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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 …

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 …

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 …

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 …

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 …

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 …

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 …