Advanced Calculus (draft version no 8)
Henri Anciaux, IME-USP
1 Review of topology
1.1 The topology of Euclidean space Notation: (e1, ..., en) is the canonical basis ofRn.
Forx=Pn
i=1xiei ∈Rn we write x= (x1, ..., xn). From the usualnormof Rn defined by
||x||:=
n
X
i=1
x2i
!1/2
,
we define the notion ofdistance, i.e. the map Rn×Rn→R defined by d(x, y) :=||x−y||.
The distance satisfies the following properties:
– d(x, x) = 0⇔x= 0;
– d(x, y) =d(y, x);
– d(x, y)≤d(x, z) +d(z, y).
Definition 1. – The open ball centered at the pointxand with radiusr >0 is the set Bx(r) :={y∈Rn| ||x−y||< r}
– a subset V ⊂ Rn is said to be a neighbourhood of x if it exists r > 0 such that Bx(r)⊂V;
– a subset U ⊂Rn is said to beopen if it is a neighbourhood of all its points;
– a subset F is said to beclosedif Fc=Rn\F is open.
Proposition 1. – If U1 and U2 are two open sets, then U1∩U2 are open as well;
– If(Ui)i∈I is a family of open sets, then ∪i∈IUi is open as well;
– IfF1 and F2 are two closed sets, then F1∪F2 are closed as well;
– If(Fi)i∈I is a family of closed sets, then ∩i∈IFi is closed as well;
Definition 2. Let U be an open subset of Rn andf :U →Rm.The functionf is said to be continuous at x if the two equivalent statements hold:
(i) ∀ >0,∃η >0 s.t. if||x−y||< η, then ||f(x)−f(y)||< ;
(ii) IfV is a neighbourhood of f(x), then f−1(V) is a neighbourhood of x.
Example 1. Let f be an affine function, i.e. f(x) = Ax+b, where A∈L(Rn,Rm) and b∈Rm.Then f is continuous.
The proof of the continuity off is an easy exercise using the next
Lemma 1. If A∈L(Rn,Rm), there exists M >0 such that ||Av|| ≤M||v||,∀v∈Rn. We shall say that a norm||.||∗ isequivalentto||.||if there exist real constantsCandC0 such thatC||.|| ≤ ||.||∗≤C0||.||.It is not difficult to check that on the one hand, all norms of Rn are equivalent, and that on the other hand, equivalent norms define the same open subsets and therefore have the same continuous functions. The concept of differentiable map function, which will be defined later, turns out to be also independent of the choice of the norm.
1.2 General topology
Definition 3. Let Ω be a set and τ a set of subsets of Ω, i.e. τ ⊂ P(Ω) which satisfies the two following properties
(i) ∅,Ω∈τ;
(ii) ∀U1, U2∈τ, U1∩U2∈τ;
(iii) If(Ui)i∈I is a subfamily of τ, then∪i∈IUi∈τ.
Then τ is said to be a topologyonΩand the pair (Ω, τ) is said to be atopological space.
The elements of τ are said to be open.
Example 2. – (Rn, τ0)whereτ0 is the set of open subsets (in the sense of Section1.1).
We call τ0 the usual topology.
– LetΩ a subset. We set τf,Ω :=P(Ω) andτg,Ω :={∅,Ω}.
– Let(Ω, τ) a topological set and A⊂Ω. We set τA:={U∩A|U ∈τ}. The topology τA is called trace topology(or induced topology).
– Let (Ω, τ) a topological set and ∼ an equivalence relation. We denote by [x] the equivalence class of x, by Ω := Ω/˜ ∼ the space of equivalence classes (quotient space) and finally by π : Ω → Ω˜ the canonical projection π(x) = [x]. We set ˜τ :=
{π(U)|U ∈τ}. The topology τ˜ is called quotient topology.
Definition 4. Let (Ω, τ) a topological space and x∈ Ω. A subset V of Ω is said to be a neighbourhood of x is there exists an open subsetU ∈τ such that x∈U ⊂V.
Definition 5. A topological space (Ω, τ) is said to be Hausdorff (or separated,or T2) if for all x, y ∈ Ω, x 6=y, there exist two neighbourhoods V and W of x and y respectively which are disjoint.
Definition 6. Let (Ω, τ) and (Ω0, τ0) be two topological spaces and f : Ω → Ω0. The functionf is said to becontinuousatx (with respect toτ andτ0) if , for all neighbourhood W of f(x), f−1(W) is a neighbourhood of x.
Proposition 2. Let(Ω, τ)and(Ω0, τ0) be two topological spaces andA⊂Ω.Iff : Ω→Ω0 is continuous, then f|A:A→Ω0 is continuous as well with respect to the trace topology.
Definition 7. A topological space (Ω, τ) is said to be compact if any open covering of Ω admits a finite subcovering;
Proposition 3. Let (Ω, τ) and (Ω0, τ0) be two topological spaces and f : Ω→ Ω0. If f is continuous and Ω is compact then Ω0 is compact as well.
Definition 8. Let (Ω, τ) and (Ω0, τ0) be two topological spaces and f : Ω → Ω0. The functionf is said to be ahomeomorphism(with respect toτ andτ0) iff is a bijection and f and f−1 are continuous at all points.
Theorem 1. Letf :Rn→Rm be a homeomorphism (with respect to the usual topologies).
Then m=n.
Definition 9. Let(Ω, τ)a connected topological space such that, for allx∈Ω,there exists an open neighbourhood U of x and a homeomorphism ϕ : U → V, where V is an open subset of Rn. Then (Ω, τ) is said to be a topological manifold. The number n (which is well defined by Theorem 1) is called the dimension of (Ω, τ). The map ϕ is called a local chart of (Ω, τ) at x.
Proposition 4. Let(Ω, τ)and (Ω0, τ0) be two topological manifolds andf : Ω→Ω0.Then f is continuous at the point x∈Ωif and only if, given local charts ϕ and ϕ0 of Ωand Ω0 at x and f(x) respectively, the mapϕ0◦f◦ϕ−1 is continuous at the point ϕ(x).
