Computational Geometry Workshop

Wichita State University, Dept. of Mathematics

Differential Forms

Last Changed: 4 Sep 2017

Authors: Justin M. Ryan,

This notebook contains sample code related to the lecture on Differential Forms. Terse lecture notes may be found at

http://geometerjustin.com/teaching/cgw/notes/notes_02.html

So future readers know what version we're using:

version()

'SageMath version 7.5.1, Release Date: 2017-01-15'

Force $\LaTeX$ display:

%display latex

Define the chart we're working in/on:

U = Manifold(3,'U',latex_name=r'\mathbb{U}',start_index=1);
coord.<x,y,z>=U.chart();
e = coord.frame();
de = coord.coframe();

We want to work with the differential $k$-forms on $\mathbb{U}$. We begin by defining two 1-forms on $\mathbb{U}$.

t1 = U.diff_form(1,latex_name=r'\theta_1',);
t1[:] =[x, 2*z-x,sin(y)];
t2 = U.diff_form(1,latex_name=r'\theta_2');
t2[:] =[0,-z,y];

We can now compute the exterior product of $\theta_1$ and $\theta_2$. The command for this is just ".wedge()". The exterior product $\theta_1 \wedge \theta_2$ is a 2-form.

t1w2 = t1.wedge(t2);
print(t1w2)
t1w2.display()

2-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\theta_1\wedge \theta_2 = -x z \mathrm{d} x\wedge \mathrm{d} y + x y \mathrm{d} x\wedge \mathrm{d} z + \left( -x y + {\left(2 \, y + \sin\left(y\right)\right)} z \right) \mathrm{d} y\wedge \mathrm{d} z$$

Exterior products are alternating.

t2w1 = t2.wedge(t1);
t2w1 == -t1w2

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}$$

The exterior derivative of a form increases the degree by 1.

dt1 = t1.exterior_derivative();
dt2 = t2.exterior_derivative();
print(dt1)
dt1.display()

2-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\theta_1 = -\mathrm{d} x\wedge \mathrm{d} y + \left( \cos\left(y\right) - 2 \right) \mathrm{d} y\wedge \mathrm{d} z$$

print(dt2)
dt2.display()

2-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\theta_2 = 2 \mathrm{d} y\wedge \mathrm{d} z$$

The exterior derivative of a smooth function $f \in \mathfrak{F} = \Omega^0$ is the usual differential of $f$ from Calc III. Recall that we need to define smooth functions on $\mathbb{U}$ as scalar fields.

f = U.scalar_field({coord: x^2 - 2*z^(sin(y))});
df = f.exterior_derivative();
print(df)
df.display()

1-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, x \mathrm{d} x -2 \, z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) \mathrm{d} y + \left( -2 \, z^{\sin\left(y\right) - 1} \sin\left(y\right) \right) \mathrm{d} z$$

df == f.differential()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}$$

Applying the exterior derivative to the same form twice always yields 0 (i.e., the form identically equal to 0).

ddf = df.exterior_derivative();
print(ddf)
ddf.display()

2-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}0$$

ddt1 = dt1.exterior_derivative();
print(ddt1)
ddt1.display()

3-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\mathrm{d}\theta_1 = 0$$

ddt2 = dt2.exterior_derivative();
print(ddt2)
ddt2.display()

3-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\mathrm{d}\theta_2 = 0$$

Differential $k$-forms map $\mathfrak{X}^k \to \mathfrak{F}$.

