The following is an excerpt from Sharpe's *Differential Geometry - Cartan's Generalization of Klein's Erlangen Program*.



> 
> Now we come to the question of higher derivatives. As usual in modern 
>  differential geometry, we shall be concerned only with the skew-symmetric 
>  part of the higher derivatives. In essence, what we shall be doing is taking 
>  the partial derivatives with respect to the base (i.e., manifold) variables 
>  and skew-symmetrizing the result, thus forgetting about the part of the 
>  higher derivatives that vanish under this procedure. However, this will not 
>  be made explicit in our treatment. The part of the higher derivative that 
>  disappears has not been studied much in differential geometry since Elie 
>  Cartan showed how useful it is to consider only the skew-symmetric part, 
>  that is, the exterior derivative. The old masters did use the symmetric part...
> 
> 
> 


"Partial derivatives with respect to the base" must be the covariant derivative of the connection. If I correctly understand what's written in [this answer](https://math.stackexchange.com/a/1980156/223002), then we have for any **torsion free** connection on a manifold the equality $ \mathrm dw=\operatorname{Alt}(\nabla w)$.


This already seems rather remarkable since the exterior derivative is intrinsic.


**Question 1.** What is the geometric meaning of the above fact for torsion-free connections? How does taking anti-symmetrization "forget" the structure of a horizontal bundle (connection)?


**Question 2.** Still for a torsion-free connection, what is the geometric meaning of "the part of the higher derivative that vanish under this procedure" of anti-symmetrization (i.e the symmetric part)?


**Question 3.** What's the conceptual picture for an arbitrary connection on a manifold (possibly with torsion)? Is the anti-symmetrization still the exterior derivative? What is the symmetric part?


*Remark on Q3.* I vaguely recall being told that torsion measures the failure of the fundamental theorem of calculus, and also that in torsion-free connections the parallel transport commutator is given by the Lie bracket (again, the latter is intrinsic).



Symmetric and anti-symmetric parts of the covariant derivative of a connection
The meaning of higher-order derivatives in differential geometry is better understood through *jet bundles*. The covariant derivative $\nabla\phi$ of (say) a smooth section $\phi:M\rightarrow E$ of a vector bundle $\pi:E\rightarrow M$, being a derivative, can be seen as a section of the so-called *first-order jet bundle* $\pi^1\_0:J^1 E\cong\mathrm{End}(TM,E)\rightarrow E$ of $E$. Likewise, derivatives of higher order (say, $k\in\mathbb{N}$) of smooth sections of $\pi$ are described in a coordinate-free fashion by the so-called *jet bundle of order* $k$ $\pi^k\_0:J^k E\rightarrow E$. This hierarchy of (vector) bundles over $E$ is endowed with vector bundle projections $\pi^k\_l:J^kE\rightarrow J^l E$, $k>l$ such that $\pi^l\_m\circ\pi^k\_l=\pi^k\_m$ for all integers $k>l>m\geq 0$. Roughly speaking, one may identify elements of $J^k E$ locally with Taylor polynomials of order $k$, and the projections $\pi^k\_l$ discard derivatives of order above $l$. 


The key point is that the leading-order derivatives are, in a sense, always symmetric: the kernel of $\pi^k\_{k-1}$ is a bundle whose total space is canonically isomorphic to $S^kT^\*M\otimes E$ = the tensor product of the *symmetrized* $k$-fold tensor power of $T^\*M$ with $E$. Particularly, any reference to a specific covariant derivative when globally defining highest-order derivatives disappears - more precisely, all yield the same result (differences only affect lower-order terms). One can exploit this fact to write smooth sections of $J^k E$ in terms of symmetrized iterated covariant derivatives (see e.g. [this MO question of mine](https://mathoverflow.net/questions/240329/jets-of-sections-of-vector-bundles-expressed-by-symmetrized-iterated-covariant-d)).


This also means that any partial antisymmetrization will always discard highest-order derivatives - this holds irrespectively of $\nabla$ being torsion-free or not (for general vector bundles, this concept usually has no meaning). Hence, if you *totally* antisymmetrize (as done with $k$-forms), only derivatives of *at most first order* survive. This is of course related to the nilpotency of the exterior derivative. 


