10 May 2014 No Comments

# Chapter 14: Lie Group Homotopy and Global Topology

We have seen that the Lie algebra $\g$ of a Lie group $\G$ encodes a great deal of information about the group. Indeed, through the Campbell Baker Hausdorff Theorem, it encodes *all* the local information about the group: two Lie groups with the same Lie algebra are locally isomorphic. However, the Lie algebra is evidently not the whole tale, for there are different Lie groups which have the same Lie algebra: the simply connected $SU(2)$ and the doubly connected $SO(3)$ for example, or the simply connected, noncompact $(\R,\,+)$ and the compact $U(1) \cong SO(2)$ whose fundamental group is $\mathbb{Z}$.

The last piece of information specifying a connected Lie group is some discrete information setting its global topology. To study this global topology, we need some basic *homotopy theory*. This is the aim of this post.

Recall that a connected Lie group is a topological group (Definition 9.30): *i.e.* a group with a topology wherein the group product is continuous with respect to this topology). A continuous $C^0$ path in such a group (indeed in any topological space) is a continuous mapping $\sigma :[0,\,1]\to \G$ (the interval is given its wonted topology in $\R$). We have seen (Theorem 9.34) that the concepts of path connectedness and connnectedness are the same in a connected Lie group, although this of course is not generally true for any topological space. Note that a subset of $\R^N$ is connected if and only if it is path connected.

**Definition 14.1****: (Path Connecting Two Points)**

Suppose $\sigma :[0,\,1]\to \G$ is a $C^0$ path in $G$ with $\sigma(0)=\gamma_0\in\G$ and $\sigma(1)=\gamma_1\in\G$. Then we say that path $\sigma$ links $\gamma_0$ to $\gamma_1$.

On the class of all $C^0$ paths between all pairs of points in $\G$ we can define a path concatenation product.

**Definition 14.2****: (Path Concatenation Product)**

Let $C^0$ path $\sigma_1:[0,\,1]\to\G$ link point $\gamma_1\in\G$ to $\gamma_2\in\G$ and suppose $C^0$ path $\sigma_2:[0,\,1]\to\G$ takes up where $\sigma_1$ left off by linking the point $\gamma_2\in\G$ to a third point $\gamma_3\in\G$, i.e. $\sigma_1(0)=\gamma_1,\,\sigma_1(1)=\sigma_2(0)=\gamma_2 ,\,\sigma_2(1)=\gamma_3$. Then we define the *path concatenation product* $\sigma_3:[0,\,1]\to\G =\sigma_1 \ast \sigma_2$ by:

\begin{equation}\label{PathConcatenationProductDefinition_1}\sigma_3(\tau)=\left\{\begin{array}{ll}\sigma_1(2\,\tau);&0\leqslant \tau \leqslant \frac{1}{2}\\\sigma_2(2\,\tau-1);&\frac{1}{2}\leqslant \tau \leqslant 1\end{array}\right.\end{equation}

In short, we begin at $\gamma_1 $ and run along path $\sigma_1 $ twice as quickly as we wontedly would to get to $\gamma_2 $, then do the same for $\sigma_2$ to reach $\gamma_3$.

**Definition 14.3****: (Homotopic Paths)**

Two paths $\sigma,\,\sigma^\prime:[0,\,1]\to \G$ with the same endpoints i.e. $\sigma(0)=\sigma^\prime(0);\,\sigma(1)=\sigma^\prime(1)$ are called *homotopic* if there is a continuous mapping $\tilde{\sigma}:[0,\,1]\times [0,\,1]\to \G$ such that $\tilde{\sigma}(\tau ,\,0)=\sigma(\tau)$ and $\tilde{\sigma}(\tau,\,1)=\sigma^\prime(\tau)$ and the function $\tilde{\sigma}$ is called *a homotopy* between the paths $\sigma$ and $\sigma^\prime$.

We note that this definition is simply a very natural the formalisation of the concept of continuously deforming one path into the other.

We now condense the (big) set of all $C^0$ paths through $\G$ by saying that two paths between the same two endpoints are equivalent if they are homotopic, *i.e. *one can be continuously deformed into the other. The homotopy relationship $\sigma \sim \sigma^\prime$ is clearly an equivalence relationship, being reflexive (every path is homotopic to itself, *i.e.* $\sigma \sim \sigma$), symmetric ($\sigma_A \sim \sigma_B \Rightarrow \sigma_B \sim \sigma_A$) and transitive ($\sigma_A \sim \sigma_B;\, \sigma_B \sim \sigma_C \Rightarrow \sigma_A \sim \sigma_C$).

**Definition 14.4****: (Homotopy Class)**

The equivalence classes of $C^0$ paths through $\G$ defined by the homotopy equivalence relationship $\sim$ are called the *homotopy classes* of $C^0$ paths through $\G$. We write the homotopy class of the path $\sigma:[0,\,1]\to\G$ as $[\sigma]$.

**Lemma 14.5****: (Well Definedness for Homotopy Class Concatenation Product)**

If $\sigma_1,\,\sigma_1^\prime$ are $C^0$ paths in $\G$ linking $\gamma_0,\,\gamma_1\in\G$ with $\sigma_1\sim\sigma_1^\prime$ and if $\sigma_2,\,\sigma_2^\prime$ are $C^0$ paths in $\G$ linking $\gamma_1,\,\gamma_2\in\G$ with $\sigma_2\sim\sigma_2^\prime$, then $\sigma_1\ast\sigma_2\sim\sigma_1^\prime\ast\sigma_2^\prime$, *i.e.* the concatenation product of homotopy classes is well defined (*i.e.* independent of the class members chosen to calculate the product of classes).

**Proof: **Show Proof

By definition, there are homotopies $\tilde{\sigma}_1:[0,\,1]\times[0,\,1]\to \G$ and $\tilde{\sigma}_2:[0,\,1]\times[0,\,1]\to \G$ continuously deforming $\sigma_1(\tau) = \tilde{\sigma}_1(\tau,\,0)$ to $\sigma_1^\prime(\tau) = \tilde{\sigma}_1(\tau,\,1)$ and $\sigma_2(\tau) = \tilde{\sigma}_2(\tau,\,0)$ to $\sigma_2^\prime(\tau) = \tilde{\sigma}_2(\tau,\,1)$, respectively. Now construct $\tilde{\sigma}:[0,\,1]\times[0,\,1]$ as follows:

\begin{equation}\label{HomotopyClassConcatenationProductWellDefinednessLemma_1}\tilde{\sigma}(\tau,\,\varsigma)=\left\{\begin{array}{ll}\tilde{\sigma}_1(2\,\tau,\,\varsigma);&0\leqslant \tau \leqslant \frac{1}{2}\\\tilde{\sigma}_2(2\,\tau-1,\,\varsigma);&\frac{1}{2}\leqslant \tau \leqslant 1\end{array}\right.\end{equation}

readily shown to be a homotopy between $\sigma_1\ast\sigma_2$ and $\sigma^\prime_1\ast\sigma^\prime_2$. $\quad\square$

Things get interesting when the beginning and end point of a path are the same, *i.e.* we talk about homotopies between loops and homotopy classes of loops. For a given point, we can consider the homotopy classes of loops that begin and end at the point. The classes together with concatenation $\ast$ is a clearly a group; the concatenation of two loops is another loop (the endpoint is the same), the concatenation product is associative and the inverse loop is defined by $\sigma^{-1}(\tau)=\sigma(1-\tau)$; the group is wontedly written $\pi_1(\G,\,\gamma)$ and called the “fundamental group of the path connected topological space $\G$ at point $\gamma\in\G$”, and was first described by Henri Poincaré. The subscript 1 stands for “first homotopy group”. Although we shall not used them, the higher homotopy group $\pi_n(\G,\,\gamma)$ is defined by considering $n-$dimensional spheres joined to a base point $\gamma$, and thinking of two spheres as homotopic if one can be continuously deformed into the other whilst keeping the base point fixed, and the group product is a higher dimensional generalisation of concatenation in $\pi_1(\G,\,\gamma)$, see [Hatcher] for tdetails. The aim here is to encode more topological information about the space in question: the the 3-ball has a trivial fundamental group, as has a 3-ball with a hollow bubble inside it but the two are clearly not homeomorphic. The second homotopy group, however, detects the difference: a sphere enclosing the hollow cannot be deformed to a point, whereas one not enclosing the hollow can. Topological spaces with different homotopy groups are never homeomorphic, but the converse is not true: non-homeomorphic spaces can have the same homotopy groups. So homotopy in general is only one tool for testing for topological differences. However, in the case of connected Lie groups, the fundamental group $\pi_1(\G,\,\gamma)$, together with the Lie algebra, encodes everything there is to know about a connected Lie group’s topology.

In a connected space, it makes no difference which point $\gamma $ is chosen as all the groups $\pi_1(\G,\,\gamma)$ for different $\gamma\in\G$ are isomorphic. For if we want to switch from base point $\gamma$ to $\gamma’$ we simply choose a path $\sigma_{\gamma ,\,\gamma^\prime} $ leading from $\gamma $ to $\gamma^\prime$ and define the isomorphism:

\begin{equation}\label{BasePointSwitchIsomorphism}\Sigma :\pi_1(\G,\,\gamma)\to \pi_1(\G,\,\gamma^\prime);\,\Sigma(\sigma)=\sigma_{\gamma ,\,\gamma^\prime}^{-1}\,\ast \,\sigma\, \ast \,\sigma_{\gamma ,\,\gamma^\prime}\end{equation}

Informally, we transform a loop at $\gamma $ to one at $\gamma’$ by running from $\gamma^\prime$ to $\gamma$, running the loop, then running back to $\gamma^\prime$ (see Figure 14.1. Hence we can write $\pi_1(\G)$ and mostly forget about base points for connected Lie groups and we confine our calculations to homotopy classes of loops beginning and ending at $\id$.

