24. Riemannian curvature tensor of type (0,4)

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'

Assume now that $M$ is a Riemannian manifold with metric $g$ and $\nabla$ is the corresponding Levi-Civita connection. We define Riemannian curvature tensor of type $(0,4)$ by

\begin{equation} R(X, Y, Z , W ) = g(R(X, Y )Z , W ),\quad \text{for } X,Y,Z,W\in \mathfrak{X}(M). \tag{24.1} \end{equation}

Since $g$ is non-singular (cf. notebook 22) the curvature (0,4)-type tensor contains the same information as the corresponding (1,3)-type tensor from the previous notebook.

In a local coordinate system $(x^1,\ldots,x^n)$ we have

\begin{equation} R_{ijkl}=g_{im}R^m_{jkl}. \tag{24.1'} \end{equation}


Symmetries of Riemannian curvature tensor


If $X, Y, Z , W$ are vector fields in $M$ and $∇$ is the Levi–Civita connection, then

\begin{equation} \begin{matrix} R(X, Y, Z , W ) + R(Y, Z , X, W ) + R(Z , X, Y, W ) = 0,\\ R(X, Y, Z , W ) = −R(Y, X, Z , W ),\\ R(X, Y, Z , W ) = −R(X, Y, W, Z ),\\ R(X, Y, Z , W ) = R(Z , W, X, Y ). \end{matrix} \tag{24.2} \end{equation}


Since the Levi-Civita connections are torsion-free, the first Bianchi identity (23.4) reduces to $R(X, Y)Z + R(Z, X)Y + R(Y, Z)X=0$ and implies the first relation.

The antisymmetry $R(X, Y) = −R(Y, X)$ implies the second equality.

The third equality is equivalent to $R(X, Y, Z , Z ) = 0.$ Indeed, if it holds then clearly $R(X, Y, Z , Z ) = 0.$ Conversely, if $R(X, Y, Z , Z ) = 0$, we have

$$R(X, Y, Z + W, Z + W ) = 0\\ ⇔R(X, Y, Z , Z ) + R(X, Y, Z, W ) +R(X, Y, W, Z ) + R(X, Y, W, W )=0\\ ⇔ R(X, Y, Z , W ) + R(X, Y, W, Z ) = 0.$$

To prove $R(X, Y, Z , Z ) = 0$, let us note that since the Levi-Civita connection is compatible with the metric, we have

$$ g(∇_X ∇_Y Z , Z ))=X ( g(∇_Y Z , Z ))-g( ∇_Y Z , ∇_X Z),\\ g(∇_Y ∇_X Z , Z ))=Y ( g(∇_X Z , Z ))-g( ∇_X Z , ∇_Y Z). $$

It is also clear, that for connections compatible with the metric, taking $V=[X,Y]$ we obtain

$$V(g(Z,Z))=2g(\nabla_V Z,Z) \Longrightarrow [X,Y](g(Z , Z))=2 g(∇_{[X,Y ]} Z , Z). $$

Therefore $$ R(X, Y, Z , Z ) = g(∇_X ∇_Y Z , Z) − g(∇_Y ∇_X Z , Z) − g(∇_{[X,Y ]} Z , Z)\\ = X(g( ∇_Y Z , Z)) − g(∇_Y Z , ∇_X Z)\\ − Y(g( ∇_X Z , Z))+ g(∇_X Z , ∇_Y Z)\\ − \frac{1}{2}[X, Y ](g( Z , Z)) $$ But we have also $$ g(\nabla_YZ,Z)=Y(g(Z,Z))-g(Z,\nabla_YZ),\\ g(\nabla_XZ,Z)=X(g(Z,Z))-g(Z,\nabla_XZ), $$

and consequently $$ g(\nabla_YZ,Z)=\frac{1}{2}Y(g(Z,Z)),\\ g(\nabla_XZ,Z)=\frac{1}{2}X(g(Z,Z)), $$

therefore

$$ R(X, Y, Z , Z )\\ =\frac{1}{2}X(Y( g(Z , Z ))-\frac{1}{2}Y(X( g(Z , Z ))-\frac{1}{2}[X,Y]( g(Z , Z ))\\ =\frac{1}{2}[X,Y]( g(Z , Z ))-\frac{1}{2}[X,Y]( g(Z , Z ))=0. $$

