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

$$\begin{gathered} R(X,Y){\mathrm{\nabla }}_{Z}W+R([X,Y],Z)W+{\mathrm{\nabla }}_{[[X,Y],Z]}W \\ +R(Y,Z){\mathrm{\nabla }}_{X}W+R([Y,Z], \end{gathered}$$