**Figure 14.1****: Changing a Base Point in a Connected Topological Space**

In any path connected topological group, therefore, *a fortiori*, in any connected Lie group, the group product and the path-concatenation product of the fundamental group are intimately related. Indeed, it is useful to specialise homotopy theory for topological groups only, because the following lemma shows that this theory is considerably simpler and more readily grasped than general homotopy theory.

**Theorem 14.6**

**: (Group and Concatenation Product The Same for Homotopy Classes)**

As binary operations defined on homotopy classes, the group and concatenation products are the same operation.

**Proof: **Show Proof

C\onsider loops with fixed point at $\id$, and let $\sigma_0 $define the identity class in $\pi_1(\G,\,\id)$; we take $\sigma_0 \left( \tau \right)=\id$. Then consider representatives $\sigma_1$ and $\sigma_2$ from any homotopy classes and the path $\sigma_3 (\tau)=(\sigma_1 \ast \sigma_0 )(\tau)\bullet (\sigma_0 \ast \sigma_2 )(\tau)$, where $\ast $ is the concatenation product and $\bullet$ the group product. Since $(\sigma_0 \ast \sigma_2 )(\tau)=\sigma_2 (0)=\id$ for $0\leqslant \tau \leqslant \frac{1}{2}$ and likewise $(\sigma_1 \ast \sigma_0 )(\tau)=\sigma_1 (1)=\id$ for $\frac{1}{2}\leqslant \tau \leqslant 1$, we have:

\begin{equation}\label{ConcatenationAndGroupProductTheorem_1}\sigma_3 (\tau)=\sigma_1 (\tau)\ast \sigma_2 (\tau)\end{equation}

But then $\sigma_1 \ast \sigma_0 $ is homotopic to $\sigma_1$: to see this, consider:

\begin{equation}\label{ConcatenationAndGroupProductTheorem_2}\sigma^\prime_1 (\tau,\,\varsigma)=\left\{\begin{array}{ll}\sigma_1 ((2-\varsigma)\,\tau );&0\leqslant \tau \leqslant \frac{1}{2-\varsigma}\\\id;&\frac{1}{2-\varsigma}\leqslant \tau \leqslant 1\end{array}\right.\end{equation}

thus $\sigma^\prime_1 (\tau,\,\varsigma)$ confines the path’s variation to the interval $0\leqslant \tau \leqslant 1/(2-\varsigma)$; thereafter the path idles at $\id$. Likewise, $\sigma_0 \ast \sigma_2 $ is homotopic to $\sigma_2$ through an analogously defined

\begin{equation}\label{ConcatenationAndGroupProductTheorem_3}\sigma^\prime_2 (\tau,\,\varsigma)\left\{\begin{array}{ll}\id;&0\leqslant \tau \leqslant 1-\frac{1}{2-\varsigma}\\\sigma_2((2-\varsigma )\,\tau+\varsigma-1);&1-\frac{1}{2-\varsigma}\leqslant \tau \leqslant 1\end{array}\right.\end{equation}

that idles at $\id$ for $0\leqslant \tau \leqslant 1-(2-\varsigma )^{-1}$ then switches on the variation defined by $\sigma_2 $ sped up by a factor of $2-\varsigma$ to complete the loop $\sigma_2 $ in the leftover time at $\tau =1$. Moreover, the group product $\bullet$ is continuous, so $\sigma_3(\tau)=(\sigma_1 \ast \sigma_0 )(\tau)\bullet (\sigma_0 \ast \sigma_2 )(\tau)=\sigma_1 (\tau)\ast \sigma_2 (\tau)$ and $\sigma_4(\tau)=\sigma_1 (\tau)\bullet \sigma_2 (\tau)$ are homotopic through $\sigma^\prime_3 (\tau,\,\varsigma)=\sigma^\prime_1 (\tau,\,\varsigma) \bullet \sigma^\prime_2 (\tau,\,\varsigma)$. Thus, $\sigma_1 (\tau)\ast \sigma_2 (\tau)\sim \sigma_1 (\tau)\bullet \sigma_2 (\tau)$, or, since concatenation is well defined at the class level (Lemma 14.5):

\begin{equation}

\label{ConcatenationAndGroupProductTheorem_4}

[\sigma_1(\tau)\ast\sigma_2(\tau)]=[\sigma_1(\tau)]\ast[\sigma_2(\tau)]=[\sigma_1(\tau)\bullet\sigma_2(\tau)]
\end{equation}

where $[\sigma]$ stands for the homotopy class of path $\sigma:[0,\,1]\to\G$. $\quad\square$

So we now use this homotopy theory to prove that:

**Theorem 14.7****: (Fundamental Group of Path Connected Topological Group is Abelian)**

The fundamental group at any base point point of a topological group is Abelian. Thus the fundamental group of a connected Lie group is Abelian.

Note: The author of this proof [Gutzwiller] has not formally published it elsewhere, thus he and I believe this is the first formal publishing of this argument.

**Proof: **Show Proof

Consider loops $\sigma_1$ and $\sigma_2 $ with fixed point at $\id$, and consider the parameterised path $\tilde {\sigma}(\tau,\,\varsigma)=\sigma_1(\varsigma\,\tau)^{-1}\,\bullet\,\sigma_1(\tau)\,\bullet\,\sigma_2 (\tau)\,\bullet\,\sigma_1 (\varsigma \,\tau)$ where $\bullet $ is the group product and the inverses are topological group ones, not path inverses in the fundamental group. We have:

\begin{equation}\label{AbelianFundamentalGroupTheorem_1}\begin{array}{lcl}\tilde{\sigma}(0,\,\varsigma)&=&\sigma_1(0)^{-1}\,\bullet\, \sigma_1(0)\,\bullet\, \sigma_2(0)\,\bullet\,\sigma_1(0)\\

&=&\sigma_1(\varsigma)^{-1}\,\bullet\, \sigma_1(1)\,\bullet\,\sigma_2 (1)\,\bullet\,\sigma_1(\varsigma)=\tilde{\sigma}(1,\,\varsigma)\end{array}\end{equation}

$\tilde{\sigma}(\tau,\,\varsigma)$ is thus a loop with endpoints fixed at $\id$ for all values of the parameter $\varsigma$. Given the continuous group product, $\tilde{\sigma}(\tau,\,\varsigma)$ is clearly a homotopy between the loops $\sigma_1(\tau)\,\bullet\,\sigma_2 (\tau)$ and $\sigma_2(\tau)\,\bullet\, \sigma_1(\tau)$. But by Theorem 14.6, these are representatives of the classes $[\sigma_1(\tau)]\,\ast\, [\sigma_2(\tau)]$ and $[\sigma_2 (\tau)]\,\ast\, [\sigma_1 (\tau)]$, respectively, which are thus shown to be homotopic. $\quad\square$

As a topological group, any connected Lie group has an Abelian fundamental group. This is one of many ways wherein Lie groups are “special”, i.e. highly specialised manifold. A general connected toplogical manifold (indeed general differentiable or analytic connected manifold) can be constructed to have *any* finitely generated group: for example, the fundamental group of a trefoil knot is the non-Abelian braid group $B_3$.

**Example 14.8****: (The Abelian Torus Group)**

The construction of Theorem 14.7 is shown for the 2-torus in Figure 14.2, where the fundamental polygon is the square ABCD: following the standard visualisation, we identify sides AB and CD (gluing them together so that the heavy arrows along them line up) to form a tube, then identify AC and BD, again so that the vertical arrows line up. We begin our path at the group identity 1, and ride to $\sigma_1 (\varsigma)^{-1}$ along path section 1 represented by $\sigma_1 (\varsigma \,\tau )^{-1}$. Then we leave from $\sigma_1 (\varsigma)^{-1}$ and run a poloidal (short loop threading the torus’s hole) loop (path section 2) ($i.e.$ the full loop $\sigma_1 )$ back to $\sigma_1 (\varsigma)^{-1}$. Then we leave from $\sigma_1 (\varsigma)^{-1}$ and run a toroidal (long loop, not threading the hole) loop (path section 3) ($i.e.$ the full loop $\sigma_2 )$ back to $\sigma_1 (\varsigma)^{-1}$. Lastly, we run the path section 4 represented by $\sigma_1 (\varsigma \,\tau )$ from $\sigma_1 (\varsigma)^{-1}$ back to the group identity. When $\varsigma =0$ path section 1 idles at the group 1, as does path section 4. Thus we have path section 2 (loop $\sigma_1 )$ followed by 3 (loop $\sigma_2 )$. When $\varsigma =1$, $\sigma_1(\varsigma)^{-1}$ has run the full loop $\sigma_1^{-1}$ to merge with the group identity: path section 1 is the full loop $\sigma_1 ^{-1}$, followed by path section 2 (loop $\sigma_1 )$, which mutually annihilate. Thus, we are left with path section 3 (loop $\sigma_2 )$ followed by path section 4, which is now the full loop $\sigma_1 $.

**Figure 14.2****: Illustration of Theorem 14.7 for the Abelian Torus Lie Group**

We now recall Lemma 8.7 and see that two Lie groups that have the same Lie algebra must be locally isomorphic. How then can they be different? Evidently there is some ambiguity in the exponential function that transforms the Campbell Baker Hausdorff product into the group product.

