), y == 2*v/(u^2 + v^2 + 1), z == -(u^2 + v^2 - 1)/(u^2 + v^2 + 1)]]"
]
},
"execution_count": 49,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"forget()\n",
"%display latex\n",
"var('u v x y z') # symbolic variables\n",
"assume(z<0) # assume z<0\n",
" # solve the system above\n",
" # with respect to x,y,z\n",
"solve([u==x/(1+z),v==y/(1+z),x^2+y^2+z^2==1],[x,y,z])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"we obtain the inverse transformation\n",
"$$x=\\frac{2u}{u^2+v^2+1},\\quad y=\\frac{2v}{u^2+v^2+1},\\quad z=-\\frac{u^2+v^2-1}{u^2+v^2+1}.$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.14**\n",
"\n",
"As we mentioned in the case of $S^1$ both the stereographic projections from North and South poles are predefined in `manifolds.Sphere`."
]
},
{
"cell_type": "code",
"execution_count": 50,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ A : & \\left(\\theta, \\phi\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\cos\\left(\\theta\\right)\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{NP}\\} : & \\left(u, v\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, u}{u^{2} + v^{2} + 1}, \\frac{2 \\, v}{u^{2} + v^{2} + 1}, \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{SP}\\} : & \\left({u'}, {v'}\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + {v'}^{2} + 1}, \\frac{2 \\, {v'}}{{u'}^{2} + {v'}^{2} + 1}, -\\frac{{u'}^{2} + {v'}^{2} - 1}{{u'}^{2} + {v'}^{2} + 1}\\right) \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\iota:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ A : & \\left(\\theta, \\phi\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\cos\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\sin\\left(\\phi\\right) \\sin\\left(\\theta\\right), \\cos\\left(\\theta\\right)\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{NP}\\} : & \\left(u, v\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, u}{u^{2} + v^{2} + 1}, \\frac{2 \\, v}{u^{2} + v^{2} + 1}, \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right) \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{SP}\\} : & \\left({u'}, {v'}\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + {v'}^{2} + 1}, \\frac{2 \\, {v'}}{{u'}^{2} + {v'}^{2} + 1}, -\\frac{{u'}^{2} + {v'}^{2} - 1}{{u'}^{2} + {v'}^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"iota: S^2 → E^3\n",
"on A: (theta, phi) ↦ (x, y, z) = (cos(phi)*sin(theta), sin(phi)*sin(theta), cos(theta))\n",
"on S^2-{NP}: (u, v) ↦ (x, y, z) = (2*u/(u^2 + v^2 + 1), 2*v/(u^2 + v^2 + 1), (u^2 + v^2 - 1)/(u^2 + v^2 + 1))\n",
"on S^2-{SP}: (up, vp) ↦ (x, y, z) = (2*up/(up^2 + vp^2 + 1), 2*vp/(up^2 + vp^2 + 1), -(up^2 + vp^2 - 1)/(up^2 + vp^2 + 1))"
]
},
"execution_count": 50,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"S2=manifolds.Sphere(2) # sphere S^2\n",
" # stereographic proj. from North pole:\n",
"stereoN.=S2.stereographic_coordinates(pole='north')\n",
" # stereographic proj. from South pole:\n",
"stereoS.=S2.stereographic_coordinates(pole='south')\n",
"Phi=S2.embedding() # embeddings\n",
"Phi.disp() # show embeddings "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.15**\n",
"\n",
"To show graphically how the projection from the North pole acts, let us extract \n",
"an appropriate part of definition from the previous cell."
]
},
{
"cell_type": "code",
"execution_count": 51,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{llcl} \\Phi_N:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{NP}\\} : & \\left(u, v\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, u}{u^{2} + v^{2} + 1}, \\frac{2 \\, v}{u^{2} + v^{2} + 1}, \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right) \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\Phi_N:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{NP}\\} : & \\left(u, v\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, u}{u^{2} + v^{2} + 1}, \\frac{2 \\, v}{u^{2} + v^{2} + 1}, \\frac{u^{2} + v^{2} - 1}{u^{2} + v^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"PhiN: S^2 → E^3\n",
"on S^2-{NP}: (u, v) ↦ (x, y, z) = (2*u/(u^2 + v^2 + 1), 2*v/(u^2 + v^2 + 1), (u^2 + v^2 - 1)/(u^2 + v^2 + 1))"
]
},
"execution_count": 51,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# continuation\n",
"E=S2.ambient() # ambient space E^3\n",
"c_cart. = E.cartesian_coordinates() # Cartesian coordinates\n",
"fun=list(Phi.coord_functions(stereoN).expr()) # embedd. functions\n",
"PhiN = S2.continuous_map(E,{(stereoN, c_cart): fun},\n",
" name='PhiN',latex_name=r'\\Phi_N') # define embedding PhiN\n",
"PhiN.display() # show PhiN"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In the figure below we show the arcs on the sphere, which correspond to the lines $v=\\text{const}$ on the $(u,v)$ plane (for the stereographic projection from the North pole)."
