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\centerline{{\large \bf Solutions to May 2010 Differential Geometry Exam}} 

\vspace{0.5cm}
\noindent
{\bf \large Question 1}

\vspace{0.5cm}

\noindent
{\bf Qu 1 i)} \\
A real manifold, $\mathcal{M}$, of dimension $m$ is a topological (Hausdorff) space with open sets $\{ U_i \}$ and a set of charts, $\{ (U_i, \phi_i ) \}$ where;
\begin{description}
\item[a)] the open sets provide an open cover; $\cup_i U_i = \mathcal{M}$
\item[b)] the $\phi_i$ are homeomorphisms from $U_i$ onto an open set $U_i' \in \mathbb{R}^m$;
\begin{eqnarray*}
\phi_i : \quad U_i & \rightarrow & \mathbb{R}^m \nonumber \\
p & \rightarrow & x(p)
\end{eqnarray*}
\item[c)] the transition functions, $t_{ij} = \phi_i \cdot \phi_j^{-1}$, defined in non-trivial overlaps, $U_i \cap U_j \ne \phi$;
\begin{eqnarray*}
t_{ij} : \quad \mathbb{R}^m & \rightarrow & \mathbb{R}^m \nonumber \\
x & \rightarrow & y = t_{ij}(x) ,
\end{eqnarray*}
are smooth $C^\infty$ functions in all their arguments.
\end{description}
A complex manifold, $\mathcal{M}$, of complex dimension $m$ has the same definition except that replacing b) and c), are;
\begin{description}
\item[b')] the $\phi_i$ are homeomorphisms from $U_i$ onto an open set $U_i' \in \mathbb{C}^m$;
\begin{eqnarray*}
\phi_i : \quad U_i & \rightarrow & \mathbb{C}^m \nonumber \\
p & \rightarrow & z(p)
\end{eqnarray*}
\item[c')] the transition functions, $t_{ij} = \phi_i \cdot \phi_j^{-1}$, defined in non-trivial overlaps, $U_i \cap U_j \ne \phi$;
\begin{eqnarray*}
t_{ij} : \quad \mathbb{C}^m & \rightarrow & \mathbb{C}^m \nonumber \\
z & \rightarrow & w = t_{ij}(z) ,
\end{eqnarray*}
are analytic (holomorphic) functions in all their arguments.
\end{description}


\noindent
{\bf Qu 1 ii)} \\
Use stereographic projection and hence with 2 charts. Describe $S^n$ as unit radius sphere in $\mathbb{R}^{n+1}$. Take coordinates $\{ x_1, x_2 , \ldots , x_{n}, z \}$ in $\mathbb{R}^{n+1}$. Then $S^n = \{ (x_i, z) \in \mathbb{R}^{n+1} | x^2_i + z^2 = 1 \}$.\\

\noindent
{\bf Chart N} Define `North' chart $(U_N,\phi_N)$ as follows. We take the open set $U_N = \{ (x_i,z) \in S^2 | -1 \le z < 1\}$ ie. we exclude only the North pole point of the sphere. The homeomorphism $\phi_N$, from $U_N$ to $\mathbb{R}^2$ is given by associating to a point $(x_i,z) \in S^2$ the $x_i$ coordinates of the intersection of the line from the North pole through the point, with the plane $z = -1$. Label the intersection point $(X_i,-1) \in \mathbb{R}^{n+1}$. Simple geometry gives,
\begin{equation*}
X_i = \frac{2}{1-z} x_i
\end{equation*}
Then we may write the map $\phi_N$ as,
\begin{eqnarray*}
\phi_N : \quad U_N & \rightarrow & \mathbb{R}^n \nonumber \\
(x_i,z) & \rightarrow & X_i
\end{eqnarray*}

