This is a very interesting result. The proof follows Marsden and Hoffman. First here is some linear
algebra.
Theorem 18.8.1Let L : ℝ^{n}→ ℝ^{N}have rank m. Then there exists a basis
{u1,⋅⋅⋅,um,um+1, ⋅⋅⋅,un}
such that a basis for ker
(L )
is
{um+1,⋅⋅⋅,un }
.
Proof: Since L has rank m, there is a basis for L
(ℝn)
which is of the form
{Lu1,⋅⋅⋅,Lum }
Then if
∑
ciui = 0
i
you can do L to both sides and conclude that each c_{i} = 0. Hence
{u1,⋅⋅⋅,um }
is linearly independent. Let
{v1,⋅⋅⋅,vk}
be a basis for ker
(L)
. Let x ∈ ℝ^{n}. Then Lx = ∑_{i=1}^{m}c_{i}Lu_{i} for some choice of scalars c_{i}.
Hence
( ∑m )
L x − ciui = 0
i=1
which shows that there exist d_{j} such that
∑m ∑k
x = ciui + djvj
i=1 j=1
It follows that span
(u1,⋅⋅⋅,um, v1,⋅⋅⋅,vk)
= ℝ^{n}. Is this set of vectors linearly independent?
Suppose
∑m ∑k
ciui + djvj = 0
i=1 j=1
Do L to both sides to get
∑m
ciLui = 0
i=1
Thus each c_{i} = 0. Hence ∑_{j=1}^{k}d_{j}v_{j} = 0 and so each d_{j} = 0 also. It follows that k = n − m and we can
let
{v ,⋅⋅⋅,v } = {u ,⋅⋅⋅,u } . ■
1 k m+1 n
Another useful linear algebra result is the following lemma.
Lemma 18.8.2Let V ⊆ ℝ^{n}be a subspace and suppose A
(x)
∈ℒ
(V,ℝN )
for x in some open setU. Also suppose x → A
(x)
is continuous for x ∈ U. Then if A
(x0)
is one to one on V for somex_{0}∈ U, then it follows that for all x close enough to x_{0}, A
(x)
is also one to one on V .
Proof: Consider V as an inner product space with the inner product from ℝ^{n} and A
(x)
^{∗}A
(x)
. Then
A
(x)
^{∗}A
(x)
∈ℒ
(V,V )
and x → A
(x)
^{∗}A
(x)
is also continuous. Also for v ∈ V,
( ∗ )
A (x) A(x)v,v V = (A (x)v,A(x)v )ℝN
If A
(x0)
^{∗}A
(x0)
v = 0, then from the above, it follows that A
(x0)
v = 0 also. Therefore, v = 0 and
so A
(x0)
^{∗}A
(x0)
is one to one on V . For all x close enough to x_{0}, it follows from continuity
that A
(x)
^{∗}A
(x )
is also one to one. Thus, for such x, if A
(x)
v = 0, Then A
(x)
^{∗}A
(x )
v = 0
and so v = 0. Thus, for x close enough to x_{0}, it follows that A
(x)
is also one to one on V .
■
Theorem 18.8.3Let f : A ⊆ ℝ^{n}→ ℝ^{N}where A is open in ℝ^{n}. Let f be a C^{r}function andsuppose that Df
(x)
has rank m for all x ∈ A. Let x_{0}∈ A. Then there are open sets U,V ⊆ ℝ^{n}withx_{0}∈ V, and a C^{r}function h : U → V with inverse h^{−1} : V → U also C^{r}such thatf ∘ hdependsonly on
(x1,⋅⋅⋅,xm )
.
Proof: Let L = Df
(x0)
, and N_{0} = kerL. Using the above linear algebra theorem, there
exists
{u1,⋅⋅⋅,um}
such that
{Lu1,⋅⋅⋅,Lum }
is a basis for Lℝ^{n}. Extend to form a basis for ℝ^{n},
{u1,⋅⋅⋅,um,um+1, ⋅⋅⋅,un}
such that a basis for N_{0} = kerL is
{um+1,⋅⋅⋅,un}
. Let
M ≡ span (u1,⋅⋅⋅,um ).
Let the coordinate maps be ψ_{k} so that if x ∈ ℝ^{n},
x = ψ1(x)u1 + ⋅⋅⋅+ ψn (x)un
Since these coordinate maps are linear, they are infinitely differentiable.
