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\begin{flushright}
{{\bf Toby Wiseman}} 
\end{flushright}

\centerline{{\large \bf Solutions to 2007 Differential Geometry Exam}} 

\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 $S^2$ as unit radius sphere in $\mathbb{R}^3$. Then $S^2 = \{ (x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = 1 \}$.\\
{\bf Chart N} Define `North' chart $(U_N,\phi_N)$ as follows. We take the open set $U_N = \{ (x,y,z) \in S^2 | -1 \le z < 1\}$ 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 $(x,y,z) \in S^2$ the $x,y$ coordinates of the intersection of the line from the North pole through the point, with the plane $z = -1$. Label the intersection point $(X,Y,-1) \in \mathbb{R}^3$. Simple geometry gives,
\begin{equation*}
X = \frac{2}{1-z} x \; , \quad Y = \frac{2}{1-z} y
\end{equation*}
Then we may write the map $\phi_N$ as,
\begin{eqnarray*}
\phi_N : \quad U_N & \rightarrow & \mathbb{R}^2 \nonumber \\
(x,y,z) & \rightarrow & (X,Y)
\end{eqnarray*}

{\bf Chart S} Define `Sorth' chart $(U_S,\phi_S)$ in an analogous manner. We take the open set $U_S = \{ (x,y,z) \in S^2 | -1 < z \le 1\}$ ie. we exclude only the Sorth pole point of the sphere. The homeomorphism $\phi_S$, from $U_S$ to $\mathbb{R}^2$ 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 $(X',Y',+1)$. Simple geometry gives,
\begin{equation*}
X' = \frac{2}{1+z} x \; , \quad Y' = \frac{2}{1+z} y
\end{equation*}
Then we may write the map $\phi_S$ as,
\begin{eqnarray*}
\phi_S : \quad U_S & \rightarrow & \mathbb{R}^2 \nonumber \\
(x,y,z) & \rightarrow & (X',Y')
\end{eqnarray*}

{\bf Transition functions} We note that in the overlap region $U_N \cap U_S = \{ -1 < z < 1 \}$;
\begin{equation*}
X^2 + Y^2 = \frac{4}{(1-z)^2} (x^2 + y^2) = \frac{4}{(1-z)^2} (1 - z^2) = 4 \frac{1+z}{1-z}
\end{equation*}
and that,
\begin{equation*}
\frac{X'}{X} = \frac{1-z}{1+z} = \frac{4}{X^2 + Y^2}
\end{equation*}
and similarly for $Y'/Y$. This yields the relation between the coordinates;
\begin{equation}
X' = \frac{4}{X^2+Y^2} X \; , \quad Y' = \frac{4}{X^2+Y^2} Y
\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, Y) & \rightarrow & (X',Y')
\end{eqnarray*}
defined above by equation \eqref{eq:qu1eq1}. In the overlap this function is smooth, $C^\infty$, in the arguments $X,Y$.\\

\noindent
{\bf Qu 1 iii)} \\
$\mathbb{CP}^1$ is the set of undirected complex lines in $\mathbb{C}^2$. Let us take `homogeneous coordinates' $(z_1,z_2) \in \mathbb{C}^2 - \{0\}$, where $\mathbb{C}^2 - \{0\}$ is $\mathbb{C}^2$ with the origin removed. Then a complex line in $\mathbb{C}^2$ labelled by $(z_1, z_2)$ is given by the set $\{ (\lambda z_1, \lambda z_2) \in \mathbb{C}^2 \; \forall \lambda \in \mathbb{C} \}$. Hence we may obtain a labelling of points in $\mathbb{CP}^1$ using the homogeneous coordinates $(z_1,z_2) \in \mathbb{C}^2 - \{0\}$ and identifying the points $(z_1,z_2) \sim (\lambda z_1,\lambda z_2)$ for $\lambda \in \mathbb{C}$. Using this labelling we can give a 2 chart altas;\\
{\bf Chart N} $U_N$ is the open set labelled by homogeneous coordinates \\$\{ (z_1,z_2) \; \forall \; z_1,z_2 \in \mathbb{C} \; \mathrm{with} \; z_1 \ne 0  \}$.  
\begin{eqnarray*}
\Phi_{N} : \quad U_N & \rightarrow & \mathbb{C} \nonumber \\
(z_1,z_2) & \rightarrow & z = z_2/z_1
\end{eqnarray*}
{\bf Chart S} $U_S$ is the open set labelled by homogeneous coordinates \\$\{ (z_1,z_2) \; \forall \; z_1,z_2 \in \mathbb{C} \; \mathrm{with} \; z_2 \ne 0  \}$.  
\begin{eqnarray*}
\Phi_{S} : \quad U_S & \rightarrow & \mathbb{C} \nonumber \\
(z_1,z_2) & \rightarrow & z' = z_1/z_2
\end{eqnarray*}
{\bf Transition functions:} On the overlap region where $z\ne 0$ and $z'\ne 0$ we have $z' = t_{SN}(z) = 1/z$. This is analytic (holomorphic) in the overlap region since $z \ne 0$. \\


