Show linearized Riemann tensor is gauge invariant I'm working in general relativity and figured this problem would be more suited to mathematics than physics. 
The linearized Riemann tensor is given by:
$$R_{\alpha \beta \mu \nu}=-\frac{1}{2}\left[h_{\alpha \mu, \beta \nu}+h_{\beta \nu, \alpha \mu}-h_{\alpha \nu, \beta \mu}-h_{\beta \mu, \alpha \nu}\right]$$
I want to show this is invariant under the gauge transformation 
$$h_{\alpha \beta} \rightarrow h_{\alpha^{\prime} \beta^{\prime}}=h_{\alpha \beta}-\xi_{\alpha, \beta}-\xi_{\beta, \alpha}$$
This seems relativitely straightforward, but I am unsure of how to plug this gauge transformation into the Riemann tensor, since the indices end up being swapped around.

The first term contributes a difference of $$ -(\xi_{\alpha,\mu\beta\nu}+\xi_{\mu,\alpha\beta\nu})$$ Similarly, the second term $$ -(\xi_{\beta,\nu\alpha\mu}+\xi_{\nu,\beta\alpha\mu})$$ and the third $$+(\xi_{\alpha,\nu\beta\mu}+\xi_{\nu,\alpha\beta\mu})$$ Hey, stop right there. The second term of the second one $-\xi_{\nu,\beta\alpha\mu}$ and the second term of the third one $\xi_{\nu,\alpha\beta\mu}$ cancel out since partial derivatives commute. Same with the first term of the first one and the first term of the third one. If you write down the fourth one surely it will cancel the remaining two.
(To get the differences I just used the fact that partial derivatives are linear, so just distribute over the terms of the $h_{\alpha'\beta'}$... you can just tack the indices from the partial derivative onto the end.)