{\bf Chart S} Define `Sorth' chart $(U_S,\phi_S)$ in an analogous manner. We take the open set $U_S = \{ (x_i,z) \in S^n | -1 < z \le 1\}$ ie. we exclude only the Sorth pole point of the sphere. The homeomorphism $\phi_S$, from $U_S$ to $\mathbb{R}^{n+1}$ is given by projecting a point in the sphere from the South pole onto the plane $z = +1$.
Label the intersection point on the plane as $(X_i',+1)$. Simple geometry gives,
\begin{equation*}
X'_i = \frac{2}{1+z} x_i
\end{equation*}
Then we may write the map $\phi_S$ as,
\begin{eqnarray*}
\phi_S : \quad U_S & \rightarrow & \mathbb{R}^n \nonumber \\
(x_i,z) & \rightarrow & X'_i
\end{eqnarray*}

{\bf Transition functions} We note that in the overlap region $U_N \cap U_S = \{ -1 < z < 1 \}$;
\begin{equation*}
X^2_i = \frac{4}{(1-z)^2} x^2_i = \frac{4}{(1-z)^2} (1 - z^2) = 4 \frac{1+z}{1-z}
\end{equation*}
and that, (with no sum over $i$)
\begin{equation*}
\frac{X'_i}{X_i} = \frac{1-z}{1+z} = \frac{4}{X^2_i}
\end{equation*}
This yields the relation between the coordinates;
\begin{equation}
X'_i = \frac{4}{X^2_j} X_i 
\label{eq:qu1eq1}
\end{equation}
Now the transition function $t_{SN} = \phi_S \cdot \phi_N^{-1}$ is the map defined in the overlap region $U_N \cap U_S$
\begin{eqnarray*}
t_{SN} : \quad \mathbb{R}^2 & \rightarrow & \mathbb{R}^2 \nonumber \\
X_i & \rightarrow & X'_i
\end{eqnarray*}
defined above by equation \eqref{eq:qu1eq1}. In the overlap this function is smooth, $C^\infty$, in the arguments $X_i$.\\

\noindent
{\bf Qu 1 iii)} \\
$\mathbb{CP}^n \equiv$ the set of undirected \emph{complex} lines through origin in $\mathbb{C}^{n+1}$. \\

\noindent
Take `homogeneous coordinates' $(z_1,\ldots,z_{n+1})$ in $\mathbb{C}^{n+1} - \{0\}$ where we identify $(z_1,\ldots,z_{n+1}) \sim (\lambda z_1,\ldots, \lambda z_{n+1})$ for any $\lambda \in \mathbb{C} - \{0\}$. \\

\noindent
Require $(n+1)$ charts:
\begin{eqnarray}
U_m : && \mathrm{all}\; p \in  \mathbb{CP}^n \; s.t. \; z_m \ne 0  \\
\psi_m : && \mathbb{CP}^n \qquad \qquad  \rightarrow   \mathbb{C}^n \\
&& p  = (z_1,\ldots,z_{n+1}) \rightarrow \left( \frac{z_1}{z_m}, \frac{z_2}{z_m}, \ldots,   \frac{z_{m-1}}{z_m} , \frac{z_{m+1}}{z_m} \ldots,  \frac{z_{n+1}}{z_m} \right) \\
&& \qquad \qquad \qquad \qquad \qquad \qquad = \left( Z^{(m)}_1, \ldots Z^{(m)}_n \right)
\end{eqnarray} \\

\noindent
Transition functions eg. for $U_1, U_2$ overlap: 
\begin{equation}
\phi_{12}: \left(Z^{(1)}_1 , \ldots, Z^{(1)}_n\right) \rightarrow  \left(Z^{(2)}_1(Z^{(1)}) , \ldots, Z^{(2)}_n(Z^{(1)})\right)
\end{equation}
with,
\begin{equation}
Z^{(2)}_1  = \frac{1}{Z^{(1)}_1} , \qquad Z^{(2)}_i  = \frac{Z^{(1)}_i}{Z^{(1)}_1}  \; \mathrm{for} \; i\ne1
\end{equation}
Hence the transition functions are indeed analytic on the overlaps (where $Z^{(1)}_1 \ne 0$). \\

