Are Finsler manifolds semi-Riemannian?  A Finsler manifold is defined as a differentiable manifold with a metric defined on so that any well-defined curves of finite arc length is given by a generalized arc length integral of an asymmetric norm over each tangent space defined at a point. This generalizes the Riemannian manifold structure since the norm is no longer required to be induced by an inner product and therefore the Finsler manifold is not necessarily Euclidean in the tangent space structure. 
A collegue of mine and I recently got into an arguement over whether or not Finsler manifolds are semi- or psuedo-Reimannian. I say no-by definition, A semi-Reimannian manifold-like an Loretzian manifold in relativity theory-is still required to have a metric tensor as it's normed structure,which is clearly an inner product. We simply weaken the condition of positive definiteness(i.e. the associated quadratic form of the norm is real valued) to nondegeneracy (i.e. the tangent space is isomorphic with its dual space). Both conditions require the distance structure to be induced by an inner product. 
My colleague's argument is that the key property of semi-Riemannian manifolds is that they admit local signed coordinate structures that allow the distinction of different kinds of tangent spaces on the manifold. Local isomorphisms can be defined-especially in infinite-dimensional extensions of classical relativistic spaces-that make certain Fisler manifolds eqivilent to relativistic models of space-time. 
I honestly don't know enough about the research that's been done on this. Is he right? This seems very bizarre to me, but it may indeed be possible to use specially constructed mappings to convert Finsler spaces to semi-Riemannian ones and vice-versa. I seriously doubt it could be done globally without running into serious topological barriers. 
I'd like the geometers to chime in on this,particularly ones who are well-versed in relativistic geometry: Am I right? Can Finsler manifolds be defined in such a manner as to be true semi-Reimannian manifolds? Can local isomorphisms or differentiomorphisms be defined to interconvert them?  
 A: No, a Finsler metric is in general not semiriemannian. As you and José indicate, a semiriemannian metric is always given by a nondegenerate quadratic form on tangent vectors at each point in the manifold. In other words, if you fix a basis of the tangent space for a given point, then the norm of that vector is given by a homogeneous quadratic polynomial in the coefficients of the vector with respect to the basis.
On the other hand, a Finsler metric is given by a norm function on the tangent space of each point in the manifold. This norm function must be convex, and additional regularity and convexity assumptions are often made. However, there is no requirement that the norm function be given by a quadratic form. It could be given by a higher even degree polynomial in the coefficients of a vector with respect to a basis. But it could be an arbitrary sufficiently smooth sufficiently convex function, too.
One way to think about this is to consider the standard flat models. The standard flat semiriemannian model is just $R^n$ with the metric given by a non-degenerate quadratic form. The standard flat Finsler model is $R^n$ with a (sufficiently smooth and convex) Banach norm, i.e. a finite dimensional Banach space. There are obviously a lot more of the latter than the former.
A: The answer is no. Any smooth manifold admits a Riemannian metric using paracompactness and partitions of unity: in short, a convex sum of positive definite symmetric matrices is positive definite symmetric. So any manifold has such a structure. But there are topological obstructions to the existence of global pseudo-Riemmannian metrics of other prescribed signatures.
http://en.wikipedia.org/wiki/Semi-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds 
EDIT, 16 July:  I was looking for an example of said topological obstructions, and with an assist fom Willie Wong it has worked out: the ordinary sphere $$ \mathbb S^2 \subseteq \mathbb R^3 $$ does not possess a signature $(+,-) $ metric which I suppose ought to be called Lorentzian for this dimension. The topological obstruction is that $ \mathbb S^2$  cannot have a smooth (tangent) "line field," just as it cannot have a smooth nonzero tangent vector field by Brouwer. Now, a Lorentzian metric would give (pointwise)  null cones, in this case a pair of distinct but intersecting lines in each tangent plane. As we are using $\mathbb R^3, $ we can cheat and define a line field from the angle bisector of the $+$ part of the cone. 
