Linear connection and covariant derivative: help needed to clear up confusion in extension of definitions

Question

Let $\nabla$ be a linear connection defined on the tangent bundle of a manifold $M$. We have, with $X(M)$ being the global sections module of $TM$, $$\nabla: X(M)\to \Omega^1(M)\otimes X(M)$$ We can extend it as a derivation of degree 1 on the $\Omega^*(M)\otimes X(M)$ complex by the formula $$\nabla(\omega\otimes X)=d\omega\otimes X+(-1)^{r}\omega\wedge\nabla(X)$$ where $\omega\in\Omega^r(M)$ and $X\in X(M)$

On the other hand, a covariant derivative defined on $TM$ extends in a unique way to the duals of vector fields (i.e., covector fields), and to arbitrary tensor fields, that ensures compatibility with the tensor product and trace operations (tensor contraction). For instance, for $\omega\in \Omega^r(M)$ and $X, Y\in X(M)$, we get $$\nabla_Y(\omega\otimes X)=\nabla_Y(\omega)\otimes X+\omega\otimes\nabla_Y(X)$$

I am confused here because I cannot reconcile the two formulas. What am I misunderstanding?

Edit: following chat discussion with @levat and his clear explanations, a quick last question: considering a connection on the tangent bundle $TM$ $$∇:Γ(TM)→Ω^1(M)⊗Γ(TM)$$ if we want to extend it on the forms with value in any type of tensorial bundle above $M$ we can do it step by step from $Ω^p(M)⊗Γ(T^{q,r}M)$ to either $Ω^{p+1}(M)⊗Γ(T^{q,r}M)$ or to $Ω^p(M)⊗Γ(T^{q+1,r}M)$ each time using the right "derivative" that preserves the respective algebra product.

Then my question is: do these two derivatives (the covariant one and the full covariant one) commute to reach $Ω^{p+1}(M)⊗Γ(T^{q+1,r}M)$?

Answer 1

Your degree one extension of $\nabla$ from $\mathcal{X}(M)$ to $$\Omega^{*}(M;TM) := \Omega^{*}(M) \otimes \mathcal{X}(M) = \Gamma(\Lambda^{*}(TM) \otimes TM)$$ is usually denoted by $d_{\nabla} \colon \Omega^{*}(M;TM) \rightarrow \Omega^{*+1}(M;TM)$ and called the covariant exterior derivative of $TM$-valued differential forms on $M$. Explicitly, $d_{\nabla}$ eats a section of $\Lambda^k(T^{*}M) \otimes TM$ and returns a section of $\Lambda^{k+1}(T^{*}M) \otimes TM$. From this description you can see that $$d_{\nabla}^k \colon \Gamma(\Lambda^k(T^{*}M) \otimes TM) \rightarrow \Gamma(\Lambda^{k+1}(T^{*}M) \otimes TM)$$ doesn't look like a connection on some bundle. Stated differently, the operator $d_{\nabla}^k$ eats a $(k,1)$ alternating tensor on $M$ and returns a $(k+1,1)$ alternating tensor on $M$.

Now,you can use your favorite identification (or even definition, depending on how you set up things) and think of elements of $\Omega^{k}(M;TM)$ as special sections of $(T^{*}M)^{\otimes k} \otimes TM$ (general $(k,1)$-tensors). Like you noted, the connection $\nabla$ induces a connection on various associated bundles so we get a connection $\nabla^{(k,1)}$ on $(T^{*}M)^{\otimes k} \otimes TM$ that allows us to differentiate $(k,1)$ tensors on $M$. The connection $$\nabla^{(k,1)} \colon \Gamma((T^{*}M)^{\otimes k} \otimes TM) \rightarrow \Gamma(T^{*}(M) \otimes ((T^{*}M)^{\otimes k} \otimes TM)) = \Gamma((T^{*}M)^{\otimes k + 1} \otimes TM)$$

(also known as the "full covariant derivative") eats a $(k,1)$ tensor on $M$ and returns a $(k+1,1)$ tensor on $M$. However, if you feed it with a $(k,1)$-alternating tensor $\omega \otimes X$, there is no reason $\nabla^{(k,1)}(\omega \otimes X)$ will be a $(k+1,1)$-alternating tensor.

Thus, $d_{\nabla}^k$ and $\nabla^{(k,1)}$ are two different operators. In fact, if $\nabla$ is symmetric then $d_{\nabla}^k$ and $\nabla^{(k,1)}$ are closely related - the operator $d_{\nabla}^k$ is obtained up to combinatorical constants depending on your conventions as the anti-symmetrization of $\nabla^{(k,1)}$. For details, see here.