I have a smooth manifold M with an affine connection with components $\Gamma^h_{ij}$. I have some conditions on the components of this connection, namely : they are symmetric $\Gamma^h_{ij}=\Gamma^h_{ji}$ and they have the same metric trace as the levi-civita connection $\hat{\Gamma}^h_{ij}$ of manifold M $g^{ij}\Gamma^h_{ij}=g^{ij}\hat{\Gamma}^h_{ij}$. Where $g_{ij}$ is the metric corresponding to the levi-civita connection.
My question is if I can derive some relation between my connection $\Gamma^h_{ij}$ and the levi-civita connection $\hat{\Gamma}^h_{ij}$ from these two conditions ? For example something like $\Gamma^h_{ij}=\hat{\Gamma}^h_{ij}+S^h_{ij}$, where $S^h_{ij}$ is some symmetric, traceless tensor or maybe that they must be the same ?
I would be glad for suggestions of books or papers where this topic or similar topic is described or maybe what part of mathematics should I look into to get the answer.
Thank you.
Yes, you can conclude $\Gamma = \hat \Gamma + S$ for some symmetric, trace-free $S.$ This is very easy: just define $S = \Gamma - \hat \Gamma$ and check that this is symmetric and trace-free using the facts you know about $\Gamma$ and $\hat \Gamma.$ Since the difference of connections is a tensor, this $S$ will be a well-defined symmetric, trace-free $(1,2)$-tensor field defined on $M$; and it can be any such tensor field, since we can add any such $S$ to the Levi-Civita connection to get a $\Gamma.$ If $M$ has dimension $n>1$, then the set of such tensor fields forms a $C^\infty(M)-$module of rank $n \binom n 2 - 1>0,$ so you can't conclude $\Gamma = \hat \Gamma;$ but you have a very good description of the difference.