(**EDIT 2**) To illustrate this, let us consider the concept of iterated covariant derivatives, mentioned in the MO question linked in the second paragraph. To that end, we combine a covariant derivative on the vector bundle $\pi$ with a covariant derivative on $TM$ in order to define covariant derivatives on the tensor bundles $\otimes^k T^\*M \otimes E\rightarrow M$ for all $k\in\mathbb{N}$ by means of Leibniz's rule. We denote all these covariant derivatives collectively by $\nabla$. Once this is done, the iterated covariant derivatives $\nabla^k\phi$ of all orders $k\in\mathbb{N}$ of smooth sections $\phi$ of $\pi$ are defined recursively as follows: $$\nabla^0\phi=\phi\ ,\,\nabla^1\phi=\nabla\phi\ ,\,\nabla^{k+1}\varphi=\nabla(\nabla^k\phi)\ ,$$ so that $\nabla^k\phi$ is a smooth section of $\otimes^k T^\*M \otimes E\rightarrow M$ for all $k\in\mathbb{N}$. Particularly, for $k=2$, one can write $$\nabla^2\phi(X,Y)=\nabla\_X(\nabla\_Y\phi)-\nabla\_{\nabla\_X Y}\phi\ .$$ If we antisymmetrize $\nabla^2\phi$, we obtain $$\nabla^2\phi(X,Y)-\nabla^2\phi(Y,X)=\nabla\_X(\nabla\_Y\phi)-\nabla\_Y(\nabla\_X\phi)-\nabla\_{\nabla\_X Y-\nabla\_Y X}\phi$$ $$=\text{Riem}\_\nabla(X,Y)\phi-\nabla\_{T\_\nabla(X,Y)}\phi\ ,$$ where $$\text{Riem}\_\nabla(X,Y)\phi=\nabla\_X(\nabla\_Y\phi)-\nabla\_Y(\nabla\_X\phi)-\nabla\_{[X,Y]}\phi$$ is the Riemann curvature tensor of the covariant derivative *on* $\pi$ and $$T\_\nabla(X,Y)=\nabla\_X Y-\nabla\_Y X-[X,Y]$$ is the torsion tensor of the covariant derivative *on* $TM$. The above expression clearly depends on derivatives of $\phi$ of at most first order. Notice that the above antisymmetrization is closely related to the operation of *exterior covariant differentiation* $d^\nabla$ of $E$-valued differential forms, as mentioned in Jesper Göransson's answer: since $\phi$ is a $E$-valued 0-form and therefore $d^\nabla\phi=\nabla\phi$, we have that $$d^\nabla(d^\nabla\phi)(X,Y)=\nabla\_X(\nabla\_Y\phi)-\nabla\_Y(\nabla\_X\phi)-\nabla\_{[X,Y]}\varphi=\text{Riem}\_\nabla(X,Y)\varphi$$ $$=\nabla^2\phi(X,Y)-\nabla^2\phi(Y,X)+\nabla\_{T\_\nabla(X,Y)}\varphi\ .$$ Particularly, if $E=M\times\mathbb{R}$ (i.e. sections of $\pi$ are just real-valued functions on $M$) and $\nabla$ is just ordinary differentiation of functions on $M$ (so that $\text{Riem}\_\nabla=0$ and $T\_\nabla=0$), we thus recover the nilpotency of the (ordinary) exterior derivative.


A similar phenomenon happens if we antisymmetrize the vector field entries of the tensor field $\nabla^k\phi$ for $k>2$ - any antisymmetrization with respect to (say) $j$ entries leaves us with an expression which depends on $\nabla^i\phi$ for $i$ only up to $k-(j-1)=k-j+1$ for all $2\leq j\leq k$.
If $j<k$, this is what is called "partial antisymmetrization" above.


Returning to covariant derivatives $\nabla$ *on* $M$ (that is, on $TM$), the equivalence of the vanishing of the torsion tensor $T\_\nabla(X,Y)$ of the connection $\nabla$ (which, by the way, simply states that $\nabla\_X Y-\nabla\_Y X$ is the Lie bracket of $X$ and $Y$ = Lie derivative of $Y$ along $X$) to the antisymmetric part of $\nabla\omega$ being equal to $d\omega$ for all 1-forms $\omega$, mentioned by Deane Yang in his comment to the OP's question, comes from Henri Cartan's formula relating Lie differentiation and exterior differentiation: $\mathscr{L}\_{\!X}\omega=d(i\_X\omega)+i\_X(d\omega)$. 


(**EDIT 1**) To see this, suppose that the torsion tensor of $\nabla$ vanishes, i.e. $$\nabla\_X Y-\nabla\_Y X=[X,Y]=\mathscr{L}\_{\!X}Y\ .$$ Recall that the action of $\nabla$ on an 1-form $\omega$ is given by Leibniz's rule$$(\nabla\_X\omega)(Y)=X(\omega(Y))-\omega(\nabla\_X Y)$$ and so is the Lie derivative of $\omega$: $$(\mathscr{L}\_{\!X}\omega)(Y)=X(\omega(Y))-\omega(\mathscr{L}\_{\!X}Y)=X(\omega(Y))-\omega([X,Y])\ .$$ The antisymmetric part of $\nabla\omega$ is then given by $$(\nabla\_X\omega)(Y)-(\nabla\_Y\omega)(X)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\ .$$
Notice that I used the vanishing of the torsion of $\nabla$ in the rhs. On the other hand, Henri Cartan's formula tells us that $$d\omega(X,Y)=(\mathscr{L}\_{\!X}\omega)(Y)-Y(\omega(X))=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\ ,$$ hence indeed $(\nabla\_X\omega)(Y)-(\nabla\_Y\omega)(X)=d\omega(X,Y)$. Conversely, if one assumes this last identity and backtracks the above calculations, one obtains that $\omega(T\_\nabla(X,Y))=0$ for all 1-forms $\omega$, hence $T\_\nabla(X,Y)=0$ for all vector fields $X,Y$.


More generally, torsion introduces a sort of "internal rotation" on the frame bundle along parallel transport which cannot be accounted for by pushing forward by the diffeomorphism flow integrating the vector field along which we are differentiating (the latter is the effect measured by Lie derivatives) - hence the failure of the fundamental theorem of calculus. Interestingly, given any covariant derivative $\nabla$ on $M$, there is a unique covariant derivative $\nabla'$ on $M$ which is *torsion-free* and *has the same geodesics as* $\nabla$, which means that this "internal rotation" effect can be cancelled by suitably redefining $\nabla$. This is not to say that torsion should be always discarded - for instance, when dealing with foliations, there is a natural connection (the so-called *Bott connection*) which is only defined along vector fields tangent to the foliation - that is, a so-called *partial connection*. This connection is flat but has nonvanishing torsion, which therefore is important in dealing with the holonomy of the foliation.