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"# continuation\n",
"p=stereoN.plot(c_cart,mapping=PhiN,ranges={u:(-4,4),v:(-4,4)},\n",
" number_values={u:0,v:14},color='black',thickness=3,\n",
" plot_points=200,label_axes=False) # plot coordinate lines\n",
"s=sphere(color='lightgrey',opacity=0.6) # plot sphere\n",
"(p+s).show(frame=False,label_axes=False) # combine plots"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.16**\n",
"\n",
"Now let us do the same for the South pole."
]
},
{
"cell_type": "code",
"execution_count": 53,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\begin{array}{llcl} \\Phi_S:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{SP}\\} : & \\left({u'}, {v'}\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + {v'}^{2} + 1}, \\frac{2 \\, {v'}}{{u'}^{2} + {v'}^{2} + 1}, -\\frac{{u'}^{2} + {v'}^{2} - 1}{{u'}^{2} + {v'}^{2} + 1}\\right) \\end{array}\\)"
],
"text/latex": [
"$\\displaystyle \\begin{array}{llcl} \\Phi_S:& \\mathbb{S}^{2} & \\longrightarrow & \\mathbb{E}^{3} \\\\ \\mbox{on}\\ \\mathbb{S}^{2}\\setminus\\{\\mathrm{SP}\\} : & \\left({u'}, {v'}\\right) & \\longmapsto & \\left(x, y, z\\right) = \\left(\\frac{2 \\, {u'}}{{u'}^{2} + {v'}^{2} + 1}, \\frac{2 \\, {v'}}{{u'}^{2} + {v'}^{2} + 1}, -\\frac{{u'}^{2} + {v'}^{2} - 1}{{u'}^{2} + {v'}^{2} + 1}\\right) \\end{array}$"
],
"text/plain": [
"PhiS: S^2 → E^3\n",
"on S^2-{SP}: (up, vp) ↦ (x, y, z) = (2*up/(up^2 + vp^2 + 1), 2*vp/(up^2 + vp^2 + 1), -(up^2 + vp^2 - 1)/(up^2 + vp^2 + 1))"
]
},
"execution_count": 53,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# continuation\n",
"fun=list(Phi.coord_functions(stereoS).expr()) # embedd. functions\n",
"PhiS = S2.continuous_map(E,{(stereoS, c_cart): fun},\n",
" name='PhiS',latex_name=r'\\Phi_S') # define embedd PhiS\n",
"PhiS.display() # show PhiS"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"In the figure below we show the arcs on the sphere, which correspond to the lines $v'=\\text{const}$ on the $(u',v')$ plane (for the stereographic projection from the South pole)."
]
},
{
"cell_type": "code",
"execution_count": 54,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\n",
"\n"
],
"text/plain": [
"Graphics3d Object"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"p=stereoS.plot(c_cart,mapping=PhiS,ranges={up:(-4,4),vp:(-4,4)},\n",
" number_values={up:0,vp:25},color='black',thickness=3,\n",
" plot_points=200,label_axes=False) # plot coordinate lines\n",
"s=sphere(color='lightgrey',opacity=0.6) # plot sphere\n",
"(p+s).show(frame=False,label_axes=False) # combine plots"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Transition map"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"From the relations $\\ (u,v)=(\\frac{x}{1-z},\\frac{y}{1-z})\\ $ and $\\ (u',v')=(\\frac{x}{1+z},\\frac{y}{1+z})\\ $ it follows\n",
"$\\ (1+z)(u',v')=(1-z)(u,v)=(x,y)\\ $, so the transition map is of the form \n",
"$$(u',v')=\\frac{1-z}{1+z}(u,v).$$\n",
"\n",
"To calculate the fraction $\\ \\frac{1-z}{1+z}\\ $ let us note that $\\ z=\\frac{u^2+v^2-1}{u^2+v^2+1}=\\frac{r^2-1}{r^2+1},\\ $ where $\\ r^2=u^2+v^2.\\ $ Accordingly $\\ \\frac{1-z}{1+z}=\\frac{1-\\frac{r^2-1}{r^2+1}}{1+\\frac{r^2-1}{r^2+1}}=\\frac{r^2+1-r^2+1}{r^2+1+r^2-1}=\\frac{2}{2r^2}=\\frac{1}{r^2}.\\ $ Consequently the transition map has the form\n",