**Example 14.9****: (Different Lie Groups with the same Lie Algebra)**

This is not surprising: let’s think about the simply connected, noncompact $(\R,\,+)$ and $(\R^+\setminus\{0\},\,\ast)$ and the compact $U(1) \cong SO(2)$ whose fundamental group $\pi_1(U(1))$ is the discrete group $\mathbb{Z} = \pi_1(U(1))$ of integers. All three have the same Lie algebra, to wit $(\R,\,+)$ and the Lie bracket is trivial ($[X,\,Y]=0\,\forall\,X,\,Y\in\R$). If $X,\,Y\in\R$, then we have three different exponential maps:

\begin{equation}\label{RealsAndU1Exponentiation}\begin{array}{ll}\exp:\R\to\R;&\exp(X) = X\\\exp:\R\to\R^+\setminus\{0\};&\exp(X) = 1+X+\frac{X^2}{2!}+\frac{X^3}{3!} +\frac{X^4}{4!} +\cdots\\\exp:\R\to U(1);&\exp(X) = 1 + i\,X – \frac{X^2}{2!} -i\,\frac{X^3}{3!} + \frac{X^4}{4!} + \cdots\end{array}\end{equation}

We can sharpen insight into the different exponential maps by describing the above three Lie groups as matrix Lie groups:

Lie Group |
Matrix Lie Group |
Lie Algebra |

$(\R,\,+)$ | $\left(\left\{\left.\left(\begin{array}{cc}1&z\\0&1\end{array}\right)\right|\;z\in\R\right\},\,\bullet\right)$ | $\left\{\left.\left(\begin{array}{cc}0&z\\0&0\end{array}\right)\right|\;z\in\R\right\}$ |

$(\R^+\setminus\{0\},\,\ast)$ | $\left(\left\{\left.\left(\begin{array}{cc}e^z&o\\0&e^z\end{array}\right)\right|\;z\in\R\right\},\,\bullet\right)$ | $\left\{\left.\left(\begin{array}{cc}z&0\\0&z\end{array}\right)\right|\;z\in\R\right\}$ |

$(\R,\,+)$ | $\left(\left\{\left.\left(\begin{array}{cc}\cosh z&\sinh z\\\sinh z&\cosh z\end{array}\right)\right|\;z\in\R\right\},\,\bullet\right)$ | $\left\{\left.\left(\begin{array}{cc}0&z\\z&0\end{array}\right)\right|\;z\in\R\right\}$ |

$U(1)$ | $\left(\left\{\left.\left(\begin{array}{cc}\cos z&-\sin z\\\sin z&\cos z\end{array}\right)\right|\;z\in\R\right\},\,\bullet\right)$ | $\left\{\left.\left(\begin{array}{cc}0&-z\\z&0\end{array}\right)\right|\;z\in\R\right\}$ |

**Table 14.1****: Matrix Realisations of Different Lie groups with the Same Lie Algebra $\R\cong\mathfrak{u}(1)$**

In all the matrix Lie groups, the group product $\bullet$ is the matrix product. The matrix Lie groups equivalents unify the exponential definition as being defined by the standard, universally convergent Taylor series for the matrix exponential. The pair comprising:

- The exponential Taylor series $e^X=1+X+\frac{X^2}{2!}+\frac{X^3}{3!} +\frac{X^4}{4!} +\cdots$ and
- When $\left\|X-\id\right\|<1$, its unique inverse $\log X = (X-\id) – \frac{(X-\id)^2}{2}+\frac{(X-\id)^3}{3}-\frac{(X-\id)^4}{4}+\cdots$ (the Mercator-Newton series) uniformly convergent $\left\|X-\id\right\|<1$

are all that are needed, in the case of linear Lie groups (as shown by Chapters 1 and 2 of [Rossmann]), to bring into being all the properties of the exponential map so far derived, such as $\d_\tau \gamma = \gamma \,X;\,\gamma(0)=\id$ for $\gamma\in\G,\,X\in\g$, analytic geodesic co-ordinates, the Campbell Baker Hausdorff series and so forth with $\G$ and $\g$ the relevant Lie group and its algebra. But we can now see the reason that there is an ambiguity in how the exponential is defined when the *same* Lie algebra is realised as different matrix Lie algebras, for example the two isomorphic realisations of $\R$ and $\mathfrak{u}(1)$ as matrix Lie algebras in Table 14.1. Specifically, we can have different sets of matrices realising the basis of a given Lie algebra such that the characteristic equations fulfilled by corresponding basis matrices are quite different. The Cayley-Hamilton theorem then interacts with the exponential series to beget matrix functions of corresponding Lie algebra members with quite different behaviours. In Table 14.1, for example, the lone basis vector for the Lie algebra realised as different matrices fulfills different characteristic equations:

\begin{equation}\label{BasisCharacteristicEquations}\begin{array}{lll}\hat{i}=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)&\hat{i}^2=-\id&\exp\left(\hat{i}\,\tau\right)=\left(\begin{array}{cc}\cos\tau&-\sin\tau\\\sin\tau&\cos\tau\end{array}\right)\\\hat{i}^\prime=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)&(\hat{i}^\prime)^2=+\id&\exp\left({\hat{i}}^\prime\,\tau\right)=\left(\begin{array}{cc}\cosh\tau&\sinh\tau\\\sinh\tau&\cosh\tau\end{array}\right)\end{array}\end{equation}

A second example is even more dramatic. The groups $SU(2)$ and $SO(3)$ have the same Lie algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ and, as matrix groups they are realised by different dimensional matrices, to wit $2\times2$ and $3\times3$ matrices, respectively.

**Example 14.10****: (3D Rotation Group $SO(3)$ and Unit Quaternion Group $SU(2)$)**

Witness the two different Rodrigues formulas mapping the same Lie algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ to the topologically different $SU(2)$ and $SO(3)$:

\begin{equation}\label{SU2SO3Exponentiation}\begin{array}{rll}

\mathfrak{su}(2)\to SU(2):& H_{2\times 2}\mapsto\exp\left(H_{2\times 2}\right)=\cos\left(||H_{2\times 2}||\right)\;I_{2\times 2} +\frac{\sin\left(||H_{2\times 2}||\right)}{||H_{2\times 2}||}\, H_{2\times 2}\\

\mathfrak{so}(3)\to SO(3):& H_{3\times 3}\mapsto\exp\left(H_{3\times 3}\right)=I_{3\times3}+\frac{\sin\left(||H_{3\times 3}||\right)}{||H_{3\times 3}||}\,H_{3\times 3} +\frac{1-\cos\left(||H_{3\times 3}||\right)}{||H_{3\times 3}||^2}\,H_{3\times 3}^2\\

\end{array}\end{equation}

where:

\begin{equation}\label{SU2SO3ExponentiationDefinitions}H_{2\times2} = \left(\begin{array}{cc}i\,z&i\,x+y\\i\,x- y&-i\,z\end{array}\right)\quad\quad H_{3\times3} = \left(\begin{array}{ccc}0&z&-y\\-z&0&x\\y&-x&0\end{array}\right)\end{equation}

and $||H_{2\times2}||=||H_{3\times3}||=\sqrt{x^2+y^2+y^2}$. (In $SO(3)$ the CBH formula has a “closed form” expression, see [Engø], and a like closed form expression for $SU(2)$ follows by the same “tricks”). The same exponential Taylor series is used in each case, it’s simply that the exponential mapping’s nonlinearity manifests itself differently in each case owing to the Cayley-Hamilton theorem’s working on different characteristic equations fulfilled by $H_{2\times2}$ and $H_{3\times3}$.

So there is further information, beyond that encoded in the Lie algebra needed to fully define a Lie group, and that information lies in the ambiguity in defining the exponential map. One could study the exponential in detail seeking what and where the ambiguity lies, but there is a much simpler approach. For evidently, since the Campbell Baker Hausdorff series shows that the group product is *wholly* defined within a small enough neighbourhood (in the group topology) of the multiplicands, the difference between different Lie groups with the same Lie algebra must lie wholly confined to the global topologies of the respective Lie groups, which are locally homeomorphic (and diffeomorphic). So the little patches in Theorem 10.5 are precisely the same for two Lie groups with the same Lie algebra; the ambiguity must lie in how we can paste these little patches together to build the different connected Lie groups. A related observation is that the adjoint representation annihilates the group’s centre (set of all elements commuting with all others, which become the representation’s kernel), thus, although the Campbell Baker Hausdorff series can “see” a continuous part of the centre through the Euclidean sum $X + Y$, the higher order terms cannot see the discrete centre, *i.e. * one of whose elements arises for example in $SU(2)$ as $\exp(i\pi\mathbf{\sigma}_3)$, where $\mathbf{\sigma}_3=\mathrm{diag}(1,\,-1)$ is the third Pauli spin matrix. So the further information is contained in the group’s discrete centre, which turns out to be the fundamental group of the connected Lie group, *i.e.* a specification of the Lie group’s global topology.

The easiest way to answer all these global topology questions turns out to be the consideration of when we can find a “converse” to Theorem 7.12; *i.e.* every continuous homomorphism $\rho:\G\to\H$ from connected Lie group $\G$ to connected Lie group $\H$ induces a corresponding homomorphism $d\rho:\g\to\h$ between the corresponding Lie algebras $\g$ and $\h$; when, then, can or do Lie algebra homomorphisms induce homomorphisms of the corresponding Lie groups? Note that a Lie algebra homomorphism, as a linear map on a finite dimensional vector space, is needfully continuous, indeed analytic (with only a linear term in the Taylor series!). The more contemporary jargon for the “converse” of Theorem 7.12 is the “*lifting*“of a Lie algebra homomorphism.

**Definition 14.11****: (Lifting of a Lie Algebra Homomorphism)**

If a homomorphism $\phi:\g\to\h$ from the Lie algebra $\g$ of the Lie group $\G$ to the Lie algebra $\h\subset\g$ of the Lie group $\H\subset\G$ induces a corresponding Lie group homomorphism $\Phi:\G\to\H$ with $\phi$ as its Lie map, the we say that the Lie algebra homomorphism $\phi:\g\to\h$ *lifts* to the Lie group homomorphism $\G\to\H$.

