23Manifold_Curvature.ipynb

Kernel: SageMath 9.6

23. Curvature

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'

Consider first the case of Euclidean connection (notebook 21).

Recall that for an open subset $U\subset R^n$, with Cartesian coordinate system $(U,x^1,\ldots,x^n)$ and vector fields $X,Y\in\mathfrak{X}(U)$, $Y=Y^k \frac{\partial}{\partial x^k}$, we defined the Euclidean connection by $D_XY=X(Y^k)\frac{\partial}{\partial x^k}$.

From this definition it follows that $$D_XD_YZ =X(Y(Z^k))\frac{\partial}{\partial x^k},\quad \text{ for } X,Y,Z\in\mathfrak{X}(U).$$ In fact, from the definition of $D_YZ$ it follows that if we put $\tilde{Y}^k=Y(Z^k)$, then $$D_XD_YZ=D_X(Y(Z^k)\frac{\partial}{\partial x^k}) =D_X(\tilde{Y}^k\frac{\partial}{\partial x^k}) =X(\tilde{Y}^k)\frac{\partial}{\partial x^k}= X(Y(Z^k))\frac{\partial}{\partial x^k}.$$

Using this representation for iterated Euclidean connection we can see that $$D_XD_YZ-D_YD_XZ =X(Y(Z^k))\frac{\partial}{\partial x^k} -Y(X(Z^k))\frac{\partial}{\partial x^k} =[X,Y](Z^k)\frac{\partial}{\partial x^k} =D_{[X,Y]}Z.$$ i.e. $$\begin{equation} D_XD_YZ-D_YD_XZ -D_{[X,Y]}Z=0. \tag{*} \end{equation}$$

Recall also that for general connection $\ \nabla\$ and general vector fields $\displaystyle X=X^i\frac{∂}{∂x^i}, Y=Y^j\frac{∂}{∂x^j}$ we have $\displaystyle ∇_X Y = (X^i\frac{∂}{∂x^i} Y^k + Γ^k_{ij} X^i Y^j )\frac{∂}{∂x^k}$, where $Γ^k_{ij}$ are defined by $\displaystyle ∇_{\frac{∂}{∂x^i}} \frac{∂}{∂x^j} = Γ^k_{ij}\frac{∂} {∂x^k}$ and consequently the equality $(*)$ is not true. Nevertheless the left hand side of $(*)$ can be used as a kind of measure of "flatness" of the manifold or a measure how much the geometry of the manifold differs from the geometry of the Euclidean space.

Curvature map

The curvature $R$, of the connection ∇ is a map that associates to each pair of vector fields an operator from $\mathfrak{X}(M)$ into itself, given by

$$\begin{equation} R(X, Y)Z = ∇_X ∇_Y Z − ∇_Y ∇_X Z− ∇_{[X,Y]}Z,\quad \text{for}\ X, Y, Z ∈ \mathfrak{X}(M). \tag{23.1} \end{equation}$$

From the properties of the covariant derivative and Lie bracket it follows

$$R(X, Y) = −R(Y, X)\\$$

To prove the tensorial property (see notebook 13) of $R$ , we need to show that $R(X, Y )Z$ is multilinear over $C^∞ (M )$ in each of the three vector fields. First we show linearity for the $X$ variable, from which linearity immediately follows for the $Y$ variable.
Let $f_1 , f_2 ∈ C^∞ (M ).$ Then

$$R(f_1 X_1 + f_2 X_2 , Y )Z \\ =\nabla_{f_1X_1+f_2X_2}\nabla_YZ-\nabla_Y\nabla_{f_1X_1+f2X_2}Z- \nabla_{[f_1X1+f_2X_2,Y]}Z \\ = (f_1 ∇_{X_1} + f_2 ∇_{X_2} )∇_Y Z -∇_Y (f_1 ∇_{X_1} + f_2 ∇_{X_2} )Z − ∇_{[f_1 X_1 ,Y ]+[f_2 X_2 ,Y ]} Z.$$

