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Homotopy theory, homological algebra, algebraic treatments of manifolds.
20
votes
Are there examples of non-orientable manifolds in nature?
Well, these microscopic examples (molecules), and small-scale examples (pulley-belts), and far-out, conjectured cosmological examples are all well and good, but as a Bostonian born and bred I am disap …
27
votes
2
answers
3k
views
Euler Characteristic of a manifold with non-vanishing vector field,
A friend of mine recently asked me if I knew any simple, conceptual argument (even one that is perhaps only heuristic) to show that if a triangulated manifold has a non-vanishing vector field, then Eu …
2
votes
Does a *topological* manifold have an exhaustion by compact submanifolds with boundary?
Doesn't this depend on the definition of "manifold". If the only condition is being locally Euclidean, then there are connected non second countable examples (e.g., the "long line") for which the answ …
44
votes
Accepted
Motivating the de Rham theorem
Here is a really "trivial" application. Since a volume form (say from a Riemannian metric) for a compact manifold $M$ is clearly closed (it has top degree) and not exact (by Stoke's Theorem), it follo …
4
votes
Does a finite-dimensional Lie algebra always exponentiate into a universal covering group
The short answer to 3. is "no". The simplest example is the circle group, $e^{it}$ of complex numbers of absolute value 1, (thought of as a $1 \times 1$ matrices), its Lie algebra $A$ consists of the …
8
votes
Is $L^p(\mathbb{R})$ minus the zero function contractible?
Here is a simple proof for case of a Hilbert space $V$. Since
$V$ minus the origin deformation retracts onto the unit sphere
$S^\infty$, it suffices to show that $S^\infty$ is contractible,
and that …
4
votes
Compactness and Covering Spaces
Well, the obvious argument that any sequence has a convergent subsequence that your three friends used for the metrizeable case generalizes easily to show that any net has a convergent subnet in the g …