1.3 Exercises
1. Prove Propositions1 and2 and Lemma1. Prove that the statements (i) and (ii) of Definition3 are equivalent.
2. Let Ω be a set and F a set of subsets of Ω, i.e. F ⊂ P(Ω) which satisfies the two following properties
(i) ∅,Ω∈ F;
(ii) ∀U1, U2 ∈ F, U1∪U2 ∈ F;
(iii) If (Ui)i∈I is a subfamily ofF, then∩i∈IUi ∈ F. Prove that τ :={Fc= Ω\F|F ∈ F }is a topology on Ω.
3. LetA be a finite set of points of Rn. Prove thatτf,A=τA.
4. Let be two topological spaces (Ω, τ) and (Ω0, τ0).Prove that ifτ0 =τg,Ω0 orτ =τf,Ω, any functionf : Ω→Ω0 is continuous. Prove that if τ =τg,Ω and τ0 =τf,Ω0, then a functionf : Ω→Ω0 is continuous if and only if it is constant.
5. Define S1 := {(x1, x2) ∈ R2|x21 +x22 = 1} endowed with the trace topology and S˜1 := R/∼, where ∼ is defined by x ∼ y ⇔ x−y ∈ 2πZ, endowed with the quotient topology. Prove thatS1and ˜S1 are homeomorphic and are a 1-dimensional topological manifold.
6. Determine which of the topological spaces of Example 2 are Haussdorff.
2 Differentiable maps
2.1 Differentiability
Remember that a functionf :I →R,whereI is an open interval, is said to bedifferentiable at the point x∈I,if
limt→0
f(x+t)−f(x)
t exists
When this limit exists, is it denoted by f0(x).
Observe that the differentiability of a function of the real variable implies continuity.
Moreover, it is worth noting that the differentiability offatxis equivalent to the existence of a real number A such that
limt→0
f(x+t)−f(x)−At
t = 0.
Of course, when this happens, the numberA is nothing butf0(x).
The definition of the differential of a function of the real variable suggests the following:
Definition 10. Let U be an open subset of Rn and f :U →Rm.Let x∈U andv ∈Rn. Thedirectional derivativeoff atxin the directionvis the following limit (when it exists):
limt→0
f(x+tv)−f(x)
t .
When this limits exists, we shall denote it by ∂f∂v(x).
It turns out that the concept of directional derivative, though quite intuitive, is not the relevant one. To be convinced of this, it is sufficient to have a close look to the following Example 3. Define f(x1, x2) by
( f(0,0) = 0 f(x1, x2) = xx421x2
1+x22 if (x1, x2)6= (0,0).
Observe first thatf is not continuous at(0,0)sincef(t, t2) = 1/2fort6= 0andf(0,0) = 0.
However, we claim that f admits directional derivative in any direction v ∈ Rn. To see this, we set v=r(cosθ,sinθ) and calculate
∂f
∂v = lim
t→0
f((0,0) +tv)−f(0,0)
t = lim
h→0
t3r3cos2θ sinθ t(t4r4cos4+t2r2sin2θ).
If sinθ= 0, the ratio above is constant and equal to zero, so its limits exists, being equal to zero. On the other hand, if sinθ6= 0, then the limit above is rcosθ cotθ.
Definition 11. Let U be an open subset of Rn and f :U →Rm.The function f is said to be differentiableat x if there exists a linear map A∈L(Rn,Rm) such that
h→0lim
||f(x+h)−f(x)−A(h)||
||h|| = 0.
In the following, the linear map Awill be denoted dfx and calledthe differential of f atx.
The first observation about this fundamental concept is that the differential is unique:
Lemma 2. If f as in the definition above is differentiable atx, then dfx is unique.
Proof. Assume there existsA, B ∈L(Rn,Rm) such that
h→0lim
||f(x+h)−f(x)−A(h)||
||h|| = 0
and
h→0lim
||f(x+h)−f(x)−B(h)||
||h|| = 0, By the triangular inequality, and setting e:= ||h||h ,
h→0lim
||A(h)−B(h)||
||h|| = lim
||h||→0||(A−B)(e)||.
It follows that (A−B)(e) = 0, for all unit vectore∈Rn,||e||= 1.By linearity, this implies that A−B = 0
The next two propositions show that the more restrictive and quite less intuitive defi- nition of differentiability is the right one: it implies continuity (this is desirable since we want to make use of topological concepts in calculus) and existence of directional derivative (hence we don’t lose the intuitive picture):
Proposition 5. If f is differentiable at x, then it is continuous at x.
Proof. We makeh=y−x, so we have
y−x→0lim
||f(y)−f(x)−A(y−x)||
||y−x|| = 0 given >0,there existsη >0 such that if||y−x||< η,then
||f(y)−f(x)−A(y−x)||< ||y−x||< η.
On the other hand, by Lemma 1, there existsM >0 such that||A(y−x)|| ≤M||x−y||<
M η. Hence, by the triangular inequality,
||f(y)−f(x)|| ≤ ||f(y)−f(x)−A(y−x)||+||A(y−x)|| ≤η(+M).
Given 0 >0, we need to prove the existence of η such that if ||y−x||< η, then||f(y)− f(x)||< 0.The inequality just obtained show that this is granted choosingη= +M0 .
Proposition 6. If f is differentiable atx, then it admits directional derivative atx in all directions v ∈Rn.Moreover, we have ∂f∂v(x) =dfx(v).
On the other hand, differentiability happens to be difficult to establish in practice (un- less directional differentiability). Fortunately, the next definition and proposition provides a simple criterion of differentiability.
Definition 12. Let U be an open subset of Rn. A function f :U → Rm is said to be of class C1 if all his partial derivatives ∂x∂f
i exist and are continuous in U.
Theorem 2. A function f :U →Rm of classC1 is differentiable on U.