In the proof of the fourth relation we use the first one to obtain

$$ R(X, Y, Z , W )+R(Y, Z , X, W )+R(Z , X, Y, W )=0,\\ R(Y, Z, W, X) +R(Z , W, Y, X )+R(W, Y, Z , X )=0,\\ R(Z , W, X, Y )+R(W, X, Z , Y )+R(X, Z , W, Y )=0,\\ R(W, X, Y, Z )+R(X, Y, W, Z )+R(Y, W, X, Z )=0. $$

Adding side by side and using the third equality we obtain

$$ R(Z , X, Y, W ) + R(W, Y, Z , X ) + R(X, Z , W, Y ) + R(Y, W, X, Z ) = 0. $$

Using the second and third relation, we have

$$2R(Z , X, Y, W ) − 2R(Y, W, Z , X ) = 0.$$


The component-wise version of (24.2) reads as follows:

\begin{equation} \begin{matrix} R_{i jkl} + R_{ikl j} + R_{il jk} = 0,\\ R_{ijkl} = −R_{jikl},\\ R_{ijkl} = −R_{ijlk},\\ R_{ijkl} = R_{klij}. \end{matrix} \tag{24.2'} \end{equation}


The case of of 2-dimensional manifolds in $R^3$


If $n=2$, then the indices take values 1 or 2 (or 0,1 if Python is used). The antisymmetry in $i,j$ and $k,l$ means that for nonzero components of of the tensor $R_{ijkl}$, we have $i\not=j$ and $k\not=l$. Furthermore changing the order of $i,j$ or $k,l$ we only change the sign of the corresponding components $R_{ijkl}$. Thus, for manifolds of dimension $n = 2,$ there is only one independent component of the Riemannian curvature tensor, namely, $R_{1212}$. The remaining ones vanish or can be obtained by permutations of indices $(i,j)=(1,2), \ (k,l)=(1,2)$ and an appropriate change of signs.

Suppose we are given a surface in Euclidean 3-space

$$ x=x(u,v),\ \ y=y(u,v),\ \ z=z(u,v),\ \ (u,v)\in V\subset R^2,$$

and a non-singular point $P=(x_0 , y_0 , z_0 )=(x(u_0,v_0),y(u_0,v_0),z(u_0,v_0))$ on it. i.e.,

$$ \text{rank}\left. \left( \begin{matrix} x_u & y_u & z_u\\ x_v & y_v & z_v \end{matrix} \right) \right|_{(u_0,v_0)}=2. $$

We shall assume that the $z$-axis is perpendicular to the tangent plane to the surface at the point $P,$ in which case the $x$-axis and $y$-axis will be parallel to it. The surface may then be given locally about (i.e. in a neighborhood of) the point $P$ by an equation of the form $z = f(x, y)$, where $z_0 = f(x_0 , y_0 )$ and

$$\frac{\partial f}{\partial x}\Big|_{\substack{x=x_0\\y=y_0}} =\frac{\partial f}{\partial y}\Big|_{\substack{x=x_0\\y=y_0}}=0, \quad \text{i.e.}\quad \text{grad}\, f\,\Big|_{\substack{x=x_0\\y=y_0}}=0. $$

Given a surface $z = f(x, y)$, and a point $P=(x_0 , y_0 , z_0 )$ on it at which grad $f = 0$, we can define the Gaussian curvature $K$ of the surface at $(x_0,y_0,z_0)$ to be the determinant of the matrix

$$K=\det (f_{x^ix^j})\Big|_{\substack{x=x_0\\y=y_0}}= \det \left( \begin{matrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{matrix} \right)\Big|_{\substack{x=x_0\\y=y_0}}. $$

The metric on such surface is defined by

$$ g_{11}=1+f^2_x,\quad g_{12}=f_xf_y,\quad g_{22}=1+f_y^2.$$

By the assumption $\text{grad}\, f\,\Big|_{\substack{x=x_0\\y=y_0}}=0$ at point $P$, the derivatives of $g_{ij}$ with respect to $x$ and $y$ vanish, so all Levi-Civita connection coefficients $\Gamma^m_{ij}=0$ at $P$.

