Michael Albanese
Reputation
45,059
86/100 score
 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 $k = n-1$? added 18 characters in body; edited tags; edited title 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 $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$ added 12 characters in body 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? added 1 character in body 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? deleted 5 characters in body; edited tags 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 $\mathbb{R}^2\setminus\{p_1,\dots,p_n\}$ added 11 characters in body; edited title 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 1 reviewed Reject Which is the difference between $P(A \mid B)$ and $P(A=t \mid B)$ in a Bayesian Network?