## When are two finite abelian groups isomorphic?

A useful method to prove that two (finite) groups are nonisomorphic is to compare the number of elements of a given order. Of course, the converse is not always true. For example, let us consider Heisenberg group

$H_p : = \left\{ \left( \begin{array}{ccc} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array} \right) \mid a,b,c \in \mathbb{Z}_p \right\}$,

with $p \geq 3$ a prime number. Noticing that for all $n \geq 0$ and $a,b,c \in \mathbb{Z}_p$,

$\left( \begin{array}{ccc} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array} \right)^n = \left( \begin{array}{ccc} 1 & na & nb + \frac{n(n-1)}{2} ac \\ 0 & 1 & nc \\ 0 & 0 & 1 \end{array} \right)$,

we deduce that $H_p$ is a group of order $p^3$ with $p^3-1$ elements of order $p$, as $G_p :=\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_p$. However, $G_p$ and $H_p$ are not isomorphic: one is abelian and not the other.

Nevertheless, it turns out that considering the number of elements of any order is sufficient for finite abelian groups:

Main Theorem: If two finite abelian groups have the same number of elements for any order, then they are isomorphic.

In order to prove our main theorem, we first prove several classical lemmas, usually introduced to show the fundamental theorem of finite abelian groups.

If $G$ is a finite abelian group and $p$ is a prime number, we define the $p$-torsion of $G$ by

$G(p)= \{ g\in G \mid \mathrm{ord}(g) \ \text{is a power of} \ p \}$.

It is straightforward to check that $G(p)$ is a subgroup of $G$, and our first lemma shows that a finite abelian group is determined by its $p$-torsions.

Lemma 1: Let $G$ be a finite abelian group. Then $G \simeq \bigoplus\limits_{p \in \mathbb{P}} G(p)$.

Proof. Let $g \in G$ and let $n=p_1^{r_1} \dots p_m^{r_m}$ denote its order. Because $\displaystyle \frac{n}{p_1^{r_1}}, \dots, \frac{n}{p_m^{r_m}}$ are coprime, there exist $u_1, \dots, u_m \in \mathbb{Z}$ such that $\displaystyle u_1 \frac{n}{p_1^{r_1}} + \dots + u_m \frac{n}{p_m^{r_m}} =1$, hence

$g = u_1 \underset{\in G(p_1)}{\underbrace{ \frac{n}{p_1^{r_1}} \cdot g }} +\dots + u_m \underset{\in G(p_m)}{ \underbrace{ \frac{n}{p_m^{r_m}} \cdot g }}$,

so $G= G(p_1)+ \dots + G(p_m)$.

Let $g \in G(p) \cap (G(p_1)+ \dots+G(p_r))$ where $p,p_1,\dots,p_r$ is a family of different primes. In particular, $g=h_1+ \dots + h_r$ for some $h_i \in G(p_i)$. Therefore, when $n_1,\dots,n_r$ are large enough,

$p_1^{n_1} \dots p_r^{n_r} \cdot g = 0$,

so either $p_1^{n_1} \dots p_r^{n_r}$ is divisible by $p$ or $g=0$. Thus, $G(p) \cap (G(p_1)+ \dots + G(p_r)=\{0\}$.

Finally, we deduce that $G= \bigoplus\limits_{p \in \mathbb{P}} G(p)$. $\square$

Afterward, we prove Cauchy’s theorem for abelian groups:

Lemma 2: Let $G$ be a finite abelian group and $p$ be a prime number. If $p$ divides $|G|$, then $G$ has an element of order $p$.

Proof. According to lemma 1, it is sufficient to prove that an abelian $p$-group has an element of order $p$. We prove it by induction on $|G|$. If $|G|=p$, then $G$ is cyclic and there is nothing to prove. If $|G|>p$, either $G$ has a nontrivial proper subgroup $H$ or $G$ is cyclic (and again there is nothing to prove). In the first case, we may apply our induction hypothesis to $H$ to deduce that $G$ has an element of order $p$. $\square$

Lemma 3: Let $G$ be a finite abelian $p$-group. If $G$ has only one subgroup $H$ of order $p$ then $G$ is cyclic.

Proof. Let us prove lemma 3 by induction on $|G|$. If $|G|=p$ then $G$ is cyclic and there is nothing to prove. If $|G|>p$, let

$\varphi : \left\{ \begin{array}{ccc} G & \to & G \\ g & \mapsto & pg \end{array} \right.$.

Because $\mathrm{ker}(\varphi)$ is the set of  elements of $G$ of order at most $p$, we deduce that $\mathrm{ker}(\varphi)=H$. In particular, $\mathrm{ker}(\varphi) \neq \{0\}$ implies that $| \varphi(G)| < |G|$ so that our induction hypothesis may be applied to $\varphi(G)$: there exists an element $g \in G$ generating $\varphi(G)$.

Because $\varphi(G) \simeq G/H$, we deduce that $\langle g \rangle +H=G$. But $\langle g \rangle$ cannot have more than one subgroup of order $p$, hence $H \leq \langle g \rangle$ and finally $G= \langle g \rangle$. Therefore, $G$ is cyclic. $\square$

Lemma 4: Let $G$ be a finite abelian $p$-group and let $C \leq G$ denote a cyclic subgroup of maximal order. Then $G$ contains a subgroup $H$ so that $G= H \oplus C$.

Proof. Let us prove lemma 4 by induction on $|G|$. If $|G|=p$, $G=C$ and there is nothing to prove. More generally, if $G$ is cyclic then $G=C$ and there is nothing to prove, so, if $|G|>p$, we may suppose that $G$ is not cyclic.

According to lemma 3, $G$ has a subgroup $K$ of order $p$ that is different from the one contained in $C$. In particular, $K \cap C= \{0\}$, so $C/K$ is a cyclic subgroup of maximal order in $G/K$, so that we may apply our induction hypothesis to $G/K$, and find a subgroup $J \leq G/K$ satisfying

$G/K=C/K \oplus J$.

Let $H$ denote the reciprocal image of $J$ in $G$ by the quotient map $G \to G/K$; in particular, $J=H/K$. Because $K$ is a subgroup of $H$, $G=C+H+K=C+H$. Because $(C/K) \cap (H/K) = \{0\}$ in $G/K$ implies $(C+K) \cap H=K$ in $G$, we deduce that $C \cap H=C \cap K =\{0\}$. Therefore, $G= H \oplus C$. $\square$

Proof of our theorem. Let $G_1,G_2$ be two finite abelian groups with the same number of elements of each order. According to lemma 1, we may suppose that they are $p$-groups for some prime $p$. If $|G_1|=1$, clearly $G_1 \simeq \{0\} \simeq G_2$.

Now suppose by induction that our theorem holds when one group has an order smaller than or equal to $n$ and suppose that $|G_1|=n+1$. According to lemma 4, there exist $H_1 \leq G_1$, $H_2 \leq G_2$ and $m \geq 1$ such that $G_1 \simeq H_1 \oplus \mathbb{Z}_m$ and $G_2 \simeq H_2 \oplus \mathbb{Z}_m$, where $m$ is the highest order of an element of $G_1$ and $G_2$.

Applying our induction hypothesis, we deduce that $H_1 \simeq H_2$ hence $G_1 \simeq G_2$. $\square$

Therefore, our theorem is an easy consequence of lemma 4. It is worth noticing that the fact that any finite abelian group is a direct sum of cyclic groups is also an easy corollary of lemma 4.

We conclude with two corollaries:

Corollary 1: If $G \oplus H \simeq K \oplus H$ then $G \simeq K$.

Corollary 2: If $G \oplus G \simeq H \oplus H$ then $G \simeq H$.

They can be shown by induction on the cardinality and using the identity

$\displaystyle n_k(G \oplus H)= n_k(G)n_k(H)+ n_k(G) \sum\limits_{j=1}^{k-1} n_j(H) + n_k(H) \sum\limits_{j=1}^{k-1} n_j(G)$,

where $n_k( \cdot)$ denotes the number of elements of order $k$.

In fact, the two corollaries above are true for any finite groups.

## SO(3) is a simple group: a geometric proof

Our aim is to show that the group of orientation-preserving isometries of $\mathbb{R}^3$, that is

$SO(3) = \{ M \in GL(3,\mathbb{R}) \mid MM^T= \mathrm{Id}, \det(M) =1 \}$,

is a simple group. Here we use a geometric point of view, mixing the proofs that can be found in Stillwell’s book, Naive Lie theory, and Perrin’s book, Cours d’algèbre. We begin with an easy structure lemma:

Lemma 1: If $M \in SO(3)$, there exists an orthonormal basis in which $M$ may be written as

$\left( \begin{matrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{matrix} \right)$

for some $\theta \in [0,2 \pi)$.

Proof. First, notice that

$\det(M- \mathrm{Id})= \det( M- \ MM^T)= \det(M) \cdot \det( \mathrm{Id}- M) = (-1)^3 \det( M- \mathrm{Id})$,

hence $\det(M- \mathrm{Id})=0$ ie. there exists $v \in \mathbb{R}^3 \backslash \{ 0 \}$ such that $Mv=v$. Let $W=v^{\bot}$. Then for all $w \in W$,

$\langle Mw,v \rangle = \langle w, Mv \rangle = \langle w,v \rangle=0$,

so $W$ is stable under $M$. Therefore, $M$ may be written as $\left( \begin{matrix} 1 & 0 \\ 0 & S \end{matrix} \right)$ in an orthonormal basis; in particular, because $MM^T= \mathrm{Id}$ and $\det(M)=1$, $S \in SO(2)$.

If $S= \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$, the equality $S^T S = \mathrm{Id}$ becomes $\left\{ \begin{array}{l} a^2+c^2=1 \\ b^2+d^2=1 \\ ab+cd=0 \\ ad-bc=1 \end{array} \right.$. The two first equations allow us to set $\left\{ \begin{array}{l} a= \cos(\theta), \ c= \sin(\theta) \\ d = \cos( \varphi), \ b= \sin(\theta) \end{array} \right.$ for some $\theta, \varphi \in [0,2\pi)$. Then, the two last equations give

$\left\{ \begin{array}{l} \cos(\theta) \sin(\varphi) + \sin(\theta) \cos(\varphi)=0 \\ \cos(\theta) \cos(\varphi)- \sin(\varphi) \sin(\theta) = 1 \end{array} \right.$, that is $\left\{ \begin{array}{l} \sin(\theta+ \varphi)=0 \\ \cos( \theta+ \varphi)=1 \end{array} \right.$.

