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15h
comment Does there exist a non-parallelisable manifold with exactly $k$ linearly independent vector fields?
This approach won't work. Note that $T((S^1)^{k-2}\times M_f) \cong \varepsilon^k\times L$ for some line bundle $L$. But $L$ is orientable, and hence trivial, so $(S^1)^{k-2}\times M_f$ is parallelisable. The same problem will occur if $M_f$ is replaced by the total space of any fibre bundle over $S^1$ with toric fibres.
1d
comment Does there exist a non-parallelisable manifold with exactly $k$ linearly independent vector fields?
Also note that this approach can be seen as a generalisation of the one you gave in your answer as the Klein bottle is the mapping torus associated to the antipodal map on $S^1$.
1d
comment Does there exist a non-parallelisable manifold with exactly $k$ linearly independent vector fields?
I was trying to see how to obtain an orientable example by looking at three manifolds $M$ which were $S^1\times S^1$ bundles over $S^1$. One way to construct such a manifold is by taking the mapping torus $M_f$ of a homeomorphism $f : S^1\times S^1 \to S^1\times S^1$. The product $(S^1)^{k-2}\times M_f$ will always have at least $k$ linearly independent vector fields by the same argument as in your answer. Maybe there is an $f$ such that $(S^1)^{k-2}\times M_f$ will have exactly $k$ linearly indepenedent vector fields. I haven't been able to establish this though.
1d
revised What can you say about the $k$-th cohomology group of a closed orientable $n$-manifold for $k = n$ and $k = n-1$?
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1d
answered What can you say about the $k$-th cohomology group of a closed orientable $n$-manifold for $k = n$ and $k = n-1$?
May
3
comment Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
Actually I meant $\mathbb{R}^3\setminus\{\ell_1, \dots, \ell_n\}$. I have edited accordingly.
May
3
revised Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
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May
3
revised Why is $\int_{X} d(\alpha \wedge *\bar{\beta})$ zero on a compact hermitian manifold?
added 135 characters in body
May
2
revised Complex manifold with subvarieties but no submanifolds
added 106 characters in body
May
2
revised Equivalent definition of metric compatibility for a connection: what does $\nabla g$ mean?
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May
2
revised Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
added 368 characters in body
May
2
revised Why is $\int_{X} d(\alpha \wedge *\bar{\beta})$ zero on a compact hermitian manifold?
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May
2
answered Why is $\int_{X} d(\alpha \wedge *\bar{\beta})$ zero on a compact hermitian manifold?
May
2
revised Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
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May
2
comment Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
This is one of those situations where visualising what's happening is much easier than writing it down explicitly.
May
2
answered Showing $\mathbb{R}^3$ minus $n$ parallel lines is homotopic to $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$
May
1
comment Does there exist a non-parallelisable manifold with exactly $k$ linearly independent vector fields?
@studiosus: Would you be willing to expand your comment into an answer? Also, $T^{k-1}\times K$ is non-orientable. Do you know how to obtain orientable examples?
May
1
reviewed Reject Dynamic Bayesian Networks without restrictions
May
1
reviewed Reject Parental Markov Condition Example
May
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reviewed Reject Which is the difference between $P(A \mid B)$ and $P(A=t \mid B)$ in a Bayesian Network?