Relationship between second derivative of a function on a manifold and its covariant Hessian I'm interested in how the covariant Hessian $\nabla^2 f :TM \oplus TM \to \mathbb{R}$ of $f$ and the second derivative of $f$, $d(df):TTM \to \mathbb{R}$ are related once we view $TTM = TM \oplus TM$ by the splitting induced by $\nabla$ I describe below. I became interested in this after looking into $C^2(M)$ as a Banach space. The definition of the norm always uses the covariant Hessian $\nabla^2f$ in place for the second derivative. This make sense in that it's easier to define the norm, but it seems strange to me that we need a connection to define the space whereas $C^1(M)$ can be defined with just $f$ and $df$ (which just needs the smooth structure of M). I suppose you need a metric to then define the norm of $df$, which would then give a connection, but still I'm still interested in the question on its own now. 
Here are the details:
Let $f:M \to \mathbb{R}$ be a smooth function on a manifold $M$. We can consider the derivative of $f$, $df:TM \to \mathbb{R}$ by $df(p,v) = df_p(v)$. Now, considering $df$ as a map between the smooth manifolds $TM$ and $\mathbb{R}$ we can take the derivative again to get a function $d^2f: TTM \to \mathbb{R}$ by $d^2f((p,v),w) = d(df)_{(p,v)}(w)$. Note that $d^2f$ is not the exterior derivative of the one form $df$, which would be zero.
Now, we can split the double tangent bundle $TTM$ as follows: For $\pi:TM \to M$ we have $d\pi : TTM \to TM$ and we can define $VTM = \ker d\pi$. A choice of complementary subspace $HTM$ such that $TTM = VTM \oplus HTM$ is equivalent(kind of) to a connection on $M$. This is because $VTM \cong TM$ by the regular value theorem, for example. So if $\nu: VTM \oplus HTM \to VTM \cong TM$ is the projection then we can define a connection by $\nabla X = \nu \circ dX$ considering $X$ as a funtion $X : M \to TM$. So, for example, $\nabla_Y X = \nu(dX(Y))$. We get a connection on $M$. 
Getting to the point, we have $HTM \cong TM$ and the restriction of $d\pi$ to $HTM$ gives an isomorphism $HTM \cong TM$. Consequently $TTM \cong TM \oplus TM$. So we can think of $d^2f$ as $d^2f : TM \oplus TM \to \mathbb{R}$.
I want to know the relationship between this $d^2f:TM \oplus TM \to \mathbb{R}$ and the covariant Hessian $\nabla^2 f: TM \oplus TM \to \mathbb{R}$. I imagine the differ somehow by something to do with the curvature of $\nabla$. All of my attemps at this have been long calulations that lead nowhere so I'm ommiting what I've tried. Thanks.
 A: Let us start by discussing the (abuse of) notation in the post. The Hessian $\nabla^2f$ is a $(0,2)$-tensor on $M$. In other words, it is a vector bundle morphism $TM\otimes TM\to\mathbb{R}$, or equivalently, a bilinear map $TM\oplus TM\to\mathbb{R}$. Of course, $TM$ here is thought of as a vector bundle over $M$.
On the other hand, the isomorphisms $VTM\cong TM$ and $HTM\cong TM$ only make sense when $TM$ denotes the pullback of $TM$ to $TM$. Explicitly, if $\pi:TM\to M$ denotes the projection, then $\pi^*TM$ is a vector bundle over $TM$, and the above two isomorphisms only hold when $TM$ is replaced by $\pi^* TM$. Hence, when we say $d(df)$ is a map $TM\oplus TM\to\mathbb{R},$ we are not referring to the same $TM$ as in the Hessian case. In particular, the two maps cannot be the same, as they are not defined on the same spaces.
However, the above does not mean that the two maps are not related to one another. Let $p\in M$, and let $v,w\in T_pM$. Extend $w$ to a vector field $X_w$ on a neighborhood of $p$, such that $\nabla_vX_w|_p=0.$ (Here, $\nabla$ can be any given connection on $M$). By definition, we have $$\nabla^2f(v,w)=v(df(X_w))-df(\nabla_vX_w)=v(df(X_w)).$$ Thinking of $df$ as a map $TM\to\mathbb{R}$, the expression on the right is the derivative of $df$ at the point $(p,w)\in TM$, in the horizontal direction over $v$. Using the notation of the post, this means that $$\nabla^2f(v,w)=d(df)_{(p,w)}(0,v).$$