Recall the formulas defining the Riemannian curvature tensor at $P$ ( here, as usual $(g_{ij})$ is the matrix of metric components and $(g^{ij})$ its inverse).

$$\displaystyle\Gamma_{rs}^q=\frac{1}{2}g^{tq}\Big(\frac{\partial g_{rt}}{\partial x^s} +\frac{\partial g_{ts}}{\partial x^r}- \frac{\partial g_{sr}}{\partial x^t}\Big),\\ R^m_{kij}=\frac{\partial \Gamma^m_{kj}}{\partial x^i} -\frac{\partial \Gamma^m_{ki}}{\partial x^j} +\Gamma^m_{li}\Gamma^l_{kj} -\Gamma^m_{lj}\Gamma^l_{ki},\\ R_{ijkl}=g_{im}R^m_{jkl} =g_{im} \Big(\frac{\partial \Gamma^m_{jl}}{\partial x^k} -\frac{\partial \Gamma^m_{jk}}{\partial x^l}\Big),\\ $$

(in the last formula we used the fact that $\Gamma^m_{ki}=0$ at $P=(x_0,y_0,z_0)$). To obtain the derivatives of Christoffel symbols we need their values in a neighborhood of $P$

$$\Gamma^m_{jl}=\frac{1}{2}g^{pm}\Big(\frac{\partial g_{jp}}{\partial x^l} +\frac{\partial g_{pl}}{\partial x^j}- \frac{\partial g_{lj}}{\partial x^p}\Big), $$

so using the fact that derivatives of $g_{ij}$ vanish at $P$ we obtain

$$\frac{\partial \Gamma^m_{jl}}{\partial x^k} =\frac{1}{2}g^{pm}\big(\frac{\partial^2 g_{jp}}{\partial x^k\partial x^l} +\frac{\partial^2 g_{pl}}{\partial x^k\partial x^j}- \frac{\partial^2 g_{lj}}{\partial x^k\partial x^p}\Big). $$

Similarly

$$\Gamma^m_{jk}=\frac{1}{2}g^{pm}\Big(\frac{\partial g_{jp}}{\partial x^k} +\frac{\partial g_{pk}}{\partial x^j}- \frac{\partial g_{kj}}{\partial x^p}\Big), $$

so $$ \frac{\partial \Gamma^m_{jk}}{\partial x^l} =\frac{1}{2}g^{pm}\big(\frac{\partial^2 g_{jp}}{\partial x^l\partial x^k} +\frac{\partial^2 g_{pk}}{\partial x^l\partial x^j}- \frac{\partial^2 g_{kj}}{\partial x^l\partial x^p}\Big). $$

Since $g_{im}g^{pm}=\delta_i^p$, then $R_{ijkl}$ is a difference of two expressions:

$$\frac{1}{2}\delta_i^p\big(\frac{\partial^2 g_{jp}}{\partial x^k\partial x^l} +\frac{\partial^2 g_{pl}}{\partial x^k\partial x^j}- \frac{\partial^2 g_{lj}}{\partial x^k\partial x^p}\big) = \frac{1}{2}\big(\frac{\partial^2 g_{ji}}{\partial x^k\partial x^l} +\frac{\partial^2 g_{il}}{\partial x^k\partial x^j}- \frac{\partial^2 g_{lj}}{\partial x^k\partial x^i}\big) $$

and

$$\frac{1}{2}\delta_i^p\big(\frac{\partial^2 g_{jp}}{\partial x^l\partial x^k} +\frac{\partial^2 g_{pk}}{\partial x^l\partial x^j}- \frac{\partial^2 g_{kj}}{\partial x^l\partial x^p}\big) = \frac{1}{2}\big(\frac{\partial^2 g_{ji}}{\partial x^l\partial x^k} +\frac{\partial^2 g_{ik}}{\partial x^l\partial x^j}- \frac{\partial^2 g_{kj}}{\partial x^l\partial x^i}\big). $$

Thus

$$R_{ijkl}=\frac{1}{2}\Big( \frac{\partial^2 g_{il}}{\partial x^k\partial x^j}+ \frac{\partial^2 g_{kj}}{\partial x^l\partial x^i}- \frac{\partial^2 g_{lj}}{\partial x^k\partial x^i}- \frac{\partial^2 g_{ik}}{\partial x^l\partial x^j}\Big). $$

