The two terms you wrote down are roughly the horizontal and vertical parts of the metric. Roughly speaking, the first part gives the metric for the first factor $T_pM$ of $T_{p,v}TM$. The second part gives the metric for the second factor $T_vT_pM$.
By the definition of the projection operator $\pi$ and the definition of the covariant derivative on $(M,\langle\cdot\rangle)$, it is clear that the expression you wrote down is coordinate independent. There are several things to check to make sure that it is a metric
- It is positive definite
- It is bilinear
- It is tensorial
Since we already have coordinate independence, it is most convenient to work over a fixed coordinate system.
Let $\{x^1,\ldots,x^n\}$ be a system of coordinates for $M$; this can be extended to a system of local coordinates $\{x^1,\ldots,x^n;y^1,\ldots,y^n\}$ for $TM$ where $(x,y)$ corresponds to the point $(p,v)\in TM$ with $p$ the point in $M$ specified by $x$, and $v\in T_pM$ given by $\sum y^i\partial/\partial x^i$.
At a fixed point $(p,v)$, an element of $T_{p,v}TM$ can be then described by
$$ V = \sum \xi^i \frac{\partial}{\partial x^i} + \sum \zeta^i \frac{\partial}{\partial y^i}$$
with the projection
$$ d\pi(V) = \sum \xi^i \frac{\partial}{\partial x^i} $$
Express the curve $\alpha(t) = (p,v)(t)$ in this coordinates we have that the condition $\alpha'(0) = V$ is simply the statement that $\frac{d}{dt}p^i(0) = \xi^i$ and $\frac{d}{dt}v^i(0) = \zeta^i$.
With this we can compute
$$ \frac{D}{dt}v^i(0) = \frac{d}{dt} v^i(0) + \Gamma^i_{jk}\left(\frac{d}{dt}p^j(0)\right)\left(v^k(0)\right) $$
using the definition, and where $\Gamma$ is the Christoffel symbol of the Riemannian metric on $M$. This we immediately see to be
$$ \frac{D}{dt}v^i(0) = \zeta^i + \Gamma^i_{jk}v^k(0)\xi^j $$
which in fact is a linear map from $T_{p,v}TM$ to $T_pM$.
Now the three properties are easily checked:
- It is tensorial because the expressions are completely independent of which curve $\alpha$ is chosen, as long as $\alpha'(0)= V$.
- It is bilinear because $T_{p,v}TM\ni V\mapsto (d\pi(V),\frac{D}{dt}v(0))\in T_pM \oplus T_pM$ is linear, and the Riemannian metric on $M$ is bilinear
- Since the Riemannian metric induced on $T_pM\oplus T_pM$ is positive definite, to prove that the new object is also positive definite, it suffices to check that the map $V\mapsto (d\pi(V),\frac{D}{dt}v(0))$ is injective. But this is true by direct inspection.
The Riemannian metric provides a natural identification of the cotangent space with the tangent space. Therefore the cotangent space is also naturally equipped with a Euclidean (or if you prefer Hilbert-space) metric. The simplest way of thinking of the identification is to send a tangent vector $v$ to the covector given by the inner product with $r$, namely $\langle v,\cdot\rangle$.
So they are naturally isomorphic as Hilbert spaces, Banach spaces, and metric spaces.
Expressed in terms of indices, the relation is simply this. If $v_i$ represents a vector and $\alpha^j$ a covector then the relation $v_i=g_{ij}\alpha^j$ (summation over a repeated index as usual) gives a way of passing between a vector and its corresponding covector. Here $g_{ij}$ is the Riemannian metric.
Best Answer
Things are much nicer i.e. more general and canonical than that!
a) Let $\pi:N\to M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0\to T^{vert}(N) \to T(N) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $\pi: N\to M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $\pi^{-1} T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]\stackrel {def}{=}\pi^{-1}(\pi (n))$ of $\pi$.
In other words at a point $n\in N$ the fibre of $T^{vert}(N)$ is $$T^{vert}_n(N)=T_n(N[n]) \quad (**)$$
b) Consider now a vector bundle $\pi:V\to M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V\;$ (yes, vector bundles on a vector bundle!) :
$$ 0\to T^{vert}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (***)$$
In this new set-up you have the interesting identification $ T^{vert}(V)=\pi^{-1}(V)$.
This boils down to the fact that the tangent space at any point $e\in E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0\to \pi^{-1}(V) \to T(V) \stackrel {d\pi}{\to} \pi^{-1}T(M) \to 0 \quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0\to V \to T(V)|M \stackrel {d\pi}{\to} T(M) \to 0 \quad \quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M \cong V \oplus T(M)$$ At a point $(m,0)\in M\subset V$ this gives the decomposition $$ T_{(m,0)}(V) \cong V_m \oplus T_m(M)$$ and for $V=T^*(M)$ this is what you were looking for.