X1 = U.vector_field(latex_name=r'X_1');
X1[:] = [x,-x,2*z];
X2 = U.vector_field(latex_name=r'X_2');
X2[:] = [cos(z),-sin(z),z];

print(df(X1))
df(X1).display()

Scalar field on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & 2 \, x z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) + 2 \, x^{2} - 4 \, z^{\sin\left(y\right)} \sin\left(y\right) \end{array}$$

$df(X_1)$ and $X_1(f)$ coincide. In Calc III lingo, they both represent the (non-normalized) directional derivative of $f$ in the direction of $X_1$.

print(X1(f))
X1(f).display()

Scalar field on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & 2 \, x z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) + 2 \, x^{2} - 4 \, z^{\sin\left(y\right)} \sin\left(y\right) \end{array}$$

df(X1) == X1(f)

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}$$

$2$-forms take two vector fields as arguments.

dt1(X1,X2).display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} \mathrm{d}\theta_1\left(X_1,X_2\right):& \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & -{\left(x \cos\left(y\right) - 2 \, x\right)} z - x \cos\left(z\right) + {\left(2 \, z {\left(\cos\left(y\right) - 2\right)} + x\right)} \sin\left(z\right) \end{array}$$

dt1(X2,X1).display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} \mathrm{d}\theta_1\left(X_2,X_1\right):& \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & {\left(x \cos\left(y\right) - 2 \, x\right)} z + x \cos\left(z\right) - {\left(2 \, z {\left(\cos\left(y\right) - 2\right)} + x\right)} \sin\left(z\right) \end{array}$$

dt1(X1,X2) == -dt1(X2,X1)

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}$$

The space $\Omega^n$ is a 1-dimensional space whose basis covector is the Euclidean volume form on $\mathbb{U}$.

vol = U.diff_form(3,latex_name=r'\omega');
vol[1,2,3] = 1;

print(vol)

3-form on the 3-dimensional differentiable manifold U

vol.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\omega = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z$$

vol(2*e[1],4*e[2],e[3])

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mbox{Scalar field on the 3-dimensional differentiable manifold U}$$

vol(2*e[1],4*e[2],e[3]).display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} & \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & 8 \end{array}$$

p=U.point((1,2,3))

vol(2*e[1],4*e[2],e[3])(p)

$$\newcommand{\Bold}[1]{\mathbf{#1}}8$$

The volume form can be used to calculate the determinant of a linear transformation of $T\mathbb{U}$. By a linear transformation of $T\mathbb{U}$, we mean a linear transformation of each tangent space that varies smoothly between tangent spaces. We call such a map a tangent-space automorphism field on $\mathbb{U}$.

A = U.automorphism_field('aut',latex_name=r'A');
A[e,:] = [[1,x,z],[0,3,y],[0,0,5]];
print(A)
A.display()

Field of tangent-space automorphisms aut on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}A = \frac{\partial}{\partial x }\otimes \mathrm{d} x + x \frac{\partial}{\partial x }\otimes \mathrm{d} y + z \frac{\partial}{\partial x }\otimes \mathrm{d} z + 3 \frac{\partial}{\partial y }\otimes \mathrm{d} y + y \frac{\partial}{\partial y }\otimes \mathrm{d} z + 5 \frac{\partial}{\partial z }\otimes \mathrm{d} z$$

A.display_comp()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} A_{ \phantom{\, x} \, x }^{ \, x \phantom{\, x} } & = & 1 \\ A_{ \phantom{\, x} \, y }^{ \, x \phantom{\, y} } & = & x \\ A_{ \phantom{\, x} \, z }^{ \, x \phantom{\, z} } & = & z \\ A_{ \phantom{\, y} \, y }^{ \, y \phantom{\, y} } & = & 3 \\ A_{ \phantom{\, y} \, z }^{ \, y \phantom{\, z} } & = & y \\ A_{ \phantom{\, z} \, z }^{ \, z \phantom{\, z} } & = & 5 \end{array}$$

A[e,:]

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & x & z \\ 0 & 3 & y \\ 0 & 0 & 5 \end{array}\right)$$

detA = vol(A(e[1]),A(e[2]),A(e[3]));
detA.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} \omega\left( A\left( \frac{\partial}{\partial x } \right) , A\left( \frac{\partial}{\partial y } \right) , A\left( \frac{\partial}{\partial z } \right) \right) : & \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & 15 \end{array}$$

Of course, in general the determinant depends on the point $p \in \mathbb{U}$.

