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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?

brunoh
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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 $d^k$ and $\nabla^{(k,1)}$ are closely related - the operator $d^k$ is obtained up to combinatorical constants depending on your conventions as the anti-symmetrization of $\nabla^{(k,1)}$. For details, see here.

brunoh
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levap
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    BTW, your formula $\nabla (\omega \otimes X) = \nabla \omega \otimes X + \omega \otimes \nabla X$ is wrong. The full covariant derivative doesn't satisfy a (aesthetically pleasing) Liebnitz rule. The correct formula is $\nabla_Y(\omega \otimes X) = \nabla'_Y \omega \otimes X + \omega \otimes \nabla_Y X$ where $\nabla'$ is the induced connection on $(T^{*}M)^{\otimes k}$. – levap Feb 17 '17 at 22:34
  • Concerning your comment, and using your own notations in your answer, it means I should have written something like $\nabla^{(k,1)}(\omega\otimes X)=\nabla^{(k,0)}\omega\otimes X+\omega\otimes\nabla^{(0,1)}X$ right? That is why I do not consider my formula wrong but just abusing notation. – brunoh Feb 17 '17 at 22:52
  • Your answer is crystal clear. Thank you very much for it, and I especially like the use of food semantics :=). I understand now that I was also confused by the use of similar notations and ... denominations! But the objects are actually quite different. – brunoh Feb 17 '17 at 22:55
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    @brunoh: That is the "abuse of notation" part but the wrong part is that the Liebnitz rule doesn't hold if you don't "plug in" a vector field (the order of the arguments is wrong). If we take your formula and plug in $X_0,\dots,X_k$ we get $(\nabla \omega \otimes X)(X_0,\dots,X_k) = (\nabla \omega \otimes X)(X_0,\dots,X_k) + (\omega \otimes \nabla X)(X_0,\dots,X_k) = (\nabla X_0 \omega)(X_1,\dots,X_k) X + \omega(X_0,\dots,X_{k-1}) \nabla_{X_k} X$ but this is wrong. The direction of differentiation in both parts of the right hand side equation should be the same. This is why you don't have – levap Feb 17 '17 at 23:00
  • a Liebitz formula for $\nabla (\omega \otimes X)$ (without permuting the arguments), only for $\nabla_Y (\omega \otimes X)$. This is another aspect in which the formula for $d_{\nabla}$ and the formula for $\nabla^{(k,1)}$ is different. – levap Feb 17 '17 at 23:01
  • (And yeah, it is very helpful in the beginning not to make the abuse of notation in denoting all the various induced connections and operators by the same symbol $\nabla$. I understand the necessity of it (otherwise, the notation becomes horrendous) but, at least in the beginning, this point should be more emphasized than it is in most textbooks). – levap Feb 17 '17 at 23:10
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    Nice point. +1 It was clear in my mind but contradicted by the notations. I have added confusion upon confusion. I have edited my question to follow your wise corrections. Thank you again. – brunoh Feb 17 '17 at 23:11
  • @brunoh: You're welcome! – levap Feb 17 '17 at 23:12
  • It is really confusing in most textbooks and in Wikipedia, since the same notations ($\nabla_Y$) is used and sometimes the same denominations! If I understand well now, there are two different extensions of the same connection on a tangent bundle to two different type of bundles, one graded-commutative and the other not. That is probably why one construction can be recovered from the other after anti-symetrization. – brunoh Feb 17 '17 at 23:31
  • Let us continue this discussion in chat. – levap Feb 18 '17 at 01:03