Therefore, $\varphi=- \theta ~ \text{mod} ~ 2\pi$ and $M$ may be written as the mentionned matrix. $\square$

Definition: If $M$ is as in the lemma, we say that $M$ is a rotation of axis $v$ and angle $\theta$.

Lemma 2: Let $M \in SO(3)$ and $P \in SO(3)$. Suppose that $M$ is a rotation of axis $v$. Then $PMP^{-1}$ is a rotation of axis $Pv$.

Proof. Let $W = v^{\bot}$. Then $PMP^{-1}(Pv)=Pv$ and $PW=(Pv)^{\bot}$ is stable under $PMP^{-1}$. Therefore, according to lemma 1, $PMP^{-1}$ is a rotation of axis $Pv$. $\square$

Lemma 3: Following the notations given on the figure below, the rotation around $P$ of angle $\theta$ composed with the rotation around $Q$ of angle $\varphi$ is a rotation around $R$ whose angle equals to twice the angle at $R$ of the spherical triangle $PQR$.

Proof. Let $H_1,H_2$ be two planes intersecting along a line $D$. Let $r_1,r_2$ be the reflections with respect to $H_1,H_2$ respectively.  Then $r_1r_2$ fixes $D$ and it is not difficult to see that the restriction of $r_1r_2$ to the plane normal to $D$ is a rotation of angle twice the angle $\theta$ between $H_1$ and $H_2$. Therefore, according to lemma 1, $r_1r_2$ is a rotation of axis $D$ and of angle $2\theta$.

Thus, if $R_{\varphi}, R_{\theta}$ denotes the rotations around $Q, P$ of angles $\varphi,\theta$ respectively, and if $r_{\mathcal{L}}, r_{\mathcal{M}}, r_{\mathcal{N}}$ denotes the reflections with respect to $\mathcal{L},\mathcal{M}, \mathcal{N}$ respectively, we can write

$R_{\varphi}=r_{\mathcal{N}} r_{\mathcal{M}}$ and $R_{\theta}= r_{\mathcal{M}} r_{\mathcal{L}}$,

hence $R_{\varphi} R_{\theta}=r_{\mathcal{N}} r_{\mathcal{L}}$, that is $R_{\varphi} R_{\theta}$ is a rotation around $R$ of angle twice the angle at $R$ of the spherical triangle $PQR$. $\square$

Now, we give two lemmas to characterize the natural action of $SO(3)$ on the sphere $\mathbb{S}^2$.

Lemma 4: The action $SO(3) \curvearrowright \mathbb{S}^2$ is transitive, ie. for every $x,y \in \mathbb{S}^2$ there exists $g \in SO(3)$ such that $g \cdot x =y$.

In fact, it is a special case of the following lemma, which states that the action $SO(3) \curvearrowright \mathbb{S}^2$ is “as $2$-transitive as possible”.

Lemma 5: The action $SO(3) \curvearrowright \mathbb{S}^2$ is isometrically 2-transitive, ie. for every $x_1,x_2,y_1,y_2 \in \mathbb{S}^2$ satisfying $\| x_1 -y_1 \| = \| x_2 - y_2 \|$ there exists $g \in SO(3)$ such that $g \cdot x_1=x_2$ and $g \cdot y_1 = y_2$.

Proof. Let $R_1$ be a rotation sending $x_1$ to $x_2$, and $R_2$ be a rotation of axis $x_2=Rx_1$ sending $R_1y_1$ to $y_2$. Then $R := R_2 \circ R_1 \in SO(3)$ and $\left\{ \begin{array}{l} Rx_1= R_2(R_1x_1)=R_2x_2=x_2 \\ Ry_1= R_2(R_1y_1)=y_2 \end{array} \right. . \square$

Theorem: $SO(3)$ is a simple group.

Proof. Let $N \lhd SO(3)$ be a non-trivial normal subgroup. We first want to show that the action $N \curvearrowright \mathbb{S}^2$ is transitive. Let $u \in N \backslash \{ \mathrm{Id} \}$ be a rotation of axis $\mathbb{R}a$ for some $a \in \mathbb{S}^2$ and let $x \in a^{ \bot} \cap \mathbb{S}^2$. Because $u \neq \mathrm{Id}$, $u(x) \neq x$, so $d:= \| x-u(x) \| >0$.

Notice that $u$ sends the meridian $L$ passing through $x$ to the meridian passing through $u(x)$, so $\|z-u(z) \|$ runs over $[0,d]$ when $z$ runs over $L$: $\|z -u(z) \|$ tends to $0$ (resp. $d$) when $z$ tends to the north pole (resp. to the equator). More precisely, if $0 \leq m \leq d$, it is possible to find so $\lambda \geq 0$ such that

$\displaystyle \frac{ \| u(x+ \lambda a) - (x+ \lambda a) \| }{ \| x+ \lambda a \| } =m$.

Because $u$ is linear and $x \in a^{\bot}$, the above equation is equivalent to

$\displaystyle d=m \| x+ \lambda a \| = \sqrt{ \|x\|^2 + \lambda^2 \| a \|^2} = \sqrt{1+ \lambda^2}$, hence $\displaystyle \lambda = \sqrt{ \frac{d^2}{m^2} -1 }$.

Let $v,w \in \mathbb{S}^2$ such that $\| v - w \| \leq d$. We just saw that there exists $y \in \mathbb{S}^2$ such that $\| u(y)-y \| = \|v-w \|$. According to lemma 5, there exists $g \in SO(3)$ such that $g(v)=y$ and $g(w)=u(y)$. Then $gug^{-1} \in N$ because $N$ is a normal subgroup and $gug^{-1}(v)=w$.

Therefore, we proved that the action $N \curvearrowright \mathbb{S}^2$ is transitive “for the small distances”. Now, if $v,w \in \mathbb{S}^2$ are any points, clearly there exists a sequence of points $v=p_0,p_1,\dots,p_n=w \in \mathbb{S}^2$ satisfying $\| p_i - p_{i+1} \| \leq d$ for all $0 \leq i \leq n-1$. In particular, for all $0 \leq i \leq n-1$, there exists $u_i \in N$ such that $u_i(p_i)=p_{i+1}$, and finally $u_{n-1} \cdots u_0$ is an element of $N$ sending $v$ to $w$. Thus, the action $N \curvearrowright \mathbb{S}^2$ is transitive.

In particular, there exist $x \in \mathbb{S}^2$ and $u \in N$ such that $u(x)=-x$; such an element of $SO(3)$, with $-1$ as an eigenvalue, is called a codimension-two reflection. Clearly, using lemmas 1 and 2, two codimension-two reflections are always conjugated. Therefore, since $N$ is a normal subgroup, any codimension-two reflection belongs to $N$.  We deduce, according to lemma 3 and using the notations of the figure below, that the rotation of angle $\theta$ around $R$ belongs to $N$.

Clearly, when $P$ runs over the equator, $\theta$ runs over $[0, \pi)$. Therefore, $N$ contains rotations of any angle. Because two rotations of the same angle are conjugated and $N$ is a normal subgroup, we deduce that any rotation belongs to $N$, that is $N= SO(3)$. $\square$

Nota Bene: The conclusion of our proof, based on lemma 3, may be replaced with an algebraic argument, stating that the codimension-two reflections generate $SO(3)$. Therefore, in order to show that a normal subgroup is $SO(3)$, it is sufficient to find only one codimension-two reflection belonging to it.

Another classical proof is to find a codimension-two reflection in a nontrivial normal subgroup is to use a connectedness argument and the continuity of the trace $M \mapsto \mathrm{tr}(M)=1+2 \cos(\theta)$.

Finally, a proof based on the isomorphism $SU(2)/ \mathbb{Z}_2 \simeq SO(3)$ exhibit in Fundamental group of SO(3) and Quaternions can be found in Artin’s book, Algebra.

## Euler-Poincaré characteristic (I): Euler’s formula and graph theory

This is the introductory note of a sequence of notes dedicated to Euler-Poincaré characteristic. Here, we deal with Euler’s formula from a graph theoritical viewpoint and exhibit several applications including a proof of Pick’s theorem and the classification of Plato’s solids.

We say that a grah is planar if it can be drawn on $\mathbb{R}^2$ (or equivalently on $\mathbb{S}^2$) without edge-crossings. Determining whether a given graph is planar or not is a difficult question in general, and Euler’s formula gives a useful necessary condition:

Theorem: (Euler’s formula) Let $\Gamma \subset \mathbb{R}^2$ be a connected planar graph. Let $v,e,f$ denote respectively the number of vertices, edges and faces (including the unbounded one) of $\Gamma$. Then

$v-e+f=2$.

Proof. Let $T \subset \Gamma$ be a spanning tree. Let $\Gamma^*$ be the dual graph of $\Gamma$ and $T^*$ be the subgraph of $\Gamma^*$ corresponding to the edges of $\Gamma \backslash T$.

Because $T$ is connected, $T^*$ is acyclic, and because $T$ is acyclic, $T^*$ is connected. Therefore, $T^*$ is a spanning tree. We deduce that

$\left\{ \begin{array}{l} v(T)-e(T)=1 \\ v(T^*)-v(T^*)=1 \end{array} \right. \Rightarrow v(T) -e(T)-e(T^*) +v(T^*)=2$.

But $v(T)=v$, $v(T^*)=v(\Gamma^*)=f$ and $e(T)+e(T^*)=e$, hence $v-e+f=2$. $\square$

Twenty proofs of Euler’s formula (including this one) can be found on Eppstein’s webpage.

Theorem: The complete graph $K_n$ is planar if and only if $n \in \{0,1,2,3,4 \}$.

Proof. We may see easily that $K_n$ is planar when $0 \leq n \leq 4$, just draw it! Then, because $K_n$ is a subgraph of $K_m$ when $n \leq m$, it is sufficient to prove that $K_5$ is not planar. Suppose by contradiction that $K_5$ can be drawn on the plane without edge-crossings.

