16Manifold_Ext_Der.ipynb

Kernel: SageMath 9.6

16. Exterior derivative

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'

If $(x^1,\ldots,x^n)$ are local coordinates on a smooth manifold $M$, then from (14.9) we know that for any differential $k$-form on $M$

$$\omega=\sum_{1\leq i_1<\ldots<i_k\leq n}ω_{i_1 ...i_k}dx^{i_1}\wedge\ldots\wedge dx^{i_k}= \frac{1}{k!}ω_{i_1 ...i_k}dx^{i_1}\wedge\ldots\wedge dx^{i_k}.$$

The exterior derivative of $\omega\$ can be defined locally by

ParseError: KaTeX parse error: Undefined control sequence: \label at position 278: …, \end{matrix} \̲l̲a̲b̲e̲l̲{}\tag{16.1} \e…

($\ d\omega_{i_1\ldots i_k}\$ denotes the differential of the scalar function $\ \omega_{i_1\ldots i_k}$).

Example 16.1

Take a 1-form and compute its exterior derivative.

%display latex
dim = 3                                   # dimension of manifold N
N = Manifold(dim, 'N')                    # manifold N
X = N.chart(' '.join(['x'+str(i)+':x^{'+str(i)+'}' for i in range(dim)]))  # chart on N
al = N.diff_form(1, name=r'\alpha')        # 1-form alpha  
def astr(i): return 'a_'+str(i)           # names of components
af = [N.scalar_field(function(astr(i))(*X), name=astr(i)) 
                     for i in range(dim)]  # component functions
al[:] = af                                 # define all components
al.disp()                                  # show al

$\displaystyle \alpha = a_{0}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}} + a_{1}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{1}} + a_{2}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{2}}$

Use the method exterior_derivative() to compute the result:

dal = al.exterior_derivative()
dal.disp()

$\displaystyle \mathrm{d}\alpha = \left( -\frac{\partial\,a_{0}}{\partial {x^{1}}} + \frac{\partial\,a_{1}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + \left( -\frac{\partial\,a_{0}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + \left( -\frac{\partial\,a_{1}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{1}}} \right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}$

Let us apply the definition (16.1) for comparison:

# continuation
dx = X.coframe()                        # (dx0,dx1,dx2)
                                        # apply the formula (16.1):
s = sum([(af[i]).differential().wedge(dx[i]) for i in range(dim)])
# SageMath automatically displays the wedge products 
# with increasing indices of variables
s.disp()                                # show the result

$\displaystyle \mathrm{d}a_0\wedge \mathrm{d} {x^{0}} + \mathrm{d}a_1\wedge \mathrm{d} {x^{1}} + \mathrm{d}a_2\wedge \mathrm{d} {x^{2}} = \left( -\frac{\partial\,a_{0}}{\partial {x^{1}}} + \frac{\partial\,a_{1}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + \left( -\frac{\partial\,a_{0}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + \left( -\frac{\partial\,a_{1}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{1}}} \right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}$

Note that for example (since $dx^i\wedge dx^i=0$)

$$da_0\wedge dx^0=(\frac{\partial a_0}{\partial x^0}dx^0+\frac{\partial a_0}{\partial x^1}dx^1+\frac{\partial a_0}{\partial x^2}dx^2)\wedge dx^0\\ =\frac{\partial a_0}{\partial x^1}dx^1\wedge dx^0+ \frac{\partial a_0}{\partial x^2}dx^2\wedge dx^0 =-\frac{\partial a_0}{\partial x^1}dx^0\wedge dx^1 -\frac{\partial a_0}{\partial x^2}dx^0\wedge dx^2,$$

and analogously for the remaining terms (in $\ \ da_2\wedge dx^2\ \$ there is no need of reordering the terms).

Example 16.2

Consider now a 2-form and its exterior differential.