Since (cf. notebook 12) $\ [f X , Y ] = f [X , Y ] − Y (f )X$, we have

$$R(f_1 X_1 + f_2 X_2 , Y )Z = f_1 ∇_{X_1} ∇_Y Z + f_2 ∇_{X_2} ∇_Y Z\\ − f_1 ∇_Y ∇_{X_1} Z − Y (f_1 )∇_{X_1}Z − f_2 ∇_Y ∇_{X_2} Z − Y (f_2 )∇_{X_2} Z\\ − ∇_{f_1 [X_1 ,Y ]−Y (f_1 )X_1} Z − ∇_{f_2[X_2 ,Y ]−Y (f_2 )X_2} Z\\ =f_1 ∇_{X_1} ∇_Y Z − f_1 ∇_Y ∇_{X_1} Z − f_1 ∇_{[X_1 ,Y ]} Z\\ + f_2 ∇_{X_2} ∇_Y Z − f_2 ∇_Y ∇_{X_2} Z − f_2 ∇_{[X_2 ,Y ]}Z\\ − Y (f_1 )∇_{X_1} Z − Y (f_2 )∇_{X_2} Z + Y (f_1 )∇_{X_1} Z + Y (f_2 )∇_{X_2} Z\\ = f_1 R(X_1 , Y )Z + f_2 R(X_2 , Y )Z.$$

Next we check the linearity for the $Z$ variable

$$R(X, Y)( f_1 Z_1+f_2Z_2)\\ = ∇_X ∇_Y (f_1 Z_1+f_2Z_2) − ∇_Y ∇_X (f_1 Z_1+f_2Z_2) − ∇_{[X,Y ]} (f_1 Z_1+f_2Z_2)\\ =∇_X (Y (f_1 )Z_1+ f_1 ∇_Y Z_1) − ∇_Y (X(f_1 )Z_1 + f_1 ∇_X Z_1) -[X,Y](f_1)Z_1-f_1 ∇_{[X,Y]}Z_1\\ +∇_X (Y (f_2 )Z_2+ f_2 ∇_Y Z_2) − ∇_Y (X(f_2 )Z_2 + f_2 ∇_X Z_2) -[X,Y](f_2)Z_2-f_2 ∇_{[X,Y]}Z_2\\ =X(Y (f_1 ))Z_1 + Y (f_1)∇_X Z_1 + X(f_1 )∇_Y Z_1 + f_1∇_X ∇_Y Z_1\\ − Y (X(f_1))Z_1 − X(f_1) ∇_Y Z_1 − Y (f_1) ∇_X Z_1 − f_1∇_Y ∇_X Z_1\\ − (X(Y (f_1))X − Y (X(f_1))Z_1 − f_1 ∇_{[X,Y ]} Z_1\\ +X(Y (f_2 ))Z_2 + Y (f_2)∇_X Z_2 + X(f_2 )∇_Y Z_2 + f_2∇_X ∇_Y Z_2\\ − Y (X(f_2))Z_2 − X(f_2) ∇_Y Z_2 − Y (f_2) ∇_X Z_2 − f_2∇_Y ∇_X Z_2\\ − (X(Y (f_2))X − Y (X(f_2))Z_2 − f_2 ∇_{[X,Y ]} Z_2\\ = f_1 (∇_X ∇_Y Z_1 − ∇_Y ∇_X Z_1 − ∇_{[X,Y ]} Z_1)\\ +f_2 (∇_X ∇_Y Z_2 − ∇_Y ∇_X Z_2 − ∇_{[X,Y ]} Z_2)\\ =f_1R(X,Y)Z_1+f_2R(X,Y)Z_2.$$

Curvature tensor of type $T^{(1,3)}M$

Since $R$ takes its values in $\mathfrak{X}(M)$, it does not satisfy the definition of tensor field (which takes its values in $C^\infty(M) )$, however, $R$ is equivalent to the tensor field of type $T^{(1,3)}M$ defined by $\tilde{R}(α, X, Y, Z) ≡ α(R(X, Y)Z).$

Remark. In SageMath Manifolds the method riemann returns this $(1,3)$-type tensor.

A connection $∇$ is flat if its curvature tensor is zero.