\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. \\

\noindent
{\bf Qu 1 v)} \\
On $m$-dimensional real manifold $\mathcal{M}$ the tangent bundle $T\mathcal{M}$ is defined as the collection of tangent spaces,
\begin{equation*}
T\mathcal{M} = \bigcup_{p\in \mathcal{M}}  T_p\mathcal{M}
\end{equation*}
It is a vector bundle, with base $\mathcal{M}$, fiber $\mathbb{R}^m$ and structure group $GL(m,\mathbb{R})$. The projection map is define by requiring the fiber at each point on the base to be the tangent vector space at that point, ie. $\pi^{-1}(p) = T_p\mathcal{M}$. A point $u \in T\mathcal{M}$ is a vector $u \in T_p\mathcal{M}$ with $\pi(u) = p$.\\
For an altas $\{ (U_i, \phi_i) \}$ with coordinates $x_{(i)} = \phi_i(p)$ for $p \in U_i$, we can write a vector $v = \left. v^\mu_{(i)} \frac{\partial}{\partial x^\mu_{(i)}} \right|_p \in T_pU_i$ and hence have the local trivialization;
\begin{eqnarray*}
\Phi_{i} : \quad U_i \times \mathbb{R}^m & \rightarrow & \pi^{-1}(U_i) \nonumber \\
p \; , \quad v^\mu & \rightarrow & v =  \left. v^\mu_{(i)} \frac{\partial}{\partial x^\mu_{(i)}} \right|_p
\end{eqnarray*}
In an overlap $U_i \cap U_j \ne \phi$, we can write a vector $v \in T_p\mathcal{M}$ for $p \in U_i \cap U_j$ as,
\begin{equation}
v =  \left. v^\mu_{(i)} \frac{\partial}{\partial x^\mu_{(i)}} \right|_p =  \left. v^\mu_{(j)} \frac{\partial}{\partial x^\mu_{(j)}} \right|_p\; , \quad v^\mu_{(i)} =  \left.  \frac{\partial x^\mu_{(i)} }{\partial x^\nu_{(j)}} \right|_p v^\nu_{(j)} 
\end{equation}
and we can express the same element of $T\mathcal{M}$ in the two trivializations as;
\begin{equation}
v = \Phi_i(p,v_{(i)}^\mu) = \Phi_j(p,v_{(j)}^\mu) = \Phi_i(p, (t_{ij} v_{(j)})^\mu)
\end{equation}
where $t_{ij}$ are the transition functions, and explicitly we see from above that,
\begin{equation*}
v_{(i)}^\mu = (t_{ij} v_{(j)})^\mu = \left.  \frac{\partial x^\mu_{(i)} }{\partial x^\nu_{(j)}} \right|_p v^\nu_{(j)} 
\end{equation*}
and hence we see $t_{ij} \in GL(m,\mathbb{R})$, since $\det{\partial x^\mu_{(i)}/\partial x^\nu_{(j)}} \ne 0$. \\




\newpage

\centerline{{\large \bf Solutions to 2007 Differential Geometry Exam }} 

\vspace{0.5cm}

\noindent
{\bf Qu 2 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  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 2 ii)} 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 2 iii)}\\
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 2 iv)}\\
Take a basis for left invariant vector fields $\{ L_a \}$ with $L_a = L_{g*} (e_a)$ where $\{e_a\}$ are a basis for  $T_eG$. Now take the set of one-forms $\{ \theta^a \}$, so,
\begin{equation*}
\delta^a_{~b} = < \theta^a , L_b > |_g= < \theta^a |_g , L_{g*} e_b > = <  L_g^{~*} (\theta^a |_g) , e_b > 
\end{equation*}
For $g=e$ we also have,
\begin{equation*}
\delta^a_{~b} = < \theta^a , L_b > |_e = < \theta^a |_e, e_b >
\end{equation*}
Hence we have $L_g^{~*} (\theta^a |_g) = \theta^a |_e = e^a$. \\

