Rank of a smooth map is lower semicontinuous?












0












$begingroup$


Let $F:Mrightarrow N$ be a smooth map between manifolds, $pin M$. Prove that if $operatorname{rank}_{p}F=r$, then there exists a neighborhood $U$ of $p$, such that for $forall qin U$, $operatorname{rank}_{q}Fgeqslant r$.



Could anyone help me? Thanks in advance.






differential-geometry manifolds smooth-manifolds







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  • 3




    $begingroup$
    This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
    $endgroup$
    – Amitai Yuval
    Dec 19 '18 at 7:32
















0












$begingroup$


Let $F:Mrightarrow N$ be a smooth map between manifolds, $pin M$. Prove that if $operatorname{rank}_{p}F=r$, then there exists a neighborhood $U$ of $p$, such that for $forall qin U$, $operatorname{rank}_{q}Fgeqslant r$.



Could anyone help me? Thanks in advance.






differential-geometry manifolds smooth-manifolds







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
    $endgroup$
    – Amitai Yuval
    Dec 19 '18 at 7:32














0












0








0





$begingroup$


Let $F:Mrightarrow N$ be a smooth map between manifolds, $pin M$. Prove that if $operatorname{rank}_{p}F=r$, then there exists a neighborhood $U$ of $p$, such that for $forall qin U$, $operatorname{rank}_{q}Fgeqslant r$.



Could anyone help me? Thanks in advance.






differential-geometry manifolds smooth-manifolds







share|cite|improve this question









$endgroup$




Let $F:Mrightarrow N$ be a smooth map between manifolds, $pin M$. Prove that if $operatorname{rank}_{p}F=r$, then there exists a neighborhood $U$ of $p$, such that for $forall qin U$, $operatorname{rank}_{q}Fgeqslant r$.



Could anyone help me? Thanks in advance.






differential-geometry manifolds smooth-manifolds




differential-geometry manifolds smooth-manifolds






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asked Dec 19 '18 at 6:51









user450201user450201

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718








  • 3




    $begingroup$
    This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
    $endgroup$
    – Amitai Yuval
    Dec 19 '18 at 7:32














  • 3




    $begingroup$
    This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
    $endgroup$
    – Amitai Yuval
    Dec 19 '18 at 7:32








3




3




$begingroup$
This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
$endgroup$
– Amitai Yuval
Dec 19 '18 at 7:32




$begingroup$
This is actually a linear-algebra fact. The map $mathrm{rank}:M_{mtimes n}(mathbb{R})tomathbb{Z}$ is lower semicontinuous.
$endgroup$
– Amitai Yuval
Dec 19 '18 at 7:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let us choose coordinate charts $(x_{1},ldots,x_{n})$ around $p$ and $(y_{1},ldots,y_{m})$ around $F(p)$. Writing $F=big(F_{1}(x_{1},ldots,x_{n}),ldots,F_{m}(x_{1},ldots,x_{n})big)$, we can express the derivative $d_{p}F$ as a matrix
$$
begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(p) & cdots & frac{partial F_{1}}{partial x_{n}}(p)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(p) & cdots & frac{partial F_{m}}{partial x_{n}}(p)
end{pmatrix}.
$$

By assumption, this matrix has rank $r$, so there is a nonzero minor of size $rtimes r$:
$$
begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(p)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(p)
end{vmatrix}neq 0.
$$
The map
$$
G:qmapsto begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(q)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(q)
end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)neq 0$ for all $qin U$. This implies that for all $qin U$
$$
d_{q}F=begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(q) & cdots & frac{partial F_{1}}{partial x_{n}}(q)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(q) & cdots & frac{partial F_{m}}{partial x_{n}}(q)
end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $rtimes r$ minor.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very clear. Really appreciate:)
    $endgroup$
    – user450201
    Dec 20 '18 at 11:11











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let us choose coordinate charts $(x_{1},ldots,x_{n})$ around $p$ and $(y_{1},ldots,y_{m})$ around $F(p)$. Writing $F=big(F_{1}(x_{1},ldots,x_{n}),ldots,F_{m}(x_{1},ldots,x_{n})big)$, we can express the derivative $d_{p}F$ as a matrix
$$
begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(p) & cdots & frac{partial F_{1}}{partial x_{n}}(p)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(p) & cdots & frac{partial F_{m}}{partial x_{n}}(p)
end{pmatrix}.
$$

By assumption, this matrix has rank $r$, so there is a nonzero minor of size $rtimes r$:
$$
begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(p)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(p)
end{vmatrix}neq 0.
$$
The map
$$
G:qmapsto begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(q)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(q)
end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)neq 0$ for all $qin U$. This implies that for all $qin U$
$$
d_{q}F=begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(q) & cdots & frac{partial F_{1}}{partial x_{n}}(q)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(q) & cdots & frac{partial F_{m}}{partial x_{n}}(q)
end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $rtimes r$ minor.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very clear. Really appreciate:)
    $endgroup$
    – user450201
    Dec 20 '18 at 11:11
















