Riemannian metric of the tangent bundle I'm trying to solve the following problem (from do Carmo's Riemannian Geometry). particularly I'm having trouble proving that the inner product defined is bilinear. 
Problem. It is possible to define a Riemannian metric in the tangent bundle $TM$ of a Riemannian manifold $M$ in the following manner. Let $(p,v)\in TM$ and $V,W$ be tangent vectors to $TM$ at $(p,v).$ Choose curves in $TM$
$$\begin{align*}& \alpha: t \rightarrow (p(t), v(t))\\
& \beta: t \rightarrow (q(s), w(s))\end{align*}$$
with $p(0)=q(0)=p$, $v(0)=w(0)=v$ and $V=\alpha'(0)$, $W=\beta'(0)$.
Define an inner product on $TM$ by
$$\langle V,W \rangle_{(p,v)} = \left\langle d\pi(V), d\pi(W)  \right\rangle_{p} + \left\langle \frac{Dv}{dt}(0) , \frac{Dw}{ds}(0) \right \rangle_{p},$$
where $\pi:TM \rightarrow M$. Prove this is a well-defined Riemannian metric on $TM$.
 A: 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. 

