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



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



Do "associative" connections exist / arise naturally in some context?
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 $X,Y,Z=f\tilde Z.$