The wording “lift, lifting” comes from the idea of *covering spaces:*

**Definition 14.12****: (Covering Space)**

Given a topological space $(\mathbb{X},\,\mathscr{T})$, a *covering space *is a bigger space $\tilde{\mathbb{X}}$ together with a projection map $p$ such that for every $x\in\mathbb{X}$ there is a neighbourhood $\U_x$ of $x$ such that the inverse image $p^{-1}(U_x)$ is a disjoint union $\bigcup\limits_{j\in \mathcal{I}} \U_{x,\,i}$ of sets $\U_{x,\,i}$ ($\U_{x,\,i}\cap\U_{x,\,j}=\emptyset$ when $j\neq i$), each of which $\U_{x,\,i}$ is open in $\tilde{\mathbb{X}}$ and such that the $\U_{x,\,i}$ are all homeomorphic to one another for some suitable index set $\mathcal{I}$; otherwise put, $p$ restricted to one and any disjoint copy $\U_{x,\,i}$ is a homeomorphism. The inverse image $p^{-1}(x)$ of a single point $x\in\mathbb{X}$ is called the *fibre* of $x$.

We think of the the copies lying over one another as in Figure 14.3, hence of properties which hold in the base neighbourhood $\U_x$ being “lifted” to the whole space. It will turn out that when a homomorphism $\phi:\g\to\h$ between Lie algebras $\g,\,\h$ can be “lifted” to a homomorphism $\Phi:\G\to\H$ between their Lie groups $\G,\,\H$, $\G$ is a covering space of $\H$ and the homomorphism i$\Phi$ is the projection $p$.

**Figure 14.3****: Relationships between Neighbourhoods in a Covering Space
**

Given a homomorphism $\phi:\g\to\h$, the obvious candidate for the Lie group homomorphism is some generalisation of $\Phi:{\Nid}_\G\to\H$ where $\Phi=\exp_\H\circ \phi \circ \log_\G$ *i.e.* $\Phi(\gamma) = \exp_\H(\phi(\log_\G(\gamma))$: we hereafter assume we use geodesic co-ordinates so that, for any Lie group, $\Nid$ is small enough that the Campbell Baker Hausdorff Theorem holds for pairs of group elements from $\Nid$ and $\exp:\V\to\Nid$ and $\log:\Nid\to\V$ are bijections. Furthermore we use the obvious notation $\exp_\G :\g\to\G$ and $\log_\G :{\Nid}_\G\to\g$ to stand for the exponential and logarithm defined between the Lie group $G$ and its algebra $\g$. So with $\Phi(\gamma) = \exp_\H(\phi(\log_\G(\gamma))$, $\log_\G$ maps $\gamma\in\G$ to the Lie algebra $\g$, whence the homomorphism maps the image to the Lie algebra $\h$ whence it can be exponentiated into $\H$ by $\exp_\H$. All looks good for the proposed homomorphism $\Phi$ locally, as the following lemma shows.

**Lemma 14.13****: (Local Lifting of Lie Algebra Homomorphism)**

Let $\phi:\g\to\h$ be a homomorphism from the Lie algebra $\g$ of the Lie group $\G$ to the Lie algebra $\h\subset\g$ of the Lie group $\H\subset\G$. Let $\K_\G$ be a nucleus of $\G$ small enough that the Campbell Baker Hausdorff Theorem holds for pairs of group elements from $\K_\G$ and $\exp:\V\subset\g\to\K_\G$ and $\log:\K_\G\to\V\subset\g$ are bijections. Then:

\begin{equation}\label{LocalHomomorphismLiftingLemma_1}\Phi:\K_\G\to\H;\,\Phi(\gamma) = \exp_\H(\phi(\log_\G(\gamma)))\end{equation}

is a local homomorphism, *i.e.* $\Phi(\gamma\,\zeta) = \Phi(\gamma\,\zeta)\,\Phi(\gamma\,\zeta)\,\forall\,\gamma,\,\zeta\in\K_\G$.

**Proof: **Let $\gamma = \exp_\G(X),\,\zeta=\exp_\G(Y)\in\K_\G$ and $X,\,Y\in\g$ then $\phi(\log_\G(\exp_\G(X)\,\exp_\G(Y)))=\phi(\varphi_{CBH}(X,\,Y)) = \varphi_{CBH}(\phi(X),\,\phi(Y))$ because $\phi$, being a Lie algebra homomorphism, respects Lie brackets and linear operations; here $\varphi_{CBH}$ is the Campbell Baker Hausdorff Product defined in Definition 8.2. Now $\varphi_{CBH}(\phi(X),\,\phi(Y)) = \log_\H(\exp_\H(\phi(X))\,\exp_\H(\phi(Y)))$, so that $\Phi(\gamma\,\zeta) = \exp_\H(\phi(X))\,\exp_\H(\phi(Y)) = \Phi(\gamma)\,\Phi(\zeta)$.$\quad\square$

So far, so good. So now, given that any element $\gamma$ in a connected Lie group $\G$ with Lie algebra $\g$ can be written as a finite product $\prod\limits_{k=1}^M \exp_\G(X_j)$ where $X_j\in\h$, the obvious definition for the global definition of a homomorphism $\Phi:\G\to\H$ is simply $\Phi(\gamma)=\prod\limits_{k=1}^M \exp_\H(\phi(X_j))$. Here, however, we strike a snag: we can write any $\gamma\in\G$ as many different equivalent products of the form $\gamma=\prod\limits_{k=1}^M \exp_\G(X_j)$ and it is not clear that the definition $\Phi(\gamma)$ is independent of exactly which of many equivalent products is chosen to represent $\gamma$. Therefore, the naïve definition is not in general sound.

However, we can specify a particular $C^0$ path from $\id$ to $\gamma\in\G$ by:

\begin{equation}\label{PathForHomomorphismDefinition}\sigma:[0,\,1]\to\G;\,\sigma(\tau) = \prod\limits_{k=1}^M\exp_\G(\mathscr{X}_j(\tau))\end{equation}

where:

\begin{equation}\label{PathForHomomorphismDefinition_2}\mathscr{X}_j(\tau) = \left\{\begin{array}{ll}0&0\leqslant\tau\leqslant\frac{j-1}{M}\\\tilde{X}_j(M\,\tau + 1-j)&\frac{j-1}{M}\leqslant\tau\leqslant\frac{j}{M}\\X_j&\frac{j}{M}\leqslant\tau\leqslant1\end{array}\right.\end{equation}

and $\tilde{X}_j(0) = 0$, $\tilde{X}_j(1) = X_j$ and $\tilde{X}_j:[0,\,1]\to\g$ is continuous. The piecewise defined exponent in $\eqref{PathForHomomorphismDefinition_2}$ “switches on” all the individual $\tilde{X}_j$ variations one after the other to link $\id$ to $\gamma=\prod\limits_{k=1}^M \exp_\G(X_j)$ by the $C^0$ path. If we specify a particular path in this way, then of course:

\begin{equation}\label{PathForHomomorphismDefinition_3}\Phi(\gamma) = \prod\limits_{k=1}^M\exp_\H(\phi(\mathscr{X}_j(1)))\end{equation}

is perfectly well defined. Moreover, we can “refine” the path specification, *i.e.* add a second by splitting a section of the path $\exp_\G(\mathscr{X}_j(\tau))$ into a concatenation of “half” paths $\exp_\G(\mathscr{X}_{j,1}(\tau))\,\exp_\G(\mathscr{X}_{j,2}(\tau))$ and then, by Lemma 14.13, the contribution $\exp_\G(\phi(\mathscr{X}_j(1)))$ is the same whether calculated as $\exp_\G(\phi(\mathscr{X}_j(1)))$ or as $\exp_\G(\phi(\mathscr{X}_{j,1}(1)))\,\exp_\G(\phi(\mathscr{X}_{j,2}(1)))$.

**Figure 14.4****: Deforming a Path in $\G$**

Furthermore we have the following theorem:

**Lemma 14.14****: (Homomorphism Definition Unaffected By Homotopy)**

The definition of $\Phi(\gamma)$ in $\eqref{PathForHomomorphismDefinition_3}$ defined for a $C^0$ path $\Gamma_1$ through $\G$ linking $\id$ and $\gamma\in\G$ is unaffected by homotopy (in $\G$’s group topology), *i.e.* the product is the same if defined in the same way for a second path $\Gamma_2$ linking $\id$ and $\gamma\in\G$ and if the $C^0$ paths $\Gamma_1,\,\Gamma_2$ are homotopic. *Otherwise put*: the definition of $\Phi(\gamma)$ in $\eqref{PathForHomomorphismDefinition_3}$ is independent of the $C^0$ path linking $\id$ and $\gamma$ for all paths through $\G$ within the same homotopy class. *Otherwise put*: the definition of $\Phi(\gamma)$ in $\eqref{PathForHomomorphismDefinition_3}$ is dependent, at most, only on the homotopy class of the path through $\G$ linking $\id$ and $\gamma$.