Proof. The idea of the proof consists of the repeated use of the mean value theorem in order to evaluate the difference f(x+h)−f(x) by the partial derivatives. We shall need a little notation. We set x0 =x andxn=x+h, and, forisuch that 1≤i≤n−1,
xi := (x1+h1, ..., xi+hi, xi+1, ..., xn).
We now apply the mean value theorem several times:
There existsc11∈(0, h1) such that f1(x1)−f1(x0) = ∂f∂x1
1(y11)h1, where we sety11:=
(c11, x2, ..., xn).
Analogously, ∀i, 2 ≤ i ≤ n, and ∀j,1 ≤ j ≤ m, there exists cij ∈ (0, hi) such that fj(xi)−fj(xi−1) = ∂f∂xj
i(yij)hi, where we setyij := (x1+h1, ..., xi−1+hi−1, cij, xi+1, ..., xn).
Observe that all the points yij belong toBx(||h||).
Therefore, we have
fj(x+h)−fj(x) =
n
X
i=1
(fj(xi)−fj(xi−1)) =
n
X
i=1
∂fj
∂xi(yij)hi. Noting that hi ≤ ||h||,it follows that
||f(x+h)−f(x)−Pn i=1
∂f
∂xi(x)hi||
||h|| ≤
Pm
j=1|fj(x+h)−fj(x)−Pn i=1
∂fj
∂xi(x)hi|
||h||
≤
Pm j=1
Pn i=1
∂fj
∂xi(yij)−∂f∂xj
i(x) hi
||h||
≤
m
X
j=1 n
X
i=1
∂fj
∂xi(yij)−∂fj
∂xi(x) .
The assumption on the continuity of the partial derivatives ∂f∂xj
i implies that lettinghtend to zero, the right hand side of the inequality above can be set arbitrarily small. This proves the claim.
Definition 13. – Let U be an open subset of Rn. A function f :U → Rm is said to be of classC2 if all his partial derivatives ∂x∂f
i exist and are of class C1;
– By induction, we way define: a functionf :U →Rm is said to be of classCk if all his partial derivatives ∂x∂f
i exist and are of class Ck−1.
– A function f :U →Rm is said to beof classC∞ if it is of class Ck,∀k.
Theorem 3 (Schwartz). Let U be an open subset of Rn and a function f :U → Rm of class C2. Then
∂2f
∂xi∂xj(x) = ∂2f
∂xj∂xi(x),∀x∈U,∀1≤i, j≤n.
Proof. The proof uses Fubini’s theorem.
So far, the discussion has been quite theoretical. The next proposition provides some concrete examples of differentials.
Proposition 7. – Let f(x) = Ax+b where A ∈ L(Rn,Rm) and b ∈ Rm. Then f is differentiable at any point x of Rn and dfx = A; In particular a constant function is differentiable, with zero differential (make A= 0) and a linear function is differentiable, with differential itself (make b= 0);
– Let U be an open subset of Rn and f1 : U → Rm1 and f2 : U → Rm2. Then the functionf :U →Rm1+m2 defined by f(x) = (f1(x), f2(x))is differentiable atx∈U if and only if both f1 andf2 are differentiable at x;
– Let f : Rn1 ×Rn2 be a bilinear map, i.e. such that the functions x 7→ f(x, y), for fixed y, and y 7→ f(x, y), for fixed x, are both linear. Then f is differentiable on Rn1+n2 and
df(x,y)(h, k) =f(x, k) +f(h, y).
Proposition 8. Let U be an open subset of Rn and f1, f2 : U → Rm two functions which are differentiable at x ∈ U. Then the functions s : U → Rm defined by s(x) = f1(x) +f2(x)and p:U →R defined by p(x) =hf1(x), f2(x)i, are differentiable atx∈U, with differential dsx=dfx1+dfx2 and dpx=hf1(x), dfx2i+hdfx1, f2(x)i.
Theorem 4 (The chain rule 1). Let f :U1 →Rn2 and g :U2 → Rn3 two maps , where U1 and U2 are two open subsets of Rn1 and Rn2 respectively, such that f(U1)⊂U2. If f is differentiable atx∈U1 andg is differentiable at f(x)∈U2,theng◦f :U1 →Rn3 is at x and we have
d(g◦f)x =dgf(x)◦dfx. Proof. Settingy:=f(x), and
ϕ(x0) :=f(x0)−f(x)−dfx(x0−x) ψ(y0) :=g(y0)−g(x)−dgy(y0−y), the differentiabily off and g atx and y respectively amounts to
(1) lim
x0→x
||ϕ(x0)||
||x0−x|| = 0 and
(2) lim
y0→y
||ψ(y0)||
||y0−y|| = 0
respectively. If we furthermore set
ρ(x0) :=g◦f(x0)−g◦f(x)−dgy◦dfx(x0−x) what we have to prove is
xlim0→x
||ρ(x0)||
||x0−x|| = 0.
We have
ρ(x0) = g(f(x0))−g(f(x))−dgy(dfx(x0−x))
= g(f(x0))−g(y)−dgy(f(x0)−f(x)−ϕ(x0))
=
g(f(x0))−g(y)−dgy(f(x0)−f(x))
+dgy(ϕ(x0))
= ψ(f(x0)) +dgy(ϕ(x0)) Hence it is sufficent to prove that
(3) lim
x0→x
||ψ(f(x0))||
||x0−x|| = 0 and
(4) lim
x0→x
||dfx(ϕ(x0))||
||x0−x|| = 0.
By (2), for > 0, there exists η > 0 such that if ||f(x0)−y|| < η, then ||ψ(f(x0))|| <
||f(x0)−y||. On the other hand, surely there exists η0 such that if||x0−x|| < η0, then
||f(x0)−y||< η.Using Lemma 1, we obtain
ψ(f(x0)) < ||f(x0)−y||
= ||ϕ(x0) +dfx(x0−x)||
≤ ||ϕ(x0)||+M||x0−x||.
