Timeline for The maximum number of vertical independent vector fields on the tangent bundle
Current License: CC BY-SA 4.0
16 events
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| May 3, 2022 at 20:29 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Dec 22, 2020 at 23:40 | comment | added | Ali Taghavi | Thank you again for your attention to my question and your interesting answer. | |
| Dec 22, 2020 at 23:34 | vote | accept | Ali Taghavi | ||
| Dec 22, 2020 at 20:57 | comment | added | Michael Albanese | @AliTaghavi: You're right, you do get canonical lifts that way. | |
| Dec 22, 2020 at 20:40 | comment | added | Ali Taghavi | I think the lift is a canonical lift and is coordinate free:Let $V_p\in TM$ is a point in the total space. Then the lift has the following formula:$X\in T_p M\mapsto d/dt(V_p+tX)\mid_{t=0}$. But each fiber $\pi^{-1}(p)$ is a linear space:Compare this construction at $V_p$ and $W_p$. I think the lifted vector field restricts to a contstant vector field on each fiber $\pi^{-1}(p)$.(The term constant vector field is a legal term in a vector space, i.e a fiber). so this canonical lifing has range in a very particular Lie subalgebra of $\chi^{\infty}(TM)$. | |
| Dec 21, 2020 at 4:12 | comment | added | Michael Albanese | @AliTaghavi: The lifts aren't canonical, but for any such lifts we have $d\pi[\tilde{X}_i, \tilde{X}_j] = [d\pi(\tilde{X}_i), d\pi(\tilde{X}_j)] = [X_i, X_j] = 0$ so $[\tilde{X}_i, \tilde{X}_j]$ is a vertical vector field, but it's not clear to me that you can choose the lifts so that the Lie bracket is zero. | |
| Dec 20, 2020 at 20:32 | comment | added | Ali Taghavi | I just realized that may be the vertical rank of $TM$ is may be equal to span of $TM$ because if we have k independent vector fields $X_1.X_2,\ldots,X_m$ on $M$ we lift them to k independent vertical vector fields $\tilde{X}_1,\ldots,\tilde{X}_k$ on $TM$ but each lifted vector field is a "constant" vector field on the fiber. So we automatically have $[\tilde{X}_i, \tilde{X}_j]=0$. Am I right? But if we pose the same question in the case of principale bundles or vector bundles, rather than $TM \to M$ we possibly get non trivial question. | |
| Dec 19, 2020 at 17:32 | comment | added | Michael Albanese | What I wrote about $d\pi$ was clearly wrong, I removed that. Thanks for pointing it out. I will be away from my computer for a while, but I will think about the rank question. | |
| Dec 19, 2020 at 17:30 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Dec 19, 2020 at 17:16 | comment | added | Ali Taghavi | Thanks again for your very interesting argument, very helpful information in the last paragraph and your helpful statement $\pi^* TM \simeq \ker d\pi$. | |
| Dec 19, 2020 at 17:07 | comment | added | Ali Taghavi | But I guess you some thing is missing in your notations when you write:$d\pi:\ker d\pi \to TM$ is an isomorphism", right? | |
| Dec 19, 2020 at 16:57 | comment | added | Ali Taghavi | A more explanation of my previuos 2 comments. let $V_p$ is given. We define a linear isomorphism from $T_p M$ to vertical vectors in $T_{V_p} TM$ with $X\mapsto d/dt V_p+tX\mid_{t=0}$. But this map preserves the Lie bracket, so commuting vector fields would be lifted tocommuting vector fields. Am I right? | |
| Dec 19, 2020 at 16:44 | comment | added | Ali Taghavi | By rank of $M$, I mean the maximum number of commuting independent vector fields. a terminology coined by, i think Milnor. So can one say that the vertical rank of TM is always equal to rank of M? | |
| Dec 19, 2020 at 16:42 | comment | added | Ali Taghavi | Thank you for your very helpful answer. To be honest i was aware of isomorphicity of a vertical space at $V_p$ with $T_p M$ but I did not pay attention to the equivalent global formulation you mentioned: The subbundle of vertical tangent vectors is isomorphic to the pull back bundle $\pi^* TM$. So by your very argument we realize that if we have $k$ commuting independent vector fields on $M$ then we have $k$ commuting independent vertical vector field on $TM$. So the vertical rank of TM is $\geq$ than the rank of $M$., | |
| Dec 19, 2020 at 14:18 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Dec 19, 2020 at 14:09 | history | answered | Michael Albanese | CC BY-SA 4.0 |