For $(i,j,k,l)=(1,2,1,2)$ we obtain $$R_{1212}=\frac{1}{2}\Big( \frac{\partial^2 g_{12}}{\partial x^1\partial x^2}+ \frac{\partial^2 g_{12}}{\partial x^2\partial x^1}- \frac{\partial^2 g_{22}}{\partial x^1\partial x^1}- \frac{\partial^2 g_{11}}{\partial x^2\partial x^2}\Big)\\ =\frac{1}{2}\Big(\frac{\partial^2 g_{12}}{\partial x\partial y}+ \frac{\partial^2 g_{12}}{\partial y\partial x}- \frac{\partial^2 g_{22}}{\partial x^2}- \frac{\partial^2 g_{11}}{\partial y^2}\Big) . $$

Since at $P$ $$(g_{11})_{yy}=2(f_{xy})^2,\quad (g_{22})_{xx}=2(f_{xy})^2,\quad (g_{12})_{xy}=f_{xx}f_{yy}+(f_{xy})^2, $$ we have at $P$ $$R_{1212}=\frac{1}{2}\big(2(f_{xx}f_{yy}+(f_{xy})^2) -2(f_{xy})^2 -2(f_{xy})^2\big)=f_{xx}f_{yy}-(f_{xy})^2=\det \left( \begin{matrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{matrix} \right)=K. $$


Ricci tensor and scalar curvature


Ricci tensor can be defined by

\begin{equation} Ric(u,v)=R(e^i,u,e_i,v), \tag{24.3} \end{equation}

for any vector fields $u,v$, where $e_i$ is any vector frame and $e^i$ the corresponding coframe.

In a Riemannian manifold with the metric $g$ the components of Ricci tensor are defined by

$$Ric_{ij}=R^k_{ikj}=g^{kl}R_{likj}=g^{lk}R_{kjli},$$

and the scalar curvature by

\begin{equation} R=g^{ij}Ric_{ij}. \tag{24.4} \end{equation}

In the case of two-dimensional manifold we have $R=2K$. In fact

$$R=g^{ij}Ric_{ij}=g^{ij}g^{lk}R_{kjli}.$$

From the symmetry properties of $R_{ijkl}$ it follows that in the obtained sum of $2^4$ elements only 4 are non-zero:

$$R=g^{22}g^{11}R_{1212}+g^{12}g^{21}R_{1221}+ g^{21}g^{12}R_{2112}+g^{11}g^{22}R_{2121}\\ =g^{22}g^{11}R_{1212}-g^{12}g^{21}R_{1212}-g^{21}g^{12}R_{1212}+g^{11}g^{22}R_{1212}\\ =2(g^{11}g^{22}-(g^{12})^2)R_{1212}=2\det (g^{ij})R_{1212}=\frac{2}{\det (g_{ij})}R_{1212}. $$

Remark in the last calculations the condition $\text{grad}\, f\,\Big|_{\substack{x=x_0\\y=y_0}}=0,$ was not used.

If as previously, the coordinates at $P$ are such that $\text{grad}\, f\,\Big|_{\substack{x=x_0\\y=y_0}}=0,$ we have $ \det (g_{ij})\Big|_{\substack{x=x_0\\y=y_0}}=1$ and $R=2K$. Since $R$ and $K$ are scalars, and consequently, are coordinate-independent, we conclude that

\begin{equation} R = 2K \tag{24.5} \end{equation}

at every (non-singular) point of the surface.


Remark. For non-singular surfaces represented locally in the form $z=f(x,y)$ (without the condition $\text{grad}\, f\,\Big|_{\substack{x=x_0\\y=y_0}}=0$) the Gauss curvature is defined by $$K=\frac{f_{xx}f_{yy}-(f_{xy})^2}{(1+(f_x)^2+(f_y)^2)^2}.$$

Remark. The calculations above give us the formula $$R_{1212}=\frac{R}{2}\det(g_{ij}).$$

