Is kinetic energy a positive definite quadratic form? Recall (Arnold, Mathematical methods of classical mechanis, 4.19, B)

Definition. Let $M$ be a riemannian manifold. The quadratic form on each tangent space
  $$ T = \frac{1}{2} \langle v, v \rangle, \qquad v \in TM_x$$
  is called kinetic energy.

In a previous page, 

Definition. A differentiable manifold with a fixed positive definite quadratic form $\langle \xi, \xi \rangle$ on every tangent space $TM_x$ is called a riemannian manifold.

It is unclear whether $T$ must be positive definite, i.e. whether the quadratic forms are the same in the two definition. It sounds like an abuse of notation, since often one finds 

$T$ is a quadratic form in $\dot q$, $T = \sum_{ij} a_{ij}(q,t)  \dot q_i \dot q_j$,

and no further specifications about the nature of the quadratic form. Moreover, I can't see any other issue with non positive definite kinetic energy, except for physical interpretation. (e.g., properties of kinetic energy with respect to Legendre transform should only rely on its being a quadratic form.)  On the other hand, I'm not aware of an example of a system with non positive definite kinetic energy. 
So, must $T$ be positive definite? Are there counter-examples?
Remark. I'm concerned with mathematical consequences of this fact, mostly with respect to possibility of construction of hamiltonian theory. Physical interpretation is not a priority at this stage.
 A: It is an interesting fact, that a big part of the mathematical theory of Lagrangian mechanics can be formulated without further specifying the form of your Lagrangian. Especially Noethers theorem does not make use of any specific form of the Lagrangian. Therefore, when sticking only to Lagrangian mechanics, you can choose the Lagrangian quite arbitrarily (of course it should be 2x differentiable), although some results will get trivial or weird. Especially, when applying Noethers Theorem, you of course will not get your usual Noether charges.
$T=\sum_{ij} a_{ij} \dot{q}_i\dot{q}_j$ is your positive definite quadratic form in local coordinates on your manifold, where $a_{ij}$ is some metric, e.g. in the euclidean case you get $a_{ij}=\delta_{ij}$. I dont see where this is abuse of notation.
As soon as it comes to Hamilton Mechanics and Legendre transform, you need the condition det$\frac{\partial^2 L}{\partial \dot{q}_i\partial \dot{q}_j}\neq0$ to make sure your Legendre Transform is unique. This is usually ensured by your kinetic energy being a positive definite quadratic form, as long as your potential is not too weird (conservative is always sufficient in this case).