1












$begingroup$

Let us choose coordinate charts $(x_{1},ldots,x_{n})$ around $p$ and $(y_{1},ldots,y_{m})$ around $F(p)$. Writing $F=big(F_{1}(x_{1},ldots,x_{n}),ldots,F_{m}(x_{1},ldots,x_{n})big)$, we can express the derivative $d_{p}F$ as a matrix
$$
begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(p) & cdots & frac{partial F_{1}}{partial x_{n}}(p)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(p) & cdots & frac{partial F_{m}}{partial x_{n}}(p)
end{pmatrix}.
$$

By assumption, this matrix has rank $r$, so there is a nonzero minor of size $rtimes r$:
$$
begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(p)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(p)
end{vmatrix}neq 0.
$$
The map
$$
G:qmapsto begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(q)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(q)
end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)neq 0$ for all $qin U$. This implies that for all $qin U$
$$
d_{q}F=begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(q) & cdots & frac{partial F_{1}}{partial x_{n}}(q)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(q) & cdots & frac{partial F_{m}}{partial x_{n}}(q)
end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $rtimes r$ minor.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very clear. Really appreciate:)
    $endgroup$
    – user450201
    Dec 20 '18 at 11:11














1












1








1





$begingroup$

Let us choose coordinate charts $(x_{1},ldots,x_{n})$ around $p$ and $(y_{1},ldots,y_{m})$ around $F(p)$. Writing $F=big(F_{1}(x_{1},ldots,x_{n}),ldots,F_{m}(x_{1},ldots,x_{n})big)$, we can express the derivative $d_{p}F$ as a matrix
$$
begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(p) & cdots & frac{partial F_{1}}{partial x_{n}}(p)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(p) & cdots & frac{partial F_{m}}{partial x_{n}}(p)
end{pmatrix}.
$$

By assumption, this matrix has rank $r$, so there is a nonzero minor of size $rtimes r$:
$$
begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(p)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(p)
end{vmatrix}neq 0.
$$
The map
$$
G:qmapsto begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(q)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(q)
end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)neq 0$ for all $qin U$. This implies that for all $qin U$
$$
d_{q}F=begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(q) & cdots & frac{partial F_{1}}{partial x_{n}}(q)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(q) & cdots & frac{partial F_{m}}{partial x_{n}}(q)
end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $rtimes r$ minor.






share|cite|improve this answer









$endgroup$



Let us choose coordinate charts $(x_{1},ldots,x_{n})$ around $p$ and $(y_{1},ldots,y_{m})$ around $F(p)$. Writing $F=big(F_{1}(x_{1},ldots,x_{n}),ldots,F_{m}(x_{1},ldots,x_{n})big)$, we can express the derivative $d_{p}F$ as a matrix
$$
begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(p) & cdots & frac{partial F_{1}}{partial x_{n}}(p)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(p) & cdots & frac{partial F_{m}}{partial x_{n}}(p)
end{pmatrix}.
$$

By assumption, this matrix has rank $r$, so there is a nonzero minor of size $rtimes r$:
$$
begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(p)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(p) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(p)
end{vmatrix}neq 0.
$$
The map
$$
G:qmapsto begin{vmatrix}
frac{partial F_{i_{1}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{1}}}{partial x_{j_{r}}}(q)\
vdots & & vdots\
frac{partial F_{i_{r}}}{partial x_{j_{1}}}(q) & cdots & frac{partial F_{i_{r}}}{partial x_{j_{r}}}(q)
end{vmatrix},
$$
is continuous and nonzero at $p$, so there exists a neighborhood $U$ of $p$ such that $G(q)neq 0$ for all $qin U$. This implies that for all $qin U$
$$
d_{q}F=begin{pmatrix}
frac{partial F_{1}}{partial x_{1}}(q) & cdots & frac{partial F_{1}}{partial x_{n}}(q)\
vdots & & vdots\
frac{partial F_{m}}{partial x_{1}}(q) & cdots & frac{partial F_{m}}{partial x_{n}}(q)
end{pmatrix}
$$
has rank at least $r$, since it has a nonzero $rtimes r$ minor.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 11:06









studiosusstudiosus

1,807713




1,807713












  • $begingroup$
    Very clear. Really appreciate:)
    $endgroup$
    – user450201
    Dec 20 '18 at 11:11


















  • $begingroup$
    Very clear. Really appreciate:)
    $endgroup$
    – user450201
    Dec 20 '18 at 11:11
















$begingroup$
Very clear. Really appreciate:)
$endgroup$
– user450201
Dec 20 '18 at 11:11




$begingroup$
Very clear. Really appreciate:)
$endgroup$
– user450201
Dec 20 '18 at 11:11


















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