It is **not** a 1-form, it is a *1-density*: a function that is continuous and homogeneous of degree 1 on the tangent space of the manifold. It also happens to be convex and positive in the complement of the zero section (actually, its restriction to each tangent space is a Euclidean norm). If the norm is not Euclidean, you have the arc-length element of a *Finsler metric.* The convexity is basically necessary and sufficient for the lower semi-continuity of the length functional (Busemann-Mayer theorem).

See my answer to this question for more on densities.

sameas the $ds$ in the expression of a Riemannian metric. As Álvarez-Paiva indicates in the currently accepted answer below, this object is an even density of rank $1$; on the one hand, such a thing may be multiplied by a scalar field to produce another even density of rank $1$, and such a thing may be integrated along any (unoriented!) $1$-dimensional submanifold (aka curve); on the other hand, such a thing may be multiplied (symmetrically!) by itself to produce (sometimes) a symmetric bilinear form. – Toby Bartels Mar 3 '13 at 8:10