It can be generalized to all components of the tensor $R_{ijkl}$ as follows:

$$R_{ijkl}=\frac{R}{2}(g_{ik}g_{jl}-g_{il}g_{jk}).$$

In fact, an explicit calculations using the symmetry properties of $R_{ijkl}$ show that

\begin{eqnarray} (i,j,k,l)=(1,2,1,2)\quad \Rightarrow & \quad \frac{R}{2}(g_{11}g_{22}-g_{12}g_{21})=\frac{R}{2}\det(g_{ij})=R_{1212},\\ (i,j,k,l)=(2,1,1,2)\quad \Rightarrow & \quad \frac{R}{2}(g_{21}g_{12}-g_{22}g_{11})=-\frac{R}{2}\det(g_{ij})=R_{2112},\\ (i,j,k,l)=(1,2,2,1)\quad \Rightarrow & \quad \frac{R}{2}(g_{12}g_{21}-g_{11}g_{22})=-\frac{R}{2}\det(g_{ij})=R_{1221},\\ (i,j,k,l)=(2,1,2,1)\quad \Rightarrow & \quad \frac{R}{2}(g_{22}g_{11}-g_{21}g_{12})=\frac{R}{2}\det(g_{ij})=R_{2121},\\ \end{eqnarray}

and for the remaining sequences of indices, both sides are zero.


Example 24.1

In $R^2$ with $g=dx\otimes dx+dy\otimes dy$, we have $R^i_{mkl}=0$ and $K=\frac{R}{2}=0.$

In [2]:
E.<x,y>=EuclideanSpace()      # Euclidean space E^2
print(E.metric().riemann())   # curvature tensor of (1,3)-type
E.metric().riemann().disp()   # show the curvature tensor
Tensor field Riem(g) of type (1,3) on the Euclidean plane E^2
Out[2]:
Riem(g) = 0

The method ricci gives the Ricci tensor:

In [3]:
print(E.metric().ricci())    # Ricci curvature tensor
E.metric().ricci().disp()    # show the result
Field of symmetric bilinear forms Ric(g) on the Euclidean plane E^2
Out[3]:
Ric(g) = 0

The Ricci scalar curvature can be obtained using ricci_scalar method:

In [4]:
E.metric().ricci_scalar().expr()  # Ricci scalar curvature
Out[4]:
0


Example 24.2

The same can be done without EuclideanSpace command.

In [5]:
%display latex
M=Manifold(2,name='R^2',start_index=1) # manifold R2
c_xy.<x,y>=M.chart()        # chart on R^2
g = M.metric('g');          # metric on R^2
g[:]=[[1,0],[0,1]]          # nonzero components
Riem=g.riemann()            # curvature (1,3)-type tensor 
print(Riem)                 # show information on curv. tensor
Riem.disp()                 # show the curvature tensor
Tensor field Riem(g) of type (1,3) on the 2-dimensional differentiable manifold R^2
Out[5]:
\(\displaystyle \mathrm{Riem}\left(g\right) = 0\)
In [6]:
print(g.ricci())            # Ricci curvature tensor information
g.ricci().disp()            # show  Ricci tensor
Field of symmetric bilinear forms Ric(g) on the 2-dimensional differentiable manifold R^2
Out[6]:
\(\displaystyle \mathrm{Ric}\left(g\right) = 0\)

The Ricci scalar curvature:

In [7]:
g.ricci_scalar().expr()     # Ricci scalar curvature 
Out[7]:
\(\displaystyle 0\)


Example 24.3

For the metric on the sphere $S^2\subset R^3$, we have $𝑔=d𝜃⊗d𝜃+sin(𝜃)^2d𝜙⊗d𝜙$, $K=R/2=1.$

In [8]:
%display latex
M = Manifold(3, 'R^3')              # R^3 ambient space for S^2
c_xyz.<x,y,z> = M.chart()           # chart on R^3
N = Manifold(2, 'N',start_index=1)  # sphere S^2
c_sph.<theta,phi>=N.chart()         # chart on S^2
psi = N.diff_map(M, (sin(theta)*cos(phi), 
                     sin(theta)*sin(phi),
                     cos(theta)),
   name='psi',latex_name=r'\psi')   # embedding S^2 -> R^3
g0=M.metric('g0')                   # metric g0 on R^3
g0[0,0],g0[1,1],g0[2,2]=1,1,1       # components of g0
g=N.metric('g')                     # metric on S^2
g.set( psi.pullback(g0) )           # g is the pullback of g0
g.disp()                            # show g
Out[8]:
\(\displaystyle g = \mathrm{d} \theta\otimes \mathrm{d} \theta + \sin\left(\theta\right)^{2} \mathrm{d} \phi\otimes \mathrm{d} \phi\)
In [9]:
Riem=g.riemann()                    # curvature (1,3)-tensor
Riem.display_comp(coordinate_labels=False)  # show components
Out[9]:
\(\displaystyle \begin{array}{lcl} \mathrm{Riem}\left(g\right)_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & \sin\left(\theta\right)^{2} \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & -\sin\left(\theta\right)^{2} \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & -1 \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & 1 \end{array}\)
In [10]:
print(g.ricci())                   # Ricci curvature tensor
g.ricci().disp()                   # show Ricci tensor
Field of symmetric bilinear forms Ric(g) on the 2-dimensional differentiable manifold N
Out[10]:
\(\displaystyle \mathrm{Ric}\left(g\right) = \mathrm{d} \theta\otimes \mathrm{d} \theta + \sin\left(\theta\right)^{2} \mathrm{d} \phi\otimes \mathrm{d} \phi\)

