# Why don't we consider "Hamiltonian vector field" on tangent bundle?

I have wondered the following for a while. Why do we consider symplectic geometry and Hamiltonian vector field on cotangent bundle but not tangent bundle? To make it clear why I have thought of finding a "Hamiltonian vector field" on tangent bundle, consider the following case:

Suppose we have a Riemannian manifold $(M,g)$ with a smooth function $V:M\to\Bbb R$ representing a potential field, and $\frac{1}{2}g(v,v)$ representing kinetic energy. Choosing an initial value $(p,v)$ on the tangent bundle $TM$, we should be able to find a path $\gamma$ on $TM$ that obeys some form of Newton's law and $\gamma(0)=(p,v)$, and there should be a vector field $X$ on $TM$ of which each of those paths described above is a flow line.

Why is that we do not consider finding a vector field induced by a function $f$ on $TM$, while we do for any smooth function $f$ on $T^*M$? Is it due to some difference in the natural structures and operations on tangent bundle and cotangent bundle? For instance, is it we always have a symplectic form on cotangent bundle, but there is no analog for tangent bundle? Is it we can do exterior differentiation on cotangent bundle naturally, while we cannot differentiate a function on tangent bundle to obtain a vector field unless we have a Riemannian metric?

• The cotangent bundle is “naturally” a symplectic manifold, while the tangent bundle only has a symplectic structure once you use a metric to identify it with the cotangent bundle. Dec 11 '17 at 15:47

In regards to your first question, you can find a vector field on $TM$ whose flow obeys Newtons law. Given $L\in C^\infty(M)$ you can lift it to $TM$. That is take $\pi^\ast L$, where $\pi:TM\to M$ is the projection map. More generally, you can start with the assumption that $L\in C^\infty(TM)$. You can then take the Legendre transform of $L$ which is a map $\Phi_L:TM\to T^\ast M$.
Letting $\omega_0=-d\theta_0$ denote the canonical $2$ and $1$-forms on $T^\ast M$, we can then let $\theta=\Phi_L^\ast\theta_0$ and $\omega=-d\theta$. This gives a $2$-form on $TM$ called the Lagrange $2$-form. Under certain conditions on $L$, the Lagrange $2$-form is symplectic.
Moreover, from $L$ we can define a new function $A\in C^\infty(TM)$ by $A(V_p)=\Phi_L(V_p)\cdot V_p$. This is referred to as the action. From $A$ we can define $E=A-L$, which is called the energy function. A Lagranginan vector field is then defined to be a vector field $X\in\Gamma(T(TM))$ satisfying $dE=\omega(X,\cdot)$. The flow of this vector field is governed exactly by the Euler-Lagrange equations, which reduces to Newtons law when $L=K-V$, the difference of the kinetic and potential energy.
As you mentioned in the question, $TM$ is not always symplectic. In order for the above to work, certain conditions need to be placed on $L$ and on the vector field $X_E$. Check out Foundations of Mechanics $2$nd addition by Abraham and Marsden for more information. Especially chapter 3.
Further to JonHerman's answer, one can define a differentiable manifold $M$, such that an autonomous second order system of ordinary differential equations on $M$ is equivalent to a certain vector field $\Gamma$ called a second order vector field on the tangent manifold $TM$.
This system of equations is Hamiltonian, and $\Gamma$ is a second order Hamiltonian vector field if there exists a Poisson structure $P$ on $TM$ and a function $f \in C^{\infty}(TM)$ such that $\Gamma$ is the Hamiltonian vector field of $f$ with respect to $P$. If this happens, the local expressions of the equations along a symplectic leaf of P are then the classical equations of Hamilton.