\noindent
{\bf Qu 2 v)}\\
So $d\omega(X,Y) = X[\omega(Y)] - Y[\omega(X)] - \omega([X,Y])$. Take $\omega = \theta^a$, $X = L_b$ and $Y = L_c$, so then,
\begin{eqnarray*}
d\theta^a(L_b,L_c) & = & L_b[\theta^a(L_c)] - L_c[\theta^a(L_b)] - \theta^a([L_b,L_c]) \\
& = & L_b[\delta^a_{~c}] - L_c[\delta^a_{~b}] - \theta^a(C_{bc}^{~~d} L_d) \\
& = & - C_{bc}^{~~d} \delta^{a}_{~d} = -C_{bc}^{~~a}
\end{eqnarray*}
So we have, 
\begin{equation*}
d\theta^a(L_b,L_c)|_g = -C_{bc}^{~~a} \quad \forall \; g \in G , \; \forall\; a, b,c
\end{equation*}
Hence we see,
\begin{equation*}
d\theta^a = -C_{bc}^{~~a} \theta^b \otimes \theta^c = -\frac{1}{2} C_{bc}^{~~a} \theta^b \wedge \theta^c
\end{equation*}
In the question we have,
\begin{equation*}
d\theta^a + K_{bc}^{~~a} \theta^b \wedge \theta^c = 0
\end{equation*}
so we see that, $K_{bc}^{~~a} = \frac{1}{2} C_{bc}^{~~a}$.\\

\noindent
{\bf Qu 2 vi)}\\
Taking the metric $g = \delta_{ab} \theta^a \otimes \theta^b$ we have the Cartan structure equation,
\begin{equation*}
d\theta^a + \omega^a_{~b} \wedge \theta^b = T^a
\end{equation*}
where $\omega^a_{~b}$ is the connection one-form and $T^a$ is the torsion two form. If we have vanishing torsion, so $T^a = 0$, then we have,
\begin{equation*}
d\theta^a + \omega^a_{~b} \wedge \theta^b = 0
\end{equation*}
and from above we see that $\omega^a_{~b} = \frac{1}{2} C_{bc}^{~~a} \theta^c$.



\newpage

\centerline{{\large \bf Solutions to 2007 Differential Geometry Exam}} 

\vspace{0.5cm}

\noindent
{\bf Qu 3 i)} Take,
\begin{equation*}
\omega = \frac{1}{r!} \omega_{\alpha_1,\ldots,\alpha_r} dx^{\alpha_1} \wedge \ldots \wedge dx^{\alpha_r}
\end{equation*}
then,
\begin{eqnarray*}
\star \omega & = & \frac{1}{r!} \omega_{\alpha_1,\ldots,\alpha_r} \star( dx^{\alpha_1} \wedge \ldots \wedge dx^{\alpha_r} ) \\
& = & \frac{1}{r!} \frac{\sqrt{|g|}}{(m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \epsilon^{\alpha_1,\ldots,\alpha_r}_{~~~~~~~\beta_{r+1},\ldots,\beta_m} dx^{\beta_{r+1}} \wedge \ldots \wedge dx^{\beta_m} 
\end{eqnarray*}
where $g = \det{g_{\mu\nu}}$ and then,
\begin{eqnarray*}
\star \star \omega & = & \frac{1}{r!} \frac{\sqrt{|g|}}{(m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \epsilon^{\alpha_1,\ldots,\alpha_r}_{~~~~~~~\beta_{r+1},\ldots,\beta_m} \star (dx^{\beta_{r+1}} \wedge \ldots \wedge dx^{\beta_m} ) \\
& = & \frac{1}{r!} \frac{|g|}{r! (m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \epsilon^{\alpha_1,\ldots,\alpha_r}_{~~~~~~~\beta_{r+1},\ldots,\beta_m} \epsilon^{\beta_{r+1},\ldots,\beta_m}_{~~~~~~~~~~\sigma_1,\ldots,\sigma_r} dx^{\sigma_1} \wedge \ldots \wedge dx^{\sigma_r} 
\end{eqnarray*}
Reordering indices we find,
\begin{eqnarray*}
\star \star \omega & = & \frac{1}{r!} \frac{|g|}{r! (m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \epsilon^{\alpha_1,\ldots,\alpha_r,\beta_{r+1},\ldots,\beta_m} \epsilon_{\beta_{r+1},\ldots,\beta_m\sigma_1,\ldots,\sigma_r} dx^{\sigma_1} \wedge \ldots \wedge dx^{\sigma_r} \\
& = & \frac{1}{r!} \frac{|g|}{r! (m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \epsilon^{\alpha_1,\ldots,\alpha_r,\beta_{r+1},\ldots,\beta_m} (-1)^{r(m-r)} \epsilon_{\sigma_1,\ldots,\sigma_r\beta_{r+1},\ldots,\beta_m} dx^{\sigma_1} \wedge \ldots \wedge dx^{\sigma_r} 
\end{eqnarray*}
where the sign arises as we moved $r$ indices past $(m-r)$ indices in the second epsilon term. Using the identity given in the question we obtain,
\begin{eqnarray*}
\star \star \omega & = & (-1)^{r(m-r)} \frac{1}{r!} \frac{|g|}{r! (m-r)!} \omega_{\alpha_1,\ldots,\alpha_r} \frac{r! (m-r)!}{\det{g_{\mu\nu}}} \delta^{[\alpha_1}_{~\sigma_1} \ldots \delta^{\alpha_r]}_{~\sigma_r}  dx^{\sigma_1} \wedge \ldots \wedge dx^{\sigma_r} \\
& = & (-1)^{r(m-r)} \frac{|g|}{g} \frac{1}{r!} \omega_{\alpha_1,\ldots,\alpha_r} \delta^{[\alpha_1}_{~\sigma_1} \ldots \delta^{\alpha_r]}_{~\sigma_r}  dx^{\sigma_1} \wedge \ldots \wedge dx^{\sigma_r} \\
& = & -(-1)^{r(m-r)} \frac{1}{r!} \omega_{\alpha_1,\ldots,\alpha_r}  dx^{\alpha_1} \wedge \ldots \wedge dx^{\alpha_r} \\
& = & -(-1)^{r(m-r)} \omega
\end{eqnarray*}
where the minus sign arises since $|g|/g = -1$ for our $(1,m-1)$ signature manifold.\\