**Proof: **Show Proof

Let the factors $\mathscr{X}_j(\tau)$ be replaced now by the continuous $\mathscr{X}_j:[0,\,1]\times[0,\,1]\to\g$ so that $\mathscr{X}_j(\tau,\,0)$ defines the path $\Gamma(0)$ in $\eqref{PathForHomomorphismDefinition}$ linking $\id$ and $\gamma\in\G$ and the factors $\mathscr{X}_j(\tau,\,1)$ define a homotopically equivalent path $\Gamma(1)$ linking linking $\id$ and $\gamma\in\G$. As the path deforms, we consider two versions at time defined by $\mathscr{X}_j(\tau,\,s)$ and $\mathscr{X}_j(\tau,\,s+\epsilon)$. By continuity of the homotopy, we can choose $\epsilon$ small enough that corresponding points $\zeta$, $\zeta^\prime$ at deformation times $s$ and $s+\epsilon$ are near enough such that $\zeta^{-1}\,\zeta^\prime\in]\K_\G$. Moreover, we can refine the sectioning of the path as in the comments above so that endpoints of sections $a,\,b,\,c$ at deformation time $s$ and their corresponding images at deformation time $s+\epsilon$ are all in the same translated copy of $\K_\G$ (see Figure 14.5). So now we think of deforming the path $\Gamma(s)$ to $\Gamma(s+\epsilon)$ one section at a time. So suppose we slide the point $b$ to $b^\prime$ in Figure 14.4 and consider the change wrought in the value of $\Phi(\gamma)$ as the path section deforms. We shift $b$ to $b^\prime$ so that **( i) **the path section stays within the translated $\K_\G$ at all times and

**(**the endpoints of the path section $a^\prime\,c$ stay the same. The former condition

*ii*)**(**means that we can use Lemma 14.13 so that the contribution to the product in $\eqref{PathForHomomorphismDefinition_3}$ of the path sections are $\Phi(\gamma_1)\,\Phi(\gamma_2) = \Phi(\gamma_1,\,\gamma_2)$ for the section $a^\prime\,b\,c$ and $\Phi(\gamma_3)\,\Phi(\gamma_4) = \Phi(\gamma_3,\,\gamma_4)$ for the deformed path sections (see Figure 14.5 for definitions of $a^\prime,\,b\,b^\prime,\,c,\,\gamma_1,\,\cdots,\,\gamma_4$). But, since the endpoints $a^\prime,\,c$ stay the same, we have $\gamma_1,\,\gamma_2=\gamma_3\,\gamma_4\,\Rightarrow\,\Phi(\gamma_1,\,\gamma_2)=\Phi(\gamma_3\,\gamma_4)$ and therefore the value of the product in $\eqref{PathForHomomorphismDefinition_3}$

*i*)*does not change as the path section $a^\prime\,c$ is deformed*. We repeat this procedure and argument for all the endpoints in the path $\Gamma(s)$ to deform it to the path $\Gamma(s+\epsilon)$ in Figure 14.4. Because the endpoints of the whole curve, namely $\id$ and $\gamma$ stay the same at all times, this means that the value of the product in $\eqref{PathForHomomorphismDefinition_3}$ is

*unchanged by the deformation of*. We can build up the whole deformation from $\Gamma(0)$ to a homotopically equivalent $\Gamma(1)$ from a finite sequence of deformations from $\Gamma(s)$ to $\Gamma(s+\epsilon)$ as in the argument above, therefore the value of hte product in $\eqref{PathForHomomorphismDefinition_3}$ is the same for all homotopically equivalent paths $\Gamma$ linking $\id$ and $\gamma$. $\quad\square$

*the path*$\Gamma(s)$ to the path $\Gamma(s+\epsilon)$**Figure 14.5****: Deforming a Path Section in a Translated $\K_\G$**

We give a simple definition first:

**Definition 14.15****: (Simple Connectedness)**

A connected topological space $(\mathbb{X},\,\mathscr{T})$ with trivial fundamental group *i.e.* $\pi_1(\mathbb{X}) = \{\id\}$ is called *simply connected*. Otherwise put: a connected topological space wherein every loop through a point can be continuously deformed (“*contracted*“) to that point (*i.e.* through a homotopy) is *simply connected*.

and now the main homomorphism lifting theorem is easily proven:

**Theorem 14.16****: (Lifting of Lie Algebra Homomorphism)**

Let $\G$ be a simply connected Lie group with Lie algebra $g$. Let $\H$ be another connected Lie group with Lie algebra $\h$. Then every homomorphism $\phi:\g\to\h$ between the Lie algebras lifts to a homomorphism $\Phi:\G\to\H$ between Lie groups and, for a general member $\gamma=\prod\limits_{k=1}^M \exp_\G(X_k)\in\G$ where $X_j \in \g$ the homomorphism is defined by:

\begin{equation}\label{HomomorphismLiftingTheorem_1}\Phi(\gamma)=\prod\limits_{k=1}^M \exp_\H(\phi(X_k))\in\G\end{equation}

**Proof: **$\G$ issimply connected, therefore there is but one, unique homotopy class for $C^0$ paths linking $\id$ and any $\gamma\in\G$; if there were more than one, say paths $\sigma_1,\,\sigma_2$ in different homotopy classes, then no homotopy could contract the loop $\sigma_1,\,\sigma_2^{-1}$ through $\id$ to the point $\id\in\G$. Therefore, by Lemma 14.14, the product in $\eqref{HomomorphismLiftingTheorem_1}$ is well defined, for there is only one homotopy class of path, therefore, a unique value of $\Phi(\gamma)$ independent of the product used to realise $\gamma\in\G$. The rule $\Phi(\gamma\,\zeta)=\Phi(\gamma)\,\Phi(\zeta)$ follows readily from the definition in $\eqref{HomomorphismLiftingTheorem_1}$. $\quad\square$

**Corollary 14.17****: (Uniqueness of Simply Connected Lie Group)**

There is at most one simply connected Lie group $\G$ with any given Lie algebra $\g$, *i.e.* any two simply connected Lie groups with the same Lie algebra must be isomorphic.

**Proof:** Show Proof

Let the two groups in question be $\G_1$ and $\G_2$, each with the same Lie algebra $\g$. So apply Theorem 14.16 to $\G_1$ with $\phi:\g\to\g;\,\phi(X)=X$ and the homomorphism $\phi$ (indeed Lie algebra isomorphism) lifts to the Lie group homomorphism $\Phi$. So now we must prove that $\Phi$ is one to one and onto.

* To Prove One-to-One*: suppose otherwise,

*i.e.*$\ker(\Phi)$ is nontrivial. Since $\Phi$ preserves Lie algebras, it is a local isomorphism and so there is a nucleus $\K\subset\G_1$ such that $\K\cap\ker(\Phi)=\{\id\}$,

*i.e.*each member $\gamma\in\ker(\Phi)$ of $\ker(\Phi)$ is confined to its own neighbourhood of the form $\gamma\,\K$. So $\ker(\Phi)$ is discrete and, by Theorem 10.5, countable. So take the identity and one other distinct kernel member $\gamma\in\G_1$:

\begin{equation}\label{UniqueSimplyConnectedLieGroupCorollary_1}\begin{array}{ll}\gamma=\prod\limits_{k=1}^M\,\exp_{\G_1}(X_{\gamma,\,k});&X_{\gamma,\,k}\in\g\\\id=\prod\limits_{k=1}^M\,\exp_{\G_1}(0\,\times\,X_{\gamma,\,k})\end{array}\end{equation}

and then their images under $\Phi$ are:

\begin{equation}\label{UniqueSimplyConnectedLieGroupCorollary_2}\Phi(\gamma)=\prod\limits_{k=1}^M\,\exp_{\G_2}(X_{\gamma,\,k})=\prod\limits_{k=1}^M\,\exp_{\G_2}(0\,\times\,X_{\gamma,\,k})=\id\end{equation}

since $\phi(X_{\gamma,\,k}) = X_{\gamma,\,k}$. But $\G_2$ is simply connected, therefore there must be continuous functions $\hat{X}_{\gamma,\,k}:[0\,\,1]\times[0\,\,1]\to\g$ such that $\hat{X}_{\gamma,\,k}(0,\,\tau)=0$ and$ \hat{X}_{\gamma,\,k}(1,\,\tau) = \tau\,X_{\gamma,\,k}$, otherwise the path $\sigma:[0,\,1]\to\G_2;\,\sigma(\tau)=\prod\limits_{k=1}^M\,\exp_{\G_2}(\tau\,X_{\gamma,\,k})$, which is a loop beginning and ending at the identity (when $\tau=0$ or $1$)in $\G_2$, would be in a distinct homotopy class from the trivial loop $\sigma^\prime(\tau) = \prod\limits_{k=1}^M\,\exp_{\G_2}(0\,\times\,X_{\gamma,\,k})$ in $\G_2$, gainsaying $\G_2$’s simple connectedness. So now, consider the trajectory $\zeta$ of the endpoint of the path in $\G_1$ (we can do this because $\phi$ is the identity Lie algebra isomorphism):

\begin{equation}\label{UniqueSimplyConnectedLieGroupCorollary_3}\zeta(\varsigma) = \prod\limits_{k=1}^M\,\exp_{\G_1}\left(\hat{X}_{\gamma,\,k}(\varsigma,\,1)\right)\in\G_1\end{equation}

which defines a $C^0$ path in $\G_1$ (since the $\hat{X}_{\gamma,\,k}:[0\,\,1]\times[0\,\,1]\to\g$ are continuous). But, by definition, $\zeta(0)=\id\in\G_1$, $\zeta(1)=\gamma\in\G_1$ and $\zeta(\varsigma)$ can only take on values in $\ker(\Phi)$ because $\Phi(\zeta(\varsigma))=\id,\,\forall\,\varsigma\in[0,\,1]$. But $\ker(\Phi)$ is discrete and countable, so, arguing as in the proof of Error: Reference not found, this is a contradition unless $\zeta(\tau)=\id,\,\forall\,\tau\in[0,\,1]$. So $\ker(\Phi) = \{\id\}$ after all.

* To Prove Onto*: suppose otherwise,

*i.e.*$\Phi(\G_1)\subset\G_2$ is a proper subset of $\G_2$. $\Phi(\G_1)$, as the image of a homomorphism that preserves Lie algebras, is a group and it contains a nucleus $\Phi(\K)\subset\G_2$ which is homeomorphic to $\K$, so therefore it must contain the smallest connected group $\Gamma_2$ of all finite products of members of $\Phi(\K)$. Arguing as in the proof of Theorem 9.31, we then prove that $\Gamma_2$ and its complement $\G_2\setminus\Gamma_2$ are open (indeed, both open and closed), so therefore $\G_2$ is disconnected, gainsaying its assumed simple connectedness unless $\Phi$ is truly onto.

