# No torsion with calculating the commutator of the covariant derivatives

For simplicity, I only calculated half of the commutator. I didn't leave everything in components because I'm uncomfortable considering (I previously messed up the indices. The following is the corrected version by Sean) $$$$\begin{split} \nabla_{V^i e_i}(\nabla_{W^je_j}X^k e_k)=\nabla_{V^i e_i}(W^j (\partial_j X^k+\Gamma_{j\rho}^k X^\rho)e_k)=V^{i}{[\partial_i(W^j(\partial_j X^k+\Gamma^k_{j\rho}X^\rho)+\Gamma^k_{i\sigma}W^j(\partial_jX^\sigma+\Gamma^\sigma_{j\rho}X^\rho)]e_k}=V^i[(\partial_i W^j)(\partial_j X^k+\Gamma^{k}_{j\rho}X^\rho)+W^j\partial_i(\partial_j W^k+\Gamma^k_{j\rho}X^\rho)+\Gamma^k_{i\sigma}W^j(\partial_jX^\sigma+\Gamma^\sigma_{j\rho}X^\rho)]e_k \end{split}$$$$ (The first term in the last line should be zero, if the vector fields in nabla are simply elements of the coordinate basis $$\partial_i$$). But this doesn't give me the torsion term, i.e. there's no minus term that arises when you consider $$\nabla_\mu X^\nu$$ as a (1,1) tensor. Like here,

we have $$-\Gamma^\lambda_{\mu\nu}\partial_\lambda V^\rho-\Gamma_{\mu\nu}^\lambda \Gamma_{\lambda\sigma}^{\rho}V^{\sigma}$$ in the third line. Why would they give different result? What's wrong with my derivation.

• Please use MathJax, not images of math. Commented Feb 18 at 21:48
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented Feb 18 at 22:04

I prefer the notation $$\nabla_X$$ to the apparently more explicit $$X^\mu\nabla_\mu$$ because in the presence of torsion there is a dangerous ambiguity lurking in the notation $$\nabla_\mu$$. To see this recall that the curvature definition $$[\nabla_X,\nabla_Y]Z- \nabla_{[X,Y]}Z= {\bf R}(X,Y) Z$$ holds even when the connection has torsion. Now take $$X\to \partial_\mu$$ and $$Y\to \partial_\nu$$, i.e. $$X^\lambda =\delta^\lambda_\mu$$, $$Y^\lambda =\delta^\lambda_\nu$$ so that we are tempted to write $$\nabla_{\partial_\mu}\stackrel{}{=}\delta^\lambda_\mu \nabla_\lambda\stackrel{?}{=} \nabla_\mu, \quad \nabla_{\partial_\nu}=\delta^\lambda_\nu \nabla_\lambda \stackrel{?}{=}\nabla_\nu.$$ Now, remembering that $$[\partial_\mu,\partial_\nu]=0$$, our curvature definition gives us
$$([\nabla_\mu,\nabla_\nu] Z)^\lambda \stackrel{?}{=} Z^\alpha{R^\lambda}_{\alpha \mu\nu}.$$ If, however, we include a Christoffel symbol $${\Gamma^\alpha}_{\nu\mu}$$ when $$\nabla_\mu$$ acts from the left on the index $$\nu$$ on $$\nabla_\nu$$ and similarly a $${\Gamma^\alpha}_{\mu\nu}$$ when $$\nabla_\nu$$ is on the left and acting on the index $$\mu$$, we end up with $$( [\nabla_\mu,\nabla_\nu] Z)^\lambda \stackrel{?}{=} Z^\alpha{R^\lambda}_{\alpha \mu\nu}- {T^\sigma}_{\mu\nu} \nabla_\sigma Z^\lambda.$$ Which is equation is correct? With the last version, remembering that
$$\nabla_X Y-\nabla_Y X= [X,Y] + T(X,Y),$$ we can use Liebnitz rule to find
$$( [X^\mu\nabla_\mu,Y^\nu \nabla_\nu]Z)^\lambda = (\nabla_X Y-\nabla_YX)^\sigma \nabla_\sigma Z^\lambda +X^\mu Y^\nu (Z^\alpha{R^\lambda}_{\alpha \mu\nu}- {T^\sigma}_{\mu\nu} \nabla_\sigma Z^\lambda)\nonumber\\ = Z^\alpha{R^\lambda}_{\alpha \mu\nu}X^\mu Y^\nu + [X,Y]^\sigma \nabla_\sigma Z^\lambda, \nonumber$$ which reproduces the curvature definition for general vector fields $$X$$, $$Y$$. Both formulae are correct in their own ways. The difference lies in whether we regard the $$\mu$$ on $$\nabla_\mu$$ as a shorthand label specifying the co-ordinate basis vector $${\partial}_\mu$$, or as an index on the numerical component of a covector.

