Identity with Riemann tensor

Is there a fast way to derive the identity

$$(\nabla_{\alpha}\nabla_{\beta}-\nabla_{\beta}\nabla_{\alpha})T_{\gamma\delta}={R_{\alpha\beta\gamma}}^{\lambda}T_{\lambda\delta}+{R_{\alpha\beta\delta}}^{\lambda}T_{\gamma\lambda}$$

where $$T$$ is a (not necessarily symmetric) $$2$$-tensor field and where I use the index and sign convention

$$(\nabla_{\alpha}\nabla_{\beta}-\nabla_{\beta}\nabla_{\alpha})v_{\gamma}={R_{\alpha\beta\gamma}}^{\lambda}v_{\lambda}?$$

Of course, one could just work out the double covariant derivatives on the left-hand side, but this procedure is rather tedious. Does anyone know a faster method to verify the identity above?

• What's your definition of the Riemann tensor? One is in terms of the action of such a commutator of covariant derivatives, albeit on vectors, but a clever argument generalizes it.
– J.G.
Commented Jan 10, 2023 at 8:02
• Maybe start with $T_{\alpha\beta}=A_{\alpha}B_{\beta}$ and use the formula for commutation of vector field. Then extend it to $\sum A_{\alpha}B_{\beta}$ by linearity
– KP99
Commented Jan 10, 2023 at 8:06
• @Ghoster Yes indeed. Fixed now Commented Jan 10, 2023 at 8:10
• Actually yes. Any Type-II tensor $T_{ab}$ is defined as sum of terms like $A_aB_a$, and this definition is completely equivalent to the multi-linear map definition, provided your underlying module has finite basis... which in physics is usually the case. You can read more about this in chapter 2 from Spinors and Space-time Volume-1
– KP99
Commented Jan 10, 2023 at 8:22
• @KP99 A yes, since the $C^{\infty}(\mathcal{M})$-module of rank two-tensor fields is isomorphic via $\Gamma^{\infty}(T^{\ast}\mathcal{M}^{\otimes 2})\cong \mathfrak{X}^{\ast}(\mathcal{M})\otimes_{C^{\infty}(\mathcal{M})}\mathfrak{X}^{\ast}(\mathcal{M})$ and the module of vector fields $\mathfrak{X}(\mathcal{M})$ is finitely-generated and projective, by Serre-Swan. Thanks! Commented Jan 10, 2023 at 8:24