Equation (3) follows. Finally, Equation (4) follows from Lemma 1 and Equation (1).
2.2 The inverse function theorem
Definition 14. Let U be a an open subset of Rn and f :U →Rm of class Ck.
– Ifn < m and dfx has maximal rankn, ∀x∈U, we will say thatf is an immersion;
– Ifn > m and dfx has maximal rankm, ∀x∈U,we will say that f is a submersion.
Definition 15. Let U be an open subset ofRn andf :U →Rm.We shall say thatf is a diffeormorphism of classCk if f :U →f(U) is a bijection and f and f−1 are of class Ck on U and f(U) respectively.
Observe that since a diffeomorphism is in particular a homeomorphism, we must have m=nby Theorem1. However this follows very easily from the chain rule: fromf−1◦f = Id, we deduce d(f−1)f(x)◦dfx = Id, we shows that dfx is invertible, and therefore that m = n. This implies also the easy-to-remember formula d(f−1)f(x) = (dfx)−1 for the differential off−1 atf(x). The Inverse Function Theorem, a strong and quite unexpected statement that we are going to see in a moment, is a kind of converse: ifdfx is invertible, then f is a diffeomorphism in a neighbourhood ofx.
Lemma 3. Let U be an open subset of Rn and f :U →Rm of classC1. Then
||f(x)−f(x0)|| ≤√
nmM||x−x0||, where M := sup1≤i≤n,1≤j≤m x∈U|∂f∂xj
i(x)|.
Proof.
Theorem 5 (The Inverse Function Theorem (IFT)). Let U be an open subset ofRn and f :U →Rnof class Ck.Let x0 ∈U and suppose that the differentialdfx0 of f at the point x0 is invertible. Then f is a local diffeomorphism at x0 of class Ck, i.e. there exists an open subset U0 of x0 in U such that the f|U0 : U0 → f(U0) is a diffeomorphism of class Ck.
Proof. The proof is not easy nor intuitive. But it is divided in a number of easy steps.
Step 1: we may assume that dfx0 =Id;
To see this, we setg(x) := (dfx0)−1.f(x) By the chain rule,g satisfies all theassumptions of the IFT, and moreover, dgx0 =Id.We claim than if g satisfies the conclusions of the IFT, then so does f. This will prove Step 1. First, if f is locally invertible, so is f and sincef−1 = (dfx0)−1◦g−1.Moreover theCk property ofg−1 implies that off−1, again by the chain rule.
Step 2: There exists a neighbourhood V of x0 such that f(x)6=f(x0),∀x∈V\{x0};
We proceed by contradiction. Assume that there exists a sequence (xn)n∈N converging to x0 such that f(xn) =f(x0),∀n∈N.Therefore, by the differentiability assumption,
0 = lim
n→∞
||f(xn)−f(x0)−dfx0(xn−x0)||
||xn−x0|| = lim
n→∞
||xn−x0||
||xn−x0|| = 1, a contradiction.
Step 3: There exists a neighbourhood V0 of x0 such that ∀x∈V0,detdfx 6= 0 and (∗) ||x−x0|| ≤2||f(x)−f(x0)||,∀x, x0 ∈V0;
By the continuity of the partial derivatives of f, and of the determinant function, there exists a neighbourhood V0 ofx0 such that
∂fj
∂xi(x)−∂fj
∂xi(x0)
< 1
2n2, ∀1≤i, j≤n,∀x∈V0 and
|detdfx−detdfx0|=|detdfx−1|<1/2,∀x∈V0.
The second inequality obviously implies that detdfx 6= 0 on V0. We will combine the first inequality with Lemma 3 applied to the map h(x) := f(x)−x. Remembering that dfx0 =Id,we have
sup
1≤i,j≤n, x∈U
∂hj
∂xi(x)
= sup
1≤i,j≤n, x∈U
∂fj
∂xi(x)−∂fj
∂xi(x0) .
Therefore, Lemma3 implies
(∗∗) ||h(x)−h(x0)||=||f(x)−f(x0)−(x−x0)|| ≤ 1
2||x−x0||.
By the triangle inequality
||x−x0||=||[f(x)−f(x0)−(x−x0)]−[f(x)−f(x0)]|| ≤ ||f(x)−f(x0)−(x−x0)||+||f(x)−f(x0)||.
Substracting, we get
||x−x0|| − ||f(x)−f(x0)|| ≤ ||f(x)−f(x0)−(x−x0)||.
Combining with Inequality (∗∗), we conclude that
||x−x0|| − ||f(x)−f(x0)|| ≤ 1
2||x−x0||, which is equivalent to Inequality (∗)
Step 4: Let r > 0 such that Br(x0) ⊂ V0, then there exists d > 0, such that f(x) >
2d, ∀x∈Sr(x0);
Since Sr(x0) is compact and f is continuous,f(Sr(x0)) is compact as well, hence closed.
Therefore its complement, which contains f(x0), is open. Hence there exists d >0 such that B2d(f(x0))∩f(Sr(x0)) =∅.Step 4 follows.
Step 5: The restriction of f to V00=f−1(Bd(f(x0))) is one-to-one;
This is equivalent to prove that ∀y ∈Bd(f(x0)), the equation f(x) =y admits a unique solution. To prove this, we introduce the function k : Br(x0) → R defined by g(x) :=
||y−f(x)||2. Since k is continuous on the compact set Br(x0), it attains its minimum.