B = U.automorphism_field('B',latex_name=r'B');
B[e,:] = [[x,0,0],[0,y,0],[0,0,exp(z)]];
B.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}B = x \frac{\partial}{\partial x }\otimes \mathrm{d} x + y \frac{\partial}{\partial y }\otimes \mathrm{d} y + e^{z} \frac{\partial}{\partial z }\otimes \mathrm{d} z$$

B.display_comp()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} B_{ \phantom{\, x} \, x }^{ \, x \phantom{\, x} } & = & x \\ B_{ \phantom{\, y} \, y }^{ \, y \phantom{\, y} } & = & y \\ B_{ \phantom{\, z} \, z }^{ \, z \phantom{\, z} } & = & e^{z} \end{array}$$

B[e,:]

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & e^{z} \end{array}\right)$$

detB = vol(B(e[1]),B(e[2]),B(e[3]));
detB.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{llcl} \omega\left( B\left( \frac{\partial}{\partial x } \right) , B\left( \frac{\partial}{\partial y } \right) , B\left( \frac{\partial}{\partial z } \right) \right) : & \mathbb{U} & \longrightarrow & \mathbb{R} \\ & \left(x, y, z\right) & \longmapsto & x y e^{z} \end{array}$$

p = U.point((1,1,1),chart=coord);
detB(p)

$$\newcommand{\Bold}[1]{\mathbf{#1}}e$$

q = U.point((-1,1,0),chart=coord);
detB(q)

$$\newcommand{\Bold}[1]{\mathbf{#1}}-1$$

Finally, we discuss how to code the results of Exercise 1.6.9 in O'Neill's Elementary Differential Geometry book. For this we need a metric (think: Riesz Representation Theorem). We'll use the Euclidean metric, a.k.a. the dot product.

g = U.riemannian_metric('g',latex_name=r'g');
g[1,1] = g[2,2] = g[3,3] =1;
print(g)
g.display()

Riemannian metric g on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}g = \mathrm{d} x\otimes \mathrm{d} x+\mathrm{d} y\otimes \mathrm{d} y+\mathrm{d} z\otimes \mathrm{d} z$$

g.display_comp()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} g_{ \, x \, x }^{ \phantom{\, x}\phantom{\, x} } & = & 1 \\ g_{ \, y \, y }^{ \phantom{\, y}\phantom{\, y} } & = & 1 \\ g_{ \, z \, z }^{ \phantom{\, z}\phantom{\, z} } & = & 1 \end{array}$$

The gradient of a function $f$ is the vector field dual to the differential $df$. Changing a covector field to a vector field is known as raising the indices, so the $\texttt{Sage}$ command is $\texttt{.up()}$.

gradf = df.up(g);
gradf.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, x \frac{\partial}{\partial x } -2 \, z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) \frac{\partial}{\partial y } + \left( -2 \, z^{\sin\left(y\right) - 1} \sin\left(y\right) \right) \frac{\partial}{\partial z }$$

gradf.display_comp()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} X_{\phantom{\, x}}^{ \, x } & = & 2 \, x \\ X_{\phantom{\, y}}^{ \, y } & = & -2 \, z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) \\ X_{\phantom{\, z}}^{ \, z } & = & -2 \, z^{\sin\left(y\right) - 1} \sin\left(y\right) \end{array}$$

df.display_comp()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\begin{array}{lcl} X_{ \, x }^{ \phantom{\, x} } & = & 2 \, x \\ X_{ \, y }^{ \phantom{\, y} } & = & -2 \, z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) \\ X_{ \, z }^{ \phantom{\, z} } & = & -2 \, z^{\sin\left(y\right) - 1} \sin\left(y\right) \end{array}$$

The curl of a vector field $X$ is the exterior derivative of the 1-form dual to $X$.

phi_X1 = X1.down(g);

To save key strokes, we can import a shortcut for $\texttt{.exterior_derivative()}$.

from sage.manifolds.utilities import xder

d_phi_X1 = xder(phi_X1);
print(d_phi_X1)
d_phi_X1.display()

