Apparently one identifies the configuration space in physics often with a manifold $M$. The tangent bundle $TM$ is then the space of all possible positions and velocities.


Furthermore, many sources seem to claim that $T^\*M$ can be regarded as the phase space, where $(q,p) \in T^\*M$ satisfies by definition that $p \in T\_q^\*M.$


Again by definition this means that $p:=\partial\_2L$ takes velocities as arguments and is linear(!) in them. Unfortunately, I don't see from the definition of the momentum by the Lagrangian why this should be a linear functional. So something is confusing me here.



Momentum a cotangent vector
The Lagrangian is a function on the tangent bundle $L:TM\rightarrow\mathbb{R}$. Given a point $q\in M$ and a Lagrangian, we can define a function $L\_q:T\_qM\rightarrow \mathbb{R}$ using the simple formula $L\_q(v\_q)=L(v\_q)$, where $v\_q\in T\_qM$ is a tangent vector at $q\in M$. Notice that $L\_q$ is a mapping between the vector spaces $T\_qM$ and $\mathbb{R}$. We may therefore consider the Frechet derivative $DL\_q:T\_qM\rightarrow L(T\_qM,\mathbb{R})$, where $L(T\_qM,\mathbb{R})$ is the space of linear maps between $T\_qM$ and $\mathbb{R}$. Notice that $DL\_q$ is essentially the partial derivative of $L$ in the velocity direction. 


Here is how the cotangent bundle comes in. The set $L(T\_qM,\mathbb{R})$ is precisely the cotangent space at $q$, i.e. $L(T\_qM,\mathbb{R})=T^\*\_qM$. Therefore the Frechet derivative of $L\_q$ is a map of the form $DL\_q:T\_qM\rightarrow T^\*\_qM$. This observation can be used to construct a mapping of $TM$ into $T^\*M$ given by $v\_q\mapsto (DL\_q)(v\_q)\equiv p\_q\in T^\*\_qM$, which is known as the Legendre transform.


Does this help?