Let $f_i$ be the corresponding faces, whose boundaries have $n(f_i)$ edges. Notice that $K_5$ is simple, so $n(f_i) \geq 3$, and that each edge belongs to two faces, hence

$\displaystyle 2e= \sum\limits_i n(f_i) \geq 3f$.

We deduce from Euler’s formula that $2=v-e+ f \leq v- \frac{1}{3}e$. But $K_5$ has five vertices and ten edges, so the relation becomes $6 \leq 5$, a contradiction. $\square$

Theorem: The complete bipartite graph $K_{m,n}$ is planar if and only if $m$ or $n \leq 2$.

Proof. If $m \leq 2$ or $n \leq 2$, then $K_{m,n}$ is planar as shown by the following figure:

Then, since $K_{m,n}$ is a subgraph of $K_{p,q}$ if $m \leq p$ and $n \leq q$, it is sufficient to prove that $K_{3,3}$ is not planar. Suppose by contradiction that $K_{3,3}$ can be drawn on the plane without edge-crossings.

Let $f_i$ be the corresponding faces, whose boundaries have $n(f_i)$ edges. Notice that any cycle in $K_{3,3}$ has even length, so $n(f_i) \geq 4$, and that each edge belongs to two faces, hence

$\displaystyle 2e= \sum\limits_i n(f_i) \geq 4f$.

We deduce from Euler’s formula that $2=v-e+f \leq v- \frac{1}{2}e$. But $K_{3,3}$ has six vertices and nine edges, so the relation becomes $4 \leq 3$, a contradiction. $\square$

Theorem: Petersen’s graph is not planar.

Proof. Suppose by contradiction that Petersen’s graph can be drawn on the plane without edge-crossings and let $f_i$ be the correponding faces. Noticing that the smallest cycle in Petersen’s graph has length five, we deduce that $n(f_i) \geq 5$, hence

$\displaystyle 2e= \sum\limits_i n(f_i) \geq 5f$.

From Euler’s formula, we deduce that $2=v-e+f \leq v- \frac{3}{5}e$. But Petersen’s graph has ten vertices and fifteen edges, so the relation becomes $2 \leq 1$, a contradiction. $\square$

Our proofs are in fact based on the corollary of Euler’s formula stating that a connected planar graph $\Gamma$ satisfies

$\displaystyle 2 \leq v- \frac{\ell-2}{\ell} e$

where $\ell$ is the length of the smallest cycle in $\Gamma$.

In fact, topological graph theory asks more generally whether, given a graph $\Gamma$ and a space $X$, there exists an embedding $\Gamma \hookrightarrow X$. The answer depends widely on the topology of $X$, so it is not surprinsing to notice topological applications of Euler’s formula.

Theorem: $\mathbb{R}^2$ and $\mathbb{R}^3$ are not homeomorphic.

Proof. We just saw that there is no embedding $K_5 \hookrightarrow \mathbb{R}^2$ whereas there is clearly an embedding $K_5 \hookrightarrow \mathbb{R}^3$. $\square$

Theorem: $\mathbb{S}^2$ and $\mathbb{R}P^2$ are not homeomorphic.

Proof. We just saw that there is no embedding $K_5 \hookrightarrow \mathbb{R}^2$. However, it is possible to draw $K_5$ on $\mathbb{R}P^2$ (or on the Möbius band) without edge-crossings, as illustrated by the following figure:

$\square$

During our proof of Euler’s formula, we implicitely used Jordan curve theorem to state that connectedness and acyclicness are dual; moreover, applying Euler’s formula for a loop gives Jordan curve theorem, so these two theorems are in fact equivalent. Therefore, while the two previous theorems follow easily from Jordan curve theorem, we did not simplify the proofs, but we formulated them as motivating examples of purely topological applications of topological graph theory.

We already gave a proof of Pick’s theorem in a previous note, here is another one based on Euler’s formula. We also use the following lemma, proved in the mentionned note:

Lemma: Let $P \subset \mathbb{R}^2$ be a polygon. Then $P$ contains a diagonal, that is an edge between two vertices lying in the interior of $P$.

Theorem: (Pick) Let $P \subset \mathbb{R}^2$ be a simple polygon whose vertices are in $\mathbb{Z}^2$. Let $i$ (resp. $b$) denote the cardinality of $\mathrm{int}(P) \cap \mathbb{Z}^2$ (resp. of $\partial P \cap \mathbb{Z}^2$). Then the area $A(P)$ of $P$ is given by the formula

$\displaystyle A(P)= i + \frac{b}{2}-1$.

Proof. According to the previous lemma, we may triangulate $P$. Moreover, if a triangle $T$ contains a point of $\mathbb{Z}^2$, we may join this point to the three vertices of $T$. Therefore, we may triangulate $P$ using triangles intersecting $\mathbb{Z}^2$ only at their vertices. For example:

Now, it is not difficult to check Pick’s formula for the previous kind of triangles, hence

$A(P)= \frac{1}{2}(f-1)$.

Let $e_b$ (resp. $e_i$) be the number of boundary edges (resp. interior edges) of the triangulation. Noticing that a boundary edge (resp. an interior edge) lies in exactly one (resp. exactly two) triangle(s), we deduce that $3(f-1)=2e_b+e_i$. Moreover, we clearly have $e_b=b$, $e=e_i+e_b$ and $v=i+b$. According to Euler’s formula:

$2 = v-e+f = v - (e_i+e_b) + \left( \frac{2}{3} e_i+ \frac{1}{3} e_b + 1 \right) =v- \frac{1}{3} e_i - \frac{2}{3} e_b +1 \\ \\ 2= (i+b) - \frac{1}{3} \left( 3 A(P) - \frac{1}{2} b \right) - \frac{2}{3} b +1 = i- A(P) + \frac{1}{2} b+1$

hence $A(P)=i + \frac{1}{2}b-1$. $\square$

Theorem: There exist only five Plato’s solids, namely the tetrahedron, the hexahedron (cube), the octahedron, the dodecahedron and the icosahedron.

Proof. If each vertex has degree $m$ and each face is a $n$-gon, then summing the number of edges adjacent to each vertex we get $2e=mv$; summing the number of boundary edges of each face we get $2e=nf$. Using Euler’s formula, we deduce the relation

$\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{1}{2} + \frac{1}{e}$.

Of course, we necessarily have $m,n \geq 3$. But if $m,n \geq 4$ then $\frac{1}{n}+ \frac{1}{m} \leq \frac{1}{2}$, which is in contradiction with the previous relation; therefore, either $m=3$ and necessarily $n \in \{ 3,4,5 \}$ so that $\frac{1}{n}+ \frac{1}{m} > \frac{1}{2}$, or $n=3$ and $m \in \{3,4,5\}$ for the same reason.

Consequently, $(m,n,e)$ is:

• either $(3,3,6)$ and the polyhedron is a tetrahedron,
• or $(3,4,12)$ and the polyhedron is a hexahedron (cube),
• or $(3,5,30)$ and the polyhedron is a dodecahedron,
• or $(4,3,12)$ and the polyhedron is an octahedron,
• or $(5,3,30)$ and the polyhedron is an icosahedron. $\square$

Theorem: Any map can be coloured so that two adjacent regions have different colors using only six colors.

Proof. First, we prove that a simple planar graph has a vertex of degree at most five. Suppose by contradiction that any vertex has degree at least six. Let $f_k$ denote the number of faces whose boundaries contain exactly $k$ edges and let $v_k$ denote the number of vertices of degree $k$. Because our graph is simple,

$\left\{ \begin{array}{l} f=f_3+f_4+ \cdots \\ 2e= 3f_3+4f_4+ \cdots \end{array} \right. \Rightarrow 2e \geq 3(f_3+f_4+ \cdots)=3f$,

and because each vertex has degree at least six,

$\left\{ \begin{array}{l} v=v_6+v_7+ \cdots \\ 2e= 6v_6+7v_7+ \cdots \end{array} \right. \Rightarrow 2e \geq 6(v_6+v_7+ \cdots) = 2v$.

Using Euler’s formula, we deduce $2 =v-e+f \leq \frac{1}{3}e- e + \frac{2}{3}e=0$, a contradiction.

Now, a map defines a planar graph; let $\Gamma$ be its dual graph. In particular, $\Gamma$ is a simple planar graph, and our theorem is equivalent to the fact that the vertices of $\Gamma$ can be coloured using only six colors so that two adjacent vertices have different colors. We show this by induction on the number $n$ of vertices of $\Gamma$. For $n\leq 6$, there is nothing to prove. Otherwise, let $v$ be a vertex of degree at most five. Thanks to our induction hypothesis, $\Gamma \backslash \{v\}$ can be coloured in a good way. Now, we can label $v$ by any color that is different from its neighbors; it is possible because $v$ has degree at most five and that we use six colors. Finally, $\Gamma$ is coloured in a good way. $\square$

Actually, the theorem is true for four colors only, but the proof is far from being as elementary as the previous one; see four color theorem. An elementary proof for five colors can be found in M. Aigner and G. Ziegler’s book, Proofs from the Book.

## Free groups acting on the circle

We present here an unusual application of Baire category theorem between topology, group theory and dynamical systems. Roughly speaking, we prove that almost all pairs of homeomorphisms of the circle are unrelated; in particular, it leads to natural occurrences of free groups.

More precisely, if $\mathrm{Homeo}^+(\mathbb{S}^1)$ denotes the set of orientation-preserving homeomorphisms of the circle, endowed with the distance

$d(f,g)= \max \left( \| f-g \|_{\infty}, \| f^{-1}-g^{-1} \|_{\infty} \right), \ \forall f,g \in \mathrm{Homeo}^+(\mathbb{S}^1)$

where $\| \cdot \|_{\infty} = \sup\limits_{\mathbb{S}^1} \| \cdot \|$, we prove

Theorem: The set of pairs $(f,g) \in \mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1)$ generating a free subgroup of rank two in $\mathrm{Homeo}^+(\mathbb{S}^1)$ is a dense $G_{\delta}$ in $\mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1)$.

Proof. With the distance given above, $\mathrm{Homeo}^+(\mathbb{S}^1)$ is a complete metric space, so Baire category theorem applies.

