The curvature tensor, $R\_{ab}{}^c{}\_d$, can be obtained from a connection which not necessarily is a metric connection.


By construction it is antisymmetric in the first two indices, since roughly speaking $[\nabla\_a , \nabla\_b] \simeq R\_{ab}{}^\bullet{}\_\bullet$. I'm assuming the connection has vanishing torsion.


**Question**


Under what conditions $R\_{ab}{}^c{}\_c$ vanishes for a general curvature?




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For example:


* If $\Gamma\_{b}{}^c{}\_c = 0$ the trace of $R$ vanishes.
* If $\partial\_a \Gamma\_{b}{}^c{}\_c = 0$ the trace of $R$ vanishes.


Is there something more general? and Could that be interpreted as a gauge fixing?



Symmetries of non-Riemannian curvature tensor
Part of the confusion is that your positional notational convention is not the standard one. In most books, what you are writing as $R\_{ab}{^c}\_d$ would be written as $R{^c}\_{dab}=-R{^c}\_{dba}$ (though the letters used would probably be different).


That said, the condition $R{^c}\_{cab}=0$ is the condition that the connection (locally) admit a parallel volume form. In other words, $R{^c}\_{cab}$ represents a $2$-form that is the curvature of the induced connection on the line bundle $\Lambda^n(T^\*M)$ (where $n$ is the dimension of the manifold). In particular, it is the exterior derivative of the $1$-form that you are writing as $\Gamma{\_b}^{c}{\_c}$. This $1$-form need not be zero (which would be $\Gamma{\_b}^{c}{\_c}=0$) nor need its coefficients be constant (which would be $\partial\_a\Gamma{\_b}^{c}{\_c}=0$). It just needs to be closed in order to have your condition on the curvature.