Therefore $\Phi$ is a one-to-one, onto homomorphism, *i.e.* a group isomorphism that preserves Lie algeras and so $G_1$ and $G_2$ are isomorphic both as abstract and Lie groups. $\quad\square$

Note that Corollary 14.17 does not say there is exactly one simply connected Lie group for every Lie algebra: as far as we know so far, there may be some Lie algebras which are not the Lie algebra of any Lie group. We shall show this, however, to be untrue and indeed *for every Lie algebra** there is precisely one simply connected Lie group*.

It’s worth taking heed that the argument in Corollary 14.17 shows that:

**Corollary 14.18****: (Lie Algebra Preserving Homomorphisms have Central Kernels)**

The kernel $\ker(\Phi)$ of any homomorphism $\Phi:\G\to\H$ between connected Lie groups $\G$ and $\H$ with the same Lie algebra is a (normal) subgroup of the discrete centre $\mathscr{Z}_D(\G)$.

**Proof: **Arguing as in Corollary 14.17, $\ker(\Phi)$ is discrete and is normal in $\G$, thus, by Error: Reference not found, is a discrete subgroup of the discrete centre $\mathscr{Z}_D(\G)$.$\quad\square$

**Example 14.19****: ($SO(3)$ is not Simply Connected)**

$SO(3)$ is not simply connected as we can understand from Figure 14.6 below.

**Figure 14.6****: ****$SO(3)$ as a Ball of Radius $\pi$ in $\R^3$**

Here we must\ imagine all the rotation operators in $SO(3)$ as points in a compactified Euclidean subspace of $\R^3$: a sphere of radius $\pi$ for which antipodean point pairs on its surface are “identified” – thought of as being the same point. A rotation of angle $\theta$ about an axis defined by the unit vector $(\hat{x},\,\hat{y},\,\hat{z})$ is drawn in the compactified space as follows: (firstly, we restrict angles to lie in the interval $(-\pi, \pi]$ so we get $\theta\mapsto\theta^\prime = \theta + 2\,k\,\pi\in(-\pi, \pi]$ by chopping off unimportant chunks of whole number $k$ multiples of $2\pi$, then we draw a vector of length $\theta^\prime$ with its tail at the origin (the group’s identity) and in the direction of $(\hat{x},\,\hat{y},\,\hat{z})$: the point at the vector’s head uniquely represents any element in $SO(3)$. Now we think about the fundamental group of $SO(3)$; there is the homotopy class of paths like $\Gamma$ which can be continuously shrunken to a point and those like $\Omega$ which cannot. Imagine following the $C^1$ path $\Omega$ through the Lie group: when we reach the point $P$ and go a distance $\epsilon$ further for any $\pi>\epsilon>0$, we find ourselves on the sphere’s “opposite side”, just beyond the point $P^\prime$ diametrically opposite to $P$. We keep going on this path $\Omega$ until we loop back to the identity. $\Omega$ cannot then be continuously shrunken back to a point at the identity. Because the path emerges from an antipodean point (actually the same point in our definition) as soon as it crosses the sphere’s surface we would need to “pull the path back though $P$” so we can loop back to origin, but we cannot do this as the path is joined beyond $P^\prime$ to the identity, so the homotopy class of $\Omega$ is an element of the fundamental group $\pi_1(SO(3))$ of $SO(3)$ distinct from the identity and so $\pi_1(SO(3))$ is not trivial. However, there is an obvious homotopy between $\Omega$ and its inverse loop $\Omega^{-1}$ (*i.e.* $\Omega$ run in the opposite sense): the homotopy is defined by a rotation of $\Omega$ through a half turn in Figure 14.6 about the origin (*i.e.* the identity $\id$). So this loop is not like winding a loop through a torus: we can continuously deform $\Omega$ into its inverse, whereas there is no way to make the arrows point the other way on a loop with arrows drawn on it threaded through a torus without first breaking the loop. So our fundamental group presentation is $\pi_1(SO(3))=\left<\Omega\,|\, \Omega^2=1\right>\cong\mathbb{Z}_2$.

**Definition 14.20****: (Universal Covering Group)**

Let $\G$ be a connected Lie group. Now defined the set:

\begin{equation}\label{UniversalCoverDefinition_1}\tilde{\G}=\{(\gamma,\,[\sigma_\gamma]);\,\gamma\in\G;\,\sigma_\gamma:[0,\,1]\to\G\ni\sigma_\gamma(0)=\id,\,\sigma_\gamma(1)=\gamma,\,\sigma_\gamma\text{ is }C^0\}\end{equation}

so that it is the set of pairs of members $\gamma$ of $\G$ together with homotopy classes $[\sigma_\gamma]$ of $C^0$ paths linking $\id$ and the respective member $\gamma\in\G$. Now define the binary operation:

\begin{equation}\label{UniversalCoverDefinition_2}\bullet:\tilde{\G}\times\tilde{\G}\to\tilde{\G};\,(\gamma,\,[\sigma_\gamma]) \bullet (\zeta,\,[\sigma_\zeta]) = (\gamma\,\zeta,\,[\sigma_\gamma\,\ast\,\gamma\,\sigma_\gamma])\end{equation}

where $\ast$ is the path concatenation product and therefore the class $[\sigma_\gamma\,\ast\,\gamma\,\sigma_\gamma]$ is well defined by Lemma 14.5. Then the pair $(\tilde{\G},\,\bullet)$, often written simply $\tilde{\G}$ when there is no likelihood of ambiguity, is the *universal covering group *of $\G$.

**Take heed:** $\sigma_\gamma:[0,\,1]\to\G$ is a $C^0$ path from $\id$ to $\gamma$ and $\sigma_\zeta:[0,\,1]\to\G$ a $C^0$ path from $\id$ to $\zeta$, with $\sigma_\gamma(0) = \sigma_\zeta(0) = \id$ and $\sigma_\gamma(1)=\gamma,\,\sigma_\zeta(1)=\zeta$. Then we use the obvious notation $\gamma\,\sigma_\zeta(\tau)$ is the group product (in $\G$) between the constant $\gamma$ and the path $\sigma_\zeta(\tau)$, so that $\gamma\,\sigma_\zeta:[0,\,1]\to\G$ is a $C^1$ path linking $\gamma$ to $\gamma\,\zeta$ with $\gamma\,\sigma_\zeta(0)=\gamma$ and $\gamma\,\sigma_\zeta(1)=\gamma\,\zeta$.

**Theorem 14.21****: Properties of the Universal Covering Group)**

For a connected Lie group $\G$, the universal cover $\tilde{\G}$ defined by Definition 14.20 has the following properties:

- $\tilde{\G}$ is a simply connected Lie group;
- The Lie algebra of $\tilde{\G}$ is $\g$, the same as that of $\G$;
- $\tilde{\G}$ is a covering space of $\G$;
- The projection $p:\tilde{\G}\to\G$ is defined by $p((\gamma,\,[\sigma_\gamma])) = \gamma$ is a homomorphism $\tilde{\G}\to\G$ and its differential $\g\to\g$ is the identity map;
- Any
*connected*covering space $\hat{\G}$ of $\G$ is itself covered by $\tilde{\G}$; - $\tilde{\G}$ has no nontrivial connected covers.

**Proof:** Show Proof

Although quite long winded, this proof is really simply checking that various definitions hold.

**$\tilde{\G}$ is a simply connected Lie group with Lie algebra $\g$:**

Products are defined and associative, because both the group multiplication in $\G$ and the path concatenation operations are also thus. The inverse of $(\gamma,\,\sigma_\gamma:[0,\,1]\to\G)$ is $(\gamma^{-1},\, (\sigma_\gamma)^{-1})$ where $(\sigma_\gamma)^{-1}:[0,\,1]\to\G;\;(\sigma_\gamma)^{-1}(\tau) = \gamma^{-1}\,\sigma_\gamma(1-\tau)$. The identity is $(\id,\,\sigma_\id)$, where $\sigma_\id:[0,\,1]\to\G;\,\sigma_\id(\tau)=\id$ is the trivial loop, or the identity in the fundamental group. So therefore $\tilde{\G}$ is a group.

Now consider what a path linking the identity to some $\tilde{\gamma}\in\tilde{\G}$ looks like. It is of the form:

\begin{equation}\label{UniversalCoveringGroupTheorem_1}\sigma_{\tilde{\gamma}}:[0,\,1]\to\tilde{G};\,\sigma_{\tilde{\gamma}}(\tau) = \left(\sigma_\gamma(\tau),\,\tilde{\sigma}_\tau:[0,\,1]\to\G\right);\, \tilde{\sigma}_\tau(u) = \sigma_\gamma(\tau\,u)\end{equation}

*i.e. *it is defined by a path of ordered pairs whose first element at the value $\tau\in[0,\,1]$ is the value at $\tau$ of the corresponding path $\sigma_\tilde{\gamma}$ through $\G$ defining the homotopy class of the link between $\id$ and $\gamma\in\G$. Now, by definition, there can only be one homotopy class of such paths in $\tilde{\G}$, because if we define the element $\tilde{\gamma}\in\tilde{\G}$ by $\gamma$ together with a linking path from a different homotopy class, we are by definition defining a distinct element $\tilde{\gamma}^\prime$ of $\tilde{\G}$. So all the possible $\tilde{\sigma}_\tau$, *i.e.* the second element in the pair in $\eqref{UniversalCoveringGroupTheorem_1}$ must be homotopic with one another at each value of $\tau\in[0,\,1]$ if we choose any other path $\sigma^\prime_\gamma(\tau)\sim \sigma_\gamma(\tau)$ to define the same element of $\tilde{\gamma}\in\tilde{\G}$. There is then precisely one homotopy class of paths linking the identity and any given element $\gamma\in\tilde{\G}$ (and thus one homotopy class for paths linking any pair of elements of $\tilde{\G}$ and one homotopy class of loops beginning and ending on any point of $\tilde{\G}$, because by assumption $\tilde{\G}$ is connected), *i.e.* *$\tilde{\G}$ is simply connected. *