2-form on the 3-dimensional differentiable manifold U

$$\newcommand{\Bold}[1]{\mathbf{#1}}-\mathrm{d} x\wedge \mathrm{d} y$$

d_phi_X1 == phi_X1.exterior_derivative()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{True}$$

In terms of the basis of $\mathfrak{X}$, the curl of $X_1$ is:

d_phi_X1_dual= d_phi_X1.up(g);
d_phi_X1_dual.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{\partial}{\partial x }\otimes \frac{\partial}{\partial y }+\frac{\partial}{\partial y }\otimes \frac{\partial}{\partial x }$$

curl_X1 = U.vector_field(latex_name=r'\mathrm{curl}(X_1)');
curl_X1[:] = [0,0,d_phi_X1[1,2]];
curl_X1.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{curl}(X_1) = -\frac{\partial}{\partial z }$$

Recall $X_1$ and verify that this corresponds with the result you'd obtain from Calc III.

X1.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}X_1 = x \frac{\partial}{\partial x } -x \frac{\partial}{\partial y } + 2 \, z \frac{\partial}{\partial z }$$

Finally, the divergence of a vector field $X$ is the coefficient of the $3$-form $d(\eta_X) = (\mathrm{div} X)\, dx \wedge dy \wedge dz$. Here, $\eta_X$ is the $2$-form corresponding to $X$ as described in O'Neill's exercise.

eta_X1 = U.diff_form(2,latex_name=r'\eta_{X_1}');
eta_X1[1,2] = X1[3];
eta_X1[1,3] = -X1[2];
eta_X1[2,3] = X1[1];
eta_X1.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\eta_{X_1} = 2 \, z \mathrm{d} x\wedge \mathrm{d} y + x \mathrm{d} x\wedge \mathrm{d} z + x \mathrm{d} y\wedge \mathrm{d} z$$

d_eta_X1 = xder(eta_X1);
d_eta_X1.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\eta_{X_1} = 3 \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z$$

div_X1 = d_eta_X1[1,2,3];
div_X1.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right) \mapsto 3$$

eta_X2 = U.diff_form(2,latex_name=r'\eta_{X_2}');
eta_X2[1,2] = X2[3];
eta_X2[1,3] = -X2[2];
eta_X2[2,3] = X2[1];
eta_X2.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\eta_{X_2} = z \mathrm{d} x\wedge \mathrm{d} y + \sin\left(z\right) \mathrm{d} x\wedge \mathrm{d} z + \cos\left(z\right) \mathrm{d} y\wedge \mathrm{d} z$$

d_eta_X2 = xder(eta_X2);
d_eta_X2.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{d}\eta_{X_2} = \mathrm{d} x\wedge \mathrm{d} y\wedge \mathrm{d} z$$

div_X2 = d_eta_X2[1,2,3];
div_X2.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right) \mapsto 1$$

The Laplacian of a function $f \in \mathfrak{F}$ is $\Delta f = \mathrm{div}(\mathrm{grad}\, f)$.

eta_f = U.diff_form(2,latex_name=r'\eta_{f}');
eta_f[1,2] = gradf[3];
eta_f[1,3] = -gradf[2];
eta_f[2,3] = gradf[1];
eta_f.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\eta_{f} = \left( -2 \, z^{\sin\left(y\right) - 1} \sin\left(y\right) \right) \mathrm{d} x\wedge \mathrm{d} y + 2 \, z^{\sin\left(y\right)} \cos\left(y\right) \log\left(z\right) \mathrm{d} x\wedge \mathrm{d} z + 2 \, x \mathrm{d} y\wedge \mathrm{d} z$$

Laplacian_f = xder(eta_f)[1,2,3];

Laplacian_f.display()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right) \mapsto -\frac{2 \, {\left(z^{\sin\left(y\right) + 2} \cos\left(y\right)^{2} \log\left(z\right)^{2} - z^{\sin\left(y\right) + 2} \log\left(z\right) \sin\left(y\right) + {\left(\sin\left(y\right)^{2} - \sin\left(y\right)\right)} z^{\sin\left(y\right)} - z^{2}\right)}}{z^{2}}$$