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Good Behaviour of Lie Bracket at a Superdensity Point of the Tangency Set of a Submanifold With Respect to a Rank 2 Distribution
Analysis Mathematica ( IF 0.6 ) Pub Date : 2021-01-05 , DOI: 10.1007/s10476-020-0063-5
S. Delladio

Let H, K be a couple of vector fields of class C 1 in an open set U ⊂ ℝ N+m , $${\cal M}$$ ℳ be a N -dimensional C 1 submanifold of U and define $${\cal T}: = \left\{{z \in {\cal M}:H\left(z \right),K\left(z \right) \in {T_z}{\cal M}} \right\}$$ T : = { z ∈ ℳ : H ( z ) , K ( z ) ∈ T z ℳ } . Then the obvious property If z 0 ∈ $${\cal M}$$ ℳ is an interior point (relative to $${\cal M}$$ ℳ )of $${\cal T}$$ T then $$\left[{H,K} \right]\left({{z_0}} \right) \in {T_{{z_0}}}{\cal M}$$ [ H , K ] ( z 0 ) ∈ T z 0 ℳ admits the following generalization: If z 0 ∈ $${\cal M}$$ ℳ is a superdensity point (relative to $${\cal M}$$ ℳ )of $${\cal T}$$ T then $$\left[{H,K} \right]\left({{z_0}} \right) \in {T_{{z_0}}}{\cal M}$$ [ H , K ] ( z 0 ) ∈ T z 0 ℳ . As a corollary we get very easily the following result of [7]: Let $${\cal D}$$ D be a C 1 distribution of rank N on an open set U ⊂ ℝ N+m and let $${\cal M}$$ ℳ be a N -dimensional C 1 submanifold of U . If z 0 ∈ $${\cal M}$$ ℳ is a superdensity point (relative to $${\cal M}$$ ℳ )of the tangency set $$\left\{{z \in {\cal M}:{T_z}{\cal M} = {\cal D}\left(z \right)} \right\}$$ { z ∈ ℳ : T z ℳ = D ( z ) } then $${\cal D}$$ D is involutive at z 0 .

中文翻译:

李括号在子流形的切集的超密度点处关于秩 2 分布的良好行为

设 H, K 是开集 U ⊂ ℝ N+m 中 C 1 类的一对向量场,$${\cal M}$$ ℳ 是 U 的 N 维 C 1 子流形并定义 $${ \cal T}: = \left\{{z \in {\cal M}:H\left(z \right),K\left(z \right) \in {T_z}{\cal M}} \right \}$$ T : = { z ∈ ℳ : H ( z ) , K ( z ) ∈ T z ℳ } 。那么显而易见的性质如果 z 0 ∈ $${\cal M}$$ ℳ 是 $${\cal T}$$ T 的内点(相对于 $${\cal M}$$ ℳ ),则 $$ \left[{H,K} \right]\left({{z_0}} \right) \in {T_{{z_0}}}{\cal M}$$ [ H , K ] ( z 0 ) ∈ T z 0 ℳ 承认以下概括:如果 z 0 ∈ $${\cal M}$$ ℳ 是 $${\cal T}$$ 的超密度点(相对于 $${\cal M}$$ ℳ ) T 然后 $$\left[{H,K} \right]\left({{z_0}} \right) \in {T_{{z_0}}}{\cal M}$$ [ H , K ] ( z 0 ) ∈ T z 0 ℳ 。作为推论,我们很容易得到 [7] 的以下结果:令 $${\cal D}$$ D 是开集 U ⊂ ℝ N+m 上秩 N 的 C 1 分布,并令 $${\cal M}$$ ℳ 是 N 维 C 1 子流形你。如果 z 0 ∈ $${\cal M}$$ ℳ 是切集 $$\left\{{z \in {\cal M 的超密度点(相对于 $${\cal M}$$ ℳ ) }:{T_z}{\cal M} = {\cal D}\left(z \right)} \right\}$$ { z ∈ ℳ : T z ℳ = D ( z ) } 然后 $${\cal D}$$ D 在 z 0 处对合。
更新日期:2021-01-05
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