We now define a neighbourhood of the identity $\id\in\tilde{\G}$ as a set of the form:

\begin{equation}\label{UniversalCoveringGroupTheorem_2}\tilde{\K}\subset\tilde{G}=\{(e^X,\,[\sigma_X:[0,\,1]\to\G;\,\sigma_X(\tau)=e{\tau\,X}])|\,X\in\g\,\left\|X\right\|<\epsilon\}\end{equation}

where $\epsilon>0$ is small enough that the Campbell Baker Hausdorff theorem applies to all pairs $e^X,\,e^Y$ when $\left\|X\right\|,\,\left\|Y\right\|<\epsilon$, *i.e.* we choose elements of a nucleus $\K =\{e^X|\,X\in\g\,\left\|X\right\|<\epsilon\}\subset\G$ in $\G$ paired with linking paths that are wholly confined within that nucleus. Then the map $p_0:\tilde{\K}\to\K;\,p_0((e^X,\,\sigma_X)) = e^X$ is clearly a bijection and a $C^\omega$ isomorphism between the sets $\K$ and $\tilde{K}$ (the path defined by $\exp(\tau\,\log(e^X\,e^Y))$ is readily shown to be homotopic to the path defined by $e^{\tau\,X}\,e^{\tau\,Y}$). The function $\log:\tilde{K}\to\g;\,\log((e^X,\,\sigma_X))=X$ is a valid logarithm function defined on $\tilde{K}$. $\tilde{G}$ together with $\tilde{K}$ and $\log$ is then readily shown to fulfill the Labeller Axiom 1, the Connectedness Axiom 2, the Group Product Continuity Axiom 3 and the Nontrivial Continuity Axiom 4. $\tilde{\G}$ is clearly path connected, therefore fulfills the Homgeneity Axiom 5 by Theorem 9.34. It is therefore a connected Lie group with the same Lie algebra $\g$ as $\G$.

**$\tilde{\G}$ is a covering space of $\G$:**

We now define the projection map $p$ very obviously:

\begin{equation}\label{UniversalCoveringGroupTheorem_3}p:\tilde{\G}\to\G;\,p((\gamma,\,\sigma_\gamma:[0,\,1]\to\G)) =\gamma)\end{equation}

This map is readily shown to make $\tilde{\G}$ into a covering space of $\G$ fulfilling Definition 14.12 when we consider the inverse images $p^{-1}(\gamma\,\tilde{K})$ of the neighbourhoods $\gamma\,\K$.

The fibres $p^{-1}(\gamma)$ of the covering are the homotopy classes of paths linking $\id\in\G$ to $\gamma\in\G$. The fibre $p^{-1}(\id)$ is the fundamental group $\pi_1(\G)$ of $\G$.

**$p:\tilde{\G}\to\G$ is a homomorphism with Lie map given by the identity map:**

This is clear from our definition of neighbourhood in $\eqref{UniversalCoveringGroupTheorem_2}$ and the discussion straight afterwards.

**Any connected covering group $\hat{\G}$ of $\G$ is itself covered by $\tilde{\G}$:**

From $\hat{\G}$ form the universal cover $\tilde{\hat{\G}}$ as defined in Definition 14.20. $\tilde{\hat{\G}}$ has the same Lie algebra as $\tilde{G}$ and has the same Lie algebra as $\tilde{G}$. So $\tilde{G}, \, \tilde{\hat{\G}}$ are isomorphic, both as abstract and as Lie groups, by Corollary 14.17.

**$\tilde{\G}$ has no nontrivial connected covers:**

Suppose there were a nontrivial cover $\hat{\G}$ of $\tilde{G}$: it is also a cover of $\G$, so argue as above, thus showing that $\hat{\G}\subseteq\G$. *Otherwise:* from first principles, show that the image inclusion map $\tilde{G}\hookrightarrow \hat{\G}$ is both open and closed in $\hat{\G}$, analogously with the method used to prove onto-ness in Corollary 14.17. $\quad\square$

Properties 5. and 6. in Theorem 14.21 justify the words “Universal Cover”: every connected cover of a Lie group is itself covered by its universal cover.

From all considerations of this post, we see how a connected Lie group’s global topology and the group’s centre are intimately related. We record the useful relationships we have proven above as follows:

**Lemma 14.22**: **(Characterisation of Global Topology of a Connected Lie Group)**

For any connected Lie group $\G$:

- Definition 14.20 uniquely defines the universal cover $\tilde{\G}$;
- $\tilde{G}$ is a discrete central extension of $\G$,
*i.e.*: $\G = \tilde{\G} / \pi_1(\G)$ and $\pi_1(\G)$ is an abelian subgroup of the abelian discrete centre $\mathscr{Z}_D(\tilde{\G})$ of the universal cover $\tilde{\G}$; - If $p:\tilde{\G}\to\G$ is the covering homomorphism from the universal cover $\tilde{\G}$ to a Lie group (
*i.e.*the function $(\gamma,\,\sigma_\gamma)\mapsto\gamma$ of $\eqref{UniversalCoveringGroupTheorem_3}$ that simply “forgets” the homotopy class of path linking the identity and the point $\gamma\in\G$), then $\G = \tilde{\G}/\ker(p)$ and $\pi_1(\G) = \ker(p)$.

$\square$

A second and highly famous demonstration of $SO(3)$’s non-simple connectedness is the following.

**Example 14.23****: (Dirac Belt Trick)**

Witness first that any member of $SO(3)$ can be represented in three dimensional space simply by the directions of the three unit vectors $\hat{\mathbf{x}},\,\hat{\mathbf{y}},\,\hat{\mathbf{z}}$ after this frame of vectors has be rotated by the member in question. We can also represent a member of the fundamental group $\pi_1(SO(3))$ of $SO(3)$ as follows. Let $\sigma:[0,\,1]\to SO(3)$ be the loop. The path is represented by a space curve for which:

- The arclength along the curve is the “time” $\tau\in[0,\,1]$;
- The tangent to the space curve at that time is $\hat{\mathbf{z}}^\prime(\tau) =\sigma(\tau)\,\hat{\mathbf{z}}$,
*i.e.*the unit vector $\hat{\mathbf{z}}$ rotated by the relevant element $\sigma(\tau)\in SO(3)$; - $\hat{\mathbf{x}}^\prime(\tau)$, $\hat{\mathbf{y}}^\prime(\tau)$, whose directions complete the definition of $\sigma(\tau)$, lie in the plane normal to the space curve;
- The pair of vectors $\hat{\mathbf{z}}^\prime(\tau)$, $\hat{\mathbf{x}}^\prime(\tau)$
**(**uniquely defines the third as $\hat{\mathbf{y}}^\prime(\tau) = \hat{\mathbf{z}}^\prime(\tau)\times\hat{\mathbf{x}}^\prime(\tau)$ and*i*)**(**defines a “ribbon” or “belt” comprising unit length fibres normal to the tangent vector to the centre of the belt.*ii*)

Thus a loop in $SO(3)$ is represented by a ribbon or belt whose two ends represent the identity element of $SO(3)$, *i.e.* the tangent to the belt’s centre is in the direction $\hat{\mathbf{z}}^\prime(0)=\hat{\mathbf{z}}^\prime(1)=(0,\,0,\,1)$ at both ends and the normal to the tangent lying along the belt’s fibre through $\tau=0,\,1$ is $\hat{\mathbf{x}}^\prime(0)=\hat{\mathbf{x}}^\prime(1)=(1,\,0,\,0)$.

If the ribbon can be continuously deformed whilst keeping the end tangents $\hat{\mathbf{z}}^\prime(0)=\hat{\mathbf{z}}^\prime(1)=(0,\,0,\,1)$ and end normals $\hat{\mathbf{x}}^\prime(0)=\hat{\mathbf{x}}^\prime(1)=(1,\,0,\,0)$ fixed, then that ribbon now represents a new loop in the same homotopy class as the former one.

A physical ribbon cannot undergo all theoretically *continuous *deformations: its mechanical elasticity and inability to pass through itself both limit the set of continuous deformations it can physically undergo. However, those it can physically undergo are accurate physical embodiments of theoretical homotopies. Otherwise put: a physical ribbon cannot represent all members of a homotopy class, but those it can physically represent are certainly in the same homotopy class.

For example, consider the ribbon with two full twists in it as in Figure 14.7. The ribbon has a marker (a card with a picture of a matryoshka doll on it) to identify orientation unambiguously. This ribbon encodes the path $\sigma:[0,\,1]\to SO(3);\,\sigma(\tau) = \exp(4\,\pi\,\hat{S}_z\,\tau)$, where $\{\hat{S}_x,\,\hat{S}_y,\,\hat{S}_z\}$ span the Lie algebra $\mathfrak{so}(3)$, $\exp(\hat{S}_x\,\theta),\,\exp(\hat{S}_y\,\theta),\,\exp(\hat{S}_z\,\theta)$ are $3\times3$ matrices of rotations through angles $\theta$ about the $x,\,y,\,z$ axes respectively and:

\begin{equation}\label{DiracBeltTrickExample_1}\hat{S}_x = \left(\begin{array}{ccc}0 & 0 & 0 \\0 & 0 & -1 \\0 & 1 & 0 \end{array}\right);\quad \hat{S}_y = \left(\begin{array}{ccc}0 & 0 & 1 \\0 & 0 & 0 \\-1 & 0 & 0\end{array}\right);\quad \hat{S}_z = \left(\begin{array}{ccc}0 & -1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)\end{equation}

**Figure 14.7**: **(Ribbon With Double Twist)**

By looping the ribbon over the doll as shown in the sequence of frames in Figure 14.8, the path $\sigma:[0,\,1]\to SO(3);\,\sigma(\tau) = \exp(4\,\pi\,\hat{S}_z\,\tau)$ encoded in the twisted ribbon of Figure 14.7 is shown to be homotopic to the trivial loop $\sigma^\prime:[0,\,1]\to SO(3);\,\sigma^\prime(\tau) = \id$.