Curvature in local coordinates

If $X=X^i\frac{\partial}{\partial x^i}, Y=Y^j\frac{\partial}{\partial x^j}, Z=Z^k\frac{\partial}{\partial x^k},$ then

$$R(X, Y)Z = X^i Y^j Z^k R^m_{kij}\frac{\partial}{\partial x^m},$$$$\begin{equation} R^m_{kij}=\frac{\partial \Gamma^m_{kj}}{\partial x^i} -\frac{\partial \Gamma^m_{ki}}{\partial x^j} +\Gamma^m_{li}\Gamma^l_{kj} -\Gamma^m_{lj}\Gamma^l_{ki} \tag{23.2} \end{equation}$$

In fact, from (23.1), the definition of Christoffel symbols and the relation $[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}]=0$ we have

$$\displaystyle R(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}) \frac{\partial}{\partial x^k}=\displaystyle =\nabla_{\frac{\partial}{\partial x^i}} \nabla_{\frac{\partial}{\partial x^j}}\frac{\partial}{\partial x^k} -\nabla_{\frac{\partial}{\partial x^j}} \nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^k}\\ =\displaystyle \nabla_{\frac{\partial}{\partial x^i}} \Big(\Gamma^m_{kj}\frac{\partial}{\partial x^m}\Big) -\nabla_{\frac{\partial}{\partial x^j}} \Big(\Gamma^m_{ki}\frac{\partial}{\partial x^m}\Big)\\ =\Gamma^m_{kj}\nabla_{\frac{\partial}{\partial x^i}} \frac{\partial}{\partial x^m} +\frac{\partial \Gamma^m_{kj}}{\partial x^i}\frac{\partial}{\partial x^m} -\Gamma^m_{ki}\nabla_{\frac{\partial}{\partial x^j}} \frac{\partial}{\partial x^m} -\frac{\partial \Gamma^m_{ki}}{\partial x^j}\frac{\partial}{\partial x^m}\\ =\Gamma^m_{kj}\Gamma^l_{mi}\frac{\partial}{\partial x^l} +\frac{\partial \Gamma^m_{kj}}{\partial x^i} \frac{\partial}{\partial x^m} -\Gamma^m_{ki}\Gamma^l_{mj}\frac{\partial}{\partial x^l} -\frac{\partial \Gamma^m_{ki}}{\partial x^j} \frac{\partial}{\partial x^m}\\ =\Big(\frac{\partial \Gamma^m_{kj}}{\partial x^i} -\frac{\partial \Gamma^m_{ki}}{\partial x^j} +\Gamma^m_{li}\Gamma^l_{kj} -\Gamma^m_{lj}\Gamma^l_{ki}\Big) \frac{\partial}{\partial x^m}.$$

Example 23.1

In notebook 21 we have noticed that all Christoffel symbols for the Euclidean connection vanish. Consequently the curvature for this connection vanishes.

R2=Manifold(2,'R^2')              # manifold R^2
c_xy.<x,y>=R2.chart()             # Cartesian coordinates
nab=R2.affine_connection('nab')   # Eucl.connection on R^2
nab[1,1,1]=0                      # all coeff. zero
R=nab.riemann();R                 # curvature of Eucl.conect.

Tensor field of type (1,3) on the 2-dimensional differentiable manifold R^2

R.disp()                          # show R

0

Example 23.2

Using the standard metric of the Euclidean space, the previous example can be simplified.

E.<x,y>=EuclideanSpace()
E.metric().riemann().disp()

Riem(g) = 0

Example 23.3

Consider the two-dimensional half-plane $y>0$ with connection coefficients defined by $\ \ \Gamma^1_{12}=\Gamma^1_{21}=\Gamma^2_{22}=-\frac{1}{y}\ \$ and $\ \ \Gamma^2_{11}=\frac{1}{y}.$

%display latex
M = Manifold(2, 'M', start_index=1)     # manifold M, y>0
c_xy.<x,y> = M.chart()                  # chart on M
                                        # affine connection on M:
nab = M.affine_connection('nabla', r'\nabla') 

                                        # Christoffel symbols:
nab[1,1,2], nab[1,2,1],nab[2,2,2],nab[2,1,1] = -1/y, -1/y,-1/y,1/y
%display latex
nab.display(coordinate_labels=False)    # show Christoffel symbols
                                        # (only non-zero ones)

