Constant Rank, Submersions, Immersions and Regular Level Sets

Suppose first that the $(I,J)$-minor is nonzero, and write $B=A[I,J]$ for the corresponding $r\times r$ submatrix.

Conversely, suppose $\mathrm{rank}A\ge r$. Choose $r$ independent columns and form the resulting $m\times r$ matrix $C$. Its column rank and hence its row rank are $r$, so its rows span ${\mathbb{R}}^r$ and contain $r$ independent rows.

By [L2], $B$ is invertible. If a linear combination of the $r$ columns of $A$ indexed by $J$ vanishes, restricting that equality to the rows in $I$ gives $Bc=0$, hence $c=0$. Those columns are independent, so $\mathrm{rank}A\ge r$.

The $r\times r$ submatrix determined by those columns and rows has rank $r$, so [L2] makes its determinant nonzero. This is an $r$-rowed minor of $A$, proving the converse and the equivalence.

If $r=0$, then $U_{\ge r}=U$; if $r>\min \{m,n\}$, then $U_{\ge r}=\mathrm{\varnothing }$. Both sets are open.

Assume $1\le r\le \min \{m,n\}$ and fix $a\in U_{\ge r}$. By [L1], one $r$-rowed minor $M(a)$ of $Jf(a)$ is nonzero.

By [L2], the same minor $M(x)$ is a continuous scalar function of $x$. Its nonzero locus contains an open neighbourhood $V$ of $a$, and [L1] gives $V\subseteq U_{\ge r}$.

Every point of $U_{\ge r}$ therefore has an open neighbourhood inside it, and the two exceptional cases were settled in step 1.1. Hence $U_{\ge r}$ is open for every $r$, proving all stated consequences.

If $r=0$, take $\mathrm{\Phi }={\mathrm{id}}_U$; it is a $C^k$ diffeomorphism on every open neighbourhood of $a$.

Suppose $r>0$. By [L1], choose a nonzero $r$-rowed minor and permute coordinates so it is the leading minor. The derivative $D\mathrm{\Phi }(a)$ is block triangular with that $r\times r$ block and an identity block of size $m-r$ on its diagonal.

Its determinant is the nonzero leading minor, including the full-rank case $r=m$ where the identity block is empty. Thus [L2] makes $D\mathrm{\Phi }(a)$ invertible and $\mathrm{\Phi }$ a local $C^k$ diffeomorphism at $a$.

Steps 1.1 and 2.1 cover every possible rank and give the asserted source coordinates.

Shrink the coordinate image to a product of open rectangles $P\times Q$ around $\mathrm{\Phi }(a)$. By [L1], $g$ is $C^k$, has constant rank $r$, and its first $r$ components are the coordinates $u$.

If $r<n$ and $r<m$, any nonzero partial derivative $\mathrm{\partial }{h}_i\mathrm{/}\mathrm{\partial }{v}_j$ would join the identity $r\times r$ block of $Dg$ to form a nonzero $(r+1)$-rowed minor, contradicting constant rank $r$ by [L1]. Hence all derivatives of $v\mapsto h(u,v)$ vanish.

For each fixed $u$, the rectangle $Q$ is connected and [L2] makes $v\mapsto h(u,v)$ constant. If $r=0$, the same argument applies to all components on the connected rectangle; if $r=m$ or $r=n$, the relevant block is empty and the conclusion is immediate.

Thus, after shrinking, $g(u,v)=(u,h(u))$ in every rank regime.

Translate the source and target distinguished points to $0$ and apply [L1], obtaining $g(u,v)=(u,h(u))$ on a product neighbourhood.

Define the target shear $\beta (z,w)=(z,w-h(z))$. It is a $C^k$ diffeomorphism with explicit inverse $(z,w)\mapsto (z,w+h(z))$ by [L2].

The composite satisfies $\beta (g(u,v))=(u,0)$. When $r=0$, [L1] makes $f$ locally constant before the target translation; when $r=m$ or $r=n$, the empty blocks make the same displayed formula literal.

Taking $\alpha$ to be the translated source coordinate map gives the asserted local normal form.

By [L1], $Df$ has rank at least $n$ on a neighbourhood of $a$; it cannot have larger rank, so its rank is constantly $n$ there.

Apply [L2]. Because $n-r=0$, its normal form is exactly the projection $(u,v)\mapsto u$.

If $m=n$, the $v$ block is also empty, so the normal form is the identity and $f$ is a local $C^k$ diffeomorphism.

By [L1], the derivative has constant rank $m$ near $a$.

By [L2], coordinate diffeomorphisms turn the restriction of $f$ into $i(u)=(u,0)$. This map is injective and its inverse on $i[U]$ is the continuous projection onto the first $m$ coordinates.

Conjugating by the coordinate diffeomorphisms shows that the restricted $f$ is an embedding. If $m=n$, the zero block is empty and the normal form is a local diffeomorphism.

If $O=\mathrm{\varnothing }$, its image is open. Otherwise fix $y\in f[O]$ and choose $x\in O$ with $f(x)=y$.

