4.6 General properties of higher homotopy groups

4.6.1 Abelian nature of higher homotopy groups

Higher homotopy groups are always Abelian; for any $n$ -loops $\alpha$ and $\beta$ at $x_0\in X$ , $[\alpha ]$ and $[\beta ]$ satisfy

$$[\alpha ]\ast [\beta ]=[\beta ]\ast [\alpha ].\text{\hspace{1em}(4.42)}$$

To verify this assertion let us observe figure 4.18. Clearly the deformation is homotopic at each step of the sequence. This shows that $\alpha \ast \beta \sim \beta \ast \alpha$ , namely $[\alpha ]\ast [\beta ]=[\beta ]\ast [\alpha ]$ .

Figure 4.18. Higher homotopy groups are always commutative, $\alpha \ast \beta \sim \beta \ast \alpha$ .

4.6.2 Arcwise connectedness and higher homotopy groups

If a topological space $X$ is arcwise connected, ${\pi }_n(X,x_0)$ is isomorphic to ${\pi }_n(X,x_1)$ for any pair $x_0,x_1\in X$ . The proof is parallel to that of theorem 4.2. Accordingly, if $X$ is arcwise connected, the base point need not be specified.

4.6.3 Homotopy invariance of higher homotopy groups

Let $X$ and $Y$ be topological spaces of the same homotopy type; see definition 4.6. If $f:X\to Y$ is a homotopy equivalence, the homotopy group ${\pi }_n(X,x_0)$ is isomorphic to ${\pi }_n(Y,f(x_0))$ ; cf theorem 4.3. Topological invariance of higher homotopy groups is the direct consequence of this fact. In particular, if $X$ is contractible, the homotopy groups are trivial: ${\pi }_n(X,x_0)=\{e\}$ , $n>1$ .

4.6.4 Higher homotopy groups of a product space

Let $X$ and $Y$ be arcwise connected topological spaces. Then

$${\pi }_n(X\times Y)\cong {\pi }_n(X)\oplus {\pi }_n(Y)\text{\hspace{1em}(4.43)}$$

cf theorem 4.6.

4.6.5 Universal covering spaces and higher homotopy groups

There are several cases in which the homotopy groups of one space are given by the known homotopy groups of the other space. There is a remarkable property

between the higher homotopy groups of a topological space and its universal covering space.

Definition 4.12

Let $X$ and $\tilde{X}$ be connected topological spaces. The pair $(\tilde{X},p)$ , or simply $\tilde{X}$ , is called the covering space of $X$ if there exists a continuous map $p:\tilde{X}\to X$ such that

(1) $p$ is surjective (onto)
(2) for each $x\in X$ , there exists a connected open set $U\subset X$ containing $x$ , such that $p^{-1}(U)$ is a disjoint union of open sets in $\tilde{X}$ , each of which is mapped homeomorphically onto $U$ by $p$ .

In particular, if $\tilde{X}$ is simply connected, $(\tilde{X},p)$ is called the universal covering space of $X$ . [Remarks: Certain groups are known to be topological spaces. They are called topological groups. For example $\mathrm{SO}(n)$ and $\mathrm{SU}(n)$ are topological groups. If $X$ and $\tilde{X}$ in definition 4.12 happen to be topological groups and $p:\tilde{X}\to X$ to be a group homomorphism, the (universal) covering space is called the (universal) covering group.]

For example, $\mathbb{R}$ is the universal covering space of $S^1$ , see section 4.3. Since $S^1$ is identified with $\mathrm{U}(1)$ , $\mathbb{R}$ is a universal covering group of $\mathrm{U}(1)$ if $\mathbb{R}$ is regarded as an additive group. The map $p:\mathbb{R}\to \mathrm{U}(1)$ may be $p:x\to {\mathrm{e}}^{\mathrm{i}2\pi x}$ . Clearly $p$ is surjective and if $U=\{{\mathrm{e}}^{\mathrm{i}2\pi x}|x\in (x_0-0.1,x_0+0.1)\}$ , then

$$p^{-1}(U)=\bigcup _{n\in \mathbb{Z}}(x_0-0.1+n,x_0+0.1+n)$$

which is a disjoint union of open sets of $\mathbb{R}$ . It is easy to show that $p$ is also a homomorphism with respect to addition in $\mathbb{R}$ and multiplication in $\mathrm{U}(1)$ . Hence, $(\mathbb{R},p)$ is the universal covering group of $\mathrm{U}(1)=S^1$ .