$\displaystyle \begin{array}{lcl} \Gamma_{\phantom{\, 1}\,1\,2}^{\,1\phantom{\, 1}\phantom{\, 2}} & = & -\frac{1}{y} \\ \Gamma_{\phantom{\, 1}\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 1}} & = & -\frac{1}{y} \\ \Gamma_{\phantom{\, 2}\,1\,1}^{\,2\phantom{\, 1}\phantom{\, 1}} & = & \frac{1}{y} \\ \Gamma_{\phantom{\, 2}\,2\,2}^{\,2\phantom{\, 2}\phantom{\, 2}} & = & -\frac{1}{y} \end{array}$

Using (23.2) with $\ (x^1,x^2)=(x,y)\$ to compute $R^1_{212}$ we obtain

$$R^1_{212}=\frac{\partial \Gamma^1_{22}}{\partial x^1} -\frac{\partial \Gamma^1_{21}}{\partial x^2} +\Gamma^1_{11}\Gamma^1_{22}+\Gamma^1_{21}\Gamma^2_{22} -\Gamma^1_{12}\Gamma^1_{21}-\Gamma^1_{22}\Gamma^2_{21}\\ =0-\frac{1}{y^2}+0\cdot 0+(-\frac{1}{y})(-\frac{1}{y}) -(-\frac{1}{y})(-\frac{1}{y})-0\cdot 0=-\frac{1}{y^2}.$$

The remaining components can be found analogously (one can use antisymmetry of $R^m_{kij}$ with respect to $i,j$).

Riem = nab.riemann();  print(Riem)         # curvature (1,3) tensor
Riem.display_comp(coordinate_labels=False) # show nonzero components

Tensor field of type (1,3) on the 2-dimensional differentiable manifold M

$\displaystyle \begin{array}{lcl} X_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{1}{y^{2}} \\ X_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{1}{y^{2}} \\ X_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{1}{y^{2}} \\ X_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{1}{y^{2}} \end{array}$

Example 23.4

Consider the plane $R^2$ with Christoffel symbols $$Γ^1_{11}= Γ^1_{22}=\frac{4u}{1+u^2+4v^2},\\ Γ^2_{11}= Γ^2_{22}=\frac{4v}{1+u^2+4v^2},$$ and the remaining symbols equal to 0.

%display latex
N=Manifold(2,name='R2',start_index=1)   # manifold M, dim=2
c_uv.<u,v>=N.chart()                    # coordinates u,v
nab=N.affine_connection('nab')          # connection on M
nab[:]=[[[4*u/(4*u^2 + 4*v^2 + 1), 0],  # Christoffel symbols
         [0, 4*u/(4*u^2 + 4*v^2 + 1)]], 
        [[4*v/(4*u^2 + 4*v^2 + 1), 0], 
         [0, 4*v/(4*u^2 + 4*v^2 + 1)]]]
nab.display(coordinate_labels=False,only_nonzero=False)

$\displaystyle \begin{array}{lcl} \Gamma_{\phantom{\, 1}\,1\,1}^{\,1\phantom{\, 1}\phantom{\, 1}} & = & \frac{4 \, u}{4 \, u^{2} + 4 \, v^{2} + 1} \\ \Gamma_{\phantom{\, 1}\,1\,2}^{\,1\phantom{\, 1}\phantom{\, 2}} & = & 0 \\ \Gamma_{\phantom{\, 1}\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 1}} & = & 0 \\ \Gamma_{\phantom{\, 1}\,2\,2}^{\,1\phantom{\, 2}\phantom{\, 2}} & = & \frac{4 \, u}{4 \, u^{2} + 4 \, v^{2} + 1} \\ \Gamma_{\phantom{\, 2}\,1\,1}^{\,2\phantom{\, 1}\phantom{\, 1}} & = & \frac{4 \, v}{4 \, u^{2} + 4 \, v^{2} + 1} \\ \Gamma_{\phantom{\, 2}\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 2}} & = & 0 \\ \Gamma_{\phantom{\, 2}\,2\,1}^{\,2\phantom{\, 2}\phantom{\, 1}} & = & 0 \\ \Gamma_{\phantom{\, 2}\,2\,2}^{\,2\phantom{\, 2}\phantom{\, 2}} & = & \frac{4 \, v}{4 \, u^{2} + 4 \, v^{2} + 1} \end{array}$

Compute components of the curvature $(1,3)$ tensor. At a first attempt we obtain some components non-simplified.