\noindent
{\bf Qu 1 iv)} \\
Recall from part ii) we have two charts; $U_N$ with coordinates $(X,Y)$ for all points with finite $X^2+Y^2$, and $U_S$ with coordinates $(X',Y')$  for all points with finite $X'^2+Y'^2$, related by the transition functions,
\begin{equation*}
X' = \frac{4}{X^2+Y^2} X \; , \quad Y' = \frac{4}{X^2+Y^2} Y
\end{equation*}
on the overlap where $X^2+Y^2 \ne 0$ and $X'^2 + Y'^2 \ne 0$.\\
Let us take complex coordinates $z = 2/(X + i Y)$ and $z' = 2/(X' - i Y')$. Then,
\begin{equation*}
z = \frac{2}{X + i Y} =  \frac{2 (X - i Y)}{X^2 + Y^2} = \frac{(X' - i Y')}{2} = \frac{1}{z'}
\end{equation*}

\noindent
The chart $U_N$ is then given by all $z \in \mathbb{C}$ excluding $z = 0$, and likewise $U_S$ is given by all $z' \in \mathbb{C}$ excluding $z' = 0$, and the transition function is $z' = 1/z$ on the overlap $z\ne 0$ and $z'\ne 0$. This is exactly our altas on $\mathbb{CP}^1$ above. \\


\newpage


\noindent
{\bf Qu 2 i)} \\

\noindent
Given an $r$-chain $c \in C_r$, it is a cycle iff $\partial c = 0$ (ie. it has no boundary), and it is a boundary iff $c = \partial a$ for some $a \in C_{r+1}$. Note that since $\partial^2 = 0$ (the boundary op is nilpotent), then a boundary is also a cycle,  $\partial c = \partial^2 a = 0$. \\

\noindent
Denote the vector space of $r$-cycles by $Z_r$ and $r$-boundaries by $B_r$. Define an equivalence of 2 $r$-cycles $z,z' \in Z_r$ by $z \sim z'$ iff $z' = z + b$ with $b\in B_r$, ie. equivalent if they differ by an $r$-boundary. \\

\noindent
Define an equivalence class of $r$-cycles by; $[ z ] = \{ z' \in Z_r | z \sim z' \}$, where $z \in Z_r$ is the representative of the class. The class $[z]$ is the set of $r$-cycles equivalent to representative $z$. If we have two $r$-cycles $a,b$ which are equivalent, $a \sim b$, then the equivalence classes defined by them are the same ie. $[a] = [b]$. \\

\noindent 
The $r^{th}$ homology vector space $H_r = \{ [z] \; \forall \; z \in Z_r \}$, ie. it is the set of equivalence classes of $r$-cycles. Its vector structure is inherited from the vector structure of the $r$-chains $C_r$, with,
\begin{equation*}
a [\alpha] + b [ \beta ]  = [ a \alpha + b \beta ] ,
\end{equation*}
for $\alpha, \beta \in Z_r$ and $a,b \in \mathbb{R}$.\\


\noindent
An $r$-form $\omega \in \Omega_r$ is closed iff $d \omega = 0$.  It is exact iff $\omega = d \beta$ for some $\beta \in \Omega^{r-1}$. Again since $d^2 = 0$  an exact form is also closed. \\

\noindent
Denote the vector space of closed $r$-forms by $Z^r$ and exact $r$-forms by $B^r$. Define equivalence for $\omega,\omega' \in Z^r$ by $\omega \sim \omega'$ iff $\omega' = \omega + b$ with $b\in B^r$.
Define an equivalence class of  by; $[ \omega ] = \{ \omega' \in Z^r | \omega \sim \omega' \}$. \\

\noindent 
The $r^{th}$ cohomology vector space $H^r = \{ [\omega] \; \forall \; \omega \in Z^r \}$. Its vector structure is inherited from the vector structure of $\Omega^r$, with,
\begin{equation*}
a [\alpha] + b [ \beta ]  = [ a \alpha + b \beta ] ,
\end{equation*}
for $\alpha, \beta \in Z^r$ and $a,b \in \mathbb{R}$.\\



\noindent
{\bf Qu 2 ii)} \\

\noindent
Simplicial complex for $S^1$ given in terms of 3 1-simplices and 3 0-simplices.\\