"$$u'=\\frac{u}{u^2+v^2},\\quad v'=\\frac{v}{u^2+v^2}.$$\n",
"\n",
"
\n",
"\n",
"**Example 4.17**\n",
"\n",
"Transition maps from one stereographic projection to the second one in `SageMath Manifolds`:"
]
},
{
"cell_type": "code",
"execution_count": 55,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.\\)"
],
"text/latex": [
"$\\displaystyle \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.$"
],
"text/plain": [
"up = u/(u^2 + v^2)\n",
"vp = v/(u^2 + v^2)"
]
},
"execution_count": 55,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"S2=manifolds.Sphere(2) # sphere S^2\n",
" # stereographic proj. from North pole:\n",
"stereoN.=S2.stereographic_coordinates(pole='north')\n",
" # stereographic proj. from South pole:\n",
"stereoS.=S2.stereographic_coordinates(pole='south')\n",
"\n",
"stereoN_to_stereoS = stereoN.transition_map(stereoS, \n",
" (u/(u^2+v^2), v/(u^2+v^2)),\n",
" intersection_name='W', restrictions1= u^2+v^2!=0,\n",
" restrictions2= up^2+vp^2!=0) # define transition map\n",
"stereoN_to_stereoS.display() # show transition map"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"`SageMath` can perform the corresponding calculation (with the help of Maxima)."
]
},
{
"cell_type": "code",
"execution_count": 56,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[2 \\, {\\left(u^{2} + v^{2}\\right)} {v'} - 2 \\, v, -2 \\, {u'} v + 2 \\, u {v'}\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[2 \\, {\\left(u^{2} + v^{2}\\right)} {v'} - 2 \\, v, -2 \\, {u'} v + 2 \\, u {v'}\\right]$"
],
"text/plain": [
"[2*(u^2 + v^2)*vp - 2*v, -2*up*v + 2*u*vp]"
]
},
"execution_count": 56,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"var('u,v,up,vp,x,y,z') # symbolic variables\n",
"\n",
"# Eliminate variables x,y,z from the system defining stereogr. proj\n",
"maxima.eliminate([u==x/(1-z),v==y/(1-z),up==x/(1+z),vp==y/(1+z),\n",
" z==(u^2+v^2-1)/(u^2+v^2+1)],[x,y,z]).sage()"
]
},
{
"cell_type": "code",
"execution_count": 57,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left[\\left[{u'} = \\frac{u}{u^{2} + v^{2}}, {v'} = \\frac{v}{u^{2} + v^{2}}\\right]\\right]\\)"
],
"text/latex": [
"$\\displaystyle \\left[\\left[{u'} = \\frac{u}{u^{2} + v^{2}}, {v'} = \\frac{v}{u^{2} + v^{2}}\\right]\\right]$"
],
"text/plain": [
"[[up == u/(u^2 + v^2), vp == v/(u^2 + v^2)]]"
]
},
"execution_count": 57,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Solve the obtained system with respect to up,vp\n",
"solve([2*(u^2 + v^2)*vp - 2*v, -2*up*v + 2*u*vp],[up,vp])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"The inverse transition map has the form\n",
"$$u=\\frac{u'}{u'^2+v'^2},\\quad v=\\frac{v'}{u'^2+v'^2}.$$"
]
},
{
"cell_type": "code",
"execution_count": 58,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right.\\)"
],
"text/latex": [
"$\\displaystyle \\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right.$"
],
"text/plain": [
"u = up/(up^2 + vp^2)\n",
"v = vp/(up^2 + vp^2)"
]
},
"execution_count": 58,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"stereoN_to_stereoS.inverse().display() # inverse transition map"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Both transformations are smooth provided $u^2+v^2$ and $u'^2+v'^2$ are nonzero."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.18**\n",
"\n",
"In fact, both transition maps are predefined in `manifolds.Sphere`."