If $w \in \mathbb{F}_2$ is a reduced word, let $X_w$ denote the set of

$(f,g,x) \in \mathrm{Homeo}^+( \mathbb{S}^1) \times \mathrm{Homeo}^+( \mathbb{S}^1) \times \mathbb{S}^1$

satisfying $w(f,g)(x) =x$. Let us show that each $X_w$ is a closed set whose interior is empty; it will allow us to deduce that the set of  $(f,g) \in \mathrm{Homeo}^+(\mathbb{S}^1) \times \mathrm{Homeo}^+(\mathbb{S}^1)$ satisfying $w(f,g)= \mathrm{Id}$ is a closed set whose interior is nonempty: indeed, if $w(f,g)= \mathrm{Id}$ and $x \in \mathbb{S}^1$, then $(f,g,x) \in X_w$, so there exist $f_0$, $g_0$ and $x_0$ arbitrarily close to $f$, $g$ and $x$ respectively, so that $(f_0,g_0,x_0) \notin X_w$ ie. $w(f_0,g_0)(x_0) \neq x_0$, hence $w(f_0,g_0) \neq \mathrm{Id}$; consequently, according to Baire category theorem, the set

$\displaystyle \{ (f,g) \mid \langle f,g \rangle \not\simeq \mathbb{F}_2 \} = \bigcup\limits_{w \in \mathbb{F}_2} \{ (f,g) \mid w(f,g)= \mathrm{Id} \}$

has empty interior and its complement  is a dense $G_{\delta}$, completing the proof.

By contradiction, let $w$ be a word of minimal length $k$ such that $X_w$ has nonempty interior. In particular, there exists a nonempty open set $U \subset X_w$. For all words $w_1,w_2 \in \mathbb{F}_2$, let

$F(w_1,w_2)= \{ (f,g,x) \mid w_1(f,g)w_2(f,g)(x)=w_2(f,g)(x) \}$

denote the image of $X_{w_1}$ by the homeomorphism  $(f,g,x) \mapsto (f,g,w_2(f,g)^{-1}(x))$; in particular, $F(w_1,w_2)$ is closed set whose interior is empty if $\mathrm{\ell g}(w_1),\mathrm{\ell g}(w_2).

Because $\bigcup\limits_{\mathrm{\ell g}(w_1),\mathrm{\ell g}(w_2) is itself a closed set whose interior is empty, there exists $(f,g,x) \in U$ such that $(f,g,x) \notin F(w_1,w_2)$ for all words satisfying $\mathrm{lg}(w_1), \mathrm{lg}(w_2).

Let $w(f,g)=h_1 \circ \cdots \circ h_k$ where $h_i \in \{ f,f^{-1},g,g^{-1}\}$ ($1 \leq i \leq k$) and set

$x_1=h_1(x), ~ \dots, ~ x_{k-1}=h_{k-1}(x_{k-2}), ~ x_k=h_k(x_{k-1})$.

First, notice that $x_k=w(f,g)(x)=x$ since $(f,g,x) \in X_w$ and that $x_i \neq x_j$ if $i \neq j$. Indeed, if it was not the case, there would exist a proper subword $w_1(f,g)$ of $w(f,g)$ fixing $x_i$, while $x_i=w_2(f,g)(x)$ for some word $w_2$, hence $(f,g,x) \in F(w_1,w_2)$, a contradiction.

To conclude, let $\tilde{f}$ and $\tilde{g}$ be two homeomorphisms arbitrarily close to $f$ and $g$ respectively, and satisfying $x_1= \tilde{h}_1(x_1), ~ \dots, ~ x_{k-1}= \tilde{h}_{k-1}(x_{k-2})$ but $x_k \neq \tilde{h}_k(x_{k-1})$.

[To do that, let $\mathbb{S}^1= \mathbb{R}/ \mathbb{Z}$ and show that each homeomorphism of the circle can be approximated by a piecewise linear homeomorphism (using a classical argument based on uniform continuity); then we deduce easily that a homeomorphism of the circle can be approximated by a homeomorphism agreeing on a finite set of points and disagreeing on another finite set of points.]

Thus, $(\tilde{f}, \tilde{g},x) \in U \subset X_w$ because $U$ is open, but $w(\tilde{f},\tilde{g})(x)=\tilde{h}_k(x_{k-1}) \neq x_k=x$ by construction, a contradiction. $\square$

We deduce from our theorem that $\mathrm{Homeo}^+(\mathbb{S}^1)$ has a lot of free subgroups, but the given proof does not provide any explicit example. In order to exhibit free subgroups, we use ping-pong lemma:

Lemma: Let $G$ be a group acting on a set $S$ such that there exist $x,y \in G$ and $X,Y \subset S$ satisfying

• $X$ and $Y$ are non-empty and disjoint,
• $x^n \cdot X \subset Y$ and $y^n \cdot Y \subset X$ for all $n \in \mathbb{Z} \backslash \{ 0\}$.

Then $\langle x,y \rangle$ is a free group of basis $\{x,y \}$.

Proof. Clearly, the second condition implies that $x$ and $y$ has infinite order. Moreover, any reduced word $w$ over $\{ x^{\pm 1}, y^{\pm 1} \}$ not in $\langle x \rangle \cup \langle y \rangle$ can be written (up to conjugacy) as

$w= y^{m_0} \cdot x^{n_1} \cdot y^{m_1} \cdots x^{n_r} \cdot y^{m_r}$ with $r \geq 1$ and $n_i,m_i \in \mathbb{S} \backslash \{ 0 \}$.

Let $a \in Y$. If $w=1$ in $G$, then $a = w \cdot a$. But $a \in Y$, hence $y^{m_r} \cdot a \in X$, hence $x^{n_r} \cdot y^{m_r} \cdot a \in Y$, and so on; finally, $w \cdot a \in X$. We deduce that $a = w \cdot a \in X \cap Y = \emptyset$, a contradiction. Therefore, $w \neq 1$ in $G$ and $\langle x,y \rangle$ is a free subgroup of basis $\{x,y \}$. $\square$

The first example of free subgroup of $\mathrm{Homeo}^+(\mathbb{S}^1)$ comes from the natural action of $GL(2, \mathbb{Z})$ on the set of lines passing through the origin, that is on $\mathbb{R}P^1$. An explicit homeomorphism between $\mathbb{R}P^1$ and $\mathbb{S}^1$ is illustrated by Figure 1, where the two blue points are identified.

Figure 1.

Now we wan to apply ping-pong lemma to $A = \left( \begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix} \right)$ and $B= \left( \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} \right)$. The actions of $A$ and $B$ on the circle are illustrated by Figures 2 and 3 respectively, through the previous homeomorphism.

Figure 2.

Figure 3.

Therefore, ping-pong lemma applies with $X$ and $Y$ as indicated:

Therefore, $\{ A, B \}$ is a free basis and generates a free subgroup of $GL(2,\mathbb{Z})$, and a fortiori of $\mathrm{Homeo}^+(\mathbb{S}^1)$.

Our second example comes from the homeomorphism $\mathbb{R} / \mathbb{Z} \simeq \mathbb{S}^1$ and is based on piecewise linear homeomorphisms:

Definition: Let $f : \mathbb{R} \to \mathbb{R}$ be a homeomorphism satisfying $f(x+1)=f(x)+1$ for all $x \in \mathbb{R}$. If there exists a sequence of reals $\{ x_i \mid i \in \mathbb{Z} \}$ satisfying $\lim\limits_{i \to \pm \infty} x_i = \pm \infty$ and such that $f$ is affine on each $[x_i,x_{i+1}]$, then the induced homeomorphism $\tilde{f} : \mathbb{R} / \mathbb{Z} \to \mathbb{R} / \mathbb{Z}$ is said piecewise linear.

The set of piecewise linear homeomorphisms of the circle is denoted by $PL_+(\mathbb{S}^1)$.

Let $f$ and $g$ be two piecewise linear homeomorphisms of the circle whose graphs are respectively

Moreover, the graphs of $f^{-1}$ and $g^{-1}$ are respectively

Easily, we deduce that for all $n \in \mathbb{Z} \backslash \{ -1,0,1 \}$,

$f^n \left( \left( \frac{1}{3}, \frac{2}{3} \right) \right) \subset \left( 0, \frac{1}{3} \right) \cup \left( \frac{2}{3}, 1 \right)$ and $g^n \left( \left( 0,\frac{1}{3} \right) \cup \left( \frac{2}{3},1 \right) \right) \subset \left( \frac{1}{3}, \frac{2}{3} \right)$.

Therefore, ping-pong lemma applies, and $\langle f^2,g^2 \rangle$ is a free subgroup of rank two of the group $PL_+(\mathbb{S}^1) \subset \mathrm{Homeo}^+(\mathbb{S}^1)$.

More information about the group of homeomorphisms of the circle can be found in Etienne Ghys’ document, Groups acting on the circle.

## Third Homotopy Group of the Sphere and Hopf Fibration

The idea of this note comes from the chapter $\pi_3(S^2)$, H. Hopf, W.K. Clifford, F. Klein written by H. Samelson in I.M. James’ book, History of Topology, and more precisely from the quoted letter from Hopf to Freudhental:

In case you are still interested in the question of the [homotopy] classes of maps of the 3-sphere $S^3$ onto the 2-sphere $S^2$ I want to tell you that I now can answer this question: there exist infinitely many classes. Namely there is a class invariant of the given map the counter image of $x$ consists of finitely many simple closed oriented polygons $P_1$, $P_2$, …, $P_a$ and likewise the counter image of $y$ consists of polygons $Q_1$, $Q_2$, …, $Q_b$. If $v_{ij}$ denotes the linking number of $P_i$ and $Q_j$, then $\sum_{i,j} v_{ij}= \gamma$ is independent of $x, ~ y$ and of the approximation and does not change under continuous change of the map. For every $\gamma$ there exists maps. Whether to every $\gamma$ there is only one map, I do not know. If the whole $S^2$ is not covered by the image, then $\gamma$ is $=0$. A consequence is that one cannot sweep the line elements on $S^2$ continuously into a point.

Our aim is to developp this nice geometric argument following problems 13, 14 and 15 of Milnor’s book, Topology from differential viewpoint. Thus, our main result is:

Theorem: The group $\pi_3(\mathbb{S}^2)$ is infinite.

In fact, it may be proved that $\pi_3(\mathbb{S}^2) \simeq \mathbb{Z}$. As a consequence that $\pi_n(\mathbb{S}^m)$ may be non-trivial when $n>m$ is that the homotopy group is not determined by the CW complex structure, unlike homology and cohomology groups.