Let us show how to apply (24.3).

First we have to define the frame and coframe fields:

In [11]:
fr=c_sph.frame()[:]                # frame
cfr=c_sph.coframe()[:]             # coframe

next we define the matrix of coefficient of the Ricci tensor field according to (24.3).

In [12]:
Ric_coeff=[[sum([Riem(cfr[i],fr[k],fr[i],fr[l]).expr() 
                 for i in range(2)])
 for k in range(2)] for l in range(2)]  # cf. (24.3)
Ric_coeff[:]                       # Ricci ten. components
Out[12]:
\(\displaystyle \left[\left[1, 0\right], \left[0, \sin\left(\theta\right)^{2}\right]\right]\)

Finally we define the Ricci tensor:

In [13]:
Ric=N.tensor_field(0,2,'Ric')     # Ricci curvature tensor
Ric[:]=Ric_coeff                  # define all components
Ric.disp()                        # show Ricci tensor
Out[13]:
\(\displaystyle Ric = \mathrm{d} \theta\otimes \mathrm{d} \theta + \sin\left(\theta\right)^{2} \mathrm{d} \phi\otimes \mathrm{d} \phi\)

Ricci scalar curvature:

In [14]:
g.ricci_scalar().expr()           # Ricci scalar curvature
Out[14]:
\(\displaystyle 2\)

Formula (24.4) can be used as follows:

In [15]:
g.inverse().contract(Ric).trace().expr()
Out[15]:
\(\displaystyle 2\)

According to (24.5), the Gaussian curvature is 1.


Example 24.4

Consider the Poincare half-plane $y>0$ with the metric defined by $g_{11}=\frac{1}{y^2},\ \ g_{22}=\frac{1}{y^2}, \ $ and remaining components equal to zero.

In [16]:
M = Manifold(2, 'M', start_index=1)  # Poincare half-plane
X.<x,y> = M.chart(coord_restrictions=lambda x,y: y>0)
g = M.metric('g')                    # metric on M
g[1,1], g[2,2] = 1/y^2, 1/y^2        # nonzero components of g

Compute (1,3) type curvature tensor:

In [17]:
Riem=g.riemann()                     # (1,3) type curvature tensor
Riem.display_comp(coordinate_labels=False)  # show nonzero comp.
Out[17]:
\(\displaystyle \begin{array}{lcl} \mathrm{Riem}\left(g\right)_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{1}{y^{2}} \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{1}{y^{2}} \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{1}{y^{2}} \\ \mathrm{Riem}\left(g\right)_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{1}{y^{2}} \end{array}\)

and next the Riemannian curvature tensor of type (0,4), using (24.1):

In [18]:
R=M.tensor_field(0,4,'R')         # R is (0,4)-type tensor field
R[:]=g.contract(Riem)[:]                # apply (24.1)
R.display_comp(coordinate_labels=False) # show (0,4) type ten. R
Out[18]:
\(\displaystyle \begin{array}{lcl} R_{\,1\,2\,1\,2}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{1}{y^{4}} \\ R_{\,1\,2\,2\,1}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{1}{y^{4}} \\ R_{\,2\,1\,1\,2}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{1}{y^{4}} \\ R_{\,2\,1\,2\,1}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{1}{y^{4}} \end{array}\)

Now compute the Ricci curvature tensor:

In [19]:
g.ricci().disp()                      # Ricci curv. tensor
Out[19]:
\(\displaystyle \mathrm{Ric}\left(g\right) = -\frac{1}{y^{2}} \mathrm{d} x\otimes \mathrm{d} x -\frac{1}{y^{2}} \mathrm{d} y\otimes \mathrm{d} y\)

and Ricci curvature scalar:

In [20]:
g.ricci_scalar().expr()               # Ricci scalar curv.
Out[20]:
\(\displaystyle -2\)


Example 24.5

Poincare disk is the set $\ \ \{(x,y)\in R^2:x^2+y^2<1\}\ \ $ with the metric defined by $g_{11}=\frac{4}{(1-x^2-y^2)^2},\ \ g_{22}=\frac{4}{(1-x^2-y^2)^2}\ \ $ and remaining components equal to zero.

