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Here is a little bit of curiousity that's been itching me, let's hope it doesn't get me killed, meow.

Definition: Let $M$ be a smooth manifold. A connection $\nabla$ on $TM$ is called associative if $\forall X,Y,Z \in \mathfrak{X}(M)$: $\nabla_{\nabla_X Y}Z = \nabla_X\nabla_Y Z$.

Using the definitions of the tensors $R$ and $T$, it is immediate that a necessary condition for $\nabla$ to be 'associative' is $$ R(X,Y)Z = \nabla_{T(X,Y)}Z $$

The motivation for this definition is the single curious observation that if $\nabla$ is 'associative', then $\mathfrak{X}(M)$ becomes an associative $\mathbb{R}$-algebra via the multiplication $X\ast Y:=\nabla_X Y$, although at the moment I am not really sure what this is good for... But one step at a time!

I've added the complex geometry tag, because I am actually more interested in the complex setting (i.e. associative $\mathbb{C}$-algebras). In other words, I look at such a pair $(M,\nabla)$, if existing, as another source for associative (complex, possibly Banach?) algebras and would like to know how sound the differential-geometric side of it is.

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No, such a connection cannot exist: Consider a function which vanish to first order at a point $p\in M$, i.e., $f(p)=0$ and $d_pf=0$, but assume that there are vector fields $X,Y$ with $(X\cdot (Y\cdot f))(p)\neq0$.Such a function clearly exist. Moreover, let $\tilde Z$ be a vector field which does not vanish at $p.$ Then, the equality does not hold for the three vector fields $X,Y,Z=f\tilde Z.$

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    $\begingroup$ Does $X\cdot Y$ denote composition as differential operators or the product in the OP? $\endgroup$ – Qfwfq Feb 8 '18 at 19:47
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    $\begingroup$ And what is $Y\cdot f$? $\endgroup$ – მამუკა ჯიბლაძე Feb 8 '18 at 21:01
  • $\begingroup$ @მამუკა ჯიბლაძე: I presume it's $Y$ acting on $f$ as a derivation (but let's Sebastian confirm or not) $\endgroup$ – Qfwfq Feb 9 '18 at 2:09
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    $\begingroup$ I mean $Y\cdot f=df(Y)$ and $X\cdot Y\cdot f=X\cdot(Y\cdot f)$. $\endgroup$ – Sebastian Feb 9 '18 at 5:59
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    $\begingroup$ It is believable but not entirely obvious: since $f(p)=d_pf=0$, then I believe also $df(Y)=0$ at $p$ for any $Y$. I understand this does not imply there is no $X$ with $X\cdot(Y\cdot f)$ nonzero at $p$ but still - to have an explicit example would be better I believe $\endgroup$ – მამუკა ჯიბლაძე Feb 11 '18 at 14:21
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Consider the associativity for $fZ$ for any $f\in \mathcal{C}^\infty(M)$. Then on the right hand side of the associativty condition you are differentiating $f$ twice, while on the left hand side you are differentiating only once. Therefore such a connection cannot exist.

(This is pretty much what Sebastian is saying in his answer. I just thought it might be worthwhile to get rid of any notation and assumptions and explain the heart of the matter.)

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