The first step is to define the linking number of two submanifolds of $\mathbb{R}^{k+1}$ with total dimension $k$. For that purpose, we use degree theory; for more information, see Milnor’s book.

Definition: Let $f : M \to N$ be a smooth map between two $n$-dimensional smooth compact oriented manifolds. If $p \in M$ is a regular point, let $\mathrm{sign} (df(p))$ be $+1$ if $df(p)$ is orientation-preserving and $-1$ otherwise. The degree of $f$ with respect to a regular value $q \in N$ is defined by

$\displaystyle \mathrm{deg}(f,q)= \sum\limits_{p \in f^{-1}(q)} \mathrm{sign}(df(p))$.

Property 1: The degree $\mathrm{deg}(f,q)$ does not depend on the regular value $q$.

In particular, property 1 allows us to define the degree $\mathrm{deg}(f)$ of a smooth map $f$ without reference to any regular value.

Property 2: If two smooth maps $f$ and $g$ are smoothly homotopic, then $\mathrm{deg}(f)= \mathrm{deg}(g)$.

Property 3: Let $M$, $N$ be two smooth manifolds and $f : M \to N$ be a smooth map. If there exists a smooth manifold $X$ whose boundary is $M$ and such that $f$ extends to a smooth map $X \to N$, then $\mathrm{deg}(f)=0$.

Properties 1 and 2 are fundamental in degree theory; property 3 is a rather technical lemma used to prove the two previous properties, but it will be useful later.

Definition: Let $M,N \subset \mathbb{R}^{k+1}$ be two submanifolds of total dimension $m+n=k$. The linking number $l(M,N)$ is defined as the degree of the linking map

$\lambda : \left\{ \begin{array}{ccc} M \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$.

Property 4: $l(M,N)= (-1)^{(m+1)(n+1)} l(N,M)$.

Proof. Let $\lambda_1, \lambda_2$ be the linking maps

$\lambda_1 : \left\{ \begin{array}{ccc} M \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a - b \| } \end{array} \right.$ and $\lambda_2 : \left\{ \begin{array}{ccc} N \times M & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$,

and let for convenience

$I : \left\{ \begin{array}{ccc} M \times N & \to & N \times M \\ (a,b) & \mapsto & (b,a) \end{array} \right.$.

Then $\lambda_1 = - \lambda_2 \circ I$. It is an easy exercice to show that $\deg(f \circ g)= \deg(f) \cdot \deg(g)$ when $g$ is a diffeomorphism, and that $\deg(-f)= (-1)^{k+1} \deg(f)$. So

$l(M,N)= \deg(\lambda_1)= (-1)^{k+1} \cdot \deg(I) \cdot \deg(\lambda_2)= (-1)^{m+n+1} (-1)^{mn} l(N,M)$. $\square$

Property 5: If $M$ bounds an oriented manifold disjoint from $N$, then $l(M,N)=0$.

Proof. If $X \subset \mathbb{R}^{k+1}$ is an oriented submanifold such that $\partial X = M$, then the map

$\Lambda : \left\{ \begin{array}{ccc} X \times N & \to & \mathbb{S}^k \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$

extends the linking map $\lambda : M \times N \to \mathbb{S}^k$, and $\partial (X \times N)= \partial X \times N= M \times N$, hence $l(M,N)= \deg(\lambda)=0$ according to property 3. $\square$

If $M$ and $N$ are two knots in $\mathbb{R}^3$, it is possible to compute $l(M,N)$ from a regular diagram.

More precisely, if $v \in \mathbb{S}^2$ and $P$ is the plane normal to $v$, then the projection of $M \cup N$ on $P$ is a regular diagram $D$ if $v$ is a regular value of $\lambda$; moreover, $\lambda^{-1}(v)$ is exactly the crossing points of $D$ when $M$ is over $N$. Therefore, $l(M,N)=r-s$ where $r$ (resp. $s$) is the number of crossing points where $\lambda$ is orientation-preserving (resp. orientation-reversing); on $D$, such crossing points correspond respectively to

For example, the linking number computed on the following diagramm is two:

Before introducing Hopf invariant, let us mention some results about cobordism, needed in the sequel:

## Some cobordism theory:

Definition: Two oriented compact manifolds $M, N$ are cobordant if there exists an oriented compact manifold $X$, called a cobordism, such that $\partial X= M \coprod (-N)$ where $-N$ denotes $N$ with the reversed orientation.

Lemma 1: Let $f : M \to \mathbb{S}^p$ be a smooth map and $y \in \mathbb{S}^p$ be a regular value. If $z \in \mathbb{S}^p$ is a regular value sufficently close to $y$, then there exists a cobordism $X \subset M \times [0,1]$ between $f^{-1}(y)$ and $f^{-1}(z)$.

Proof. If $C$ is the set of singular points of $f$, $f(C)$ is compact so there exists an open neighborhood $V$ of $y$ containing only regular values. Let $z \in V$.

Let $\gamma : [0,1] \to \mathbb{S}^p$ be a great circle from $y$ to $z$, and let $r_t$ be a rotation whose restriction to $\mathrm{span}(y,z)$ is a rotation sending $y$ to $\gamma(t)$. In particular, if $r_0= \mathrm{Id}$, $(r_t)$ defines an isotopy between $\mathrm{Id}$ and $r_1$. Let

$F : \left\{ \begin{array}{ccc} M \times [0,1] & \to & \mathbb{S}^p \\ (x,t) & \mapsto & r_t \circ f(x) \end{array} \right.$.

Because $r_t^{-1}(z) \in V$, $z$ is a regular value of $F(t, \cdot)$ and finally to $F$ since

$\displaystyle dF(m,s)= r_s \circ df(m) + \frac{\partial r_t}{\partial t}_{|t=s} \circ f(m) dt$.

Thus, $F^{-1}(z) \subset M \times [0,1]$ defines a cobordism between $f^{-1}(z)$ and $f^{-1}(r_1^{-1}(z))=f^{-1}(y)$. $\square$

Lemma 2: If $f,g : M \to \mathbb{S}^p$ are smoothly homotopic and $y \in \mathbb{S}^p$ is a regular value for both, then there exists a cobordism $X \subset M \times [0,1]$ between $f^{-1}(y)$ and $g^{-1}(y)$.

Proof. Let $H : M \times [0,1] \to \mathbb{S}^p$ be a homotopy between $f$ and $g$. Let $z$ be a regular value of $H$ close to $y$ so that $f^{-1}(y)$ and $f^{-1}(z)$, and $g^{-1}(y)$ and $g^{-1}(z)$, are cobordant (using lemma 1). Then $H^{-1}(z)$ defines a cobordism between $f^{-1}(z)$ and $g^{-1}(z)$. Lemma 2 follows since cobordism is an equivalent relation. $\square$

## Hopf invariant:

From now on, let $f : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ be a smooth map and $y \neq z$ be two regular values. Applying a stereographic projection, we may view $f^{-1}(y)$ and $f^{-1}(z)$ as submanifolds of $\mathbb{R}^{2p-1}$. Then the linking number $l(f^{-1}(y),f^{-1}(z))$ is well-defined.

Claim 1: The linking number $l(f^{-1}(y),f^{-1}(z))$ does not depend on the stereographic projection.

Proof. A stereographic projection is an orientation-preserving diffeomorphism, so if $p$ and $q$ are two stereographic projections, then $q \circ p^{-1} \times q \circ p^{-1}$ induces an orientation-preserving diffeomorphism between $pf^{-1}(y) \times pf^{-1}(z)$ and $qf^{-1}(y) \times qf^{-1}(z)$, hence

$l(pf^{-1}(y),pf^{-1}(z))= l(qf^{-1}(y),qf^{-1}(z))$. $\square$

Claim 2: The linking number $l(f^{-1}(y),f^{-1}(z))$ is locally constant as a function of $y$.

Proof. Let the mapping maps

$\lambda_1 : \left\{ \begin{array}{ccc} f^{-1}(y) \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$ and $\lambda_2 : \left\{ \begin{array}{ccc} f^{-1}(x) \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$.

According to lemma 1, there exists a cobordism $X \subset \mathbb{R}^{2p-1} \times [0,1]$ between $f^{-1}(y)$ and $f^{-1}(x)$ when $x$ is close to $y$. Let

$\Phi : \left\{ \begin{array}{ccc} X \times f^{-1}(z) & \to & \mathbb{S}^{2p-2} \times [0,1] \\ (a,t,b) & \to & \left( \frac{a-b}{ \| a-b \| },t \right) \end{array} \right.$.

For convenience, let $\partial \Phi$ be the restriction of $\Phi$ on $\partial X \times f^{-1}(z)$, and $\partial_1\Phi$ (resp. $\partial_2 \Phi$) be the restriction of $\partial \Phi$ on $f^{-1}(x) \times \{ 0 \} \times f^{-1}(z)$ (resp. on $f^{-1}(y) \times \{ 1 \} \times f^{-1}(z)$). According to property 3,

$0 = \deg(\partial \Phi)= \deg( \partial_1 \Phi)+ \deg(\partial_2 \Phi)$.

If $\varphi_1, \varphi_2$ are the obvious diffeomorphisms

$\varphi_1 : f^{-1}(x) \times \{ 0 \} \times f^{-1}(z) \to f^{-1}(x) \times f^{-1}(z)$

and

$\varphi_2 : f^{-1}(y) \times \{1\} \times f^{-1}(z) \to f^{-1}(y) \times f^{-1}(z)$,

then $\partial_i \Phi= \lambda_i \circ \varphi_i$ ($i=1,2$), hence

$0= \deg( \lambda_1 ) \cdot \deg(\varphi_1)+ \deg( \lambda_2) \cdot \deg(\varphi_2)$.

But $\varphi_1$ is orientation-preserving whereas $\varphi_2$ is not. Therefore,

$l( f^{-1}(y), f^{-1}(z))= \deg( \lambda_1)= \deg(\lambda_2)= l(f^{-1}(x), f^{-1}(z))$. $\square$

Claim 3: If $y \neq z$ are regular values of $g : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ and $\| f-g \|_{\infty} <1$, then

$l(f^{-1}(y),f^{-1}(z)) = l(g^{-1}(y), f^{-1}(z))$.

