#!/usr/bin/env python
# coding: utf-8

# # 16. Exterior derivative
# 
# 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).

# In[1]:


version()


# 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 $\ \omega_{i_1\ldots i_k}$).

# <br>
# 
# **Example 16.1**
# 
# Take a 1-form and compute its exterior derivative.

# In[2]:


get_ipython().run_line_magic('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


# Use the method `exterior_derivative()` to compute the result:

# In[3]:


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


# 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


# 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).
# 
# <br>
# 
# **Example 16.2**
# 
# Consider now a 2-form and its exterior differential.

# In[5]:


get_ipython().run_line_magic('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


# The exterior derivative:

# In[6]:


a.exterior_derivative().disp()


# 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.

# <br>
# 
# ### Exterior derivative is antiderivation
# 
# <br>
# 
# \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).
# 

# <br>
# 
# ### Exterior derivative is nilpotent
# 
# <br> 
# 
# \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
# $$ω = dη_{i_1 ...i_m} ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m} =
# \Big(\frac{\partial}{\partial x^j}η_{i_1 ...i_m}\Big)dx^j ∧ dx^{i_1} ∧ · · · ∧ dx^{i_m}.$$
# <br>
# 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
# $$\frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}dx^1\wedge dx^2
# +\frac{∂^2 \eta_{...} }{ ∂ x^2 ∂ x^1}dx^2\wedge dx^1\\
# =\frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}dx^1\wedge dx^2+
# \frac{∂^2 \eta_{...} }{ ∂ x^1 ∂ x^2}(-dx^1\wedge dx^2)=0.$$
# 

# <br>
# 
# ### Pullback and wedge product
# 
# <br>
# 
# \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).
# 

# <br>
# 
# ### Pullback and exterior derivative
# 
# <br>
# 
# 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}
# 

# <br> 
# 
# **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]:


get_ipython().run_line_magic('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


# Pullback of $\alpha$:

# In[8]:


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


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

# <br>
# 
# **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]:


get_ipython().run_line_magic('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


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

# In[10]:


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


# 
# <br>
# 
# **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


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

# In[12]:


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


# <br>
# 
# **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]:


get_ipython().run_line_magic('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


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

# In[14]:


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


# <br>
# 
# **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


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

# In[16]:


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


# 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).
# $

# <br>
# 
# **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]:


get_ipython().run_line_magic('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


# In[18]:


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


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

# <br>
# 
# ### Global formula for exterior differentials of 1-forms
# 
# <br>
# 
# 
# 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}
# 
# <br>
# 
# 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
# 
# $$
# vα(w) − wα(v) − α ([v, w])\\
# = v(f)w(x^i) + f\,v(w(x^i)\\
# − w(f)v(x^i) -f\, w(v(x^i))\\
# − f\, v(w(x^i)) + f \,w(v(x^i))\\
# = v(f)w(x^i) − w(f)v(x^i)
# = dα(v, w).
# $$
# 
# 

# <br>
# 
# **Example 16.9**
# 
# Check  (16.6) on selected $\alpha, v, w$

# In[19]:


get_ipython().run_line_magic('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)


# ## What's next?
# 
# Take a look at the notebook [One-parameter groups of transformations](https://nbviewer.org/github/sagemanifolds/IntroToManifolds/blob/main/17Manifold_One_Parameter.ipynb).