Problems with Leibniz rule in calculating the covariant derivative of a $(1,1)-$ tensor. Where is my mistake?

Question

Let be $$R=\sum _{\alpha, \beta} R^\alpha_\beta \frac{\partial}{\partial x^\alpha} \otimes dx^\beta. $$

I want to calculate $\nabla_\gamma(R)=\nabla_{\frac{\partial}{\partial x^\gamma}}(R).$

My book gives me this first result: $$\nabla_\gamma(R)=\sum_{\beta}\big(\nabla_\gamma\sum_\alpha\big[R^\alpha_\beta\frac{\partial}{\partial x^\alpha}\big]\big)\otimes dx^\beta+\sum_\alpha \frac{\partial}{\partial x^\alpha}\otimes\big(\nabla_\gamma\sum_\beta\big[R^\alpha_\beta dx^\beta\big]\big).$$

I have tried to obtain this expression using the Leibniz rule for tensors but I have found something different for the second addendum: $$\nabla_\gamma(R)=\sum_\beta\big( \nabla_\gamma\big[\sum_\alpha R^\alpha_\beta\frac{\partial}{\partial x^\alpha}\otimes dx^\beta\big]\big)=\\\sum_\beta\big(\nabla_\gamma\sum_\alpha\big[R^\alpha_\beta\frac{\partial}{\partial x^\alpha}\big]\big)\otimes dx^\beta+\sum_\beta\big(\sum_\alpha R^\alpha_\beta\frac{\partial}{\partial x^\alpha}\big)\otimes \nabla_\gamma dx^\beta=\\\\\sum_\beta\big(\nabla_\gamma\sum_\alpha\big[R^\alpha_\beta\frac{\partial}{\partial x^\alpha}\big]\big)\otimes dx^\beta+\sum_\alpha\big(\frac{\partial}{\partial x^\alpha}\otimes\sum_\beta\big(R^\alpha_\beta\nabla_\gamma dx^\beta \big) \big). $$

Where is my mistake? Why is this last expression not correct?I have only used the rules of the covariant derivative but my result is different.

How can I obtain the first formula?

Thanks in advance for the help!

Answer 1

Are you sure about the formula from the book? Because it involves twice the derivative of $R^\alpha_\beta$, which shouldn't be the case, while your result seems correct.

Answer 2

If we use the Einstein's sum convention and some basic calculations like $$\nabla_{\partial_{\gamma}}(X^s\partial_s)= \left(X^s{}_{,\gamma}+X^t\Gamma^s{}_{\gamma t}\right)\partial_s,$$ and $$\nabla_{\partial_{\gamma}}(\omega_sdx^s)= \left(\omega_{s,\gamma}-\omega_t\Gamma^t{}_{s\gamma}\right)dx^s,$$ then $$\nabla_{\partial_{\gamma}}(R^{\alpha}{}_{\beta}\partial_{\alpha}\otimes dx^{\beta})= \left(R^{\alpha}{}_{\beta,\gamma} +R^s{}_{\beta}\Gamma^{\alpha}{}_{s\gamma}-R^{\alpha}{}_s\Gamma^s{}_{\beta \gamma}\right) \partial_{\alpha}\otimes dx^{\beta}. $$

Take into account that $X^s{}_{,\gamma}=\dfrac{\partial X^s}{\partial x^{\gamma}}$ and $\omega_{s,\gamma}=\dfrac{\partial\omega_s}{\partial x^{\gamma}}$.