Shrink the local source neighbourhood from [L1] so that it lies in $O$. In the local coordinates, [L2] shows that its image contains an open neighbourhood of $y$ lying in $f[O]$.

Every point of $f[O]$ is therefore interior, so $f[O]$ is open.

Apply [L2] and restrict to its source coordinate neighbourhood $W$.

By [L1], a point in $W$ represents a point of $f^{-1}(c)$ exactly when $(u,0)=(0,0)$, which is exactly the condition $u=0$; the $v\in {\mathbb{R}}^{m-r}$ coordinates are free.

Pulling this slice back by ${\alpha }^{-1}$ gives the stated local description. The formula also covers $r=0$ and $r=m$ through the empty-block convention.

By [L1], $f$ has constant rank $n$ near $a$, and its fibre is a $C^k$ coordinate slice. By [L2], put $E=K^{\perp }$, so ${\mathbb{R}}^m=K\oplus E$ and $\dim K=m-n$.

The projection of that slice to $K$ along $E$ has derivative equal to the identity at $a$: its tangent there is $K$, because differentiating the normal-form slice and undoing the source coordinates gives $\ker Df(a)$. By [L2], this projection is a local $C^k$ diffeomorphism.

Inverting the projection writes the slice uniquely as $a+u+g(u)$. Its derivative at $0$ takes values both in $E$ and in the tangent $K$, so $Dg(0)=0$; the zero-dimensional case is the same statement with $P=\{0\}$.

This gives the asserted graph and dimension at every point of a nonempty regular fibre, while the empty-fibre case is vacuous.

For the forward direction, suppose $\gamma$ lies in the fibre and $\gamma (0)=a$. Then $f\circ \gamma$ is constant, so [L1] gives $Df(a){\gamma }^{\mathrm{\prime }}(0)=0$ and hence ${\gamma }^{\mathrm{\prime }}(0)\in T_a(f^{-1}(c))$.

For the reverse direction, suppose $v\in T_a(f^{-1}(c))=K$. Using [L2], define $\gamma (t)=a+tv+g(tv)$ for sufficiently small $\mathrm{\mid }t\mathrm{\mid }$. This curve lies in the fibre, satisfies $\gamma (0)=a$, and has ${\gamma }^{\mathrm{\prime }}(0)=v+Dg(0)v=v$.

The two implications are independent and exhaustive. In particular $v=0$ is realized by the same construction, or by the constant curve.

For each standard basis vector $e_j$, choose $w_j\in {\mathbb{R}}^m$ with $A{w}_j=e_j$, and define ${\lambda }_j=\mathrm{\ell }(w_j)$. These are finitely many choices.

For $v\in {\mathbb{R}}^m$, put $w={\sum }_{j<n}(Av{)}_j{w}_j$. Then $Aw=Av$ by [L1] and linearity, so $v-w\in \ker A$ and $\mathrm{\ell }(v)=\mathrm{\ell }(w)={\sum }_{j<n}(Av{)}_j{\lambda }_j=⟨\lambda ,Av⟩$.

If both $\lambda$ and $\mu$ work, surjectivity gives $v$ with $Av=\lambda -\mu$; then $0=⟨\lambda -\mu ,Av⟩=\mathrm{\parallel }\lambda -\mu {\mathrm{\parallel }}_2^2$, so $\lambda =\mu$. This proves existence and uniqueness.

By [L4] the set $W$ of points of $U$ at which $DG$ is surjective is open, and $a\in W$; on $W$ every point of every fibre is a submersion point, so $c$ is a regular value of $G{\mathrm{\mid }}_W$ and $a$ is a local extremum of $f$ subject to $G{\mathrm{\mid }}_W=c$. Fix $v\in \ker DG(a)$. By [L1] applied to $G{\mathrm{\mid }}_W$, choose a level-set curve $\gamma$ with $\gamma (0)=a$ and ${\gamma }^{\mathrm{\prime }}(0)=v$ inside $G{\mathrm{\mid }}_W^{-1}(c)\subseteq G^{-1}(c)$. The constrained local extremum condition in [L2] makes $f\circ \gamma$ locally extremal at $0$.

By [L2], $Df(a)v=0$. Since $v$ was arbitrary, the functional $Df(a)$ vanishes on $\ker DG(a)$.

Apply [L3] to the surjection $DG(a)$. It gives a unique $\lambda$ with $Df(a)v=⟨\lambda ,DG(a)v⟩$ for all $v$, and the gradient representation turns this equality of functionals into $\mathrm{\nabla }f(a)=DG(a{)}^{\mathsf{T}}\lambda$.

The argument derives the multiplier equation from a constrained extremum and makes no converse assertion, as claimed.

By [L1], the nonzero-gradient hypothesis makes $DG(a)$ surjective.

Apply [L2]. Since the transpose of the one-row matrix $DG(a)$ sends $\lambda$ to $\lambda \mathrm{\nabla }G(a)$, its conclusion is the displayed equation.

Uniqueness is part of [L2] and also follows directly from $\mathrm{\nabla }G(a)\mathrm{\ne }0$.