%display latex
N=3                                    # dimension of the manifold M
M = Manifold(N, 'M')                   # manifold M
X = M.chart(' '.join(['x'+str(i)+':x^{'+str(i)+'}' for i in range(N)]))  # chart on M
a = M.diff_form(2, name='a')           # 2-form a
                                       # components of the 2-form a:
x0, x1, x2 = X[:]  # coordinates x^0, x^1, x^2 of chart X as the Python variables x0, x1, x2
a01 = M.scalar_field(function('a01')(x0,x1,x2), name='a01')
a02 = M.scalar_field(function('a02')(x0,x1,x2), name='a02')
a12 = M.scalar_field(function('a12')(x0,x1,x2), name='a12')
a[0,1] = a01; a[0,2] = a02             # define components of a
a[1,2] = a12                           # (up.triangle of comp.matr.)
a.disp()                               # show a

$\displaystyle a = a_{01}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + a_{02}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + a_{12}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}$

The exterior derivative:

a.exterior_derivative().disp()

$\displaystyle \mathrm{d}a = \left( \frac{\partial\,a_{01}}{\partial {x^{2}}} - \frac{\partial\,a_{02}}{\partial {x^{1}}} + \frac{\partial\,a_{12}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}$

In this case for each differential $da_{ij}$ only one term of the form $\frac{\partial a_{ij}}{\partial x^i}dx^i$ (no summation) gives a nonzero contribution to the final result. For example, for the second term of $a$ we have

$$da_{02}\wedge dx^0\wedge dx^2=\Big(\frac{\partial a_{02}}{\partial x^0}dx^0+\frac{\partial a_{02}}{\partial x^1}dx^1+\frac{\partial a_{02}}{\partial x^2}dx^2\Big)dx^0\wedge dx^2\\ =\frac{\partial a_{02}}{\partial x^1}dx^1\wedge dx^0\wedge dx^2 =-\frac{\partial a_{02}}{\partial x^1}dx^0\wedge dx^1\wedge dx^2.$$

The calculations for the remaining terms are analogous.

Exterior derivative is antiderivation

ParseError: KaTeX parse error: Undefined control sequence: \label at position 53: …(−1)^k ω ∧ dη, \̲l̲a̲b̲e̲l̲{}\tag{16.2} \e…

for $\omega\in\Omega^k(M)$ and $\eta \in\Omega^m(M)$.

Using the linearity of the operation $d$, it is enough to prove the antiderivation property for single terms $ω = ω_{i_1 ...i_k} dx^{i_1} ∧ · · · ∧ dx^{i_k}$ and $η = η_{j_1 ... j_m }dx^{j_1} ∧· · · ∧ dx^{j_m}$ (no summation here). From (16.1) and (13.4) we obtain

$$d(ω ∧ η) = d( ω_{i_1 ...i_k} η_{j_1 ... j_m} dx^{i_1} ∧ · · · ∧ dx^{i_k} ∧ dx^{j_1} ∧ · · · ∧ dx^{j_m})\\ = d(ω_{i_1 ...i_k} η_{j_1 ... j_m} ) ∧ dx^{i_1} ∧ · · · ∧ dx^{i_k} ∧ dx^{j_1} ∧ · · · ∧ dx^{j_m}\\ =[(dω_{i_1 ...i_k} )η_{j_1 ... j_m} + ω_{i_1 ...i_k} dη_{j_1 ... j_m}] ∧ dx^{i_1} ∧ · · · ∧ dx^{i_k} ∧ dx^{j_1} ∧ · · · ∧ dx^{j_m}\\ =(dω_{i_1 ...i_k} ∧ dx^{i_1} ∧ · · · ∧ dx^{i_k} ) ∧ (η_{j_1 ... j_m} dx^{j_1} ∧ · · · ∧ dx^{j_m} )\\ +(−1)^k (ω_{i_1 ...i_k} dx^{i_1} ∧ · · · ∧ dx^{i_k} ) ∧ (dη_{j_1 ... j_m} ∧ dx^{j_1} ∧ · · · ∧ dx^{j_m}) \\ = dω ∧ η + (−1)^k ω ∧ dη.$$

We have proved (16.2).

Exterior derivative is nilpotent

ParseError: KaTeX parse error: Undefined control sequence: \label at position 38: …ta=d(d\eta)=0. \̲l̲a̲b̲e̲l̲{}\tag{16.3} \e…

Again, it is enough to consider forms $\eta= η_{i_1 ...i_m} ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m}$ (no summation here). If $ω = dη$ with $η ∈ \Omega^m (M)$, we have $$ω = dη_{i_1 ...i_m} ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m} = \Big(\frac{\partial}{\partial x^j}η_{i_1 ...i_m}\Big)dx^j ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m}.$$
Computing the exterior derivative of $\omega$ we obtain $$d\omega=d\Big(\frac{\partial}{\partial x^j}η_{i_1 ...i_m}\Big)dx^j ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m}\\ =\Big(\frac{\partial}{\partial x^p}\frac{\partial}{\partial x^j}η_{i_1 ...i_m}\Big)dx^p\wedge dx^j ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m}.$$ In the last sum if $p = j$, then $dx^p ∧ dx^j = 0,$ if $p\not= j$, then $\frac{∂^2 f }{ ∂ x^p ∂ x^j}$ is symmetric in $p$ and $j$, but $dx^p ∧ dx^j$ is alternating in $p$ and $j$, so the terms with $p \not= j$ pair up and cancel each other. For example $$\frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}dx^1\wedge dx^2 +\frac{∂^2 \eta_{...} }{ ∂ x^2 ∂ x^1}dx^2\wedge dx^1\\ =\frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}dx^1\wedge dx^2+ \frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}(-dx^1\wedge dx^2)=0.$$