In [21]:
M = Manifold(2, 'M', start_index=1)  # Poincare disc
                # Cartesian coordinates on the Poincaré disk:
X.<x,y> = M.chart('x:(-1,1) y:(-1,1)', 
    coord_restrictions=lambda x,y: x^2+y^2<1)
g = M.metric('g')                    #  metric on M
                                     # components of g
g[1,1], g[2,2] = 4/(1-x^2-y^2)^2,4/(1-x^2-y^2)^2

Compute (1,3)-type curvature tensor. Since g.riemann() gives an unsimplified result, we simplify the components one by one.

In [22]:
Riem = g.riemann();           # curvature (1,3) type tensor
                              # factor components
Riem2=[[[[Riem[a,b,c,d].factor() for a in [1,2]] 
         for b in [1,2]] for c in [1,2]] for d in[1,2]];
R=M.tensor_field(1,3,'R')     #  tensor field of type (1,3)
R[X.frame(),:]=Riem2          # simplified components
R.display_comp(coordinate_labels=False)  # show R
Out[22]:
\(\displaystyle \begin{array}{lcl} R_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \\ R_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \\ R_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \\ R_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \end{array}\)

Ricci curvature tensor:

In [23]:
Ric=g.ricci()                # Ricci curvature tensor
Ric.apply_map(factor)        # factor components
Ric.disp()                   # show Ricci curv.tensor  
Out[23]:
\(\displaystyle \mathrm{Ric}\left(g\right) = -\frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \mathrm{d} x\otimes \mathrm{d} x -\frac{4}{{\left(x^{2} + y^{2} - 1\right)}^{2}} \mathrm{d} y\otimes \mathrm{d} y\)

Ricci curvature scalar:

In [24]:
g.ricci_scalar().expr()     # Ricci scalar curvature 
Out[24]:
\(\displaystyle -2\)


Example 24.6

Consider $R^2$ with the matrix of metric components:

In [25]:
matrix([[1+4*x^2,4*x*y],[4*x*y,1+4*y^2]])
Out[25]:
\(\displaystyle \left(\begin{array}{rr} 4 \, x^{2} + 1 & 4 \, x y \\ 4 \, x y & 4 \, y^{2} + 1 \end{array}\right)\)
In [26]:
N=Manifold(2,name='R^2',start_index=1) # Manifold R^2
X.<x,y>=N.chart()                      # chart on R^2
g = N.metric('g');                     # metric g on R^2
g[:]=[[1+4*x^2,4*x*y],[4*x*y,1+4*y^2]] # components of g
Riem = g.riemann();                    # (1,3) type curv. tens.
R=N.tensor_field(0,4,'R')              # (0,4) type curv. tens.

Compute Riemannian (0,4)-type tensor.

In [27]:
R[:]=g.contract(Riem)[:]                   # use (24.1)
R.display_comp(coordinate_labels=False)    # show (0,4) t. R
Out[27]:
\(\displaystyle \begin{array}{lcl} R_{\,1\,2\,1\,2}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & \frac{4}{4 \, x^{2} + 4 \, y^{2} + 1} \\ R_{\,1\,2\,2\,1}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{4}{4 \, x^{2} + 4 \, y^{2} + 1} \\ R_{\,2\,1\,1\,2}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{4}{4 \, x^{2} + 4 \, y^{2} + 1} \\ R_{\,2\,1\,2\,1}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & \frac{4}{4 \, x^{2} + 4 \, y^{2} + 1} \end{array}\)

Ricci curvature tensor:

