整体秩定理 global rank theorem

Machine-translated from Chinese.

定理 1. 光滑映射$F:M^m\rightarrow N^n$秩为常数, 若$F$单则$F$为浸入; 若$F$满则$F$为淹没. 若$F$为双射则$F$为微分同胚.

记$r(F)=r.$ 若$F$满, 但$r<n,$ 则由秩定理有坐标系使得$F(\overline U)\subset V$为紧切片. 它在$N$中是闭的, 内部为空, 是疏朗集. 由于$M$上可以找到可列个这样的局部坐标系, 那么由于$F$是满射. 可列个疏朗集将并出$N.$ 由Baire纲定理, 这是不可能的.

若$r<m,$ 那么由秩定理$F$局部为投影, 显然不是单射. $F$为双射的情形是上面的自然推论.

参考: Introduction to Smooth Manifolds by John M.Lee

文章最后更新于 2022-09-27 23:31:00

Theorem 1. The smooth map $F:M^m\rightarrow N^n$ has a constant rank. If $F$ is simple, $F$ is immersed; if $F$ is full, $F$ is submerged. If $F$ is a bijection, $F$ is a diffeomorphism.

Remember $r(F)=r.$ if $F$ is full, but $r<n,$ Then according to the rank theorem, there is a coordinate system such that $F(\overline U)\subset V$ is a compact slice. It is closed in $N$, empty inside, and is a sparse set. Since such local coordinate systems can be found on $M$, then $F$ is surjective. It is possible to list a sparse set that will be combined $N.$ according to Baire's theorem, which is impossible.

If $r<m,$ then according to the rank theorem $F$ is locally a projection, which is obviously not an injective. The case where $F$ is a bijection is a natural corollary of the above.

Reference: Introduction to Smooth Manifolds by John M.Lee

The article was last updated on 2022-09-27 23:31:00