%DIAGRAM
\includegraphics[width=3in,height=1.5in]{complexes_S1.jpg}


\noindent
The boundary operator $\partial_1 : C_1 \rightarrow C_{0}$ is then given as: \\

\begin{eqnarray}
\partial_1 = \left( \begin{array}{ccc}
-1 & 0 & +1\\
+1 & -1 & 0 \\
0 & +1 & -1
\end{array} \right) \qquad \dim \mathrm{Ker}(\partial_1) = 1 \; , \quad \dim \mathrm{Im}(\partial_1) = 2 
\end{eqnarray}
We may directly verify that $\partial_1 \cdot \partial_1 = 0$ and hence $\partial_1$ is nilpotent. \\

\noindent
Betti number $b_i$ is the dimension of the $i$-th cohomology class $H_i$, and is given as,
\begin{eqnarray}
b_r  = \dim \mathrm{Ker}(\partial_{r}) - \dim  \mathrm{Im}(\partial_{r+1})
\end{eqnarray}\\

\noindent
The kernel of $\partial_1$ is 1-dimensional and spanned by the vector;
\begin{eqnarray}
\mathrm{Ker}(\partial_1) = \left\{ \left( \begin{array}{c}
+1 \\
+1 \\
+1
\end{array} \right) \right \}
\end{eqnarray}
So then,
\begin{eqnarray}
b_1 & = & \dim \mathrm{Ker}(\partial_1) - \dim \mathrm{Im}(\partial_2)  =  \dim \mathrm{Ker}(\partial_1) =  1 \nonumber \\
b_0 & = & \dim \mathrm{Ker}(\partial_0) - \dim \mathrm{Im}(\partial_1)  = \dim C_0 - \dim \mathrm{Im}(\partial_1) = 3 - 2 = 1
\end{eqnarray}
where in the first line we have used the fact that $\partial_2$ is trivial since there are no 2-simplices, and in the second line we have used the fact that since there are no $-1$-simplices, then the kernel of $\partial_0$ has dimension $\dim C_0$.



\noindent
{\bf Qu 2 iii)} \\

We can embed $S^1$ in $\mathbb{C}$ as $S^1 = \{ e^{i \theta} \; \forall \; \theta \in [0 , 2 \pi ) \}$.\\

\noindent
We have $\Omega^0(S^1) = \mathcal{F}(S^1)$. For $f \in \mathcal{F}(S^1)$ we have $df = \frac{\partial f(\theta)}{\partial \theta} d\theta$ and so if $df = 0$ then $\frac{\partial f(\theta)}{\partial \theta} = 0$  over $S^1$ and hence $f$ is constant. 
\\

\noindent
Thus the space of closed $0$-forms, $Z^0 =\mathbb{R}$, ie. the coefficient of the constant function. Since for $0$-forms $B^0(S^1) = 0$, then $H^0 = Z^0 = \mathbb{R}$, with the representative for a basis being any constant function.\\



\noindent
Now every $1$-form is closed. However, now not every $1$-form is exact. As above we can use our coordinate $\theta$ in the chart $\theta \in (0, 2\pi)$ to write a $1$-form $\omega \in \Omega^1(S^1)$ as,
\begin{eqnarray}
\omega = f(\theta) d\theta = d\left( \int^\theta_0 f(\theta') d\theta' \right) = d F(\theta)
\end{eqnarray}
Now the question is whether $F(\theta)$ defined locally over the chart $\theta \in (0, 2\pi)$ can be extended to a smooth function globally on the $S^1$. If it can, then $\omega$ is exact. Otherwise it is only closed. Clearly $F(\theta)$ can be extended to a smooth function only if $F(0) = F(2 \pi)$. Note that $F(0) = 0$, hence our condition is that;
\begin{eqnarray}
\mathrm{if} \quad \int_{S^1} \omega = \int_0^{2 \pi} f(\theta) d\theta = 0 \quad  \mathrm{then} \quad \omega \in B^r(S^1)
\end{eqnarray}

\noindent
Take two forms $\omega, \omega' \in Z^1(S^1)$ such that $\int \omega \ne 0$ ie. $\omega \notin B^1(S^1)$. Then take $a = (\int \omega')/(\int \omega) \in \mathbb{R}$. Now $\omega' - a \, \omega \in B^1$. Hence $[ \omega' ] = a [\omega]$ for any $\omega'$. Hence $H^1(S^1) = \mathbb{R}$ and $\omega$ gives a representative for the basis. \\

\noindent
For example $\omega = d\theta$ which is a volume form on $S^1$, gives a representative.



\noindent
{\bf Qu 2 iv)} \\

Poincare's theorem states that for a compact manifold then the Betti numbers $b_i \equiv \dim H_i$ are equal to the dimension of the cohomology vector spaces - so $b_i = \dim H^i$. We see this is true in our example of $S^1$.




\newpage


\noindent
{\bf Qu 3 i)} A Lie group $G$ is a manifold with a group structure defined by the following \emph{smooth} maps;
\begin{description}
\item[a)] Multiplication map $\cdot$ so; 
\begin{equation*}
\cdot : G \times G \rightarrow G \nonumber \\
\end{equation*}
\item[b)] Inverse map ${}^{-1}$ so; 
\begin{equation*}
{}^{-1} : G \rightarrow G \nonumber \\
\end{equation*}
\end{description}
where $\cdot$ is a smooth map from the product manifold $G\times G$ into $G$.\\

\noindent A left invariant vector field $V \in \mathcal{J}^1_0$ obeys the relation; $V |_g = L_g* (V |_e)$.\\


\noindent
{\bf Qu 3 ii)} 
Recall that for a diffeomorphism $f \in \mathrm{Diff}(G)$ that,
\begin{eqnarray}
f_* \left( [ X, Y ] |_p \right) = \left[ f_* X, f_* Y \right] |_{f(p)}
\end{eqnarray}
Consider two left invariant vector fields $X_{V_1}|_h = L_{h*} V_1, X_{V_2}|_h = L_{h*} V_2$. Then since $L_g$ is a diffeomorphism,
\begin{eqnarray}
L_{g*} \left( [ L_{h*} V_1, L_{h*} V_2] |_{h=e} \right) = \left[ L_{g*} L_{h*} V_1, L_{g*} L_{h*} V_2 \right]|_{gh} = \left[ L_{gh*} V_1, L_{gh*} V_2 \right]|_{gh}
\end{eqnarray}
Take $h = e$, and then,
\begin{eqnarray}
\left[  X_{V_1},  X_{V_2} \right]|_{g} = L_{g*} \left( [ X_{V_1}, X_{V_2} ] |_{e} \right) 
\end{eqnarray}
and hence the vector field $\left[  X_{V_1},  X_{V_2} \right]$ is left invariant. \\


\noindent  For a basis of left invariant vector fields $\{ L_i \}$, the structure constants $C_{ab}^{~~c}$ are given by $[ L_a, L_b ] = C_{ab}^{~~c} L_c$. Since the Lie bracket of two left invariant vector fields is also a left invariant vector field the structure constants are constant ie. independent of the point $g \in G$ that the defining expression is evaluated at.\\



\noindent
{\bf Qu 3 iii)} Let,
\begin{eqnarray*}
\mathcal{M}_{g'} & = & \left( \begin{matrix} a' & b' \\ - \bar{b'} & \bar{a'} \end{matrix} \right) , \quad |a'|^2 + |b'|^2 = 1 \nonumber \\
\mathcal{M}_g & = & \left( \begin{matrix} a & b \\ - \bar{b} & \bar{a} \end{matrix} \right) , \quad |a|^2 + |b|^2 = 1 \nonumber \\
\mathcal{M}_h & = & \left( \begin{matrix} c & d \\ - \bar{d} & \bar{c} \end{matrix} \right) , \quad |c|^2 + |d|^2 = 1 \nonumber \\
\end{eqnarray*}
Now take, $g' = L_g h$ so that,
\begin{equation*}
\left( \begin{matrix} a' & b' \\ - \bar{b'} & \bar{a'} \end{matrix} \right)  = \left( \begin{matrix} a & b \\ - \bar{b} & \bar{a} \end{matrix} \right)  \left( \begin{matrix} c & d \\ - \bar{d} & \bar{c} \end{matrix} \right)
\end{equation*}
where expanding the right-hand side,
\begin{equation*}
a' = a c - b \bar{d} , \quad b' = a d + b \bar{c}
\end{equation*}
Now we take,
\begin{eqnarray*}
a' & = & t' + i x' , \quad b' = y' + i z' \\
a & = & t + i x , \quad b = y + i z \\
c & = & \tau + i \rho , \quad d = \sigma + i \lambda
\end{eqnarray*}
and we have,
\begin{equation*}
t'^2 + x'^2 + y'^2 + z'^2 = t^2 + x^2 + y^2 + z^2 = \tau^2 + \rho^2 + \sigma^2 + \lambda^2 = 1
\end{equation*}
So we have,
\begin{eqnarray*}
t' & = & t \tau - x \rho - y \sigma - z \lambda \\
x' & = & t \rho + x \tau + y \lambda - z \sigma \\
y' & = & t \sigma - x \lambda + y \tau + z \rho \\
z' & = & t \lambda + x \sigma - y \rho + z \tau \\
\end{eqnarray*}

\noindent
{\bf Qu 3 iv)}\\
Left invariant vector fields satisfies $X |_{g'} = L_{g*} (X |_h)$ where $g' = g h$. Take $X |_h = v^i \frac{\partial}{\partial x^i(h)}$ with $x^i = \{ x,y,z \}$. Then,
\begin{equation*}
X |_{g'} = v^j \frac{\partial x^i(g')}{\partial x^j(h)} \frac{\partial }{\partial x^i(g')}
\end{equation*}
and now we have,
\begin{eqnarray*}
\frac{\partial x^i(g')}{\partial x^j(h)}  & = & \left( \begin{matrix} 
 \frac{\partial x'}{\partial \rho} &  \frac{\partial x'}{\partial \sigma} &  \frac{\partial x'}{\partial \lambda} \\  
 \frac{\partial y'}{\partial \rho} &  \frac{\partial y'}{\partial \sigma} &  \frac{\partial y'}{\partial \lambda} \\ 
 \frac{\partial z'}{\partial \rho} &  \frac{\partial z'}{\partial \sigma} &  \frac{\partial z'}{\partial \lambda} 
 \end{matrix} \right)  \\
 & = & \left( \begin{matrix} 
+t + x  \frac{\partial \tau}{\partial \rho} & - z + x  \frac{\partial \tau}{\partial \sigma} & + y +  x  \frac{\partial \tau}{\partial \lambda} \\  
+ z + y  \frac{\partial \tau}{\partial \rho} &  +t + y \frac{\partial \tau}{\partial \sigma} &  - x + y \frac{\partial \tau}{\partial \lambda} \\ 
- y + z  \frac{\partial \tau}{\partial \rho} & + x + z  \frac{\partial \tau}{\partial \sigma} & +t + z \frac{\partial \tau}{\partial \lambda} 
 \end{matrix} \right)  
\end{eqnarray*}
and $\frac{\partial \tau}{\partial \rho} = - \rho / \tau$, $\frac{\partial \tau}{\partial \sigma} = - \sigma / \tau$ and $\frac{\partial \tau}{\partial \lambda} = - \lambda / \tau$. \\
We are interested in the case $h = e$, the identity, so $g' = L_g e = g$. So we substitute $\rho = 0, \sigma = 0, \lambda = 0$ and so $\tau = 1$ into the above to find,
\begin{eqnarray*}
\frac{\partial x^i(g)}{\partial x^j(e)} & = & \left( \begin{matrix} 
+t  & - z  & + y \\  
+ z  &  +t  &  - x  \\ 
- y  & + x  & +t 
 \end{matrix} \right)  
\end{eqnarray*}
where $t^2 = 1 - x^2 - y^2 - z^2$. 
Take a basis for the left invariant vector fields as $\{ L_i \}$ with $i=1,2,3$ so that $(L_i)|_e = \frac{\partial}{\partial x^i(e)}$. Then,
\begin{eqnarray*}
L_1 & = & + t \frac{\partial}{\partial x} + z \frac{\partial}{\partial y} - y \frac{\partial}{\partial z} \\
L_2 & = & -z  \frac{\partial}{\partial x} + t \frac{\partial}{\partial y} + x \frac{\partial}{\partial z} \\
L_3 & = & +y  \frac{\partial}{\partial x} - x  \frac{\partial}{\partial y} + t \frac{\partial}{\partial z} 
\end{eqnarray*} \\


\noindent
{\bf Qu 3 v)}\\
For example, consider the Lie bracket,
\begin{eqnarray*}
[ L_1 , L_2 ] & = & [ t \frac{\partial}{\partial x} , L_2 ] + [  z \frac{\partial}{\partial y} , L_2 ] + [   - y \frac{\partial}{\partial z}  , L_2 ] \nonumber \\
& = & \left( + t \frac{\partial t}{\partial x} \frac{\partial}{\partial y}  + t \frac{\partial}{\partial z} + z  \frac{\partial t}{\partial x} \frac{\partial}{\partial x} - t  \frac{\partial t}{\partial y} \frac{\partial}{\partial x} - x  \frac{\partial t}{\partial z} \frac{\partial}{\partial x} \right) \nonumber \\
&& + \left( + z \frac{\partial t}{\partial y} \frac{\partial}{\partial y} - x  \frac{\partial}{\partial y} \right) \nonumber \\
&& + \left( + y \frac{\partial}{\partial x} - y \frac{\partial t}{\partial z} \frac{\partial}{\partial y} + t  \frac{\partial}{\partial z} \right)\nonumber \\
& = & - x \frac{\partial}{\partial y} + t \frac{\partial}{\partial z} - \frac{z x}{t} \frac{\partial}{\partial x} + y \frac{\partial}{\partial x} + \frac{x z}{t}  \frac{\partial}{\partial x} - \frac{z y}{t} \frac{\partial}{\partial y} - x  \frac{\partial}{\partial y} + y \frac{\partial}{\partial x} + \frac{ y z}{t} \frac{\partial}{\partial y} + t  \frac{\partial}{\partial z} \nonumber \\
& = & 2 \left( + t \frac{\partial}{\partial z} + y \frac{\partial}{\partial x}  - x  \frac{\partial}{\partial y} \right) = 2 L_3
\end{eqnarray*}
Hence the Lie bracket of $L_1$ and $L_2$ is left invariant (being proportional to $L_3$).


\newpage


\noindent
{\bf Qu 4 i)}\\
The equations of electromagnetism are given by,
\begin{eqnarray}
d F = 0 \, , \qquad d^\dagger F = j
\end{eqnarray}
for the 2-form field strength $F$ and 1-form current $j$. In particular we see that since $d F = 0$ then the 2-form $F$ is \emph{closed}. \\

Since $d^\dagger$ is nilpotent, acting it on the second equation above yields,
\begin{eqnarray}
d^\dagger j = 0
\end{eqnarray}
which when written out in components is the usual current conservation condition $\nabla^\mu j_\mu = 0$.\\

\noindent
{\bf Qu 4 ii)} 
Poincar\'e's Lemma states that a closed $r$-form defined on an open set $U_i$ of a manifold $\mathcal{M}$ that can be continuously contracted to a point, can be written as an exact form over $U_i$ (for $r>0$). So if $U_i$ is contractible and $\omega \in Z^r(U_i)$ then the Lemma states $\omega \in B^r(U_i)$ so $\omega = d a$ with $a \in \Omega^{r-1}(U_i)$. \\

Consider $\mathcal{M} = \mathbb{R}^n$. Since we can cover the whole of $\mathbb{R}^n$ with one open set, and this set can also be smoothly contracted to a point, then any closed form in $\mathbb{R}^n$ is also exact. Hence $H^i(\mathbb{R}^n) = 0$ for $i=1,\ldots,n$. In electromagnetism $d F = 0$ so $F$ is a closed 2-form. Hence for $\mathcal{M} = \mathbb{R}^4$ it is also exact and may be written as $F = d A$ for some $A \in \Omega^1(\mathbb{R}^n)$.\\


\noindent
{\bf Qu 4 iii)} 
Take the field strength $F = q \sin{\theta} \,d \theta \wedge d \phi$. Now we must check the vacuum EM equations $dF = d^\dagger F = 0$.  \\

\noindent
Firstly we will check $F$ is closed ie. $d F = 0$.
\begin{equation*}
d F = q \frac{\partial \sin{\theta} }{ \partial \theta} d\theta \wedge d \theta \wedge d \phi = 0
\end{equation*}

\noindent
Now we check $d^\dagger F = 0 \implies d\star F = 0$. So,
\begin{equation*}
\star F = + q \sin{\theta} \, \star (d \theta \wedge d \phi) = q \sin{\theta} \,\frac{\sqrt{|g|}}{2!} \epsilon^{\theta\phi}_{~~\mu\nu} dx^\mu \wedge dx^\nu = q \sin{\theta} \sqrt{|g|} g^{\theta\theta} g^{\phi\phi} dt \wedge dr
\end{equation*}
and $\sqrt{|g|} = r^2 \sin{\theta}$, $g^{\theta\theta} = (r^2)^{-1}$ and $g^{\phi\phi}= (r^2 \sin^2{\theta})^{-1}$, so we have,
\begin{equation*}
\star F = + \frac{q}{r^2} dt \wedge dr 
\end{equation*}
Hence $d\star F = q \frac{\partial}{\partial r} \left( \frac{1}{r^2} \right) dr \wedge dt \wedge dr +  \frac{q}{r^2} d^2t \wedge dr +  \frac{q}{r^2} dt \wedge dr^2  = 0$.\\


\noindent
{\bf Qu 4 iv)} 
Consider the integral of $F$ over a 2-sphere at $t = 0, r = r_0$. Then,
\begin{equation*}
\int_{S^2} F = + q \int_{S^2}  \sin{\theta} d\theta \wedge d\phi = - q \int_{S^2} d(\cos{\theta}) \wedge d\phi =  - q \int_{-1}^{+1} d(\cos{\theta}) \int_0^{2 \pi} d\phi = - 4 \pi q
\end{equation*} \\

\noindent
If we integrate an exact 2-form $\omega = d a$ over a 2-sphere and use Stokes's theorem we obtain zero,
\begin{equation*}
\int_{S^2} \omega = \int_{S^2} d a = \int_{\partial(S^2)} a = 0
\end{equation*}
since $S^2$ has no boundary.\\

\noindent
As the integral of $F$ over the $S^2$ does not vanish it follows that $F$ cannot be exact.\\


\noindent
{\bf Qu 4 v)} 
Suppose we could find a field strength $F'$ on $\mathbb{R}^4$ such that $F' = q \sin \theta d \theta \wedge d \phi$ for $r>1$. Now since it solves the EM vacuum equations it must be a closed 2-form. However in part ii) we argued that any closed 2-form on $\mathbb{R}^4$ must be exact. Hence we see that now $F'$ would have to be exact.\\

\noindent
We may again perform the same integration as in part iv) which doesn't care about whether we have removed the line $r=0$ or not, and so for a 2-sphere with $r_0 \ge 1$ gives the same answer on $\mathcal{M}$ or $\mathbb{R}^4$ for any 2-form equal to $q \sin \theta d \theta \wedge d \phi$ for $r > 1$. Since this answer is non-zero, the 2-form $F'$ cannot be exact. \\

\noindent
Hence we conclude that it is impossible to find a closed form $F'$ on $\mathbb{R}^4$ satisfying $F' = q \sin \theta d \theta \wedge d \phi$ for $r>1$. \\


\end{document}