]
},
{
"cell_type": "code",
"execution_count": 59,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.\\right)$"
],
"text/plain": [
"(up = u/(u^2 + v^2)\n",
" vp = v/(u^2 + v^2),\n",
" up = u/(u^2 + v^2)\n",
" vp = v/(u^2 + v^2))"
]
},
"execution_count": 59,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"%display latex\n",
"dim=2 # dim of the sphere\n",
"Sph=manifolds.Sphere(dim) # sphere S^2 \n",
"spher = Sph.spherical_coordinates() # spherical coord.\n",
"stereoN, stereoS = Sph.coordinate_charts('stereographic',\n",
" names=['u','v']) # sterogr. projections\n",
"A=stereoN.domain() # domain of stereoN\n",
" # intersection of domains of stereoN and stereoS:\n",
"W = Sph._stereoN_dom.intersection(Sph._stereoS_dom)\n",
" # transition from stereoN to stereoS:\n",
"FNS=Sph.coord_change(stereoN.restrict(W),stereoS.restrict(W))\n",
" # transition from stereoS to stereoN:\n",
"FSN=Sph.coord_change(stereoS.restrict(W),stereoN.restrict(W))\n",
"FNS.disp(),FSN.disp() # show transitions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.19**\n",
"\n",
"The transitions from stereoN \n",
"to stereoS and the inverse can be also obtained as compositions."
]
},
{
"cell_type": "code",
"execution_count": 60,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right.\\right)$"
],
"text/plain": [
"(u = up/(up^2 + vp^2)\n",
" v = vp/(up^2 + vp^2),\n",
" up = u/(u^2 + v^2)\n",
" vp = v/(u^2 + v^2))"
]
},
"execution_count": 60,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"dim=2 # dimension of sphere\n",
"Sph=manifolds.Sphere(dim) # manifold S^2\n",
"spher = Sph.spherical_coordinates() # spherical coordinates\n",
"stereoN,stereoS=Sph.coordinate_charts('stereographic',\n",
" names=['u','v']) # stereographic coordinates\n",
"A = spher.domain() # domain of stereogr. coord.\n",
" # intersection of stereoN and stereoS coordinates domains:\n",
"W = Sph._stereoN_dom.intersection(Sph._stereoS_dom)\n",
"V=W.intersection(A) # intersection with spher. coord. domain \n",
" # transition spher->stereoN:\n",
"F1=Sph.coord_change(spher.restrict(V), stereoN.restrict(V))\n",
" # transition stereoN->spher:\n",
"F2=Sph.coord_change( stereoN.restrict(V),spher.restrict(V))\n",
" # transition spher->stereoS:\n",
"F3=Sph.coord_change(spher.restrict(V), stereoS.restrict(V))\n",
" # transition stereoS->spher\n",
"F4=Sph.coord_change( stereoS.restrict(V),spher.restrict(V))\n",
"(F1*F4).disp(),(F3*F2).disp() # show compositions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"
\n",
"\n",
"**Example 4.20**\n",
"\n",
"Knowing the transition maps we are ready to an independent of `manifolds.Sphere` definition of $S^2$ as a manifold with two maps. "
]
},
{
"cell_type": "code",
"execution_count": 61,
"metadata": {},
"outputs": [
{
"data": {
"text/html": [
"\\(\\displaystyle \\left(\\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right.\\right)\\)"
],
"text/latex": [
"$\\displaystyle \\left(\\left\\{\\begin{array}{lcl} {u'} & = & \\frac{u}{u^{2} + v^{2}} \\\\ {v'} & = & \\frac{v}{u^{2} + v^{2}} \\end{array}\\right., \\left\\{\\begin{array}{lcl} u & = & \\frac{{u'}}{{u'}^{2} + {v'}^{2}} \\\\ v & = & \\frac{{v'}}{{u'}^{2} + {v'}^{2}} \\end{array}\\right.\\right)$"
],
"text/plain": [
"(up = u/(u^2 + v^2)\n",
" vp = v/(u^2 + v^2),\n",
" u = up/(up^2 + vp^2)\n",
" v = vp/(up^2 + vp^2))"
]
},
"execution_count": 61,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"S2 = Manifold(2, 'S^2') # S^2-sphere\n",
"U = S2.open_subset('U') # complement of the North pole\n",
"V = S2.open_subset('V') # complement of the South pole\n",
"S2.declare_union(U,V)\n",
"c_uv. = U.chart() # stereogr. coordinates from the North pole\n",
"c_upvp. = V.chart() # stereogr. coord. from the South pole\n",
"\n",
"uv_to_upvp = c_uv.transition_map(c_upvp, (u/(u^2+v^2), v/(u^2+v^2)),\n",
" intersection_name='W', # transition map\n",
" restrictions1= u^2+v^2!=0, \n",
" restrictions2= up^2+vp^2!=0)\n",
"upvp_to_uv = uv_to_upvp.inverse() # inverse transition\n",
"uv_to_upvp.disp(),upvp_to_uv.disp() # show transitions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## What's next?\n",
"\n",
"Take a look at the notebook [Spheres and spherical coordinates in higher dimensions](https://nbviewer.org/github/sagemanifolds/IntroToManifolds/blob/main/05Manifold_Spher_Higher_Dim.ipynb)."
]
}
],
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