In this paper a Weil approach to quasijets is discussed. For given manifolds $M$ and $N$, a quasijet with source $x\in M$ and target $y\in N$ is a mapping $T_x^{r}M\to T_y^{r}N$ which is a vector homomorphism for each one of the $r$ vector bundle structures of the iterated tangent bundle $T^r$ [A. Dekrét, Casopis Pest. Mat. 111, No. 4, 345-352 (1986; Zbl 0611.58004)]. Let us denote by $Q{J}^r(M,N)$ the bundle of quasijets from $M$ to $N$; the space ${\tilde{J}}^r(M,N)$ of non-holonomic $r$-jets from $M$ to $N$ is embeded into $Q{J}^r(M,N)$. On the other hand, the bundle $Q{T}_m^{r}N$ of $(m,r)$-quasivelocities...

Let $G_{n,k}$ (${\tilde{G}}_{n,k}$) denote the Grassmann manifold of linear $k$-spaces (resp. oriented $k$-spaces) in ${ℝ}^n$, $d_{n,k}=k(n-k)=\text{dim}{G}_{n,k}$ and suppose $n\ge 2k$. As an easy consequence of the Steenrod obstruction theory, one sees that $(d_{n,k}+1)$-fold Whitney sum $(d_{n,k}+1){\xi }_{n,k}$ of the nontrivial line bundle ${\xi }_{n,k}$ over $G_{n,k}$ always has a nowhere vanishing section. The author deals with the following question: What is the least $s$ ($=s_{n,k}$) such that the vector bundle $s{\xi }_{n,k}$ admits a nowhere vanishing section ? Obviously, $s_{n,k}\le d_{n,k}+1$, and for the special case in which $k=1$, it is known that $s_{n,1}=d_{n,1}+1$. Using results...

This paper has two parts. Part one is mainly intended as a general introduction to the problem of sectioning vector bundles (in particular tangent bundles of smooth manifolds) by everywhere linearly independent sections, giving a survey of some ideas, methods and results.Part two then records some recent progress in sectioning tangent bundles of several families of specific manifolds.

We show that in dimensions higher than two, the popular "red refinement" technique, commonly used for simplicial mesh refinements and adaptivity in the finite element analysis and practice, never yields subsimplices which are all acute even for an acute father element as opposed to the two-dimensional case. In the three-dimensional case we prove that there exists only one tetrahedron that can be partitioned by red refinement into eight congruent subtetrahedra that are all similar to the original...