Riem = nab.riemann();                  # curvature (1,3) tensor
Riem.display_comp(coordinate_labels=False)

$\displaystyle \begin{array}{lcl} X_{\phantom{\, 1}\,1\,1\,2}^{\,1\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & \frac{16 \, u v}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 1}\,1\,2\,1}^{\,1\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{16 \, u v}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & \frac{4 \, {\left(4 \, v^{2} + 1\right)}}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{4 \, {\left(4 \, v^{2} + 1\right)}}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{4 \, {\left(4 \, u^{2} + 1\right)}}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & \frac{4 \, {\left(4 \, u^{2} + 1\right)}}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 2}\,2\,1\,2}^{\,2\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{16 \, u v}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \\ X_{\phantom{\, 2}\,2\,2\,1}^{\,2\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & \frac{16 \, u v}{16 \, u^{4} + 16 \, v^{4} + 8 \, {\left(4 \, u^{2} + 1\right)} v^{2} + 8 \, u^{2} + 1} \end{array}$

so we decided to simplify each component separately and to introduce a new tensor with simplified components.

# simplification of components one by one
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]];
Riem1 = N.tensor_field(1, 3, name='Riem1')  # new tensor (1,3)type
Riem1[c_uv.frame(), :] = Riem2              # with simplified comp.
Riem1.display_comp(coordinate_labels=False)

$\displaystyle \begin{array}{lcl} Riem1_{\phantom{\, 1}\,2\,1\,1}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 1}} & = & -\frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 1}\,2\,1\,2}^{\,1\phantom{\, 2}\phantom{\, 1}\phantom{\, 2}} & = & \frac{4 \, {\left(4 \, u^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 1}\,2\,2\,1}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 1}} & = & -\frac{4 \, {\left(4 \, v^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 1}\,2\,2\,2}^{\,1\phantom{\, 2}\phantom{\, 2}\phantom{\, 2}} & = & \frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 2}\,1\,1\,1}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 1}} & = & \frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 2}\,1\,1\,2}^{\,2\phantom{\, 1}\phantom{\, 1}\phantom{\, 2}} & = & -\frac{4 \, {\left(4 \, u^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 2}\,1\,2\,1}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 1}} & = & \frac{4 \, {\left(4 \, v^{2} + 1\right)}}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \\ Riem1_{\phantom{\, 2}\,1\,2\,2}^{\,2\phantom{\, 1}\phantom{\, 2}\phantom{\, 2}} & = & -\frac{16 \, u v}{{\left(4 \, u^{2} + 4 \, v^{2} + 1\right)}^{2}} \end{array}$

Extension of curvature map to covariant tensor fields

If we extend the definition $R(X, Y)t ≡ ∇_X ∇_Y t − ∇_Y ∇_X t − ∇_{[X,Y]} t$, to all tensor fields $t\in T^{(0,k)}M,$ then one can check that for covariant tensor fields $\ t,s$, vector fields $\ X,Y$ and smooth function $f$

$$\begin{equation} \begin{matrix} R(X, Y)(t + s) = R(X, Y)t + R(X, Y)s,\\ R(X, Y)( f t) = f R(X, Y)t,\\ R(X, Y)(t ⊗ s) = R(X, Y)t ⊗ s + t ⊗ R(X, Y)s . \end{matrix} \tag{23.3} \end{equation}$$

The first relation follows from

$$\nabla_X\nabla_Y(t+s)-\nabla_Y\nabla_X(t+s)-\nabla_{[X,Y]}(t+s)\\ =\nabla_X\nabla_Yt+\nabla_X\nabla_Ys- \nabla_Y\nabla_Xt+\nabla_Y\nabla_Xs-\nabla_{[X,Y]}t-\nabla_{[X,Y]}s,$$

the second from

$$\nabla_X\nabla_Y(ft)-\nabla_Y\nabla_X(ft)-\nabla_{[X,Y]}(ft)\\ =\nabla_X(f\nabla_yt+(Yf)t)-\nabla_Y(f\nabla_Xt+(Xf)t) -f\nabla_{[X,Y]}t-([X,Y]f)t\\ =f\nabla_X\nabla_Yt+(Xf)\nabla_Y t +(Yf)\nabla_Xt+(X(Yf))t\\ -f\nabla_Y\nabla_Xt-(Yf)\nabla_Xt-(Xf)\nabla_Y t-(X(Yf))t\\ -f\nabla_{[X,Y]}t-([X,Y]f)t,$$