Proof. Let $H : \mathbb{S}^{2p-1} \times [0,1] \to \mathbb{S}^p$ be the homotopy

$\displaystyle H(x,t)= \frac{(1-t)f(x)+tg(x)}{ \| (1-t)f(x)+g(x) \| }$.

$H$ is well-defined since $(1-t)f(x)+tg(x)=0$ implies

$1= \| f(x) \| =t \| f(x)-g(x) \| \leq \| f-g \|_{\infty} <1$,

a contradiction. Therefore, $f$ and $g$ are smoothly homotopic. Using lemma 2, we prove in the same way as for claim 2 that the degrees of the associated linking maps are equal so that

$l(f^{-1}(y) , f^{-1}(z))= l(g^{-1}(y),f^{-1}(z))$. $\square$

Property 6: The linking number $l(f^{-1}(y),f^{-1}(z))$ depends only on the homotopy class of $f$.

Proof. According to claim 1, $l(f^{-1}(y),f^{-1}(z))$ does not depend on the stereographic projection. According to claim 2 and property 4, the linking number does not depend on the regular values $y$ and $z$ by connectedness of $\mathbb{S}^{2p-1}$. Finally, if $H$ is a smooth homotopy between $f$ and $g$, then we deduce that

$l(f^{-1}(y),f^{-1}(z))= l(g^{-1}(y),g^{-1}(z))$

from claim 3 and property 4, taking a sequence $0=t_0 < t_1 < \cdots < t_k=1$ satisfying $\| H(t_i , \cdot ) - H ( t_{i+1} , \cdot ) \|_{\infty} < 1$ for all $0 \leq i \leq k-1$. $\square$

Definition: We define the Hopf invariant $H(f)$ of $f$ as the linking number $l(f^{-1}(y),f^{-1}(z))$ for some regular values $y \neq z$.

We just showed that Hopf invariant is a homotopic invariant.

Property 7: If $f : \mathbb{S}^{2p-1} \to \mathbb{S}^p$ and $g : \mathbb{S}^p \to \mathbb{S}^p$, then $H(g \circ f )= H(f) \cdot \deg(g)^2$.

Proof. Let $y \neq z$ be two regular values of $g \circ f$ and $g$, and let $g^{-1}(y)= \{ y_1, \dots, y_r \}$ and $g^{-1}(z)= \{ z_1, \dots, z_s \}$. The inclusions

$f^{-1}(y_i) \times f^{-1}(z_i) \hookrightarrow (g \circ f)^{-1}(y) \times (g \circ f)^{-1} (z)$

induce an orientation-preserving diffeomorphism between

$\displaystyle \coprod\limits_{i=1}^r \coprod\limits_{j=1}^s \left( \mathrm{sign}(dg(y_i)) f^{-1}(y_i) \right) \times \left( \mathrm{sign}(dg(z_j)) f^{-1}(z_j) \right)$

and

$\displaystyle (g \circ f)^{-1}(y) \times (g \circ f )^{-1}(z)$,

so, if $\lambda_{ij} : \left\{ \begin{array}{ccc} f^{-1}(y_i) \times f^{-1}(z_j) & \to & \mathbb{S}^{2p-2} \\ (a,b) & \mapsto & \frac{a-b}{ \| a-b \| } \end{array} \right.$, we have:

$\begin{array}{lcl} \deg(\lambda) & = & \displaystyle \sum\limits_{i=1}^r \sum\limits_{j=1}^s \mathrm{sign}(dg(y_i)) \mathrm{sign}(dg(z_j)) \cdot \underset{= H(f)}{\underbrace{\deg(\lambda_{ij})}} \\ \\ & = & \displaystyle H(f) \left( \sum\limits_{i=1}^r \mathrm{sign}(dg(y_i)) \right) \left( \sum\limits_{j=1}^s \mathrm{sign}(dg(z_j)) \right) \\ \\ & = & H(f) \cdot \deg(g)^2 \hspace{1cm} \square \end{array}$

## Hopf fibration:

The sphere $\mathbb{S}^3$ can be viewed as the unit sphere $\{ (z_1,z_2) \mid |z_1|^2+ |z_2|^2=1 \}$ in $\mathbb{C}^2 \simeq \mathbb{R}^4$. Noticing that the intersection between $\mathbb{S}^3$ and any complexe line is a circle, we can say that $\mathbb{S}^3$ is covered by a family of circles $\mathbb{S}^1$ indexed by $\mathbb{C}P^1 \simeq \mathbb{S}^2$.

More precisely, if $(x,y,z,t) \in \mathbb{S}^3 \subset \mathbb{R}^4$ then $(x+iy,z+it) \in \mathbb{S}^3 \subset \mathbb{C}^2$ belongs to the complex line $(z+it)z_2= (x+iy)z_1$; to this line (in $\mathbb{C}P^1$) is associated the complex number $\frac{x+iy}{z+it} \in \mathbb{C} \cup \{ \infty\}$. Finally, if $h : \mathbb{S}^2 \to \mathbb{C} \cup \{\infty\}$ denotes the diffeomorphism induced by the stereographic projection, we get a point $h^{-1} \left( \frac{x+iy}{z+it} \right)$ of the sphere $\mathbb{S}^2$.

Definition: Hopf map $\pi : \left\{ \begin{array}{ccc} \mathbb{S}^3 & \to & \mathbb{S}^2 \\ (x_1,x_2,x_3,x_4) & \mapsto & h^{-1} \left( \frac{x_1+ix_2}{x_3+ix_4} \right) \end{array} \right.$ induces the Hopf fibration $\mathbb{S}^1 \to \mathbb{S}^3 \overset{\pi}{\longrightarrow} \mathbb{S}^2$.

It is possible to visualize the decomposition of $\mathbb{S}^3$ by circles in $\mathbb{R}^3$ using stereographic projection. It gives something like that, a decomposition in concentric torii each covered by their Villarceau circles:

Very nice animations about Hopf fibration can be found in Etienne Ghys, Joe Leys and Aurélien Alvarèz’s movie, Dimensions, and in Niles Johnson’s lecture, Visualizations of Hopf fibration. Now,

$\pi^{-1}(1,0,0)= \{ (x,y,x,y) \mid x^2+y^2= \frac{1}{2} \}$ and $\pi^{-1}(-1,0,0) = \{ (x,y,-x,-y) \mid x^2+y^2= \frac{1}{2} \}$

are two circles in $\mathbb{S}^3$. If $\varphi : \mathbb{S}^3 \subset \mathbb{R}^4 \to \mathbb{R}^3$ is the stereographic projection with respect to $(0,0,0,1)$, then $\varphi(x,y,z,t)= \left( \frac{x}{1-t}, \frac{y}{1-t}, \frac{z}{1-t} \right)$, so the previous circles become

$\left\{ \left( \frac{x}{1-y}, \frac{y}{1-y}, \frac{x}{1-y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ and $\left\{ \left( \frac{x}{1+y}, \frac{y}{1+y}, - \frac{x}{1+y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$.

In fact, it is just a Hopf link; precisely, if $C_1$ and $C_2$ denotes the projections of the two previous circles on the plane $z=0$, that is

$C_1=\left\{ \left( \frac{x}{1-y}, \frac{y}{1-y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$ and $C_2= \left\{ \left( \frac{x}{1+y}, \frac{y}{1+y} \right) \mid x^2+y^2= \frac{1}{2} \right\}$,

then $C_1 \cap C_2= \left\{ \left( \pm \frac{1}{\sqrt{2}}, 0 \right) \right\}$ where $C_1$ is over $C_2$ at one point, and vice-versa at the other. Therefore, $H(\pi)=1$ (be careful to the orientations!).

On the other hand, $H(f)=0$ for every constant map $f : \mathbb{S}^3 \to \mathbb{S}^2$. Indeed, let

$p : \left\{ \begin{array}{ccc} \mathbb{S}_3 & \to & D^2 \\ (x,y,z,t) & \mapsto & (z,t) \end{array} \right.$

where $D^2$ is the unit two-dimensional disk viewed as a subspace of $\mathbb{S}^2$; in particular, since $D^2$ is contractible, $p$ is homotopic to a constant map $f$ and $H(p)=H(f)$. But the circles $p^{-1}(0,0)= \{ (x,y,0) \mid x^2+y^2=1 \}$ and $p^{-1}(1/2,0)= \{ (x,y,1/2) \mid x^2+y^2= 3/4 \}$ (viewed in $\mathbb{R}^3$ thanks to the stereographic projection with respect to $(0,0,0,1)$) are clearly unlinked, hence $H(p)=0$ according to property 5.

Moreover, using property 7, we deduce that the applications

$p_n : \left\{ \begin{array}{ccc} \mathbb{S}^{3} \subset \mathbb{C}^2 & \to & \mathbb{S}^2= \mathbb{C} \cup \{ \infty \} \\ (z_1,z_2) & \mapsto & \left( z_1/z_2 \right)^n \end{array} \right.$

define a family of pairwise non-homotopic maps, since $H(p_n)= n^2$. We just proved our main theorem!

## Hopf fibration in higher dimensions:

Hopf fibration $\mathbb{S}^1 \to \mathbb{S}^3 \to \mathbb{S}^2$ is obtained using Hopf map $\left\{ \begin{array}{ccc} \mathbb{S}^3 \subset \mathbb{C}^2 & \to & \mathbb{S}^2= \mathbb{C} \cup \{ \infty \} \\ (z_1,z_2) & \mapsto & z_1/z_2 \end{array} \right.$. But the same thing can be done by replacing $\mathbb{C}$ with any real division algebra, for example the quaternions $\mathbb{H}$ or the octonions $\mathbb{O}$.

The associated maps $p : \mathbb{S}^7 \subset \mathbb{H}^2 \to \mathbb{S}^4 = \mathbb{H} \cup \{ \infty \}$ and $q : \mathbb{S}^{15} \subset \mathbb{O}^2 \to \mathbb{S}^8 = \mathbb{O} \cup \{ \infty \}$ induce respectively the Hopf fibrations $\mathbb{S}^3 \longrightarrow \mathbb{S}^7 \overset{p}{\longrightarrow} \mathbb{S}^4$ and $\mathbb{S}^7 \longrightarrow \mathbb{S}^{15} \overset{q}{\longrightarrow} \mathbb{S}^8$.

