Geometric Topology 19

Lifting Loops and Homotopies

Theorem: Let $Y = I$. Fix $x_0 \in X$ and $z \in p^{-1}(x_0)$. Let $\alpha,\beta : I \rightarrow X$ be equivalent loops based at $x_0$, meaning $\alpha \approxeq \beta$.

1. $\alpha$ has a unique lift $\tilde{\alpha}_z : I \rightarrow Z$ (a path, occasionally a loop) such that $\tilde{\alpha}(0) = z$.
2. If $F : I^2 \rightarrow X$ is the path-homotopy realizing $\alpha \approxeq \beta$, it has a unique lift $\tilde{F}: I^2 \rightarrow Z$ such that $\tilde{F}_0 = \tilde{\alpha}_z$ (the bottom edge of $I^2$).

Proof:

1. One can partition $I$ into closed intervals $[a_j,a_{j+1}]$ such that $\alpha([a_j,a_{j+1}])$ lies in an open set $U$ covered by sheets. $\tilde{\alpha}$ is defined on each in succession by bijectivity.
2. This follows from a similar argument by dividing the square $I^2$ into rectangles $R_{jk}$, each mapped into a $U$, then working upwards from $\tilde{F}_0$ to define $\tilde{F}$ on each column. $\blacksquare$

$\tilde{F}$ must be constant on the vertical lines $s=0,1$ so $\tilde{F}(0,t) = z$ and $\tilde{F}_1 = \tilde{\beta}_z$. Thus:

Corollary: The endpoint $\tilde{\alpha}_z(1)$ of $Z$ depends only on the class $[\alpha]$ of $\alpha$ in $\pi_1(X,x_0)$, so fixing $z$ determines a map $\pi_1(X,x_0) \rightarrow p^{-1}(x_0)$.

Groups Acting on Sets

Let $G$ be a group (finite or infinite, with identity element $e$), and $\Omega$ any set.

Definition: A right action of $G$ on $\Omega$ is a mapping:

such that:

1. $z \cdot e = z$,
2. $z \cdot (gh) = (z \cdot g)\cdot h$ for all $z \in \Omega$ and $g, h \in G$.

If $\Omega$ is finite of size $n$, this simply means that there is a group homomorphism $G \rightarrow S_n$.

Examples:

• Let $\Omega = \mathbb{R}^3 = \{(x_1,x_2,x_3)\}, \ G = SO(3) = \{A \in \mathbb{R}^{3,3}: A^TA = I, \det A = 1\}$. $G$ acts on $\Omega$ on the right. It also acts on $S^2$ on the right. Let $C$ be a cube centred at the origin. The subgroup of $G$ mapping $C$ onto $C$ is known to be isomorphic to $S_4$; it contains $2$-cycles, $3$-cycles, $4$-cycles (how do each of these rotate the cube?).

Lemma: Fix $z \in \Omega$. Then $G_z = \{g \in G: z \cdot g = z\}$ is a subgroup of $G$, called the stabilizer of $z$, and $g \mapsto z \cdot g$ identifies the set $\{G_z g : g \in G\}$ of right cosets with $\Omega$. Note that $G$ acts on this set of cosets on the right, so we can dispense with $\Omega$.

Action of $\pi_1$ on a Fibre

Let $p : Z \rightarrow X$ be a covering map. Fix $x_0 \in X$. Set:

Theorem:

1. Setting $z \cdot [\alpha] = \tilde{\alpha}_z(1)$ defines a right action of $G$ on $\Omega$.
2. If $z \in \Omega$, the induced homomorphism $p_{*} : \pi_1 (Z,z) \rightarrow \pi_1(X,x_0)$ is injective.
3. The stabilizer $\{[\alpha] : z \cdot [\alpha] = z\}$ is precisely the subgroup $p_{*}(\pi_1 (Z,z))$ of $G$.

Proofs:

1. It is obvious that $z \cdot [\varepsilon] = z$. To prove that $z \cdot ([\alpha][\beta]) = (z \cdot [\alpha]) \cdot [\beta]$, observe that $\tilde{\alpha \beta}_z = \tilde{\alpha}_z \tilde{\beta}_{z'}$, where $z' = \tilde{\alpha}_z(1)$ (since $\tilde{\alpha}_z \tilde{\beta}_{z'}$ is a lift of $\alpha \beta$ from $z$).
2. If $\gamma$ is a loop at $z$ such that $p \circ \gamma \approxeq_F \varepsilon$, then the unique lift $\tilde{F}$ gives $\gamma \approxeq \varepsilon_z$.
3. If $\tilde{\alpha}_z = z$, then $\tilde{\alpha}_z$ is a loop and:
   $$[\alpha] = [p \circ \tilde{\alpha}_z] = p_{*}[\tilde{\alpha}_z].$$
   Conversely, if $[\alpha] = p_{*}[\gamma]$ with $\gamma$ a loop based at $z$, then $\gamma = \tilde{\alpha}_z$ and $z \cdot [\gamma] = z$. $\blacksquare$

Fundamental Group of the Circle

Let $p : Z \rightarrow X$ be a covering map, fix $x_0 \in X$ and $z \in p^{-1}(x_0)$. Set $G = \pi_1(X,x_0)$ and $\Omega = p^{-1}(x_0)$ as before.