Clear the minimum is not attained on the boundary Sr(x0) since k(x0) =||f(x0)−y||<
d < 2d < ||f(x)−f(y)||,∀x ∈ Sr(x0). Hence there exists ¯x ∈ Br(x0) minimizing k and therefore satisfying dkx¯ = 0. We have ∂kx
i(x) = 2Pn
j=1
∂fj
xi (x)(yj −fj(x), which can be viewed as the matricial product [dkx] = 2[dfx][y−f(x)]. Since ¯x ∈V00, df¯x is invertible, the condition dkx¯ = 0 implies y = f(¯x). The fact that ¯x is the unique solution of the equation y=f(x) follows from Inequality (∗) of Step 3: if ¯xand ¯x¯are two such solutions, we have ||¯x−¯x|| ≤¯ 2||y−y||= 0, so ¯x= ¯x.¯
Step 6: The inverse map f−1 is continuous;
It follows from Inequality (∗) of Step 3: set x = f−1(y) and x0 = f−1(y0), we have
||x−x0|| ≤2||f(x)−f(x0)||, so that ||f−1(y)−f−1(y0)|| ≤2||y−y0||.
Step 7: The inverse map f−1 is differentiable;
Since f is differentiable at a pointx∈V00, we havef(x0) =f(x) +dfx(x0−x) +ϕ(x0−x) with
lim
x0→x
||ϕ(x0−x)||
||x0−x|| = 0.
Hence
(dfx)−1(f(x0)−f(x)) =x0−x+ (dfx)−1(ϕ(x0−x)).
Since for any y ∈ f(V00) there exists x such that y = f(x), we can replate x and x0 by f−1(y) and f−1(y0), getting
f−1(y0)−f−1(y) = (dfx)−1(y0−y)−(dfx)−1
ϕ(f−1(y0)−f−1(y))
.
Therefore, to prove that f−1 is differentiable at the point y with differential dfy−1 = (dff−1(y))−1 amounts to prove that
ylim0→y
(dfx)−1
ϕ(f−1(y0)−f−1(y))
||y0−y|| = 0.
By Lemma1 and then Inequality (∗), there existsM >0 such that
(dfx)−1
ϕ(f−1(y0)−f−1(y))
||y0−y|| ≤ M
ϕ(f−1(y0)−f−1(y))
||y0−y||
= M
ϕ(f−1(y0)−f−1(y))
||f−1(y0)−f−1(y)||
||f−1(y0)−f−1(y)||
||y0−y||
≤ 2M||ϕ(f−1(y0)−f−1(y))||
||f−1(y0)−f−1(y)||
By the continuity of f−1,
ylim0→y
||ϕ(f−1(y0)−f−1(y))||
||f−1(y0)−f−1(y)|| = lim
f−1(y0)→f−1(y)
||ϕ(f−1(y0)−f−1(y))||
||f−1(y0)−f−1(y)|| = 0.
The claim follows.
Step 8: The inverse map f−1 is of class Ck;
We need to prove that the partial derivatives of f−1 are of class Ck−1. The formula dfy−1= (dff−1(y))−1 expresses the partial derivatives of f−1 as the composition of
– The operation consisting of taking the inverse of a matrix with non-vanishing de- terminant. This operation involves the determinant function and the real function x→x−1, and is therefore of classC∞;
– The functionf−1, which has been proved to be continuous and differentiable in the previous steps;
– The partial derivatives of f, which are assumed to be of class Ck−1 (f being itself of classCk).
Hence, by the chain rule, the partial derivatives of f−1 are continuous and differentiable.
Repeating the argument, and using the fact that now we know that the function f−1 is of class C1, we deduce that the partial derivatives of f−1 are of class C1 themselves, so that, by the chain rule again, f−1 is of class C2. We may repeat the argument until we are stopped by the maximum regularity of the partial derivatives of f, i.e. Ck−1. Hence the the partial derivatives of f−1 are of classCk−1 and thenf−1 is of classCk.
Theorem 6 (The implicit function theorem). Let U be an open subset of Rn=Rp×Rq and f :U → Rq a function of class Ck. Let x0 = (x00, x000) ∈ U and c := f(x0). Assume that the matrix
h∂fp+j
∂xp+i(x0) i
1≤i,j≤qis invertible. Then there exists an open neighbourhood V of x00 and a functiong:V →Rq of class Ck such that f(x0, g(x0)) =c,∀x∈V.
Proof. We define the mapF :U →Rp×Rq by F(x) := (f(x),0) + (x0,0).It follows that dF(x0 is invertible ...
2.3 Exercises
1. Prove Proposition6.
2. Define f(x1, x2) by
( f(0,0) = 0
f(x1, x2) = x1x2x21−x22
x21+x22 if (x1, x2)6= (0,0).
Prove that the second derivatives of f are not continuous in (0,0) and that
∂2f
∂x1∂x2(0,0)6= ∂2f
∂x2∂x1(0,0).
3. Let f : Rn → Rm of class C1 in Rn and define g : Rn×Rn → Rm by g(x, y) = f(3x−y).
- Prove thatg is of classC1 inRn×Rn and calculatedg(x,y);
- Calculate dg(x,y) in the case n = 3, m = 1, f(x1, x2, x3) = x1 +x22 + 2x3 e (x, y) = ((1,0,2),(0,3,0)).
4. Let g1, g2, g3 and g4 real differentiable functions of the real variable. Calculate the partial derivatives of
f(x1, x2) :=
g1(x1)g2(x2), g3(x1+x2), Z x2
x1
g4(t)dt
5. Letf :Rn→R differentiable in Rn and m∈N such that
∀(t, x)∈R×Rn, f(tx) =tmf(x).
Prove that
∀x∈Rn,
n
X
i=1
xi
∂f
∂xi(x) =mf(x).
6. Let U be an open subset of R2 and f : U → R a continuous function such that
∀(x1, x2)∈U,tem-se (x21+x42)f(x1, x2) + (f(x1, x2))3 = 1. Prove thatf is of class C∞.
7. Exercices 2-35, 2-37 and 2-39 of Spivak’s book.
8. Prove that there exists > 0 such that if p
y21+y22 < , the following system of equations has a unique solution
e2x1 + sin2(x2)−1 = y1 sin3(x1)−e2x2+ 1 = y2.
3 Submanifolds of Euclidean space
3.1 What is a submanifold?
In all this section we fix three integers numbers p, q and n such that p+q = n. We furthermore introduce the canonical injectionι and projectionπ defined respectively by
ι: Rp → Rn=Rp×Rq x0 7→ (x0,0) and
π : Rn=Rp×Rq → Rp x= (x0, x00) 7→ x0.