Pullback and wedge product

ParseError: KaTeX parse error: Undefined control sequence: \label at position 51: … ω) ∧ (ψ^∗ η), \̲l̲a̲b̲e̲l̲{}\tag{16.4} \e…

for $\omega\in\Omega^k(N),\ \eta\in\Omega^m(N)$ and a smooth map $\psi:M\to N.$

In fact, we have

$$\psi^*(\omega\wedge\eta)(v_1,\ldots,v_{k+m})= \omega\wedge\eta(d\psi(v_1),\ldots,d\psi(v_{k+m}))\\ =\frac{(k+m)!}{k!m!}\mathrm{Alt}(\omega\otimes\eta)(d\psi(v_1),\ldots,d\psi(v_{k+m}))\\ =\frac{1}{k!m!}\sum_{\sigma\in S_{k+m}}\mathrm{sign}\,\sigma\, \omega(d\psi(v_{\sigma(1)},\ldots,d\psi(v_{\sigma(k)})\eta(d\psi(v_{\sigma(k+1)}),\ldots,d\psi(v_{\sigma(k+m)}))\\ =\frac{1}{k!m!}\sum_{\sigma\in S_{k+m}}\mathrm{sign}\,\sigma\, \psi^*\omega(v_{\sigma(1)},\ldots,v_{\sigma(k)}\psi^*\eta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+m)})\\ =\frac{(k+m)!}{k!m!}\mathrm{Alt}((\psi^*\omega)\otimes(\psi^*\eta))(v_1,\ldots,v_{k+m})\\ =(\psi^*(\omega))\wedge(\psi^*(\eta))(v_1,\ldots,v_{k+m}).$$

We have checked (16.4).

Pullback and exterior derivative

Using the relation between the differential of a scalar function and the pullback (cf. (15.2)) one can check that for $\omega\in\Omega^k(N)$ of the special form $\ \omega=ω_{i_1 ...i_k}dy^{i_1}\wedge\ldots\wedge dy^{i_k}\$ (no summation) and a smooth map $\psi:M\to N$

$$ψ^∗ (dω) = \psi^*(d\omega_{i_1...i_k}\wedge dy^{i_1}\wedge ...dy^{i_k})\\ =\psi^*(d\omega_{i_1...i_k})\wedge \psi^*(dy^{i_1})... \wedge\psi^*(dy^{i_k})\\ =d(ψ^∗ ω_{i_1 ...i_k} ) ∧ d(ψ^∗ y^{i_1} ) ∧ · · · ∧d(ψ^*y^{ i_k} )\\ = d [(ψ^∗ ω_{i_1 ...i_k} ) d(ψ^∗ y^{i_1} ) ∧ · · · ∧ d(ψ^∗ y^{i_k} )]\\ = d[ (ψ^∗ ω_{i_1 ...i_k} )ψ^∗ (dy^{i_1} ) ∧ · · · ∧ ψ^∗ (dy^{i_k} )]\\ = d [ψ^∗ (ω_{ i_1 ...i_k} dy^{i_1} ∧ · · · ∧ dy^{i_k} )] = d(ψ^∗ ω).$$

Due to linearity, we have proved ParseError: KaTeX parse error: Undefined control sequence: \label at position 81: …\Omega^k (N ). \̲l̲a̲b̲e̲l̲{}\tag{16.5} \e…

Example 16.3

Compute the pullback of $\ \alpha=dx\wedge dy\ \$ under the map $\Phi: R_{r,\phi}\to R^2_{x,y},$ $\Phi(r,\phi)=(r\cos\phi,r\sin\phi)$.