and the third from

$$\nabla_X\nabla_Y(t\otimes s)-\nabla_Y\nabla_X(t\otimes s)-\nabla_{[X,Y]}(t\otimes s)\\ =\nabla_X(t\otimes\nabla_Ys+(\nabla_Yt)\otimes s) -\nabla_Y(t\otimes\nabla_X s+(\nabla_X t)\otimes s)\\ -t\otimes(\nabla_{[X,Y]}s)-(\nabla_{[X,Y]}t)\otimes s\\ =t\otimes\nabla_X\nabla_Y s+(\nabla_Xt)\otimes(\nabla_Y s) +(\nabla_X\nabla_Y t)\otimes s+(\nabla_Y t)\otimes(\nabla_X s)\\ -t\otimes\nabla_Y\nabla_Xs-(\nabla_Y t)\otimes(\nabla_X s) -(\nabla_Y\nabla_X t)\otimes s-(\nabla_X t)\otimes(\nabla_Y s)\\ -t\otimes\nabla_{[X,Y]}s-(\nabla_{[X,Y]}t)\otimes s.$$

Bianchi identities

If $T$ is the torsion and $R$ the curvature, then for $X, Y, Z ∈ \mathfrak{X}(M)$, the following first Bianchi identity holds true.

$$\begin{equation} \begin{matrix} R(X, Y)Z + R(Z, X)Y + R(Y, Z)X\\ = ∇_X (T (Y, Z)) + ∇_Y (T (Z, X)) + ∇_Z (T (X, Y))\\ + T (X, [Y, Z]) + T (Y, [Z, X]) + T (Z, [X, Y]) \end{matrix} \tag{23.4} \end{equation}$$

In fact, from the definition of torsion we have $$\nabla_XY=\nabla_YX+[X,Y]+T(X,Y),\\ \nabla_ZX=\nabla_XZ+[Z,X]+T(Z,X),\\ \nabla_YZ=\nabla_ZY+[Y,Z]+T(Y,Z),$$ and therefore $$R(X, Y)Z + R(Z, X)Y + R(Y, Z)X\\ =∇_X ∇_Y Z − ∇_Y ∇_X Z − ∇_{[X,Y]} Z\\ +∇_Y ∇_Z X − ∇_Z ∇_Y X − ∇_{[Y,Z]} X\\ +∇_Z ∇_X Y − ∇_X ∇_Z Y − ∇_{[Z,X]} Y\\ =∇_X(\nabla_ZY+[Y,Z]+T(Y,Z))− ∇_Y ∇_X Z − ∇_{[X,Y]} Z\\ +∇_Y(\nabla_XZ+[Z,X]+T(Z,X))− ∇_Z ∇_Y X − ∇_{[Y,Z]} X\\ +∇_Z(\nabla_YX+[X,Y]+T(X,Y))− ∇_X ∇_Z Y − ∇_{[Z,X]} Y$$ $$\begin{equation} \begin{matrix} =∇_X([Y,Z]+T(Y,Z)) − ∇_{[X,Y]} Z\\ +∇_Y([Z,X]+T(Z,X)) − ∇_{[Y,Z]} X\\ +∇_Z([X,Y]+T(X,Y)) − ∇_{[Z,X]} Y. \end{matrix} \tag{23.5} \end{equation}$$

If we use the definition of torsion in the following form

$$T(X,[Y,Z])=\nabla_X[Y,Z]-\nabla_{[Y,Z]}X-[X,[Y,Z]],\\ T(Y,[Z,X])=\nabla_Y[Z,X]-\nabla_{[Z,X]}Y-[Y,[Z,X]],\\ T(Z,[X,Y])=\nabla_Z[X,Y]-\nabla_{[X,Y]}Z-[Z,[X,Y]],\\$$

then the subexpressions of (23.5) which do not contain $T$ take the form

$$∇_X([Y,Z]− ∇_{[Y,Z]} X = T(X,[Y,Z])+[X,[Y,Z]],\\ ∇_Y([Z,X]− ∇_{[Z,X]} Y = T(Y,[Z,X])+[Y,[Z,X]],\\ ∇_Z([X,Y]− ∇_{[X,Y]} Z = T(Z,[X,Y])+[Z,[X,Y]].\\$$

Using the Jacobi identity (notebook 12) we obtain (23.4).