We furthermore setH:=ι(Rp). Observe thatHis ap-dimensional linear subspace ofRn. Lemma 4. The maps ι and π are of class C∞. Moreover, the map π is open, i.e. the image of an open subset of Rn by π is an open subset of Rp.
Definition 16. Let U be an open subset ofRp andϕ:U →Rnan immersion of classCk such that ϕ:U →ϕ(U) is a homeomorphism,ϕ(U) being equipped with the trace topology defined in Example 2. Then ϕ is called an embedding.
Remark 1. Observe that the condition of being an embedding is stronger than just the immersion being one to one: we require that the inverse of the immersion is continuous (see Exercise 3 for a counter-example).
Proposition 9. Let U be an open subset of Rp and ϕ :U → Rn an immersion of class Ck and x00 ∈U.There exists an open neighbourhoodV of x00 such that the restriction ofϕ to V is an embedding.
Proof. Sincedϕx0
0 has rankp, the matrix [∂ϕ∂xj
i(x00)]1≤i≤p,1≤j≤nhasplines which are linearly independent. Without loss of generality, we may assume that these are the first p ones, i.e. the matrix [∂ϕ∂xj
i(x00)]1≤i,j≤p is invertible. We define
¯
ϕ: U ×Rq → Rn (x0, x00) 7→ ϕ(x) + (0, x00).
Since ∂∂xϕ¯j
i(x) = ∂ϕ∂xj
i(x0) if 1≤i≤p and ∂∂xϕ¯j
i(x) =δij if p+ 1≤i, j ≤n, we deduce that dϕ¯(x0
0,0) is invertible. By the IFT, there exists a neighbourhood W of (x00,0) inU ×Rq such that the restriction of ¯ϕ toW is a diffeomorphism. We claim that the restriction of ϕ toV :=π(W) (which, by Lemma 4, is an open subset), is an embedding. To see this, observe that
∀x0 ∈V,((π◦ϕ¯−1)◦ϕ)(x0) = (π◦ϕ¯−1)◦( ¯ϕ◦ ι)(x0) = (π◦ ι)(x0) =x0.
Hence ϕ is one to one on V and ϕ−1 = (π◦ϕ¯−1)|ϕ(V), which is continuous by Lemma 4 and Proposition 2. The conclusion follows.
Definition 17. A subset S of Euclidean space Rn is said to be a submanifold of Rn of class Ck and dimension p if, for every point x of S, there exists a open neighbourhood of x in Rn and a diffeomorphism f of classCk such that
f(U∩ S) =f(U)∩ H.
The integer number q :=n−p is called the co-dimension of S.
A submanifold of dimensionp= 1 is called a curve.
A submanifold of dimensionp= 2 is called a surface.
A submanifold of co-dimension q = 1 is called a hypersurface (in general one also assume implicitely that p >2).
Proposition 10. Submanifolds of co-dimension0 are open subsets of Rn. Submanifolds of dimension 0 are discrete set of points of Rn.
The proposition above dealt with two extreme (but still important) cases of submani- folds. The two next theorems will provide easy ways to produce more interesting examples.
Theorem 7. Let U be an open subset of Rp and ϕ:U → Rn an embedding of class Ck. Then,ϕ(U) is a submanifold ofRn of classCk.Conversely, ifS is a submanifold ofRn of class Ck and dimensionp,then for every point xof S, there exists an open neighbourhood V of x and an embedding ϕ:U →Rn of classCk, where U is an open subset of Rp, such that S ∩V =ϕ(U).
Proof. Letϕ(x00)∈ S :=ϕ(U). As in the proof of Proposition9, we may assume that the matrix [∂ϕ∂xj
i(x00)]1≤i,j≤p is invertible and we define
¯
ϕ: U ×Rq → Rn (x0, x00) 7→ ϕ(x) + (0, x00).
By the IFT, there exists a neighbourhood W of (x00,0) inU×Rq such that the restriction of ¯ϕ toW is a diffeomorphism. We now use the crucial fact (see Exercise3 thatϕ is an homeomorphism: since ϕ−1 is continuous, given an open neighbourhood U1 of x00, since ϕ(U1) is an open neighbourhood ofϕ(x00) inS, there exists an open neighbourhoodW1 of ϕ(x00) inRn such thatϕ(U1) =W1∩ S. We now definef := ¯ϕ−1 on ¯ϕ(U1×Rq)∩W and satisfies the properties that must be fulfilled by the diffeomorphism f of the definition of a submanifold.
Conversely let S be a submanifold of class Ck and x ∈ S. Let f : U → Rn like in the definition of a submanifold. We define ϕ := f−1 ◦ ι. Observe that ϕ is a bijection from (π◦f)(U) to its image and that ϕ−1 =π◦f. Hence ϕ is a homeomorphism. It is straightforward to check that ϕ(U) =S ∩V.
Definition 18. The maps ϕ:U → S whose existence is guaranted by the Theorem above are called local parametrizations of S. Furthermore, the inverse maps ϕ−1 : ϕ(U) → U are called local charts of S.
Remark 2. Observe that the existence of these local charts, since they are homeomor- phisms, prove that a submanifold is, in particular, a topological manifold (Definition 9).
Theorem 8. Let U be an open subset of Rn and g:U →Rq a submersion of class Ck. Then, for allc∈Rq, the setg−1({c}), if non empty, is a submanifold ofRnof classCkand dimension p. Conversely, if S is a submanifold of Rn of class Ck and dimension p, then for every point x ofS, there exists an open neighbourhoodU, a submersion g:U →Rq of class Ck and c∈Rq such that S ∩U =g−1({c}).
