# Riemann tensor definition for non vanishing torsion

From the definition of Riemann tensor we have:

$$\mathbf{R}\left( \mathbf{z},\mathbf{v},\mathbf{w}\right)=\nabla_{\mathbf{[v}}\nabla_{\mathbf{w}]}\mathbf{z}-\nabla_{[\mathbf{v},\mathbf{w}]}\mathbf{z} \label{riemannnew}$$

and computing the coordinates of $\mathbf{R}$ in a coordinate basis we obtain:

$$R^a_{\hphantom{a}bcd}=\partial_c\Gamma^a_{\hphantom{a}bd}-\partial_{d}\Gamma^a_{\hphantom{a}bc}+\Gamma^a_{\hphantom{a}\mu c}\Gamma^\mu_{\hphantom{a}bd}-\Gamma^a_{\hphantom{a}\mu d}\Gamma^\mu_{\hphantom{a}bc}$$

I find another way to compute the coefficient fo Riemann tensor with not vanishing torsion:

$$[\nabla_c,\nabla_d]V^a=2\nabla_{[c}\nabla_{d]}V^a = 2\partial_{[c}\nabla_{d]}V^a-2\Gamma^e_{\hphantom{e}[dc]}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}\nabla_{d]}V^e \nonumber \\ = 2\partial_{[c}(\partial_{d]}V^a+\Gamma^a_{\hphantom{a}|e|d]}V^e)+2S^e_{\hphantom{a}cd}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}(\partial_{d]}V^e+\Gamma^e_{\hphantom{a}|b|d]}V^b) \nonumber \\ = 2 \partial_{[c}\Gamma^a_{\hphantom{a}|b|d]}V^b-2 \Gamma^a_{\hphantom{a}e[c}\partial_{d]}V^e+2S^e_{\hphantom{a}cd}\nabla_eV^a+2\Gamma^a_{\hphantom{a}e[c}\partial_{d]}V^e+2\Gamma^a_{\hphantom{a}e[c}\Gamma^e_{\hphantom{a}|b|d]}V^b= \nonumber \\ =2(\partial_{[c}\Gamma^a_{\hphantom{a}|b|d]}+\Gamma^a_{\hphantom{a}e[c}\Gamma^e_{\hphantom{a}|b|d]})V^b + 2S^e_{\hphantom{a}cd}\nabla_eV^a \tag{1}$$

where the first bracket is the Riemann-Cartan tensor and second term is the part due to the non vanishing torsion tensor.

My question is:

The first term of the first definition $\nabla_{\mathbf{[v}}\nabla_{\mathbf{w}]}\mathbf{z}$ is the second equation (1) but only the first term of the second equation is the Riemann tensor. How can I solve this problem? Is the definition of the Riemann tensor incomplete?

In the invariant notation $\nabla_X\nabla_Y$ corresponds to $X^a\nabla_a(Y^b\nabla_b)$, not $X^a Y^b\nabla_a\nabla_b$, eg. the vector field $Y$ also gets differentiated.

We can define $\nabla^2_{X,Y}Z=i_Xi_Y\nabla\nabla Z$, where here $i$ means "insert into the last empty argument", then we have $$X^a\nabla_a(Y^b\nabla_b)Z^c=X^a\nabla_aY^b\nabla_bZ^c+X^aY^b\nabla_a\nabla_bZ^c,$$ so $$\nabla^2_{X,Y}Z=\nabla_X\nabla_YZ-\nabla_{\nabla_XY}Z.$$

This gives then $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z=\nabla^2_{X,Y}Z+\nabla_{\nabla_XY}Z-\nabla^2_{Y,X}Z-\nabla_{\nabla_YX}Z-\nabla_{[X,Y]}Z \\ =\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{\nabla_XY-\nabla_YX-[X,Y]}Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{T(X,Y)}Z.$$

As you can see the $[\nabla_a,\nabla_b]X^c=R^c_{\ dab}X^d$ Ricci-identity corresponds to $R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z$, which is certainly true in absence of torsion.

In the presence of torsion, this gets modified to $$R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{T(X,Y)}Z,$$ but the definition of the curvature tensor, $$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$$ doesn't depend on torsion at all.

• Ok, I can see that from the definition in presence of torsion we get that the curvature tensor is $$R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z+\nabla_{T(X,Y)}Z,$$ but only the first two terms correspond to the Riemann tensor? Curvature tensor and Riemann tensor differs in presence of torsion? – raskolnikov Jul 2 '17 at 10:47
• @raskolnikov No, all three terms correspond to the curvature tensor. The more fundamental definition of the curvature tensor is $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$, if you write this in terms of the second covariant derivative, you get the expression you wrote in the comment. – Bence Racskó Jul 2 '17 at 11:22
• Ok but $R^a_{\ bcd}$ is equal to which terms in presence of torsion? – raskolnikov Jul 2 '17 at 15:29
• @raskolnikov $R^a_{bcd}X^b=[\nabla_c,\nabla_d]X^a+T^b_{\ cd}\nabla_bX^a$. – Bence Racskó Jul 2 '17 at 15:49
• Sorry but I struggle with this problem again. In your last comment you said: $R^a_{\ bcd}X^b=[\nabla_c,\nabla_d]X^a+T^b_{\ cd}\nabla_b X^a$ but this is incompatible with my calculation of $[\nabla_c, \nabla_d]$ I wrote in my question since I can't eliminate the torsion tensor $T^b_{\ cd}$ putting my formula in your one. – raskolnikov Jul 6 '17 at 13:50