For arbitrary vectors $$V,W$$ and $$X$$, we have $$$$\begin{split} \nabla_V \nabla_W X &= \nabla_{V^\lambda \partial_\lambda}(\nabla_{W^\tau \partial_\tau}X^k e_k) \\ &=\nabla_{V^\lambda \partial_\lambda}(W^\tau (\partial_\tau X^k+\Gamma_{\tau j}^k X^j)e_k) \\ &=V^\lambda[\partial_\lambda (W^\tau(\partial_\tau X^k+\Gamma^k_{\tau j}X^j))+\Gamma^k_{\lambda i}W^\tau(\partial_\tau X^i+\Gamma^i_{\tau j}X^j)]e_k \\ &=V^\lambda[ (\partial_\lambda W^\tau)(\partial_\tau X^k+\Gamma^k_{\tau j}X^j) + W^\tau \partial_\lambda (\partial_\tau X^k+\Gamma^k_{\tau j}X^j) \\ &\qquad +\Gamma^k_{\lambda i}W^\tau(\partial_\tau X^i+\Gamma^i_{\tau j}X^j)]e_k \end{split}$$$$ Now if $$V=\partial_\mu$$, $$W=\partial_\nu$$, we have $$V^\lambda=\delta^\lambda_\mu$$, $$W^\tau=\delta^\tau_\nu$$. Substituting $$V^\lambda$$ and $$W^\tau$$ into the above equation, we get $$$$\begin{split} \nabla_{\partial_\mu} \nabla_{\partial_\nu} X &\equiv \nabla_{\delta^\lambda_\mu \partial_\lambda}(\nabla_{\delta^\tau_\nu \partial_\tau}X^k e_k) \\ &= [ \partial_\mu \partial_\nu X^k + \partial_\mu \Gamma^k_{\nu j}X^j + \Gamma^k_{\nu j}\partial_\mu X^j + \Gamma^k_{\mu i}\partial_\nu X^i + \Gamma^k_{\mu i}\Gamma^i_{\nu j}X^j]e_k. \end{split}$$$$ Then $$$$\begin{split} \nabla_{\partial_\mu} \nabla_{\partial_\nu} X - \nabla_{\partial_\nu} \nabla_{\partial_\mu} X &= [\partial_\mu \partial_\nu X^k + \partial_\mu \Gamma^k_{\nu j}X^j + \Gamma^k_{\nu j}\partial_\mu X^j + \Gamma^k_{\mu i}\partial_\nu X^i + \Gamma^k_{\mu i}\Gamma^i_{\nu j}X^j]e_k \\ &\quad - [ \partial_\nu \partial_\mu X^k + \partial_\nu \Gamma^k_{\mu j}X^j + \Gamma^k_{\mu j}\partial_\nu X^j + \Gamma^k_{\nu i}\partial_\mu X^i + \Gamma^k_{\nu i}\Gamma^i_{\mu j}X^j]e_k \\ &= ( \partial_\mu \Gamma^k_{\nu j} - \partial_\nu \Gamma^k_{\mu j} + \Gamma^k_{\mu i}\Gamma^i_{\nu j} - \Gamma^k_{\nu i}\Gamma^i_{\mu j} ) X^j e_k \\ &= {R^k}_{j\mu\nu} X^{j} e_k. \end{split}$$$$

$$\nabla_\mu \nabla_\nu X^k$$ is defined as the component of the $$(1,2)$$-tensor: $$$$\nabla\nabla X = (\nabla_\mu \nabla_\nu X^k) \mathrm{d}x^\mu \otimes \mathrm{d}x^\nu \otimes \partial_k.$$$$ We have $$$$\nabla_\mu \nabla_\nu X^k - \nabla_\nu \nabla_\mu X^k = {R^k}_{j\mu\nu} X^j - 2 \Gamma^\lambda_{[\mu\nu]} \nabla_\lambda X^k.$$$$ But $$\nabla_{\partial_\mu} \nabla_{\partial_\nu} X$$ is a vector. So $$(\nabla_{\partial_\mu} \nabla_{\partial_\nu} X)^k \neq \nabla_\mu \nabla_\nu X^k$$. And $$$$[\nabla_\mu, \nabla_\nu] X^k \neq ([\nabla_{\partial_\mu}, \nabla_{\partial_\nu}] X)^k.$$$$

• Thank you! But that still wouldn't give the minus terms right? Commented Feb 19 at 4:42
• $(\nabla_\mu X^\nu) dx^\mu \otimes \partial_\nu (V,\omega) = (\nabla_\mu X^\nu) dx^\mu(V) \omega(\partial_\nu) = (\nabla_\mu X^\nu) V^\mu \omega_\nu$.