Proof. Since dgx has rank q, the matrix [∂g∂xj
i(x)]1≤i≤n,1≤j≤q has q columns which are linearly independent. Without loss of generality, we may assume that these are the lastq ones, i.e. the matrix [∂g∂xj
i(x)]p+1≤i,j≤nis invertible. We define
f : U → Rp×Rq
x= (x0, x00) 7→ (x0, g(x0, x00)−c).
Since ∂f∂xj
i(x) =δij if 1≤i≤pand ∂f∂xj
i(x) = ∂g∂xj
i(x) ifp+1≤i, j≤n, we deduce thatdfxis invertible. By the IFT, there exists a neighbourhoodV ofxsuch that the restriction offto V is a diffeomorphism. It is then straightforward to check thatf(V∩g−1({c})) =f(V)∩H.
Conversely, let S be a submanifold of class Ck. Given x ∈ S, let U be an open neighbourhood of x and a diffeomorphism f : U → f(U) such that f(U ∩ S) = f(U)∩ π(Rn).Define g:U →Rq by g=π00◦f, where
π00 : Rn=Rp×Rq → Rq x= (x0, x00) 7→ x00. Clearly, the rank of g is maximal, andg−1({0}) =S ∩U.
3.2 Differentiable maps and tangent spaces
Lemma 5. Let S a submanifold of class Ck of Euclidean space Rn and ϕ:U → S and ϕ0 : ˜U → S two local parametrizations of S such that V := ϕ(U)∩ϕ0(U0)6= ∅. Then the map ϕ−1◦ϕ0 :ϕ0−1(V)→ϕ−1(V) is a diffeomorphism of class Ck.
Proof. We fix a point x ∈ V. Exactly as in the proof of Proposition 9, we may assume without loss of generality that the first lines of the matrices [∂ϕ∂xj
i(ϕ−1(x))]1≤i≤p,1≤j≤n are linearly independent and we define
¯
ϕ: U ×Rq → Rn (x0, x00) 7→ ϕ(x) + (0, x00).
We then have
ϕ−1◦ϕ0= ( ¯ϕ−1◦π)◦ϕ0,
which is of classCk according to the chain rule. Inverting the roles ofϕand ϕ0, we obtain that ϕ0−1◦ϕ= (ϕ−1◦ϕ0)−1 is of classCk as well, which completes the proof.
Definition 19. The diffeomorphisms ϕ−1◦ϕ0 of the lemma above are called transition functions of S.
Definition 20. Let S be a submanifold of Euclidean space Rn of class Ck. A map f : S →Rm is said to be differentiable at xif, given a local parametrization ϕ:U →Rn of S such that x∈ϕ(U),the map f ◦ϕ is differentiable at ϕ−1(x).
Remark 3. The definition above is, a priori ambiguous, since it seems to rely on a particular choice of local parametrizationϕ. However, the chain rule applied to the identity f ◦ ϕ0 = (f ◦ϕ) ◦(ϕ−1 ◦ϕ0), as well as Lemma 5 show that, if ϕ and ϕ0 two local parametrizations, thenf◦ϕ0 is differentiable atϕ0−1(x)if and only iff◦ϕis differentiable at ϕ−1(x). Hence the definition is independent of the choice of parametrization.
Definition 21(”Meta-definition”). Definitions15and14extend to submanifolds, replac- ing the words ”open subset U of Rn” by ”open subset U of a submanifold of dimension n of Rn
0.”
Theorem 9 (”Meta-theorem”). Theorems 4 and 5 extend to submanifolds, replacing the words ”open subsetU of Rn” by ”open subsetU of a submanifold of dimension nof Rn
0.”
We have defined the concept of differentiable map on submanifolds, but so far we haven’t defined what should be the differential of such a map. Since we want this differ- ential to be still a linear map, some natural vector space must be attached to a point of a submanifold. This is the concept of tangent space, that we define now:
Definition 22. Let S be a submanifold of Euclidean space Rn of class Ck, k ≥ 1. The tangent space ofS atx∈ S is the set of all the velocity vectors att= 0 of the differentiable parametrized curves γ:I →Rn such thatγ(0) =x and γ(t)∈ S,∀t∈I:
TxS :={γ0(0)|γ :I →Rn, γ(0) =x, γ(t)∈ S,∀t∈I}.
Theorem 10. The tangent spaceTxS ofx toS is a linear subspace ofRn of dimensionp.
Proof. Letf :U →Rn be a diffeomorphism such thatf(U ∩ S) =f(U)∩ H, whereU is an open neighbourhood of x. We are going to prove that TxS = (dfx)−1(H), proceeding by proving the double inclusion. Since H is a linear subspace of dimension p and df is a linear isomorphism, the claim will follow.
We first consider a differentiable parametrized curve γ : I → Rn such that γ(0) = x and γ(t)∈ S,∀t∈I.It follows that∀t∈I, (f◦γ)(t)∈ H.Differentiating this identity at t= 0, using the chain rule, we get dtd(f◦γ)(t)
t=0 =dfx(γ0(0)) ∈ H. Since the choice of γ is arbitrary, we deduce thatdfx(TxS)⊂ H, so that TxS ⊂(dfx)−1(H).
Conversely, letv ∈ H. Then f(x) +tv ∈ H,∀t ∈R, and moreover, there exists >0 such thatf(x) +tv∈f(U)∩ H,∀t∈(−, ).It follows that the curve γ(t) :=f−1(f(x) + tv), t∈ (−, ) is contained in S ∩U and, sinceγ(0) =x, we have that γ0(0) ∈TxS.On the other hand, by the chain rule again, γ0(0) = (dff(x)−1 )(v) = (dfx)−1(v)∈TxS.Since the choice of vis arbitrary, it follows that (dfx)−1(H)⊂TxS.
Proposition 11. – Let U an open subset of Rp and ϕ:U → Rn an embedding. Let x∈ S :=ϕ(U). ThenTxS =dϕϕ−1(x)(Rp).
– Let U a open subset of Rn and g:U →Rq a submersion and c∈Rq. Let x∈ S :=
g−1({c}).Then TxS =Ker dgx.