%display latex
M = Manifold(2, 'R^2')           # manifold M
c_xy.<x,y> = M.chart()           # Cartesian coordinates on M
N = Manifold(2, 'R^2p')          # manifold N
c_rp.<r,phi>=N.chart()           # polar coordinates on N
Phi = N.diff_map(M, (r*cos(phi), r*sin(phi)), name=r'\Phi') # N-> M
alpha = M.diff_form(2,r'\alpha')   # 2-form alpha on M 
alpha[0,1] = 1                     # only one nonzero comp. of alpha
alpha.disp()                       # show alpha

$\displaystyle \alpha = \mathrm{d} x\wedge \mathrm{d} y$

Pullback of $\alpha$:

plb = Phi.pullback(alpha)       # pullback of alpha
plb.display()                   # show pullback

$\displaystyle {\Phi}^*\alpha = r \mathrm{d} r\wedge \mathrm{d} \phi$

According to the remark after formula (15.6) one can check that if we replace $\ x\$ by $\ \ r\cos\phi\ \$ and $\ \ y\ \$ by $\ \ r\sin\phi\ \$ in $dx\wedge dy\ \$ and compute the differentials (w.r.t $(r,\phi)$) of the obtained functions, then we get

$$d(r\cos\phi)\wedge d(r\sin\phi)=(\frac{\partial (r\cos\phi)}{\partial r}dr+ \frac{\partial (r\cos\phi)}{\partial \phi}d\phi)\wedge (\frac{\partial (r\sin\phi)}{\partial r}dr+ \frac{\partial (r\sin\phi)}{\partial \phi}d\phi)\\ =(\cos\phi dr-r\sin\phi d\phi)\wedge (\sin\phi dr+r\cos\phi d\phi)=\\\\ r\cos^2\phi dr\wedge d\phi -r\sin^2\phi d\phi\wedge dr = r(\cos^2\phi+\sin^2\phi)dr\wedge d\phi =rdr\wedge d\phi.$$

Example 16.4

Compute the pullback of $\ \alpha=dx\wedge dy\wedge dz\ \$ under the map $\Phi: R^3_{r,\phi,\theta}\to R^3_{x,y,z}$, $\ \ \Phi(r,\phi,\theta)=(r\cos\phi\sin\theta,r\sin\phi\sin\theta,r\cos\theta).$

%display latex
M = Manifold(3, 'R^3')             # manifold M=R^3 (Cart. coord.)
c_xyz.<x,y,z> = M.chart()          # Cartesian coordinates on M
N = Manifold(3, 'R^3p')            # manifold N=R^3 (spher.coord.)
c_rpt.<r,theta,phi>=N.chart()      # spherical coordinates on N
Phi = N.diff_map(M, (r*cos(phi)*sin(theta),   # Phi N -> M
                     r*sin(phi)*sin(theta),
                     r*cos(theta)),name=r'\Phi')
alpha = M.diff_form(3,r'\alpha')   # 3-form alpha on M 
alpha[0,1,2] = 1                   # only one nonzero comp. of alpha
alpha.disp()                       # show alpha

$\displaystyle \alpha = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z$

Pullback of $\alpha$ under $\Phi$:

plb = Phi.pullback(alpha)         # pullback of alpha
plb.display()                     # show pullback

$\displaystyle {\Phi}^*\alpha = r^{2} \sin\left(\theta\right) \mathrm{d} r\wedge \mathrm{d} \theta\wedge \mathrm{d} \phi$

Example 16.5

Compute $\Phi^*\alpha\ \$ for $\Phi:R\to R^2,\ \ \Phi(t)=(\cos(t),\sin(t))\$ and $\ \alpha= \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$.

M = Manifold(2, 'R^2')            # manifold M= R^2
c_xy.<x,y> = M.chart()            # Cartesian coordinates on M
N = Manifold(1, 'R^1')            # manifold N=R^1
c_t.<t> = N.chart()               # coordinate t
Phi = N.diff_map(M, (cos(t), sin(t)), name=r'\Phi')  # Phi: N -> M
alpha = M.diff_form(1,r'\alpha')     # 1-form alpha on M
alpha[:] = -y/(x^2+y^2), x/(x^2+y^2) # components of alpha
alpha.disp()                         # show alpha

$\displaystyle \alpha = \left( -\frac{y}{x^{2} + y^{2}} \right) \mathrm{d} x + \left( \frac{x}{x^{2} + y^{2}} \right) \mathrm{d} y$

Pullback $\ \Phi^*\alpha\ \$

plb = Phi.pullback(alpha)        # pullback Phi^*alpha
plb.apply_map(factor)            # factor all components
plb.display()                    # show the result