Theorem 4.10

Let $(\tilde{X},p)$ be the universal covering space of a connected topological space $X$ . If $x_0\in X$ and ${\tilde{x}}_0\in \tilde{X}$ are base points such that $p({\tilde{x}}_0)=x_0$ , the induced homomorphism

$$p_{\ast }:{\pi }_n(\tilde{X},{\tilde{x}}_0)\to {\pi }_n(X,x_0)\text{\hspace{1em}(4.44)}$$

is an isomorphism for $n\ge 2$ . [Warning: This theorem cannot be applied if $n=1$ ; ${\pi }_1(\mathbb{R})=\{e\}$ while ${\pi }_1(S^1)=\mathbb{Z}$ .]

The proof is given in Croom (1978). For example, we have ${\pi }_n(\mathbb{R})=\{e\}$ since $\mathbb{R}$ is contractible. Then we find

$${\pi }_n\left(S^1\right)\cong {\pi }_n(\mathrm{U}(1))=\{e\}n\ge 2.\text{\hspace{1em}(4.45)}$$

Example 4.12

Let $S^n=\{x\in {\mathbb{R}}^{n+1}||x{|}^2=1\}$ . The real projective space $\mathbb{R}{P}^n$ is obtained from $S^n$ by identifying the pair of antipodal points $(x,-x)$ . It is easy to

see that $S^n$ is a covering space of $\mathbb{R}{P}^n$ for $n\ge 2$ . Since ${\pi }_1(S^n)=\{e\}$ for $n\ge 2$ , $S^n$ is the universal covering space of $\mathbb{R}{P}^n$ and we have

$${\pi }_n\left(\mathbb{R}{P}^m\right)\cong {\pi }_n\left(S^m\right).\text{\hspace{1em}(4.46)}$$

It is interesting to note that $\mathbb{R}{P}^3$ is identified with SO(3). To see this let us specify an element of SO(3) by a rotation about an axis $n$ by an angle $\theta$ $(0<\theta <\pi )$ and assign a ‘vector’ $\Omega \equiv \theta n$ to this element. $\Omega$ takes its value in the disc $D^3$ of radius $\pi$ . Moreover, $\pi n$ and $-\pi n$ represent the same rotation and should be identified. Thus, the space to which $\Omega$ belongs is a disc $D^3$ whose anti-podal points on the surface $S^2$ are identified. Note also that we may express $\mathbb{R}{P}^3$ as the northern hemisphere $D^3$ of $S^3$ , whose anti-podal points on the boundary $S^2$ are identified. This shows that $\mathbb{R}{P}^3$ is identified with SO(3).

It is also interesting to see that $S^3$ is identified with $\mathrm{SU}(2)$ . First note that any element $g\in \mathrm{SU}(2)$ is written as

$$g=\left(\begin{array}{cc}a& -\bar{b}\\ b& \bar{a}\end{array}\right)|a{|}^2+|b{|}^2=1.\text{\hspace{1em}(4.47)}$$

If we write $a=u+\mathrm{i}v$ and $b=x+\mathrm{i}y$ , this becomes $S^3$ ,

$$u^2+v^2+x^2+y^2=1.$$

Collecting these results, we find

$${\pi }_n(\mathrm{SO}(3))={\pi }_n\left(\mathbb{R}{P}^3\right)={\pi }_n\left(S^3\right)={\pi }_n(\mathrm{SU}(2))n\ge 2.\text{\hspace{1em}(4.48)}$$

More generally, the universal covering group $\mathrm{Spin}(n)$ of $\mathrm{SO}(n)$ is called the spin group. For small $n$ , they are

$$\mathrm{Spin}(3)=\mathrm{SU}(2)\text{\hspace{1em}(4.49)}$$

$$\mathrm{Spin}(4)=\mathrm{SU}(2)\times \mathrm{SU}(2)\text{\hspace{1em}(4.50)}$$

$$\mathrm{Spin}(5)=\mathrm{USp}(4)\text{\hspace{1em}(4.51)}$$

$$\mathrm{Spin}(6)=\mathrm{SU}(4).\text{\hspace{1em}(4.52)}$$

Here $\mathrm{USp}(2N)$ stands for the compact group of $2N\times 2N$ matrices $A$ satisfying $A^{†}JA=J$ , where

$$J=\left(\begin{array}{cc}0& I_N\\ -I_N& 0\end{array}\right).$$

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