Let $T$ be a $(1, 1)$ tensor field, $\lambda$ a covector field and $X, Y$ vector fields. We may define $\nabla_X T$ by requiring the ‘inner’ Leibniz rule, $$\nabla_X[T(\lambda, Y )] = (\nabla_XT)(\lambda, Y ) + T(\nabla_X \lambda, Y ) + T(\lambda, \nabla_X Y ) . $$
(a) Prove that $\nabla_XT$ defines a $(1, 1)$ tensor field.
(b) Prove that the components of the $(1, 2)$ tensor $\nabla T$ are, $$\nabla_cT^{a}_{\,\,\,b} = e_c(T^{a}_{b}) + \Gamma^{a}_{\,\,dc}T^{d}_{b} − \Gamma^{d}_{ bc}T^{a}_{d}$$ (c) Deduce that the Kronecker delta tensor is covariantly constant $\nabla \delta = 0.$
Attempt:
Really just need to check what I am going to do is correct. For a), $\nabla_X T$ defines a (1,1) tensor if it is linear in the arguments $X$ and $Y$? Linear in $X$ by definition of a connection and for $Y$;
$$\nabla_X (T (\lambda, fY)) = (\nabla_X T)(\lambda fY) + T(\nabla_X \lambda, fY) + T(\lambda, \nabla_X (fY)) $$ which can be written using the axioms of a connection $$ f(\nabla_X T)(\lambda, Y) + fT(\nabla_X \lambda, Y) + fT(\lambda, f \nabla_X Y) + T(\lambda, X(f)Y)$$ It doesn't seem to be linear in Y at all?
Thanks!
From your definition,
$$ (\nabla_XT)(\lambda, Y ) = \nabla_X[T(\lambda, Y )] - T(\nabla_X \lambda, Y ) - T(\lambda, \nabla_X Y ) . $$
To calculate $\nabla_i T^j_{\ k}$, note (We use the simplified notation $\partial_i = \frac{\partial}{\partial x^i}$)
$$\nabla_i T^j_{\ k} = (\nabla_{\partial _j} T)(dx^j , \partial_k) = \nabla_{\partial_i}[T(dx^j , \partial_k )] - T(\nabla_{\partial_i} dx^j,\partial_k ) - T(dx^j, \nabla_{\partial_i} \partial _k ) $$
$$ = \partial_i (T^j_{\ k}) + T(\Gamma_{il}^j dx^l , \partial_k) - T(dx^j , \Gamma_{ik}^l \partial_l) $$
$$ = \partial_i (T^j_{\ k}) + \Gamma_{il}^j T( dx^l , \partial_k) - \Gamma_{ik}^l T(dx^j , \partial_l) = \partial_i (T^j_{\ k}) + \Gamma_{il}^j T^l_{\ k} - \Gamma_{ik}^l T^j_{\ l}. $$
Remark: First of all $T^j_{\ k} = T(dx^j , \partial_k)$ by definition, so $\nabla_{\partial_i}[T(dx^j , \partial_k )]= \partial_i (T^j_{\ k})$. Secondly, we define the covariant differentiation of a covector field $\alpha$ so that
$$ \nabla_{X} \big( \alpha(Y)\big) = (\nabla_X \alpha) (Y) + \alpha (\nabla_XY)$$
is satisfied. In particular, written in local coordinate,
$$0 = \nabla_{\partial_i} (dx^j (\partial_k)) = \nabla_{\partial_i} dx^j (\partial_k) + dx^j (\nabla_{\partial_i} \partial_k) = \nabla_{\partial_i} dx^j (\partial_k) + \Gamma_{ik}^j$$
$$\Rightarrow \nabla_{\partial_i} dx^j = -\Gamma_{ik}^j dx^k.$$