Lemma: The map $G \rightarrow \Omega$ defined by $[\alpha] \mapsto z \cdot [\alpha]$ is:

1. surjective if $Z$ if path-connected,
2. bijective if $Z$ is simply-connected (meaning path-connected and $\pi_1(Z) = \{e\}$).

Proof of Surjectivity: Given $z' \in \Omega$, choose a path $\sigma$ from $z$ to $z'$ in $Z$. Then $\alpha = p \circ \sigma$ is a loop based at $x_0$. By uniqueness, $\tilde{\alpha}_z = \sigma$, so $\tilde{\alpha}_z(1) = z'$. Proof of Injectivity: If $z \cdot [\alpha] = z \cdot [\beta]$ then $z \cdot g = z$ where $g = [\alpha][\beta]^{-1}$. But the stabilizer of $z$ is trivial by part (3) of the previous theorem, so $[\alpha] = [\beta]$.

We can apply the lemma to $X = S^1 \subset \mathbb{C}$, $x_0 = 1, Z = \mathbb{R}, p(z) = e^{2\pi i s}$. Then $\Omega = \mathbb{Z}$. The path $s \mapsto sn$ in $\mathbb{R}$ projects to the loop $\alpha_n : s \mapsto \exp (2 \pi i sn)$ in $S^1$, and $0 \cdot [\alpha_n] = n$.

Corollary: $\pi_1 (S^1,1)$ is isomorphic to $(\mathbb{Z},+)$, i.e. the infinite cyclic group $F_1$.

Double Coverings

We previously defined $\mathbb{RP}^n$ to be the quotient of $S^n$ obtained by identifying each point $v = (x_0,x_1,...,x_n)$ with $-v=(-x_0,-x_1,...,-x_n)$, its antipodal point.

Recall that $\mathbb{RP}^2$ is homeomorphic to the surface $P$ originally defined from a square. The square model showed that $P$ is also a quotient of a circular disk $D$ in which opposite points on the boundary $\partial D$ are identified. Now deform $D$ into the southern hemisphere of $S^2$ and retain $\partial D$ as its equator. Then any equivalence class $\{v,-v\}$ (defining a point on $\mathbb{RP}^2$) has a unique representative in $P$.

Let's explain why the group $SO(3)$ of rotations fixing the origin is homeomorphic to $\mathbb{RP}^3$. Any non-identity rotation is determined by an axis (i.e. a unit vector $v$), and an angle $\theta \in [0,\pi]$. Map this to the point $(\frac{\theta}{\pi})v$ in $\mathbb{R}^3$. The only ambiguity is that an angle $\pi$ around $v$ is the same rotation as $\pi$ around $-v$. So we are identifying antipodal points of the boundary $S^2 = \partial B$ of the solid unit ball $B$. Since $B$ is just a $3$-dimensional analogue of $D$ above, the resulting space is $\mathbb{RP}^3$ by the same argument.

Torus to Klein Bottle in Squares

Represent the torus $T$ by the square $I^2$ with boundary $aba^{-1}b^{-1}$, giving $q_T : I^2 \rightarrow T$. Represent the Klein bottle $K$ by $I^2$ with boundary $cd^{-1}c^{-1}d^{-1}$, giving $q_K : I^2 \rightarrow K$.

Lemma: The mapping $p : T \rightarrow K$ defined by:

is a 2:1 covering map (the quotient map $\mathscr{P} \rightarrow \hat{\mathscr{P}}$ is denoted by $q$ rather than $\pi$).

If we regard $a,c$ as loops in $T,K$ respectively, then $p \circ a = cc$, whilst $p \circ b = d$.

$T$ to $K$ in Letters and Words

Fix $z = q_T(0,1)$ as a basepoint $z \in T$, and $x_0 = q_K(0,1) \in K$. The interior of $q_K(I^2)$ allows us to define a homotopy $\boxed{cd^{-1}c^{-1}d^{-1}} \approxeq \varepsilon$ with basepint $x_0$.

Abusing notation to identify loops with classes in $\pi_1 (K,x_0)$, write $cd^{-1}c^{-1}d^{-1} = e$:

If we now set $a=c^2$ and $b=d^{-1}$ (justified since $p_{*} : \pi_1 (T,x_0) \rightarrow \pi_1(K,x_0)$ is injective) we recover the relation $ab=ba$ or $aba^{-1}b^{-1} = e$ in $\pi_1(T,z)$.

We shall see that $\pi_1 (K,x_0)$ is the group generated by $c$ and $b$ subject only to the relation $cbc^{-1}b = e$, induced by the boxed homotopy that shrinks the loop to a point. It follows that $\pi_1 (T,z)$ is isomorphic (via $p_{*}$) to the abelian subgroup generated by $a=c^2$ and $b$. Each element of $\pi_1(T,z)$ can be written uniquely as $a^mb^n$ for some $m,n \in \mathbb{Z}$. Converting to additive notation:

Summary

Topologically the $2$-torus $T = T^2$ is the product $S^1 \times S^1$, and there is a 'combined' covering map $\mathbb{R}^2 \rightarrow T$ defined by:

A theorem about $\pi_1$ of a product of spaces confirms the isomorphism $\pi_1(T) \cong \mathbb{Z} \times \mathbb{Z}$.

Covering Spaces:

Fundamental Groups: