Chapter 3 Lie Groups

1.	Let $e\in U\subset G$ be a locally Euclidean open neighborhood of $e$. Then by Theorem 3.18, $G={\bigcup }_{n=1}^{\infty }{U}^n$. Now, fix a countable basis of $U$, say $\{B_i\}$. Then for all open set $V\subset U^n$ and for all points $g\in V$, $g$ can be expressed as $g=g_1\dots g_n$ by definition. Also, since multiplication on $G$ is continuous, we can find open neighborhoods of $g_k$, say $V_k$, such that $V_1\dots V_n\subset V$. Then, we can find $B_{i_k}$’s such that $g_k\in B_{i_k}\subset V_k$, and as a result $g=g_1\dots g_n\in B_{i_1}\dots B_{i_n}\in V$. Therefore it follows from this that $\{B_i^n\}$ becomes a basis of $U^n$ and $U^n$ is second countable, and so is $G$.
2.	(a) It is definitely a differential manifold. Also, subtraction is continuous under standard topology. (b) It is also almost obvious. (c) Since it is a subgroup of ${\mathbb{C}}^{\ast }$, and the zero set of $|z|-1$, which is nonsingular. (d) It is also a manifold, and the operation is continuous by definition of product topology. (e) It directly follows from (c) and (d). (f) Since matrix multiplication is defined by a polynomial, and inverse by a polynomial divided by determinant which is continuous. (g) Since it is the zero set of the elements below the diagonal of $Gl(n,\mathbb{R})$. (h) It is surely a manifold. Also, $(s,t)(s_1,t_1{)}^{-1}=(\frac{s}{s_1},-\frac{s{t}_1}{s_1}+t)$ is continuous. (i) It is also a manifold, and $(A,v)(A_1,v_1{)}^{-1}=(A{A}_1^{-1},-A{A}_1^{-1}{v}_1+v)$ is continuous.
3.	(a) It is a vector space, and Lie bracket operation satisfies the given conditions by Proposition 1.45. (b) Trivial Lie bracket trivially satisfies the conditions. (c)$$[B,A]=BA-AB=-[A,B],$$and$$\begin{array}{rl}[[A,B],C]+[[B,C],A]+[[C,A],B]& =ABC-BAC-CAB+CBA\\ & +BCA-CBA-ABC+ACB\\ & +CAB-ACB-BCA+BAC=0.\end{array}$$(d) $[ax+by,cx+dy]=(ad-bc)y$. It is a Lie algebra by direct calculation. (e) Cross product also satisfies the conditions. Note that $(a\times b)\times c=b(a\cdot c)-a(b\cdot c)$.
4.	(a) It suffices to check that for all smooth vector fields $X_i$ $(1\le i\le r)$ on $G$, $\omega (X_1,\dots ,X_r)$ is smooth. But$$\omega (X_1,\dots ,X_r)(\sigma )={\omega }_{\sigma }(X_{1\sigma },\dots ,X_{r\sigma })=(\delta l_{{\sigma }^{-1}}{\omega }_e)(X_{1\sigma },\dots ,X_{r\sigma })={\omega }_e(d{l}_{{\sigma }^{-1}}{X}_{1\sigma },\dots ,d{l}_{{\sigma }^{-1}}{X}_{r\sigma }),$$so we need to check that ${\omega }_e(d{l}_{{\sigma }^{-1}}{X}_{1\sigma },\dots ,d{l}_{{\sigma }^{-1}}{X}_{r\sigma })$ is smooth. Now let $\tilde{\omega }$ be a smooth form such that ${\tilde{\omega }}_e={\omega }_e$. Then for $\phi :G\times G\to G,(a,b)\mapsto ab$, $\delta \phi \tilde{\omega }$ is smooth. Also, $\left[0,X_i\right]$ is a smooth vector field on $G\times G$, and $\iota :G\to G\times G,\tau \mapsto ({\tau }^{-1},\tau )$ is smooth. Therefore, $\delta \phi \tilde{\omega }([0,X_1],\dots ,[0,X_r])(\iota (\sigma ))$ is smooth with respect to $\sigma$. But,$$\begin{array}{rl}& \delta \phi \tilde{\omega }([0,X_1],\dots ,[0,X_r])(\iota (\sigma ))\\ & \\ & =\delta \phi {\tilde{\omega }}_{({\sigma }^{-1},\sigma )}([0,X_1{]}_{({\sigma }^{-1},\sigma )},\dots ,[0,X_r{]}_{({\sigma }^{-1},\sigma )})\\ & ={\tilde{\omega }}_e(d\phi [0,X_1{]}_{({\sigma }^{-1},\sigma )},\dots ,d\phi [0,X_r{]}_{({\sigma }^{-1},\sigma )}).\end{array}$$Furthermore,$$\begin{array}{rl}d\phi [0,X_i{]}_{(a,b)}(f)& =[0,X_i{]}_{(a,b)}(f\circ \phi )\\ & =0_a(f\circ \phi \circ {\iota }_b^1)+X_{ib}(f\circ \phi \circ {\iota }_a^2)\\ & =X_{ib}(f\circ l_a)=d{l}_a{X}_{ib}(f).\end{array}$$Therefore$${\tilde{\omega }}_e(d\phi [0,X_1{]}_{({\sigma }^{-1},\sigma )},\dots ,d\phi [0,X_r{]}_{({\sigma }^{-1},\sigma )})={\omega }_e(d{l}_{{\sigma }^{-1}}{X}_{1\sigma },\dots ,d{l}_{{\sigma }^{-1}}{X}_{r\sigma }),$$and the result follows. (b) $\delta l_{\sigma }$ is an algebra endomorphism and $E_{\text{linv}}^{\ast }(G)={\bigcap }_{\sigma \in G}\{\delta l_{\sigma }$-invariant elements$\}$, so $E_{\text{linv}}^{\ast }(G)$ is a subalgebra. Also, $\epsilon :E_{\text{linv}}^{\ast }(G)\to \Lambda (G_e^{\ast }),\omega \mapsto \omega (e)$ is a homomorphism. And since$$\omega (e)={\omega }^{\prime }(e)\Rightarrow {\omega }_{\sigma }=\delta l_{{\sigma }^{-1}}{\omega }_e=\delta l_{{\sigma }^{-1}}{\omega }_e^{\prime }={\omega }_{\sigma }^{\prime }\Rightarrow \omega ={\omega }^{\prime },$$$ϵ$ is injective. Furthermore, for all $\chi \in \Lambda (G_e^{\ast })$, if we set $\omega (a)=\delta l_{a^{-1}}\chi$, then$$(\delta l_{\tau }\omega )(\sigma )=\delta l_{\tau }(\omega (\tau \sigma ))=\delta l_{\tau }\delta l_{{\sigma }^{-1}{\tau }^{-1}}\chi =\delta l_{{\sigma }^{-1}}\chi =\omega (\sigma ),$$so $\omega \in E_{\text{linv}}^{\ast }(G)$. Also, $\epsilon (\omega )=\chi$, so $ϵ$ is surjective. (c)$$\omega (X)(\sigma )={\omega }_{\sigma }(X_{\sigma })=(\delta l_{{\sigma }^{-1}}\omega {)}_{\sigma }(X_{\sigma })={\omega }_e(d{l}_{{\sigma }^{-1}}{X}_{\sigma })={\omega }_e(X_e)=\omega (X)(e),$$so $\omega (X)$ is constant on $G$. Also, if ${\alpha }^{\ast }:G_e^{\ast }\to {\mathfrak{g}}^{\ast }$ is the dual map of $\alpha$ defined on Proposition 3.7, then$${\alpha }^{\ast }(\epsilon (\omega ))(X)=\alpha (\omega (e))(X)=\omega (e)(X(e))=\omega (X)(e).$$(d)$$d\omega (X,Y)=X(\omega (Y))-Y(\omega (X))-\omega ([X,Y])=-\omega ([X,Y])$$since $\omega (X),\omega (y)$ are constants. (e) Since $\mathfrak{g}$ is closed under the Lie bracket, such $c_{ijk}$’s exist, and the relation (3) is directly followed by properties of the Lie bracket, i.e., anti-symmetry and the Jacobi identity. Also, since $\delta l_{\sigma }$ commutes with $d$, $d{\omega }_i\in E_{\text{linv}}^2(G)$ and$$d{\omega }_i(X_j,X_k)=-{\omega }_i[X_j,X_k],$$so we can write$$d{\omega }_i=\sum _{j<k}{c}_{jki}{\omega }_k\wedge {\omega }_j+{\omega }^{\prime }$$where ${\omega }^{\prime }(X,Y)=0,\forall X,Y\in (g)$. But, since$$\epsilon :E_{\text{linv}}^2(G)\to {\Lambda }_2(G_e)\cong \mathfrak{g}\wedge \mathfrak{G}$$is an isomorphism, ${\omega }^{\prime }$ must be zero. Hence the Maurer–Cartan equations are derived.
5.	Consider $\mathrm{id},{\pi }_2:\mathbb{Z}\times \mathbb{R}\to \mathbb{Z}\times \mathbb{R}$.
6.	($\Leftarrow$) part is obvious since $d\omega (X,Y)=-\omega [X,Y]$. For ($\Rightarrow$) part, $d\omega (X,Y)=-\omega [X,Y]=0$ by assumption, so by using the Maurer–Cartan equation, $d\omega$ should be in $\mathcal{I}$.
7.	(a) See [1], pp. 61–62, or follow the sketch given in the textbook. (b) See [1], pp. 64–65, or follow the sketch given in the textbook. (c) If $\pi :\tilde{X}\to X$ is a covering and $X$ is simply connected, then by (a), there is a unique continuous map $\tilde{\alpha }:X\to \tilde{X}$ s.t. $\pi \circ \tilde{\alpha }={\mathrm{id}}_X$. Therefore, $\pi \circ (\tilde{\alpha }\circ \pi )=\pi$. Then by uniqueness, $\tilde{\alpha }\circ \pi ={\mathrm{id}}_{\tilde{X}}$ also. Therefore, $\tilde{\alpha }$ and $\pi$ are inverses to each other, which means $\pi$ is a homeomorphism.
8.	(Following the proof of [4], Prop. 1.9.) There is a countable open cover $\mathcal{U}$ which elements are homeomorphic to open ball in Euclidean space. For all $U,U^{\prime }\in \mathcal{U},U\cap U^{\prime }$ has at most countably many components, each of which is path connected. Let $\mathcal{X}$ be a countable set containing one point from each component of $U\cap U^{\prime }$ (including $U=U^{\prime }$). For each $U\in \mathcal{U}$ and each $x,x^{\prime }\in \mathcal{X}$ s.t. $x,x^{\prime }\in U$, let $p_{x,x^{\prime }}^U$ be a path joining $x$ and $x^{\prime }$ and contained in $U$. Since the fundamental groups based at any two points in the same component of $M$ are isomorphic, and $\mathcal{X}$ contains at least one point in each component of $M$, we may as well choose a point $q\in \mathcal{X}$ as base point. Define a “special loop” to be a loop based at $q$ that is equal to a finite product of paths of the form $p_{x,x^{\prime }}^U$. Clearly, the set of special loops is countable, and each special loop determines an element of ${\pi }_1(M,q)$. To show that ${\pi }_1(M,q)$ is countable, therefore, it suffices to show that every element of ${\pi }_1(M,q)$ is represented by a special loop. Suppose $f:[0,1]\to M$ is any loop based at $q$. The collection of components of sets of the form $f^{-1}(U)$ as $U$ ranges over $\mathcal{U}$ is an open cover of $[0,1]$, so by compactness it has a finite subcover. Thus there are finitely many numbers $0=a_0<a_1<\dots <a_k=1$ such that $\left[a_{i-1},a_i\right]\subset f^{-1}(U)$ for some $U\in \mathcal{U}$. For each $i$, let $f_i$ be the restriction of $f$ to the interval $[a_{i-1},a_i]$, reparametrized so that its domain is $[0,1]$, and let $U_i\in \mathcal{U}$ be a coordinate ball containing the image of $f_i$. For each $i$, we have $f(a_i)\in B_i\cap B_{i+1}$, and there is some $x_i\in \mathcal{X}$ that lies in the same component of $B_i\cap B_{i+1}$ as $f(a_i)$. Let $g_i$ be a path in $B_i\cap B_{i+1}$ from $x_i$ to $f(a_i)$, with the understanding that $x_0=x_k=q$, and $g_0$ and $g_k$ are both equal to the constant path $c_q$ based at $q$. Then, because $g_i^{-1}\cdot g_i$ is path homotopic to a constant path,$$\begin{array}{rl}f& \sim f_1\cdot \dots \cdot f_k\\ & \sim g_0\cdot f_1\cdot g_1^{-1}\cdot g_1\cdot f_2\cdot g_2^{-1}\cdot \dots \cdot g_{k-1}^{-1}\cdot g_{k-1}\cdot f_k\cdot g_k^{-1}\\ & \sim {\tilde{f}}_1\cdot {\tilde{f}}_2\cdot \dots \cdot {\tilde{f}}_n\end{array}$$where ${\tilde{f}}_i=g_{i-1}\cdot f_i\cdot g_i^{-1}$. For each $i$, ${\tilde{f}}_i$ is a path in $U_i$ from $x_{i-1}$ to $x_i$. Since $B_i$ is simply connected, ${\tilde{f}}_i$ is path homotopic to $p_{x_{i-1},x_i}^{U_i}$. It follows that $f$ is path homotopic to a special loop, as claimed.
9.	$(\Rightarrow )$ part is easy since$$\delta l_{\sigma }\delta \phi \omega =\delta (\phi \circ l_{\sigma })\omega =\delta (l_{\phi (\sigma )}\circ \phi )\omega =\delta \phi \omega .$$For the converse, let $\mathcal{I}$ be an ideal generated by$$\delta {\pi }_1\delta \phi ({\omega }_i)-\delta {\pi }_2({\omega }_i)$$for projections ${\pi }_1:G\times H\to G,{\pi }_2:G\times H\to H$. Then by direct calculation similar to 3.15, (6), it is a differential ideal. Therefore, there exists $I$, maximal connected integral manifold of $\mathcal{I}$ through $(e,e)$. But since $\delta {\pi }_1,\delta {\pi }_2$, and $\delta \phi$ pull left invariant forms back to left invariant forms, by Thm 3.19, Corollary (c), $I$ is a Lie subgroup of $G\times H$. Meanwhile, let $(G,g)$ be a graph of $\phi$ where $g(\sigma )=(\sigma ,\phi (\sigma ))$. Then also by direct calculation (following 2.33), it is an integral manifold of $\mathcal{I}$. Since $G$ is connected, $g(G)\subset I$ by maximality. Now, let ${\pi }_1{|}_I:I\to G$ be a restriction map which is a homomorphism. Then, $d{\pi }_1:I_{(e,e)}\to G_e$ is an isomorphism since for$$(X,Y)\in I_{(e,e)}\iff \forall i,[\delta {\pi }_1\delta \phi ({\omega }_i)-\delta {\pi }_2({\omega }_i)](X,Y)=0\iff Y=d\phi (X).$$Therefore by Theorem 3.26, ${\pi }_1{|}_I$ is a covering map. If $(\sigma ,\tau )\in I$, then by path–connectedness, there is a path $\ell =({\ell }_G,{\ell }_H)$ from $(e,e)$ to $(\sigma ,\tau )$ in $I$. Then $({\ell }_G^{-1},\phi ({\ell }_G^{-1}))\cdot \ell$ is also a path in $I$ which joins $(e,e)$ and $(e,\phi ({\sigma }^{-1})\cdot \tau )$ since $g(G)\subset I$. However, since the “$G$ coordinate” of this path is constantly $e$ and there is a local isomorphism from a neighborhood of $(e,e)\in I$ to a neighborhood of $e\in G$, it should be a constant path, which means $\phi ({\sigma }^{-1})\cdot \tau =e$. Therefore, since $I$ is a Lie subgroup, $(\sigma ,\tau )\in I\Rightarrow ({\sigma }^{-1},{\tau }^{-1})\in I\Rightarrow \phi (\sigma )=\tau \Rightarrow I\subset g(G)$. Thus $I=g(G)$ and it means that $\phi$ is a homomorphism. Connectedness is necessary: Consider $G=\mathbb{Z},H=\mathbb{R}$, $\phi :\mathbb{Z}\to \mathbb{R},x\mapsto x^2$.
10.	If there exist $A\in \mathfrak{gl}(2,\mathbb{R})$ s.t.$$\left(\begin{array}{cc}-2& 0\\ & \\ 0& -1\end{array}\right)=e^A.$$Then regarding $A$ as an element $\mathfrak{gl}(2,\mathbb{C})$, one can find a matrix $L=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \mathfrak{gl}(2,\mathbb{C})$ s.t. $LA{L}^{-1}$ is upper triangular. Then$$L\left(\begin{array}{cc}-2& 0\\ & \\ 0& -1\end{array}\right)L^{-1}=L{e}^A{L}^{-1}=e^{LA{L}^{-1}}$$is also upper triangular.$$L\left(\begin{array}{cc}-2& 0\\ & \\ 0& -1\end{array}\right)L^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}-2ad+bc& ab\\ & \\ -cd& -ad+2bc\end{array}\right),$$thus $cd=0$. If $c=0$,$$L\left(\begin{array}{cc}-2& 0\\ & \\ 0& -1\end{array}\right)L^{-1}=\frac{1}{ad}\left(\begin{array}{cc}-2ad& ab\\ & \\ 0& -ad\end{array}\right)=\left(\begin{array}{cc}-2& b\\ & \\ 0& -1\end{array}\right).$$If $d=0$,$$L\left(\begin{array}{cc}-2& 0\\ & \\ 0& -1\end{array}\right)L^{-1}=\frac{1}{-bc}\left(\begin{array}{cc}bc& ab\\ & \\ 0& 2bc\end{array}\right)=\left(\begin{array}{cc}-1& -\frac{a}{c}\\ & \\ 0& -2\end{array}\right).$$Therefore, the diagonal elements of $e^{LA{L}^{-1}}$ should be $-2$ and $-1$. Meanwhile, if we let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $A$ (and $LA{L}^{-1}$), since $A$ is real, two eigenvalues are either both real or complex conjugates. But since $LA{L}^{-1}$ is upper triangular, the diagonal elements of $e^{LA{L}^{-1}}$ are $e^{{\lambda }_1}$ and $e^{{\lambda }_2}$, which should not be $-2$ and $-1$ in either case. Therefore, there is no $A$ satisfying the condition.
11.	For $\eta$ defined as the text, using Exercise 1.24(d),$$d\eta (X,Y)=d\eta d{\iota }_e^1(X)+d\eta d{\iota }_e^2(Y)=X+Y$$for ${\iota }_e^1:G\to G\times G,\sigma \mapsto (\sigma ,e)$ and ${\iota }_e^2:G\to G\times G,\sigma \mapsto (e,\sigma )$. Therefore,$$d\alpha =d\eta (d\sigma ,d\beta )=d\sigma +d\beta .$$Evaluating both sides at $\frac{\partial }{\partial t}{|}_{t=0}$, we obtain the given result.
12.	For any $\sigma \in G$, a path $\ell$ from $e$ to $\sigma$, and $\tau \in \ker \phi$, $\ell \cdot \tau \cdot {\ell }^{-1}$ is a curve in $\ker \phi$ also. Since $\ker \phi$ is discrete, the curve should be constant, which means $\sigma \cdot \tau \cdot {\sigma }^{-1}=\tau$. Therefore $\ker \phi \subset Z(G)$. For the second statement, for any Lie group $G$, let $\pi :\tilde{G}\to G$ is the universal covering. Then since $\pi$ is a local homeomorphism, $\ker \pi$ is discrete, so $\ker \pi \in Z(\tilde{G})$, which says that $\ker \pi$ is abelian. But by the lifting property, the fundamental group of $G$ can be identified with a subgroup of $\ker \pi$, so it is also abelian.
13.	Let the Lie algebra of 3.5(d) be $L=({\mathbb{R}}^2,[\cdot ,\cdot ])$. For any non-abelian Lie algebra $\bar{L}=({\mathbb{R}}^2,\overline{[\cdot ,\cdot ]})$ which is defined by $\overline{[e_1,e_2]}=s{e}_1+t{e}_2$, let$$f:\bar{L}\to L,a{e}_1+b{e}_2\mapsto (at-bs)e_1+(as+bt)e_2.$$It is by definition a linear isomorphism, and$$\begin{array}{rl}f(\overline{[a{e}_1+b{e}_2,c{e}_1+d{e}_2]})& =(ad-bc)f(s{e}_1+t{e}_2)=(ad-bc)(s^2+t^2)e_2,\\ [f(a{e}_1+b{e}_2),f(c{e}_1+d{e}_2)]& =[(at-bs)e_1+(as+bt)e_2,(ct-ds)e_1+(cs+dt)e_2]\\ & =((at-bs)(cs+dt)-(as+bt)(ct-ds))e_2\\ & =(ad-bc)(s^2+t^2)e_2.\end{array}$$Therefore $f$ is a Lie algebra isomorphism. For the statement below, consider $({\mathbb{R}}^{>0},\cdot )\times (\mathbb{R},+)$ instead of 3.3(h). Then it can be easily checked that $\{x\frac{\partial }{\partial x},x\frac{\partial }{\partial y}\}$ is left invariant tangent fields that satisfies$$\left[x\frac{\partial }{\partial x},x\frac{\partial }{\partial y}\right]=x\frac{\partial }{\partial y}.$$Therefore, by Theorem 3.27, Corollary, it is (up to isomorphism) the unique simply connected $2$-dimensional non-abelian Lie group.
14.	Let $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),B=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. Then$$e^A{e}^B=\left(\begin{array}{cc}1& e\\ 0& e\end{array}\right),e^{A+B}=\left(\begin{array}{cc}1& e-1\\ 0& e\end{array}\right).$$
15.	If suffices to check that all Jordan canonical forms are in the image, since basis change and the exponential function commute, and the exponent of block diagonal matrices is block diagonal matrices of which blocks consist of the exponent of the original blocks. For $A$ an elementary Jordan matrix, let $A=\lambda I\cdot N$, where the diagonal entries of $N$ are $1$’s. Since $(I-N{)}^k=0$ for $k>n$,$$\log N=-\sum _{k=1}^{\infty }\frac{(I-N{)}^K}{k}$$exists and $e^{\log N}=N$ by calculation dealing with formal series. Also, $\lambda I=e^{\log \lambda I}$ for any $\log \lambda$ regardless of the choice of the branch of $\log$. Since $\log \lambda I$ is in the center, $e^{\log \lambda I+\log N}=e^{\log \lambda I}{e}^{\log N}=A$.
16.	It is a Lie algebra because of Proposition 1.55 and if we denote it by $\mathfrak{h}$, $\mathfrak{h}\to G_e,X\mapsto X(e)$ is an isomorphism. Also, for $X$ a left invariant vector field on $G$,$$d{r}_{\sigma }d\phi (X)=d\phi d{l}_{{\sigma }^{-1}}(X)=d\phi (X),$$so $d\phi (X)\in \mathfrak{h}$. Also, for $\eta :G\times G\to G,(\sigma ,\tau )\to \sigma \tau$, $\eta ({\mathrm{id}}_G,\phi )$ is a constant function to $e$. Thus $0=d\eta (d{\mathrm{id}}_G,d\phi )=d{\mathrm{id}}_G+d\phi$ and $d\phi (X)(e)=-X(e)$. Also, by the same argument, $d\phi (\mathfrak{h})\subset \mathfrak{g}$. Then, since $\phi$ is an involution, $d\phi d\phi =\mathrm{id}$, so $\mathfrak{g}$ and $\mathfrak{h}$ are isomorphic via $d\phi$.
17.	Consider ($\mathbb{Q}$, discrete topology) $\hookrightarrow$ ($\mathbb{R}$, standard topology).
18.	By Theorem 3.50, Corollary (b) and Theorem 3.28, ${\mathbb{R}}^n$ is the unique simply connected abelian Lie group up to isomorphism. Then, since $\pi :{\mathbb{R}}^n\to G$ is the universal cover of $G$, $G\simeq {\mathbb{R}}^n/\ker \pi$. Also, the kernel of a covering homomorphism is discrete. Then following the argument in the text, $\ker \pi ={⨁}_{i=1}^k\mathbb{Z}{v}_i$ for some linearly independent elements $v_1,\dots ,v_k$. Therefore, $G\simeq {\mathbb{R}}^n/{⨁}_{i=1}^k\mathbb{Z}{v}_i\simeq {\mathbb{R}}^{n-k}\times T^k$.
19.	Clearly $A_B(V)$ is an abstract subgroup of $\mathrm{Aut}(V)$ and ${\mathfrak{d}}_B$ is a subalgebra of $\mathrm{End}(V)$. According to Theorem 3.42, $A_B(V)$ is a closed Lie subgroup of $\mathrm{Aut}(V)$ since$$A_B(V)=\bigcap _{v,w\in V}\ker (\alpha \mapsto (\alpha (v),\alpha (w))-(v,w))$$is closed. $\mathfrak{a}$ be the Lie algebra of $A_B(V)$. If $l\in \mathfrak{a}$, $\exp tl\in A_B(V)$. Therefore$$((\exp tl)(v),(\exp tl)(w))=(v,w).$$Taking their derivatives at $t=0$,$$(lv,w)+(v,lw)=0.$$Therefore $l\in {\mathfrak{d}}_B$. Conversely, suppose that $l\in {\mathfrak{d}}_B$. Then $B(l\otimes 1+1\otimes l)=0$ by definition. Thus, $B(l\otimes 1+1\otimes l{)}^n=0$ for $n\ge 1$ and $B\circ e^{t(l\otimes 1+1\otimes l)}=B$. But since$$e^{t(l\otimes 1+1\otimes l)}=e^{t(l\otimes 1)}{e}^{t(1\otimes l)}=(e^{tl}\otimes 1)(1\otimes e^{tl})=e^{tl}\otimes e^{tl},$$$\exp tl\in A_B(V)$, and $l\in \mathfrak{a}$. Therefore, $\mathfrak{a}={\mathfrak{d}}_B$.
20.	Thanks to the outline, we only need to prove that $A(G)$ is naturally isomorphic with the closed subgroup of $A(\tilde{G})$ consisting of those automorphisms of $\tilde{G}$ which map $D$ onto $D$. However, if $g:G\to G\in A(G)$, then by the lifting property, there is a unique map $\tilde{g}:\tilde{G}\to \tilde{G}$ s.t. $\pi \circ \tilde{g}=g\circ \pi$, and since $\tilde{g}\circ g^{-1}={\mathrm{id}}_{\tilde{G}}$ by the uniqueness, $\tilde{g}\in A(\tilde{G})$. By definition $\tilde{g}(D)\subset D$ and $\tilde{g}(D^c)\subset D^c$, thus $\tilde{g}(D)=D$ since it is an automorphism. Conversely, if $h\in A(\tilde{G})$ and $h(D)=D$, there exists a map $\hat{h}:G\to G$ s.t. $\hat{h}\circ \pi =\pi \circ h$ and it is also an automorphism since $\ker (\pi \circ h)=\ker \pi$. Then it is easy to see that $\hat{\tilde{g}}=g$ and $\tilde{\hat{h}}=h$.
21.	$\phi :U(n)\to S^1\times SU(n),A\mapsto (\det A,\frac{A}{\det A})$ is an diffeomorphism.
22.	(a) First, find an eigenvalue $\lambda$ and a corresponding unit eigenvector $v$. Since$$\lambda \Vert v\Vert =\lambda {\bar{v}}^{t}v={\bar{v}}^{t}Av={\bar{v}}^t{\bar{A}}^{t}v=\bar{\lambda }{\bar{v}}^{t}v=\bar{\lambda }\Vert v\Vert ,$$$\lambda$ is real. Then if $v^t\bar{w}=0$,$$⟨v,Aw⟩=v^t\overline{Aw}=v^t\bar{A}\bar{w}=v^t{A}^t\bar{w}=\lambda v^t\bar{w}=0.$$Therefore $⟨v⟩$ and $⟨v{⟩}^{\perp }$ are both $A$-stable. Then we can use induction on the rank of $A$ to find orthonormal basis which consists of $\left\{v_i\right\}$ eigenvectors of $A$. Set $B={\left(v_1{v}_2\dots v_n\right)}^{-1}$. For the real symmetric case, first find an eigenvalue $\lambda$ (which is real) and a corresponding eigenvector $v$ which can be complex valued. Since$$A\bar{v}=\overline{Av}=\overline{\lambda v}=\lambda \bar{v},$$$\bar{v}$ is also an eigenvector of $\lambda$. Therefore, $\mathrm{Re}(v)$ and/or $\mathrm{Im}(v)$ is an real eigenvector of $\lambda$ and after scaling, we can find an unit real eigenvector. Now use the same induction as above. (b) For a hermitian matrix $A$, by (a) we can find $B\in U(n)$ s.t. $BA{B}^{-1}$ is diagonal. Then $e^A=B^{-1}{e}^{B{A}^{-}1}B$. Since eigenvalues of $A$ is real, $e^{BA{B}^{-1}}$ is a diagonal matrix whose diagonal entries are all positive. Therefore, since $B\in U(n)$, $e^A$ is also positive definite hermitian. Meanwhile, for a positive definite hermitian matrix $H$, we can find $B\in U(n)$ as above. Then since eigenvalues of $H$ are all positive, we can find a real diagonal matrix $A$ s.t. $e^A=BH{B}^{-1}$. Then $e^{B^{-1}AB}=H$, and $B^{-1}AB$ is a hermitian matrix. If $e^A=e^{A^{\prime }}$ for two hermitian matrices $A$ and $A^{\prime }$, find $B\in U(n)$ s.t. $BA{B}^{-1}$ is real diagonal, and note that $e^{BA{B}^{-1}}=e^{B{A}^{\prime }{B}^{-1}}$ is positive definite diagonal. Also, there exists $C\in U(n)$ s.t. $CB{A}^{\prime }{B}^{-1}{C}^{-1}$ is real diagonal, and $e^{CBA{B}^{-1}{C}^{-1}}=e^{CB{A}^{\prime }{B}^{-1}{C}^{-1}}$ is also positive definite diagonal. But since $CBA{B}^{-1}{C}^{-1}$ is also real diagonal by direct calculation, it means $CBA{B}^{-1}{C}^{-1}=CB{A}^{\prime }{B}^{-1}{C}^{-1}$, which says $A=A^{\prime }$. Therefore $\exp$ is well-defined one-to-one onto. For a real symmetric $A$, $e^A$ is positive definite hermitian by above and real, so positive definite symmetric. Injectivity is explained above. For a positive definite symmetric matrix $H$, we can find $B\in O(n)$ as (a), and there exists a real diagonal matrix $A$ s.t. $e^A=BH{B}^{-1}$. Then note that $e^{B^{-1}AB}=H$ and $B^{-1}AB$ is real symmetric.
23.	If $\sigma =PR=P_1{R}_1$, then$$P^2=PR{R}^t{P}^t=P_1{R}_1{R}_1^t{P}_1^t=P_1^2.$$Find $A,B$ real symmetric matrices satisfying $e^A=P$ and $e^B=P_1$, and we see that$$e^{2A}=(e^A{)}^2=P^2=P_1^2=(e^B{)}^2=e^{2B},$$so by injectivity, $2A=2B$ or $A=B$. It follows that $P=P_1$. Therefore, $R=R_1$ also.
24.	$\sigma {\bar{\sigma }}^t$ is positive definite hermitian, so $\exists B\in U(n)$ s.t. $B\sigma {\bar{\sigma }}^t{B}^{-1}$ is positive definite diagonal. Set $P=B^{-1}(B\sigma {\bar{\sigma }}^t{B}^{-1}{)}^{\frac{1}{2}}B$, which is positive definite hermitian, and $R=P^{-1}\sigma$ a unitary matrix.
25.	$U(n)$ is connected, and there exists a path $tI+(1-t)P$ joining $I$ and $P$ a positive definite hermitian matrix. Thus by similar argument of Theorem 3.68 and using Exercise 24, $Gl(n,\mathbb{C})$ is connected.
26.	Follow the argument and use actions $SU(n)\times X\to X$ for (b) and $SO(n)\times {\mathbb{P}}^{n-1}\to {\mathbb{P}}^{n-1}$ for (c).

References

[1]	Allen Hatcher, Algebraic Topology, Cambridge University Press, 2002.
[2]	Daniel Murfet, Automorphisms of Power Series Rings, in his blog, 2005.
[3]	Frank W. Warner, Foundations of Differentiable Manifolds and Lie Groups, Springer, 1983.
[4]	John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Springer, 2010.
[5]	Paul Loya, An introductory course in differential geometry and the Atiyah–Singer index theorem, Binghamton University, 2005.
[6]	C. H. Dowker, Lectures on Sheaf Theory, Tata Institute of Fundamental Research, 1956.