Difference tensor between two connections

Question

I am using these supergravity lecture notes by Gary W. Gibbons. On page 18, the author claims that geodesics and autoparallels coincide for a theory with totally antisymmetric torsion, and proves it by using the following identity (which he states without proof):

$$\Gamma_{\mu\;\;\nu}^{\;\;\rho}=\{_{\mu\;\;\nu}^{\;\;\rho}\}+K_{\mu\;\;\nu}^{\;\;\rho},$$

where

$$K_{\mu\rho\nu}=-\frac{1}{2}(T_{\mu\rho\nu}+T_{\rho\mu\nu}+T_{\rho\nu\mu}).$$

Here, $\Gamma_{\mu\;\;\nu}^{\;\;\rho}$ are the coefficients of an arbitrary connection with totally antisymmetric torsion and $\{_{\mu\;\;\nu}^{\;\;\rho}\}$ are the coefficients for the Levi-Civita connection, while $T_{\rho\nu\mu}$ is the torsion. Is there a proof of this identity somewhere?

Answer 1

OP's question is probably spurred by the fact that Ref. 1 forgets to mention that:

1. The other connection $\nabla:\Gamma(TM)\times \Gamma(TM)\to \Gamma(TM) $ with lower Christoffel symbols $$\Gamma_{ij,k}~:=~ \Gamma_{ij}^{\ell}~g_{\ell k}\tag{1}$$ is still assumed to be compatible with the metric $$\nabla g = 0\qquad\Leftrightarrow\qquad \partial_kg_{ij}=\Gamma_{ki,j}+\Gamma_{kj,i}.\tag{2}$$

2. The contorsion tensor $K~\in~\Gamma\left(T^{\ast}M\otimes \bigwedge^2(T^{\ast}M)\right)$ is defined$^1$ as the difference between the metric-compatible $\nabla$ and the Levi-Civita connection $\nabla^{LC}$, i.e. $$K_{i,jk}~:=~\Gamma_{ij,k}-\Gamma^{LC}_{ij,k} ~\stackrel{(2)}{=}~-K_{i,kj}.\tag{3} $$

3. The torsion tensor $T~\in~\Gamma\left(\bigwedge^2(T^{\ast}M)\otimes T^{\ast}M \right)$ is defined$^1$ as $$ T(X,Y)~:=~\nabla_XY-\nabla_YX-[X,Y] $$ $$\quad\Leftrightarrow\quad T_{ij,k}~:=~\Gamma_{ij,k}-\Gamma_{ji,k}~\stackrel{(3)}{=}~K_{i,jk}-K_{j,ik}.\tag{4} $$

4. One may show that the inverse relation to (4) is $$ K_{i,jk}~\stackrel{(4)}{=}~ \frac{1}{2}\left(T_{ij,k}-T_{jk,i}+T_{ki,j}\right),\tag{5}$$ and vice-versa.

5. Let us decompose the contorsion tensor $$K_{i,jk} ~=~ \frac{1}{2}(K^+_{i,jk}+K^-_{i,jk}), \tag{7}$$ in components $$ K^{\pm}_{i,jk}~:=~K_{i,jk}\pm K_{j,ik}, $$ $$ K^+_{i,jk}~=~T_{ki,j}+T_{kj,i}, \qquad K^-_{i,jk}~=~T_{ij,k}, \tag{8} $$ that are symmetric/antisymmetric wrt. the first two indices $i\leftrightarrow j$.

6. The geodesic equation reads $$0~=~\nabla^{LC}_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{LC,\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma^{LC}_{ij,k}\dot{\gamma}^i\dot{\gamma}^j.\tag{9} $$

7. In contrast, the auto-parallel equation $$0~=~\nabla_{\dot{\gamma}}\dot{\gamma} \qquad\Leftrightarrow\qquad \ddot{\gamma}^{\ell}~=~-\Gamma^{\ell}_{ij}\dot{\gamma}^i\dot{\gamma}^j$$ $$\qquad\Leftrightarrow\qquad -g_{k\ell}\ddot{\gamma}^{\ell}~=~\Gamma_{ij,k}\dot{\gamma}^i\dot{\gamma}^j~=~\left(\Gamma^{LC}_{ij,k}+\frac{1}{2}K^+_{i,jk}\right)\dot{\gamma}^i\dot{\gamma}^j\tag{10} $$ can detect the symmetric part $K^+_{i,jk}$ of the contorsion.

8. Observation: $$T_{ij,k} \text{ is totally antisymmetric} \qquad\Leftrightarrow\qquad K_{i,jk} \text{ is totally antisymmetric} $$ $$\qquad\Leftrightarrow\qquad K^+_{i,jk}~=~0.\tag{11}$$ In the affirmative case, we have $T_{ij,k}=2K_{i,jk}$, and the geodesic & auto-parallel equations (9) & (10) are the same.

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$^1$ Pertinent applications of the musical isomorphism are implicitly implied from now on.

Answer 2

On the page 17-18 of this paper has told how to proof. It say " It is possible to invert the torsion equation to ‘solve for’ the spin connection in terms of the tetrad, its derivatives and the torsion tensor. All one needs to do is manipulate the combination $\left( T_{\nu\lambda\mu} + T_{\lambda\mu\nu} - T_{\mu\nu\lambda} \right)$ and use the invertibility of the (co-)tetrad. The result is: \begin{eqnarray} % \omega_{\mu}\,^{IJ} & := & \hat{\omega}_{\mu}\,^{IJ}(e) + K_{\mu}\,^{IJ}(e, T) \\ % \hat{\omega}_{\mu}\,^{IJ} & := & \frac{1}{2}\left[ e^{\nu I}\left(\partial_{\mu} e_{\nu}^J - \partial_{\nu} e_{\mu}^J\right) - e^{\nu J}\left(\partial_{\mu} e_{\nu}^I - \partial_{\nu} e_{\mu}^I\right) - e^{\nu I} e^{\lambda J} \left(\partial_{\nu}e^K_{\lambda} - \partial_{\lambda}e^K_{\nu} \right) e_{\mu K} \right] \\ % K_{\mu}\,^{IJ} & := & - \frac{1}{2} e^{\nu I} e^{\lambda J} \left( T_{\nu\lambda\mu} + T_{\lambda\mu\nu} - T_{\mu\nu\lambda} \right) % \end{eqnarray}

The $K$ is called the con-torsion tensor and $\hat{\omega}$ is the torsion-free spin connection which is explicitly determined by the tetrad. The Affine connection equation is the corresponding inversion of the metric compatibility condition (covariant constancy of the metric) to express the general affine connection in terms of the torsion free Christoffel connection plus the torsion combinations."