$\displaystyle {\Phi}^*\alpha = \mathrm{d} t$

Example 16.6

Compute the pullback of $\ \ \alpha=a_0(x,y)dx+a_1(x,y)dy\ \$ under the map $\ \psi:R^2_{u,v}\to R^2_{x,y},$ $\ \psi(u,v)=(\psi_1(u,v),\psi_2(u,v)).$

%display latex
M = Manifold(2, r'R^2_{uv}')         # manifold M=R^2_uv
c_uv.<u,v>=M.chart()                 # coordinates u,v
N = Manifold(2, r'R^2_{xy}')         # manifold N=R^2_xy
c_xy.<x,y> = N.chart()               # Cartesian coord. on N
psi1 = M.scalar_field(function('psi1')(u,v), name=r'\psi1') # first comp. of psi
psi2 = M.scalar_field(function('psi2')(u,v), name=r'\psi2') # second comp. of psi
psi = M.diff_map(N,(psi1.expr(),psi2.expr()), name=r'\psi') # psi: M -> N
al = N.diff_form(1,name=r'\alpha')   # 1-form on N
astr = ['a'+str(i) for i in range(2)]  # names of components of al
                                       # component functions:
af = [N.scalar_field(function(astr[i])(x,y), name=astr[i]) for i in range(2)]
al[:] = af                           # define all components
al.disp()                            # show al

$\displaystyle \alpha = a_{0}\left(x, y\right) \mathrm{d} x + a_{1}\left(x, y\right) \mathrm{d} y$

Pullback $\ \ \psi^*\alpha$:

alplb = psi.pullback(al)            # pullback of al
alplb.disp()                        # show pullback

$\displaystyle {\psi}^*\alpha = \left( a_{0}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{1}}{\partial u} + a_{1}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{2}}{\partial u} \right) \mathrm{d} u + \left( a_{0}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{1}}{\partial v} + a_{1}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{2}}{\partial v} \right) \mathrm{d} v$

Example 16.7

For $\ \psi\ \$ as above compute $\ \ \psi^*(dx\wedge dy)$.

# continuation
beta = N.diff_form(2, r'\beta')      # 2-form beta on N
beta[0,1] = 1                        # only one nonzero component
beta.disp()                          # show beta

$\displaystyle \beta = \mathrm{d} x\wedge \mathrm{d} y$

$\ \ \psi^*(dx\wedge dy)$:

betaplb = psi.pullback(beta)        # pullback psi^*beta
betaplb.disp()                      # show pullback

$\displaystyle {\psi}^*\beta = \left( -\frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial u} + \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial v} \right) \mathrm{d} u\wedge \mathrm{d} v$

Note that the unique component of the obtained pullback is the determinant of the Jacobian matrix $\left(\begin{matrix} \frac{\partial \psi_1}{\partial u} & \frac{\partial \psi_1}{\partial v}\\ \frac{\partial \psi_2}{\partial u} & \frac{\partial \psi_2}{\partial v} \end{matrix} \right).$

Example 16.8

For the map $\psi:R^3_{u,v,w}\to R^3_{x,y,z}$, defined by $\ \psi(u,v,w)=(\psi_1(u,v,w),\psi_2(u,v,w),\psi_3(u,v,w))\ \$ compute $\ \psi^*(dx\wedge dy\wedge dz).$

%display latex
M = Manifold(3, r'R^3_{uvw}')       # manifold M=R^3_uvw
c_uvw.<u,v,w>=M.chart()             # coordinates on M
N = Manifold(3, r'R^3_{xyz}')       # manifold N=R^3_xyz  
c_xy.<x,y,z> = N.chart()            # Cartesian coord on N
psi1 = M.scalar_field(function('psi1')(u,v,w), name=r'\psi1') # first component of psi
psi2 = M.scalar_field(function('psi2')(u,v,w), name=r'\psi2') # second component of psi
psi3 = M.scalar_field(function('psi3')(u,v,w), name=r'\psi3') # third component of psi
psi = M.diff_map(N,(psi1.expr(),psi2.expr(),psi3.expr()) ,name=r'\psi') # psi: M->N
beta = N.diff_form(3,r'\beta')      # 3-form beta on N
beta[0,1,2] = 1                     # only one nonzero component of beta 
beta.disp()                         # show beta

$\displaystyle \beta = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z$

betaplb=psi.pullback(beta)          # pullback psi*beta
betaplb.disp()                      # show pullback