Next I will define coordinate maps for x ∈ ℝ^{N}. Then by the above construction,
{Lu1,⋅⋅⋅,Lum }
is a
basis for L
(ℝn )
. Let a basis for ℝ^{N} be
{ }
Lu1,⋅⋅⋅,Lum,vm+1, ⋅⋅⋅,vN
(Note that, since the rank of Df
(x)
= m you must have N ≥ m.) The coordinate maps ϕ_{i} will be defined
as follows for x ∈ ℝ^{N}.
Now define two infinitely differentiable maps G : ℝ^{n}→ ℝ^{n} and H : ℝ^{N}→ ℝ^{n},
G(x) ≡ (0,⋅⋅⋅,0,ψm+1 (x),⋅⋅⋅,ψn (x))
H (y) ≡ (ϕ1 (y ),⋅⋅⋅,ϕm (y),0,⋅⋅⋅,0)
For x ∈ A ⊆ ℝ^{n}, let
g (x) ≡ H (f (x))+ G(x) ∈ ℝn
Thus the first term picks out the first m entries of f
(x )
and the second term the last n−m entries of x. It
is of the form
(ϕ1 (f (x )),⋅⋅⋅,ϕm (f (x)),ψm+1 (x),⋅⋅⋅,ψn (x ))
Then
Dg (x0)(v) = HL (v)+ Gv =HLv + Gv (18.24)
(18.24)
which is of the form
Dg (x0)(v) = (ϕ1(Lv),⋅⋅⋅,ϕm (Lv ),ψm+1 (v ),⋅⋅⋅,ψn(v))
If this equals 0, then all the components of v, ψ_{m+1}
(v)
,
⋅⋅⋅
,ψ_{n}
(v)
are equal to 0. Hence
m
v = ∑ ciui.
i=1
But also the coordinates of Lv,ϕ_{1}
(Lv )
,
⋅⋅⋅
,ϕ_{m}
(Lv)
are all zero so Lv = 0 and so 0 =∑_{i=1}^{m}c_{i}Lu_{i} so by
independence of the Lu_{i}, each c_{i} = 0 and consequently v = 0.
This proves the conditions for the inverse function theorem are valid for g. Therefore, there is an
open ball U and an open set V , x_{0}∈ V , such that g : V → U is a C^{r} map and its inverse
g^{−1} : U → V is also. We can assume by continuity and Lemma 18.8.2 that V and U are small
enough that for each x ∈ V,Dg
(x)
is one to one. This follows from the fact that x → Dg
(x )
is
continuous.
Since it is assumed that Df
(x)
is of rank m,Df
(x)
(ℝn )
is a subspace which is m dimensional, denoted
as P_{x}. Also denote L
(ℝn)
= L
(M )
as P.
PICT
Thus
{Lu1,⋅⋅⋅,Lum }
is a basis for P. Using Lemma 18.8.2 again, by making V,U smaller if necessary,
one can also assume that for each x ∈ V, Df
(x)
is one to one on M (although not on ℝ^{n}) and HDf
(x)
is
one to one on M. This follows from continuity and the fact that L = Df
(x0)
is one to one on
M. Therefore, it is also the case that Df
(x)
maps the m dimensional space M onto the m
dimensional space P_{x} and H is one to one on P_{x}. The reason for this last claim is as follows: If
Hz = 0 where z ∈ P_{x}, then HDf
(x)
w = 0 where w ∈ M and Df
(x)
w = z. Hence w = 0
because HDf
(x)
is one to one, and so z = 0 which shows that indeed H is one to one on
P_{x}.
Denote as L_{x} the inverse of H which is defined on ℝ^{m}× 0, L_{x} : ℝ^{m}× 0 → P_{x}. That 0 refers to the
N − m string of zeros in the definition given above for H.
Define h ≡ g^{−1} and consider f_{1}≡f ∘ h. It is desired to show that f_{1} depends only on x_{1},
has rank m.)
Thus it suffices to consider only y ∈ M in the right side of the above. However, for such y,π_{2}Gy = 0
because to be in M,ψ_{k}
(y )
= 0 if k ≥ m + 1, and so the left side of the above equals 0. Thus it appears
this term on the left is 0 for any y chosen. How can this be so? It can only take place if
D_{2}f_{1}
(g(x))
= 0 for every x ∈ V . Thus, since g is onto, it can only take place if D_{2}f_{1}
(x)
= 0 for all
x ∈ U. Therefore on U it must be the case that f_{1} depends only on x_{1},