It can be shown (for example in Hatcher’s book, Algebraic topology, using a homological interpretation of Hopf invariant) that $H(p)=H(q)=1$, hence:

Theorem: $\pi_3(\mathbb{S}^2)$, $\pi_7(\mathbb{S}^4)$ and $\pi_{15}(\mathbb{S}^8)$ are infinite.

However, Hopf fibrations do not exist in other dimensions since a finite-dimensional real division algebra has dimension one, two, four or eight.

## Rado graph and 0-1 law for finite graphs

Given a set of vertices $X$, a random graph is given by deciding, independently with probability $1/2$, whether each unordered pair of vertices should be joined by an edge or not. Following Peter Cameron’s book, Permutation Groups, we prove here that there exists a countable graph $R$ (unique up to isomorphism) such that each random countable graph is isomorphic to $R$ with probability $1$. Moreover, we shall see that the existence of such a graph implies a 0-1 law for first-order properties of finite graphs.

Definition: We say that a graph $\Gamma$ has the property $\mathcal{P}$ if for all disjoint finite subsets of vertices $U,V$ there exists a vertex $z \notin U \cup V$ joined to every vertex of $U$ and to no vertex of $V$.

The two following lemmas show that, up to isomorphism, there exists only one countable graph satisfying property $\mathcal{P}$: the so-called Rado graph, denoted $R$.

Lemma 1: Two countable graphs satisfying property $\mathcal{P}$ are isomorphic.

Proof. Let $\Gamma_1, \Gamma_2$ be two countable graphs satisfying property $\mathcal{P}$. For convenience, let $(f_1,f_2, \dots)$ and $(g_1,g_2, \dots)$ be an enumeration of $\Gamma_1$ and $\Gamma_2$ respectively.

Let $\varphi_0 : \emptyset \to \Gamma_2$ be the empty morphism. Then, by induction:

1) If $n$ is odd: Let $f$ be the first vertex in the enumeration of $\Gamma_1$ not in the domain $S_{n-1}$ of $\varphi_{n-1}$. If $U$ (resp. $V$) denotes the set of vertices of $S_{n-1}$ joined (resp. not joined) to $f$ by an edge, according to property $\mathcal{P}$, there exists a vertex $z \notin \varphi_{n-1}(U) \cup \varphi_{n-1}(V) = \varphi_{n-1}(S_{n-1})$ joined by an edge to any vertex of $\varphi_{n-1}(U)$ and to no vertex of $\varphi_{n-1}(V)$. Therefore, $\varphi_{n-1}$ can be extended to $\varphi_n : S_{n-1} \cup \{f\} \to \Gamma_2$ by setting $\varphi_n(f)=z$.

2) If $n$ is even: Let $g$ be the first vertex in the enumeration of $\Gamma_2$ not in the range $R_{n-1}$ of $\varphi_{n-1}$. If $U$ (resp. $V$) denotes the set of vertices of $R_{n-1}$ joined (resp. not joined) to $g$ by an edge, according to property $\mathcal{P}$, there exists a vertex $z \notin \varphi_{n-1}^{-1}(V) \cup \varphi_{n-1}^{-1}(U)= \varphi_{n-1}^{-1}(R_{n-1})$ joined by an edge to any vertex of $\varphi_{n-1}^{-1}(U)$ and to no vertex of $\varphi_{n-1}^{-1}(V)$. Therefore, $\varphi_{n-1}$ can be extended to $\varphi_n : S_{n-1} \cup \{z\} \to R_{n-1} \cup \{g \}$ by setting $\varphi_n(z)=g$.

Finally, $\varphi := \bigcup\limits_{n \geq 0} \varphi_n$ defines an isomorphism between $\Gamma_1$ and $\Gamma_2$. $\square$

It is worth noticing that the previous proof gives an isomorphism as soon as $\varphi_0$ is an isomorphism between two subgraphs. Therefore, we have:

Property 1: $R$ is homogenous, ie. every isomorphism between two finite induced subgraphs can be extented to an automorphism of $R$.

Moreover, using only the first step of our induction, we have:

Property 2: $R$ is universal, ie. every at most countable graph is isomorphic to a subgraph of $R$.

In fact, it can be shown that $R$ is the only countable, universal and homogenous graph (up to isomorphism); see Peter Cameron’s book for more details.

Lemma 2: A countable random graph has property $\mathcal{P}$ with probability $1$.

Proof. Because there exist only countably many pairs of finite subsets of vertices $(U,V)$, it is sufficient to show that the event that the property $\mathcal{P}$ for $U$ and $V$ fixed fails has probability $p$ zero. But the probability that a point $z \notin U \cup V$ be joined by an edge to any vertex of $U$ and to no vertex of $V$ is $1/2^k$, where $k= |U \cup V|$. Therefore,

$\displaystyle p = \lim\limits_{n \to + \infty} \left( 1- \frac{1}{2^k} \right)^n =0$. $\square$

In particular, lemma 2 shows that there exists a countable graph satisfying property $\mathcal{P}$, justifying our definition of $R$. We deduce the following theorem from lemmas 1 and 2:

Theorem 1: A countable random graph is isomorphic to $R$ with probability $1$.

From now on, let us consider graph theory from model theory viewpoint, that is we view a graph as a model of the theory

$\mathcal{T}_{\text{graph}} = \left\{ \forall x, \neg (x \sim x) ; \forall x,y (x \sim y \Leftrightarrow y \sim x) \right\}$

over the langage $\mathcal{L}= \{ \sim, =\}$. Here, the relation $x \sim y$ means that the vertices $x$ and $y$ are joined by an edge; therefore, in $\mathcal{T}_{\text{graph}}$ the graphs are unoriented and do not contain loops. It is worth noticing that a graph satisfies property $\mathcal{P}$ if and only if it satisfies the first-order sentence

$\Phi_{m,n} = \forall x_1, \dots, x_m \forall y_1, \dots, y_n \exists z \Phi_{m,n} ( \overline{x} , \overline{y},z)$

for all $m,n \geq 1$, where

$\Phi_{mn} (\overline{x}, \overline{y},z) = \left[ \bigwedge\limits_{i,n} (x_i \neq y_j ) \Rightarrow \left( \bigwedge\limits_{i,j} (z \neq x_i,y_j ) \wedge \bigwedge\limits_i (z \sim x_i ) \wedge \bigwedge\limits_{j} \neg (z \sim y_j) \right) \right]$.

Therefore, $\mathcal{P}$ is a first-order property. We now are able to deduce a 0-1 law for finite graphs:

Definition: A property is true in almost all finite graphs if the proportion of graphs of cardinality $n$ satisfying the given property tends to $1$ as $n \to + \infty$.

Theorem 2: Let $\sigma$ be a sentence in the first-order langage of graph theory. If $R \models \sigma$ (resp. if $R \not\models \sigma$) then $\sigma$ is true (resp. false) in almost all finite graphs. In particular, a first-ordre sentence is either true in almost all finite graphs or false in almost all finite graphs.

Proof. Suppose that $R \models \sigma$. By our previous remark, the sentence $\sigma$ can be logically deduced from the sentences $\Phi_{m,n}$, ie. $\{\Phi_{m,n}, m,n \geq 1 \} \vdash \sigma$. Then, it is a consequence of compactness theorem that $\sigma$ can be logically deduced from only finitely many sentences $\Phi_{m,n}$. Therefore, it is sufficient to prove that each $\Phi_{m,n}$ fails in a random graph of cardinality $k$ with probability $p_k$ such that $p_k \underset{k \to + \infty}{\longrightarrow} 0$.

Clearly, $p_k$ is smaller than the product of the number of choices of a subset of $m+n$ vertices by the probability that the other vertices are not correctly joined (so that $\Phi_{m,n}$ be not satisfied), that is

$\displaystyle p_k \leq C_{k}^{m+n} \left( 1- \frac{1}{2^{m+n}} \right)^{k-m-n} \underset{k \to + \infty}{\longrightarrow} 0$.

If $R \not\models \sigma$, then $R \models \neg \sigma$ and $\neg \sigma$ is true in almost all finite graphs. $\square$

It can be unsatisfactory that we only give a non-constructive proof for the existence of $R$; to conclude, we give an explicit construction:

Let $\mathbb{N}$ be the set of vertices; two integers $n < m$ are joined by an edge if and only if the $n$-th digit (from right to left) of $m$, written in binary, is one. Then, the constructed graph has property $\mathcal{P}$: Let $U, V$ be two finite disjoint subsets of vertices and $p \geq 1$ be such that $2^p>x$ for all $x \in U \cup V$. Now, let $z$ be the binary number $\epsilon_p \cdots \epsilon_0$, where $\epsilon_p=1$, and for every $0 \leq i \leq p-1$, $\epsilon_i=1$ if $i \in U$ and $0$ otherwise (in particular, if $i \in V$). Then $z \notin U \cup V$ is joined by an edge to every vertex of $U$ and to no vertex of $V$.

## Hairy ball theorem without degree theory

Often, hairy ball theorem is mentioned as an application of degree theory derived from singular homology in algebraic topology or from de Rham cohomology in differential topology; we also mentionned it as an application of Brouwer’s degree in Brouwer’s Topological Degree (III): First Applications. Here we describe an argument due to Milnor, exposed in Gallot, Hullin and Lafontaine’s book Riemannian Geometry, using only integration of differential forms.

For convenience, we begin by exposing some basic definitions and first properties about integration on manifolds.

If $M$ is a smooth manifold, let $TM$ and $\Omega^k(M)$ denote respectively its tangent bundle and its space of differential forms of degree $k$ (ie. the vector space $\Gamma( \Lambda^k T^*M)$ of sections of the $k$-th extorior power of the cotangent bundle).  We say that $M$ is orientable if there exists an atlas $(U_i,\varphi_i)$, said oriented, such that $\varphi_i \circ \varphi_j^{-1}$ is orientation-preserving for all $i,j$.

Property: A smooth manifold $M$ is orientable if and only if there exists $\omega \in \Omega^n(M)$ satisfying $\omega_m \neq 0$ for all $m \in M$. Such a differential form is a volume form.

