16. Exterior derivative

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

In [1]:
version()
Out[1]:
'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

\begin{equation} \begin{matrix} dω = \sum_{1\leq i_1<\ldots<i_k\leq n} dω_{i_1 ...i_k} ∧ dx^{i_1} ∧ · · · ∧ dx^{i_k}\\ =\sum_{1\leq i_1<\ldots<i_k\leq n} \sum_{m=1}^n \Big(\frac{\partial}{\partial x^m}ω_{i_1 ...i_k}\Big) ∧ dx^m\wedge dx^{i_1} ∧ · · · ∧ dx^{i_k}, \end{matrix} \label{}\tag{16.1} \end{equation}

($\ 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.

In [2]:
%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
Out[2]:
\(\displaystyle \alpha = a_{0}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}} + a_{1}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{1}} + a_{2}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{2}}\)

Use the method exterior_derivative() to compute the result:

In [3]:
dal = al.exterior_derivative()
dal.disp()
Out[3]:
\(\displaystyle \mathrm{d}\alpha = \left( -\frac{\partial\,a_{0}}{\partial {x^{1}}} + \frac{\partial\,a_{1}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + \left( -\frac{\partial\,a_{0}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + \left( -\frac{\partial\,a_{1}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{1}}} \right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}\)

Let us apply the definition (16.1) for comparison:

In [4]:
# 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
Out[4]:
\(\displaystyle \mathrm{d}a_0\wedge \mathrm{d} {x^{0}} + \mathrm{d}a_1\wedge \mathrm{d} {x^{1}} + \mathrm{d}a_2\wedge \mathrm{d} {x^{2}} = \left( -\frac{\partial\,a_{0}}{\partial {x^{1}}} + \frac{\partial\,a_{1}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + \left( -\frac{\partial\,a_{0}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + \left( -\frac{\partial\,a_{1}}{\partial {x^{2}}} + \frac{\partial\,a_{2}}{\partial {x^{1}}} \right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}\)

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.

In [5]:
%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
Out[5]:
\(\displaystyle a = a_{01}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}} + a_{02}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{2}} + a_{12}\left({x^{0}}, {x^{1}}, {x^{2}}\right) \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}\)

The exterior derivative:

In [6]:
a.exterior_derivative().disp()
Out[6]:
\(\displaystyle \mathrm{d}a = \left( \frac{\partial\,a_{01}}{\partial {x^{2}}} - \frac{\partial\,a_{02}}{\partial {x^{1}}} + \frac{\partial\,a_{12}}{\partial {x^{0}}} \right) \mathrm{d} {x^{0}}\wedge \mathrm{d} {x^{1}}\wedge \mathrm{d} {x^{2}}\)

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


\begin{equation} d(ω ∧ η) = dω ∧ η + (−1)^k ω ∧ dη, \label{}\tag{16.2} \end{equation}

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


\begin{equation} d^2\eta=d(d\eta)=0. \label{}\tag{16.3} \end{equation}

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


\begin{equation} ψ^∗ (ω ∧ η) = (ψ^∗ ω) ∧ (ψ^∗ η), \label{}\tag{16.4} \end{equation}

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 \begin{equation} ψ^∗ (dω) = d(ψ^∗ ω),\quad \mathrm{for}\quad ω ∈ \Omega^k (N ). \label{}\tag{16.5} \end{equation}


Example 16.3

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

In [7]:
%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
Out[7]:
\(\displaystyle \alpha = \mathrm{d} x\wedge \mathrm{d} y\)

Pullback of $\alpha$:

In [8]:
plb = Phi.pullback(alpha)       # pullback of alpha
plb.display()                   # show pullback
Out[8]:
\(\displaystyle {\Phi}^*\alpha = r \mathrm{d} r\wedge \mathrm{d} \phi\)

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 $\Phi: R^3_{r,\phi,\theta}\to R^3_{x,y,z}$, $\ \ \Phi(r,\phi,\theta)=(r\cos\phi\sin\theta,r\sin\phi\sin\theta,r\cos\theta).$

In [9]:
%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
Out[9]:
\(\displaystyle \alpha = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z\)

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

In [10]:
plb = Phi.pullback(alpha)         # pullback of alpha
plb.display()                     # show pullback
Out[10]:
\(\displaystyle {\Phi}^*\alpha = r^{2} \sin\left(\theta\right) \mathrm{d} r\wedge \mathrm{d} \theta\wedge \mathrm{d} \phi\)


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$.

In [11]:
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
Out[11]:
\(\displaystyle \alpha = \left( -\frac{y}{x^{2} + y^{2}} \right) \mathrm{d} x + \left( \frac{x}{x^{2} + y^{2}} \right) \mathrm{d} y\)

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

In [12]:
plb = Phi.pullback(alpha)        # pullback Phi^*alpha
plb.apply_map(factor)            # factor all components
plb.display()                    # show the result
Out[12]:
\(\displaystyle {\Phi}^*\alpha = \mathrm{d} t\)


Example 16.6

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

In [13]:
%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
Out[13]:
\(\displaystyle \alpha = a_{0}\left(x, y\right) \mathrm{d} x + a_{1}\left(x, y\right) \mathrm{d} y\)

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

In [14]:
alplb = psi.pullback(al)            # pullback of al
alplb.disp()                        # show pullback
Out[14]:
\(\displaystyle {\psi}^*\alpha = \left( a_{0}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{1}}{\partial u} + a_{1}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{2}}{\partial u} \right) \mathrm{d} u + \left( a_{0}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{1}}{\partial v} + a_{1}\left(\psi_{1}\left(u, v\right), \psi_{2}\left(u, v\right)\right) \frac{\partial\,\psi_{2}}{\partial v} \right) \mathrm{d} v\)


Example 16.7

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

In [15]:
# 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
Out[15]:
\(\displaystyle \beta = \mathrm{d} x\wedge \mathrm{d} y\)

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

In [16]:
betaplb = psi.pullback(beta)        # pullback psi^*beta
betaplb.disp()                      # show pullback
Out[16]:
\(\displaystyle {\psi}^*\beta = \left( -\frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial u} + \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial v} \right) \mathrm{d} u\wedge \mathrm{d} v\)

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 $\ \psi^*(dx\wedge dy\wedge dz).$

In [17]:
%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
Out[17]:
\(\displaystyle \beta = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z\)
In [18]:
betaplb=psi.pullback(beta)          # pullback psi*beta
betaplb.disp()                      # show pullback
Out[18]:
\(\displaystyle {\psi}^*\beta = \left( -{\left(\frac{\partial\,\psi_{1}}{\partial w} \frac{\partial\,\psi_{2}}{\partial v} - \frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial w}\right)} \frac{\partial\,\psi_{3}}{\partial u} + {\left(\frac{\partial\,\psi_{1}}{\partial w} \frac{\partial\,\psi_{2}}{\partial u} - \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial w}\right)} \frac{\partial\,\psi_{3}}{\partial v} - {\left(\frac{\partial\,\psi_{1}}{\partial v} \frac{\partial\,\psi_{2}}{\partial u} - \frac{\partial\,\psi_{1}}{\partial u} \frac{\partial\,\psi_{2}}{\partial v}\right)} \frac{\partial\,\psi_{3}}{\partial w} \right) \mathrm{d} u\wedge \mathrm{d} v\wedge \mathrm{d} w\)

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 $\alpha, v, w$

In [19]:
%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)
Out[19]:
\(\displaystyle \mathrm{True}\)

What's next?

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