The proposition above implies that given a local parametrization ϕ : U → Rn of a submanifold S of Rn, its differential dϕ induces a linear isomorphism between the lin- ear spaces Rp and TxS. This, and the next lemma allow to define the differential of a differentiable map between manifolds:
Definition 23. LetS be a submanifold of classCk of Rn andRn andf :S →Rn
0. Then if f is differentiable at x ∈ S, its differential is the linear map dfx ∈L(TxS,Rn
0) defined by dfx =d(f◦ϕ)ϕ−1(x)◦(dϕϕ−1(x))−1, where ϕis a local chart of S.
Remark 4. Again, it must be checked that the definition above is independent of a partic- ular choice of parametrization ϕ. This task is left to the reader as an (excellent) exercise.
Proposition 12. LetS andS0 be a submanifold ofRnandRn
0 respectively. If f :S → S0 is differentiable at x∈ S,thendfx(TxS)⊂Tf(x)S0.
Theorem 11 (The chain rule 2). Let f : S1 → S2 and g : S2 → S3 two maps, where S1,S2 and S3 are three submanifolds of class Ck of Rn1, Rn2 and Rn3 respectively. If f is differentiable at x ∈ S1 and g is differentiable at f(x) ∈ S2, then g◦f : S1 → S3 is differentiable at x and we have
d(g◦f)x =dgf(x)◦dfx. 3.3 Exercises
1. Prove Propositions 10,11 and 12. Prove that Definition23 is unambiguous, i.e. it does not depend on the the particular choice of parametrization ϕ.
2. Let x ∈ Rn and y ∈ Rm and define the matrice x⊗y := [aij]1≤i≤n,1≤j≤m, where aij =xiyj. Prove that the rank of the matrix is 0 if x or y vanishes, and 1 in the other cases. Conversely, prove that if M has rank 1, then there exists x ∈ Rn and y∈Rm such thatM =x⊗y.
Deduce that there is no square matrix which (i) has rank 1, (ii) is symmetric (i.e.
aij =aji and (iii) is traceless (i.e.Pn
i=1aii= 0).
3. Let
ϕ: (−1,∞) → R2 t 7→ (t3−t, t2).
Prove that ϕ is an immersion, is one-to-one, but not an embedding (Hint: to prove that ϕ is not an embedding, set xn = ϕ(n1 −1) and x0 = ϕ(1) = (0,1).
Clearly limn→∞xn = x0 in R2, and therefore for the trace topology as well, while limn→∞ϕ−1(xn) =−1,which does not belong to the open interval (−1,∞). Hence ϕ−1 cannot be continuous.)
4. We identify the set of square matrices n×n with Rn
2. Prove that the following subsets of matrices are submanifolds of Rn
2 a determine their dimension:
GL(n,R) :={M ∈M at(n×n)|detM 6= 0}
SL(n,R) :={M ∈M at(n×n)|detM = 1}
SO(n,R) :={M ∈GL(n,R)| hM x, M yi=hx, yi, ∀x, y∈Rn} Prove that
TIdSL(n,R) ={N = [aij]1≤i,j≤n|trN =
n
X
i=1
aii= 0}
TIdSO(n,R) ={N = [aij]1≤i,j≤n|aij +aji= 0,∀i, j,1≤i, j≤n}.
5. Fori= 1,2, letSi be a submanifold ofRni of dimensionpi and classCk. Prove that S1× S2:={(x1, x2)∈Rn1×Rn2|x1 ∈ S1, x2 ∈ S2}
is a submanifold of Rn1+n2 of dimension p1+p2 and class Ck. Let ¯x1 ∈ S1 and
¯
x2 ∈ S2; prove that the maps ι1 : S1 → S1 × S2 and ι2 : S2 → S1 × S2 defined by ι1(x1) = (x1,x¯2) and ι2(x2) = (¯x1, x2) respectively are immersions. Describe a natural linear isomorphism between T(x1,x2)S1× S2 and Tx1S1⊕Tx2S2.
6. Let S be a submanifold ofRn of dimension p and class Ck, k >1. Prove that the tangent bundle of S
TS:={(x, v)∈Rn×Rn|x∈ S, v∈TxS}
is a submanifold of R2n of dimension 2p and class Ck−1. Prove that the map π : TS → S defined byπ(x, v) =x is a submersion.
7. Fori= 1,2, letSibe a submanifold ofRnof dimensionpiand classCk. Assume that
∀x∈ S1∩ S2, the dimension of TxS1∩TxS2 is minimal, i.e.p1+p2−n. Prove that S1∩ S2 is a submanifold of dimensionp1+p2−nand classCk, and that∀x∈ S1∩ S2, we have
Tx(S1∩ S2) =TxS1∩TxS2.
4 Differentiable manifolds
4.1 What is a manifold?
Definition 24. A differentiable manifoldMof classCkand of dimensionnis a topogical set equipped with a family of homeomorphismsϕi :Ui→ M,where theUi are open subsets of Rn,such that
1. ϕi(Ui) is a covering ofM;
2. ∀i, j ∈I, the map ϕ−1i ◦ϕj is a diffeomorphism of class Ck whenever it is defined, i.e. when the subset W =ϕi(Ui)∩ϕj(Uj)6=∅;
The maps ϕi are called local parametrizations, their inverse ϕ−1i :ϕ(Ui)→ Ui are called local charts, and the diffeomorphisms ϕ−1i ◦ϕj are called transition maps. .
Example 4. A submanifold of dimension nand of class Ck of Rn
0 is also a manifold of dimension n of classCk. This comes from Proposition7 and Lemma 5.
There is a converse of this elementary fact, whose proof uses partition of unity, a technical tool that we shall introduce later.
Example 5 (The real projective space). Given a non-vanishing vector v of Rn+1, we define the vector line spanned byv:
[v] :={w∈Rn+1|w is collinear tov}.
The real projective space of dimensionn is the set of such vector lines:
RPn={[v], v∈Rn+1}.