**Figure 14.8****: A Homotopy Showing Equivalence of The Double Twist and The Identity Loop**

In motion, this sequence looks like Figure 14.9

**Figure 14.9****: Unwinding the Two-Twist from the Dirac Belt** (*Click here to download high resolution animation *(*approx 10MB)*)

The homotopy imparted to the path $\sigma$ in **Figure 14.8** can be qualitatively described as follows (see Figure 14.10 for definitions). Although qualitative, the description allows the full functional homotopy to be derived and so this can indeed be readily made into a full and precise description of the homotopy.

**Figure 14.10****: Definitions for Describing the Homotopy between Double Twist and Identity Loop**

The ribbon begins at the path beginning $\tau=0$, marked point *A* in Figure 14.10 and section *AB* is straight, rigid and still throughout the whole homotopy. The tangent to its centreline is in the $\hat{\mathbf{z}}$ direction and the normal in the ribbon plane is in the $\hat{\mathbf{x}}$ direction, so that $\sigma(0)$ defines $\id\in SO(3)$. Likewise for the last section *GH *of the ribbon: this is straight, rigid through out the homotopy and is only translated so that the centreline always points in the $\hat{\mathbf{z}}$ direction and the normal in the ribbon plane is in the $\hat{\mathbf{x}}$ direction, so that $\sigma(1)$ too defines $\id\in SO(3)$ at all times. Sections *BC, DE* and *FG* bend in the $y-z$ plane (*i.e. *the unit normal $\hat{\mathbf{n}}=\kappa \d_s \hat{\mathbf{t}}$ in the Frenet Serret frame defined by the ribbon’s centreline is in the $y-z$ plane). Sections *BC* and *FG *bend through angle $\theta$ and *DE* bends in the opposite sense through angle $2\,\theta$.

So the inverse homotopy deforming the identity loop into the double twist is defined by the following steps:

- The ribbon begins straight, so that $\theta=0$, and also untwisted, so it encodes the identity loop;
- $\theta$ smoothly increases from $0$ to $\pi/2$, so that the ribbon “droops” and sections
*CD*and*EF*hang “vertically” (in the $-\hat{\mathbf{y}}$ direction) as in Frame 6 of Figure 14.8; - The doll then undergoes pure translation within the $x-z$ plane so that it traces frames 5, 4, 3, 2, 1 of Figure 14.8. Sections
*AB*and*BC*are still throughout this stage, sections*FG*and*GH*undergo, like the doll, pure translation in the $x-z$ plane, the semicircular section*DE*undergoes pure, rigid rotation about the centreline of section*CD*, which hangs still and vertical throughout this stage. However, the ribbons of the now vertical sections*CD*and*EF*twist uniformly and work as swivvels whilst the rest of the ribbon moves rigidly, so that they each take up one full twist as the doll, undergoing pure translation, fares the path shown in Figure 14.8 backwards; - Now $\theta$ decreases smoothly from $\pi / 2$ to $0$, so that the twisted ribbon straightens out;
- The straightened ribbon now has two full twists, one full twist of each confined to sections
*CD*and*EF*. The twist is then allowed to smoothly ungather and spread itself uniformly along the ribbon’s whole length, thus reaching the double twist configuration shown in Figure 14.7; - To undo a double twist as in Figure 14.7, we of course simply do the above backwards: smoothly gather the double twist into sections
*CD*and*EF*with one full twist in each of these sections, increase $\theta$ from $0$ to $\pi/2$, translate the doll as shown in frames 1 through 6 of Figure 14.8 so that sections*CD*and*EF*work as swivvels, thus untwisting themselves, then decrease $\theta$ from $\pi/2$ back to $0$ again so that ribbon encodes the identity loop.

However, if we begin with one rather than two twists in Figure 14.7 we encode the path $\sigma:[0,\,1]\to SO(3);\,\sigma(\tau) = \exp(2\,\pi\,\hat{S}_z\,\tau)$ and the sequence shown in **Figure 14.11** follows when we impart the same homotopy to the path as we did in Figure 14.8.

**Figure 14.11****: Trying to Undo the Lone Twist**

In the sequence shown by Figure 14.11 one full twist in the belt is converted to a full twist in the opposite sense when we impart the homotopy described above. In motion, this sequence looks like Figure 14.12.

**Figure 14.12****: Evolution of the One-Twist Dirac Belt **(*Click here to download high resolution animation *(*approx 10MB*))

Similar homotopies can be imparted to the doll whilst two belts are attached to her, one leading from the doll towards $z\to -\infty$, the other leading from the doll towards $z\to +\infty$. Thus the sequence sketched in Figure 14.13 shows that a double twist by a doll in the middle of an endlessly long ribbon can be undone by looping both halves of the ribbon at once as shown without cutting the ribbon, nor with the ribbon’s needing to pass through itself.

**Figure 14.13****: Untwist Sequence for a Doll at the Middle of an InfiniteRibbon**

In motion, the untwist sequence for an infinitely long Dirac belt looks like Figure 14.14.

**Figure 14.14****: Untwisting the Two-Twist Infinite Dirac Belt **(*Click here to download high resolution animation *(*approx 10MB*))

whilst the evolution of an infinitely long Dirac belt representing the one-twist (-1) homotopy class is as shown in Figure 14.15.

**Figure 14.15****: Evolution of the One-Twist Infinite Dirac Belt **(*Click here to download high resolution animation *(*approx 10MB*))

The Dirac Belt trick is almost a precise analogy. That the double twist can be deformed by looping the ribbon as just described sketches a proof for the homotopic equivalence between the doubly twisted and untwisted state. That is, the fact that the physical object can undergo this homotopy is a precice analogy for the mathematica proof of homotopic equivalence. However, the Dirac Belt is embedded in three-dimensional Euclidean space, whereas a homotopy in $SO(3)$ is not. In particular this means that some valid homotopies are encoded by a transformation of the Dirac Belt where the ribbon passes through itself. So we need to be careful in saying that the fact that the physical object cannot undo a lone twist does not prove the existence of a second homotopy class in $SO(3)$. Cartainly, the inference the other way around is true: the existence of the separate homotopy class shows why the lone twist cannot be undone. However, there is a way to make the analogy complete. Suppose we consider a homotopy where the ribbon does pass through itself. We can easily transform this into a homotopy into one where the ribbon does not pass through itself by looping the end of the ribbon behind the self-intersection point, so the ribbon immediately before passing through itself and the ribbon straight afterwards are homotopically equivalent by translation of one of the ribbon’s ends that does not need the ribbon to pass through itself. That is, even though the physical ribbon cannot pass through itself, there is no homotopy involving self intersection that can untwist a lone twist in the ribbon if there is no homotopy that the physical ribbon can undergo. Thus the configuration spaces of the half-infinite Dirac Belts of Figure 14.9 and Figure 14.12 have the same homotopy classes as does $SO(3)$. This argument breaks down for the infinite Dirac Belts of Figure 14.14 and Figure 14.15, so, whilst the existence of two homotopy classes in $SO(3)$ shows that there must be *at least* two homotopy classes in the infinite ribbon configuration space, the converse is not true: the fact that there happen to be two homotopy classes in the latter does not prove that there are two distinct classes in the former.

Excellent intuitive references for further discussion of the Dirac Belt Trick can be found in [Hart, Francis and Kauffman] and [Bolker] The above homotopy sequences can be simulated in detail with the Mathematica simulation presented below. If for some reason this is not working on your browser, visit the copy I have uploaded to the Wolfram Demonstrations Project Site. Click here to go to the Wolfram Demonstrations version of the simulator

Use the control “homotopy” to set whether the doll begins her untwist sequence with two full twists in the ribbon (as in Figure 14.8: set “homotopy” to 1) or with one full twist in the ribbon (as in Figure 14.11: set “homotopy” to -1). Click the “+” symbol next to the main “time” slider to expand the controls: you can use these to slow the simulation down (double down arrow) or speed it up (double up arrow), begin, halt or one-step (“+” and “-” buttons) the simulation. You can also use the “belts” control to put one or two belts on on the doll: the two belt simulation shows that the untwisting sequence works whether there are one or two belts and thus that the doll’s behaviour is the same even if she is at the middle of an endlessly long ribbon, this is even though we might otherwise think that a “free” end is needed to allow the doll to pass under the ribbon. We can see with the two belt trick that a free end is not needed.

You can also grab the Doll-and-Ribbon object, strech and shrink it and re-orient it in space to get a better view – it has full 3D controls. The simulation’s size can also be zoomed by grabbing the edge of its window.

**References:**

- Hatcher, Allen, “Algebraic topology” Cambridge University Press, 2002
- L. Gutzwiller, Private correspondence with Luke Gutzwiller, Department of Mathematics, Northwestern University, 2033 Sheridan Road, Evanston, IL 60208-2730, 2011
- Wulf Rossmann: “
*Lie Groups: An Introduction through Linear Groups (Oxford Graduate Texts in Mathematics)”* - K. Engø, “On the BCH-formula in so(3)”,
*BIT Numerical Mathematics,***41**, number 3, 2001,*pp*629-632 - John C. Hart, George K. Francis and Louis H. Kauffman, “Visualizing Quaternion Rotation”, Assoc. Comp. Machinery (ACM) Trans. on Graphics,
**13**, No. 3, July 1994, Pages 256-276. - Ethan D. Bolker, “The Spinor Spanner”,
*The American Mathematical Monthly*,**80**, No. 9. November, 1973, pp. 977-984

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