Sketch of proof. Let $\omega \in \Omega^n(M)$ be a volume form and let $(U_i,\varphi_i)$ be a family of charts covering $M$. Because $(\varphi_i)_* \omega_{|U_i} \in \Omega^n (\varphi_i(U_i))$ is a volume form, we can write

$(\varphi_i)_* \omega_{|U_i} = f_i ~ dx_1 \wedge \cdots \wedge dx_n$

where $f_i \in C(\varphi_i(U_i), \mathbb{R})$ does not vanish. Without loss of generality, we may suppose that $f_i > 0$ (otherwise, switch two coordinates of the chart). Then $(U_i, \varphi_i)$ is an oriented atlas.

Conversely, let $(U_i, \varphi_i)$ be an oriented atlas and $(\chi_i)$ be an associated partition of unity. Then

$\displaystyle \omega = \sum\limits_i \chi_i \varphi_i^* (dx_1 \wedge \cdots \wedge dx_n)$

is a volume form. $\square$

As an example, we may mention the differential form

$\displaystyle \omega = \sum\limits_{i=0}^n (-1)^i x^i dx_0 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n \in \Omega^n(\mathbb{R}^{n+1})$.

Then $\omega$ induces a volume form on $\mathbb{S}^n$ viewed as the unit sphere in $\mathbb{R}^{n+1}$. [In fact, $\omega$ is the canonical volum form of $\mathbb{S}^n$ as a Riemannian manifold. In particular, we define $\displaystyle \mathrm{Vol}(\mathbb{S}^n) = \int_{\mathbb{S}^n} \omega$.]

An interesting property of volum forms is that, if $\omega \in \Omega^n(M)$ is a volume form and $\alpha \in \Omega^n(M)$, then there exists $f \in C(M, \mathbb{R})$ such that $\alpha= f \omega$. It is clear locally, and then it is sufficient to extend it globally using a partition of unity.

Definition: Let $M$ be an orientable compact $n$-dimensional smooth manifold. There exists one and only one linear map $\int_{M} : \Omega^n(M) \to \mathbb{R}$ such that for every $\omega \in \Omega^n(M)$ supported in an open chart $(U,\varphi)$,

$\displaystyle \int_M \alpha = \int_{\varphi(U)} (\varphi^{-1})^* \alpha$.

[Where the right-hand side is the usual integral of a differential form defined on an open subspace in $\mathbb{R}^n$.]

Sketch of proof. First, suppose that such a linear map exists. Let $(U_i, \varphi_i)$ be a finite family of charts covering $M$ (such a family exists by compactness) and $(\chi_i)$ be an associated partition of unity. Then necessarily:

$\displaystyle \int_M \omega= \int_M \sum\limits_{i} \chi_i \omega_{|U_i} = \sum\limits_{i} \int_M \chi_i \omega_{|U_i} = \sum\limits_{i} \int_{\varphi(U_i)} (\varphi_i^{-1})^* \chi_i \omega_{|U_i}$,

hence $\int_M$ is uniquely defined by the mentionned properties. Conversely, it is sufficient to show that the above equality does not depend on the family of charts and on the partition of unity chosen. An argument for that is to notice that if $\varphi : U \to V$ is a diffeomorphism between two open subsets $U,V \subset \mathbb{R}^n$, and $\omega \in \Omega^n(\mathbb{R}^n)$, then

$\displaystyle \int_U \varphi^* \omega= \epsilon \int_V \omega$,

where $\epsilon=1$ if $\varphi$ is orientation-preserving and $-1$ otherwise. We can write $\omega= f dx_1 \wedge \cdots \wedge dx_n$ for some $f \in C(M, \mathbb{R})$, and

$\begin{array}{lcl} \displaystyle \int_U \varphi^* \omega & = & \displaystyle \int_U f \circ \varphi ~ \varphi^* (dx_1 \wedge \cdots \wedge dx_n) \\ \\ & = & \displaystyle \int_U f \circ \varphi ~ \mathrm{Jac} (\varphi) ~ dx_1 \wedge \cdots \wedge dx_n \\ \\ & = & \displaystyle \epsilon \int_U f \circ \varphi ~ | \mathrm{Jac} ( \varphi ) | ~ dx_1 \wedge \cdots \wedge dx_n \\ \\ & = & \displaystyle \epsilon \int_{\varphi(U)} f = \epsilon \int_V \omega \hspace{1cm} \square \end{array}$

Using the proof above, we may notice:

Property: Let $M,N$ be two compact orientable $n$-dimensional smooth manifolds and $\omega \in \Omega^n(N)$. If $f : M \to N$ is a diffeomorphism then

$\displaystyle \int_N \omega= \epsilon \int_M f^*\omega$,

where $\epsilon=1$ if $f$ is orientation-preserving and $\epsilon=-1$ otherwise.

Theorem: (Hairy ball theorem) If $n$ is even, any continuous vector field on $\mathbb{S}^n$ has a zero.

Proof. Let $X$ be a continuous vector field on $\mathbb{S}^n$, that is $X : \mathbb{S}^n \to \mathbb{R}^{n+1}$ is a map satisfying $X(p) \cdot p =0$ for every $p \in \mathbb{S}^n$. By contradiction, suppose that $X$ does not vanish.

According to Stone-Weierstrass theorem, it is possible to approximate $X$ by a sequence of smooth maps $X_k : \mathbb{S}^n \to \mathbb{R}^{n+1}$. Then, if $X_k(p)^{\top}$ denotes the orthogonal projection of $X_k(p)$ on $T_p \mathbb{S}^n$, $X_k^{\top}$ is a sequence of smooth vector fields on $\mathbb{S}^n$ converging uniformely to $X$. Therefore, we may suppose without loss of generality that $X$ is smooth.

[Such an argument may be generalized to other manifolds; see Tubular neighborhood.]

Let $\omega \in \Omega^n(\mathbb{S}^n)$ be the volume form as defined above by the restriction on $\mathbb{S}^n$ of

$\displaystyle \omega = \sum\limits_{i=0}^n (-1)^i x^i \underset{ := \omega_i }{ \underbrace{ dx_0 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n }} \in \Omega^n ( \mathbb{R}^{n+1})$

and let $f_{\epsilon} : S(1) \to S( \sqrt{1+\epsilon^2}), \ x \mapsto x + \epsilon X(x)$. For every $p \in S(1)$ and $v_0, \dots, v_n \in \mathbb{R}^{n+1}$, we may write

$\begin{array}{lcl} ( f_{\epsilon}^* \omega)_p(v_0, \dots, v_n) & = & \sum\limits_{i=0}^n (-1)^i (p^i+ \epsilon X(p)^i) \omega_i (v_0+ \epsilon dX(p) \cdot v_0, \dots, v_n+ \epsilon dX(p) \cdot v_n) \\ & = & \sum\limits_{i=0}^n \sum\limits_{k=0}^{n+1} (-1)^i (p^i+ \epsilon X(p)^i) \epsilon^k \omega_{ik} (v_0, \dots, v_n) \end{array}$

where $\omega_{ik} \in \Omega^n(\mathbb{R}^{n+1})$ is such that $\omega_{i0}=\omega_i$ and $\omega_{ik}(v_0, \dots, v_n)= \omega_i (\tilde{v_0}, \dots, \tilde{v_n})$ with $\tilde{v_j} = v_j$ or $dX(p) \cdot v_j$. Therefore, there exist $\alpha_k \in \Omega^n (\mathbb{R}^{n+1})$ such that

$\displaystyle f_{\epsilon}^* \omega= \omega+ \sum\limits_{k=1}^{n+2} \epsilon^k \alpha_k \hspace{1cm} (1)$.

First, let us show how conclude the proof if $f_{\epsilon}$ is a diffeomorphism:

Let $r= \sqrt{1+ \epsilon^2}$ for convenience, $\psi : x \mapsto rx$ be a diffeomorphism from $S(1)$ to $S(r)$ and  as mentioned above $\displaystyle \mathrm{Vol}(\mathbb{S}^n)= \int_{\mathbb{S}^n} \omega$. Then

$\displaystyle \mathrm{Vol}(\mathbb{S}^n) (1+ \epsilon^2)^{\frac{n+1}{2}} = r^{n+1} \int_{\mathbb{S}^n} \omega = \int_{S(1)} \psi^* \omega= \int_{S(r)} \omega= \int_{S(1)} f_{\epsilon}^* \omega$,

so according to $(1)$, the expression $\displaystyle (1+\epsilon^2)^{\frac{n+1}{2}}$ is a polynomial in $\epsilon$. Of course, it is only possible when $n$ is odd.

To conclude the proof, we shall see that $f_{\epsilon}$ is a diffeomorphism when $\epsilon$ is sufficiently small. From $(1)$, if $g_k \in C(\mathbb{S}^n,\mathbb{R})$ is such that $\alpha_k = g_k \omega$,

$\displaystyle f_{\epsilon}^* \omega= \left( 1+ \sum\limits_{k=1}^{n+1} \epsilon^k g_k \right) \omega$.

By compactness, the $g_k$ are bounded so the quantity $1+ \sum\limits_{k=1}^{n+1} \epsilon^k g_k$ does not vanish when $\epsilon$ is small enough. Therefore, $f_{\epsilon}^*\omega$ is a volume form and $f_{\epsilon}$ is an immersion.

Moreover, if $f_{\epsilon}$ were not injective for $\epsilon$ small enough, there would exist sequences $(\epsilon_k)$, $(x_k)$ and $(y_k)$ such that $\epsilon_k \to 0$, $x_k \neq y_k$ and $f_{\epsilon_k}(x_k)= f_{\epsilon_k}(y_k)$, hence

$\displaystyle \frac{x_k-y_k}{\| x_k-y_k \|} = - \epsilon_k \frac{X(x_k)-X(y_k)}{ \| x_k-y_k \| }$.

The first term has norm one, $\epsilon_k \to 0$ and the last term is bounded according to mean value theorem: a contradiction.

Therefore, for $\epsilon$ small enough, $f_{\epsilon}$ is an injective immersion, and consequently a local diffeomorphism according to inverse function theorem. In particular, $\mathrm{Im}(f_{\epsilon})$ is open but also closed by compactness of $S(1)$; by connectedness, $\mathrm{Im}(f_{\epsilon})= S(r)$ and $f_{\epsilon}$ is a diffeomorphism. $\square$