$\displaystyle {\psi}^*\beta = \left( -{\left(\frac{\partial\,\psi_{1}}{\partial w} \frac{\partial\,\psi_{2}}{\partial v} - \frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial w}\right)} \frac{\partial\,\psi_{3}}{\partial u} + {\left(\frac{\partial\,\psi_{1}}{\partial w} \frac{\partial\,\psi_{2}}{\partial u} - \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial w}\right)} \frac{\partial\,\psi_{3}}{\partial v} - {\left(\frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial u} - \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial v}\right)} \frac{\partial\,\psi_{3}}{\partial w} \right) \mathrm{d} u\wedge \mathrm{d} v\wedge \mathrm{d} w$

The unique component of the pullback is the Laplace expansion of the determinant of the Jacobian matrix $$\left(\begin{matrix} \frac{\partial \psi_1}{\partial u} & \frac{\partial \psi_1}{\partial v} & \frac{\partial \psi_1}{\partial w} \\ \frac{\partial \psi_2}{\partial u} & \frac{\partial \psi_2}{\partial v} & \frac{\partial \psi_2}{\partial w} \\ \frac{\partial \psi_3}{\partial u} & \frac{\partial \psi_3}{\partial v} & \frac{\partial \psi_3}{\partial w} \end{matrix} \right).$$

Global formula for exterior differentials of 1-forms

For smooth differential 1-form $\ \alpha\$ and smooth vector fields $\ v,w\$ on the manifold $M$ we have

$$\begin{equation} dα(v, w) = v\,α(w) − w\,α(v) − α ([v, w]). \tag{16.6} \end{equation}$$

Both sides of this equation are linear in the sense that if $\alpha =\sum f_idx^i$, then

$$d(\sum f_idx^i)(v,w)=\sum d(f_idx^i)(v,w),\\ v[(\sum f_idx^i)(w)]=\sum v[f_idx^i(w)],\\ w[(\sum f_idx^i)(v)]=\sum w[f_idx^i(v)],\\ (\sum f_idx^i)([v,w])=\sum(f_idx^i)([v,w]),$$

therefore we only need to prove the equation for a single term $f dx^i$.

For the left hand side of (16.6) we obtain

$$dα(v, w) = d(f dx^i)(v, w) = (df ∧ dx^i )(v, w)\\ = df (v)dx^i(w) − df (w)dx^i(v) = v(f)\,w(x^i) − w(f)\,v(x^i).$$

For the first term on the right hand side we have

$$v\,α(w) = v (f dx^i) (w) = v (f dx^i(w))\\ = v (f · w(x^i)) = v(f)w(x^i) + f\,v(w(x^i)),$$

and similarly

$$w\,α(v) = w(f)v(x^i) + f\, w(v(x^i)).$$

Since $[v, w] = vw − wv$

$$α ([v, w]) = (f dx^i) ([v, w]) = f · [v, w](x^i)\\ = f ·( vw − wv)(x^i) = f\, v(w(x^i)) − f \,w(v(x^i)).$$

Combining the obtained equalities we have

$$vα(w) − wα(v) − α ([v, w])\\ = v(f)w(x^i) + f\,v(w(x^i)\\ − w(f)v(x^i) -f\, w(v(x^i))\\ − f\, v(w(x^i)) + f \,w(v(x^i))\\ = v(f)w(x^i) − w(f)v(x^i) = dα(v, w).$$

Example 16.9

Check (16.6) on selected $\alpha, v, w$

%display latex
dim=2                                     # dimension of manifold N
N = Manifold(dim, 'N')                    # manifold N
X = N.chart(' '.join(['x'+str(i)+':x^{'+str(i)+'}' for i in range(dim)]))  # chart on N
x0, x1 = X[:]  # coordinates x^0, x^1 of chart X as the Python variables x0, x1
al = N.diff_form(1, name=r'\alpha')          # 1-form alpha  
def astr(i): return 'a_'+str(i)           # names of components
af = [N.scalar_field(function(astr(i))(x0, x1), name=astr(i)) 
                     for i in range(dim)]    # component functions
al[:] = af                                   # define all components
v = N.vector_field(x0, x1)                   # vector field v
w = N.vector_field(x1, x0)                   # vector field w
L = al.exterior_derivative()(v, w)           # Left hand side of (16.6)
R = v(al(w)) - w(al(v)) - al(v.bracket(w))   # Right hand side of (16.6)
L == R                                       # check (16.6)

$\displaystyle \mathrm{True}$

What's next?

Take a look at the notebook One-parameter groups of transformations.