In [28]:
Ric=g.ricci()                          # Ricci curvature tensor
Ric.apply_map(factor)                  # factor components
Ric.disp()                             # show the result
Out[28]:
\(\displaystyle \mathrm{Ric}\left(g\right) = \frac{4 \, {\left(4 \, x^{2} + 1\right)}}{{\left(4 \, x^{2} + 4 \, y^{2} + 1\right)}^{2}} \mathrm{d} x\otimes \mathrm{d} x + \frac{16 \, x y}{{\left(4 \, x^{2} + 4 \, y^{2} + 1\right)}^{2}} \mathrm{d} x\otimes \mathrm{d} y + \frac{16 \, x y}{{\left(4 \, x^{2} + 4 \, y^{2} + 1\right)}^{2}} \mathrm{d} y\otimes \mathrm{d} x + \frac{4 \, {\left(4 \, y^{2} + 1\right)}}{{\left(4 \, x^{2} + 4 \, y^{2} + 1\right)}^{2}} \mathrm{d} y\otimes \mathrm{d} y\)

and Ricci scalar curvature:

In [29]:
factor(g.ricci_scalar().expr()/2)      # Ricci curvature scalar
Out[29]:
\(\displaystyle \frac{4}{{\left(4 \, x^{2} + 4 \, y^{2} + 1\right)}^{2}}\)


Example 24.7

Consider the embedding of the hyperboloid into $R^3$ defined by $\ \ \psi(u,v)= (u, v,u^2-v^2)$. The metric on the hyperboloid can be defined as the pullback $\psi^*g_0\ \ $ under $\psi$ where $g_0$ is the standard metric on $R^3$.

In [30]:
%display latex
M = Manifold(3, 'R3')                  # manifold R^3
c_xyz.<x,y,z> = M.chart()              # Cart. coord. on R^3
N=Manifold(2,name='R2',start_index=1)  # hyperboloid N
c_uv.<u,v>=N.chart()                   # coordinates on N
g0 = M.metric('g0');                   # standard metr, of R^3
g0[:]=[[1,0,0],[0,1,0],[0,0,1]];       # components of g0
psi = N.diff_map(M, (u, v,u^2-v^2) ,name='psi')  # embedding
g=N.metric('g')                        # metric on N
g.set(psi.pullback(g0))                # g is pullback of g0
g.disp()                               # show g
Out[30]:
\(\displaystyle g = \left( 4 \, u^{2} + 1 \right) \mathrm{d} u\otimes \mathrm{d} u -4 \, u v \mathrm{d} u\otimes \mathrm{d} v -4 \, u v \mathrm{d} v\otimes \mathrm{d} u + \left( 4 \, v^{2} + 1 \right) \mathrm{d} v\otimes \mathrm{d} v\)

Compute the Riemannian curvature tensor of type (0,4).

In [31]:
Riem = g.riemann();                    # curv. (1,3) tensor
R=N.tensor_field(0,4,'R')              # tensor field (0,4)-type
R[:]=g.contract(Riem)[:]               # use (24.1)
R.display_comp(coordinate_labels=False) # show (0,4) type ten. R
Out[31]:
\(\displaystyle \begin{array}{lcl} R_{\,1\,2\,1\,2}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{4}{4 \, u^{2} + 4 \, v^{2} + 1} \\ R_{\,1\,2\,2\,1}^{\phantom{\, 1}\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{4}{4 \, u^{2} + 4 \, v^{2} + 1} \\ R_{\,2\,1\,1\,2}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{4}{4 \, u^{2} + 4 \, v^{2} + 1} \\ R_{\,2\,1\,2\,1}^{\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{4}{4 \, u^{2} + 4 \, v^{2} + 1} \end{array}\)

Ricci tensor:

In [32]:
Ric=g.ricci()                         # Ricci curvature tensor
Ric.apply_map(factor)                 # factor components
Ric.disp()                            # show the Ricci tensor
Out[32]:
\(\displaystyle \mathrm{Ric}\left(g\right) = -\frac{4 \, {\left(4 \, u^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \mathrm{d} u\otimes \mathrm{d} u + \frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \mathrm{d} u\otimes \mathrm{d} v + \frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \mathrm{d} v\otimes \mathrm{d} u -\frac{4 \, {\left(4 \, v^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \mathrm{d} v\otimes \mathrm{d} v\)

Ricci scalar curvature:

In [33]:
factor(g.ricci_scalar().expr())      # Ricci scalar curvature 
Out[33]:
\(\displaystyle -\frac{8}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}}\)

What's next?

Take a look at the notebook Torsion and curvature forms.