TITLE: Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
QUESTION [6 upvotes]: Given a smooth manifold $M$, there is a vector bundle over $M$, denoted $\tau M$, known as the second-order tangent bundle. The fiber $\tau_mM$ at $m\in M$ is the collection of linear operators $A_m:C^\infty(M)\rightarrow\mathbb{R}$ that satisfy
$$ A_m(f^3)=3f(m)A_m(f^2)-3f^2(m)A_m(f) $$
for each $f\in C^\infty(M)$. It isn't too hard to show that each section of $\tau M$ is locally of the form
$$A_m=a^i(m)\partial_i+b^{ij}(m)\partial^2_{ij},$$
where $a^i$ and $b^{ij}$ are smooth functions, $b^{ij}=b^{ji}$, $\partial_i=\frac{\partial}{\partial x^i}$, $\partial^2_{ij}=\frac{\partial^2}{\partial x^i\partial x^j}$, and I'm using the Einstein summation convention.
For each $m\in M$ There is a short exact sequence
$$T_mM\rightarrow \tau_m M\rightarrow T_mM\odot T_mM$$
where $\odot$ denotes the symmetric tensor product. The first arrow is the inclusion map that sends a vector at $m$ to its corresponding directional derivative operator $C^\infty(M)\rightarrow\mathbb{R}$. The second arrow is given by $A_m\mapsto \hat{A}_m$, where $\hat{A}_m$ is defined by the formula
$$ \hat{A}_m(\mathbf{d}_mf,\mathbf{d}_mg)=A_m(fg)-f(m) A_m(g)-g(m) A_m(f), $$
and $f,g\in C^\infty(M)$. 
My question: One way to split the sequence, and thereby identify $\tau M$ with $TM\oplus(TM\odot TM)$, is to assign to each $m\in M$ a subspace $S_m\subset \tau_mM$ that is complementary to $T_mM\subset \tau_m M$. Is there a name for such an assignment?
I'd just like to know if there is an established name so that I can more easily search for what people already know about such things.

REPLY [6 votes]: Given a connection on the tangent bundle, you can define the second covariant derivative $\nabla^2f$ by
$$ \nabla^2f[X, Y] := \partial_X\partial_Y f - \partial_{\nabla_X Y} f.$$
Then $\nabla^2f$ is a symmetric tensor provided that $\nabla$ was torsion-free, which Robert pointed out but I missed in the first moment.
(This is because
$$\nabla^2f[Y, X] = \partial_X\partial_Y f + \partial_{[Y, X]} f - \partial_{\nabla_Y, X} = \partial_X\partial_Y f  - \partial_{\nabla_X, Y}+ T(X, Y),$$
where $T$ is the torsion tensor of nabla.)
Now given such a torsion-free connection, you can associate to an element polynomial $p \in TM \odot TM$ the operator 
$$Pf = \langle p, \nabla^2 f\rangle.$$
This gives a splitting of your sequence, since the principal symbol of $P$ will be $p$ again.
Conversely, if $S: TM \odot TM \longrightarrow \tau M$ is such a splitting, set $\Gamma$ to be the projection onto $TM$ along the image of $S$ (i.e. $\Gamma = \iota^{-1}(\mathrm{id} - S \circ \pi)$ with $\pi:\tau M \longrightarrow TM \odot TM$ the projection). Then 
$$\nabla_X Y := \Gamma(\partial_X \partial_Y)$$
is a torsion-free connection.