The second Bianchi identity reads as follows.

For $X, Y, Z, W ∈ \mathfrak{X}(M)$

$$\begin{equation} \begin{matrix} ∇_X( R(Y, Z)W) + ∇_Y( R(Z, X)W )+ ∇_Z( R(X, Y)W)\\ = R(Y, Z)∇_X W + R(Z, X)∇_Y W + R(X, Y)∇_Z W\\ + R([Y, Z], X)W + R([Z, X], Y)W + R([X, Y], Z)W. \end{matrix} \tag{23.6} \end{equation}$$

To check the identity let us note that from (23.1) it follows

$$\begin{equation} \begin{matrix} R([X, Y], Z)W+\nabla_{[[X,Y],Z]}W= \nabla_{[X,Y]}\nabla_ZW-\nabla_Z\nabla_{[X,Y]}W,\\ R([Y, Z], X)W+\nabla_{[[Y,Z],X]}W= \nabla_{[Y,Z]}\nabla_XW-\nabla_X\nabla_{[Y,Z]}W,\\ R([Z, X], Y)W+\nabla_{[[Z,X],Y]}W= \nabla_{[Z,X]}\nabla_YW-\nabla_Y\nabla_{[Z,X]}W, \end{matrix} \tag{23.7} \end{equation}$$

therefore (we changed the order in the second column of expressions)

$$∇_X( R(Y, Z)W) + ∇_Y( R(Z, X)W )+ ∇_Z( R(X, Y)W)\\ =∇_X(\nabla_Y\nabla_ZW-\nabla_Z\nabla_YW-\nabla_{[Y,Z]}W)\\ +∇_Y(\nabla_Z\nabla_XW-\nabla_X\nabla_ZW-\nabla_{[Z,X]}W)\\ +∇_Z(\nabla_X\nabla_YW-\nabla_Y\nabla_XW-\nabla_{[X,Y]}W)\\ =\nabla_X\nabla_Y\nabla_ZW-\nabla_Y\nabla_X\nabla_ZW-\nabla_X\nabla_{[Y,Z]}W\\ +\nabla_Y\nabla_Z\nabla_XW-\nabla_Z\nabla_Y\nabla_XW-\nabla_Y\nabla_{[Z,X]}W\\ +\nabla_Z\nabla_X\nabla_YW-\nabla_X\nabla_Z\nabla_YW-\nabla_Z\nabla_{[X,Y]}W.\\$$

In all three rows of the obtained sum we can recognize incomplete curvature tensors computed for $\nabla_ZW, \nabla_XW, \nabla_YW$ respectively. If we subtract and add the lacking terms with covariant derivatives along the corresponding Lie brackets, we obtain the following form of the last sum

$$R(X,Y)\nabla_ZW+\nabla_{[X,Y]}\nabla_ZW-\nabla_X\nabla_{[Y,Z]}W\\ +R(Y,Z)\nabla_XW+\nabla_{[Y,Z]}\nabla_XW-\nabla_Y\nabla_{[Z,X]}W\\ +R(Z,X)\nabla_YW+\nabla_{[Z,X]}\nabla_YW-\nabla_Z\nabla_{[X,Y]}W\\$$

In all three lines, the last two expressions are equal to the right hand sides of (23.7), therefore the obtained sum is equal

$$R(X,Y)\nabla_ZW+R([X, Y], Z)W+\nabla_{[[X,Y],Z]}W\\ +R(Y,Z)\nabla_XW+R([Y, Z], X)W+\nabla_{[[Y,Z],X]}W\\ +R(Z,X)\nabla_YW+R([Z, X], Y)W+\nabla_{[[Z,X],Y]}W.$$

The sum of the last column is zero according to Jacobi identity (notebook 12), so (23.6) is proved.

Bianchi identities will be used in the proof of (24.2) in the next notebook.

What's next?

Take a look at the notebook Riemannian curvature tensor of type (0,4).