\noindent
{\bf Qu 3 ii)} The 2-form $F \in \Omega^2$ can be written in components as $F = \frac{1}{2} F_{\mu\nu} dx^\mu \wedge dx^\nu$, where $F_{\mu\nu}$ give the components of the field strength tensor. The 1-form $j \in \Omega^1$ in components is $j = j_\mu dx^\mu$ with $j^\mu$ (note the raised index) giving the components of the charge current vector. \\

\noindent
Now $d^\dagger F = j$ but $(d^\dagger)^2 = \star d \star \star d \star = -(-1)^{r(m-r)} \star d^2 \star = 0$ since $d^2 = 0$, so we have,
\begin{equation*}
0 = (d^\dagger)^2 F = d^\dagger j
\end{equation*}
The condition $d^\dagger j = 0$ gives the local current conservation condition (and in components is $\nabla_\mu j^\mu = 0$). \\

\noindent
Take an $(m-1)$-chain $V \in C_{m-1}$ and $S = \partial V \in C_{m-2}$. Then $Q = \int_S \star F = \int_V d\star F$ using Stoke's thm. Using the defn of $d^\dagger$,
\begin{equation*} 
d^\dagger F  = (-1)^{2 m + m} \star d \star F = (-1)^{m} \star d \star F =  j
\end{equation*} 
Thus,
\begin{equation*} 
 \star j = (-1)^{m} ( \star \star d \star F ) =  (-1)^{m} ( -(-1)^{m-1} d \star F ) =  -(-1)^{2 m -1} d \star F =   d \star F 
\end{equation*} 
So,
\begin{equation*} 
Q = \int_V d\star F = \int_V \star j
\end{equation*} 
\\

\noindent
{\bf Qu 3 iii)} 
Poincar\'e's Lemma states that a closed $r$-form (for $r > 0$) defined on an open set $U_i$ of a manifold $\mathcal{M}$ that can be smoothly contracted to a point, can be written as an exact form over $U_i$. So if $U_i$ is smoothly 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 $\mathcal{M} = \mathbb{R}^4$. Since we can cover the whole of $\mathbb{R}^4$ with one open set, and this set can also be smoothly contracted to a point, then any closed form in $\mathbb{R}^4$ is also exact. (Hence $H^i(\mathbb{R}^4) = 0$ for $i=1,2,3,4$). 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 $A \in \Omega^1$ - now $A = A_\mu dx^\mu$ where $A^\mu$ is the `4-vector' potential.\\

\noindent
{\bf Qu 3 iv)} We take the metric $g = dt \otimes dt + dr \otimes dr + r^2 d\theta \otimes d\theta + r^2 \sin^2{\theta} d\phi \otimes d\phi$ on $\tilde{\mathcal{M}}$. Now $F = dA$ with $A = \frac{q}{r} dt$ for constant $q$. So,
\begin{equation*}
dA = d\left( \frac{q}{r} \right) \wedge dt = q d\left(\frac{1}{r} \right) \wedge dt = - \frac{q}{r^2} dr \wedge dt
\end{equation*}
Hence $F = + \frac{q}{r^2} dt \wedge dr$. Now we must check the vacuum EM equations $dF = d^\dagger F = 0$. Clearly the first condition is true, $d F = 0$, since $F$ is exact. Thus we must check $d^\dagger F = 0 \implies d\star F = 0$. So,
\begin{equation*}
\star F = + \frac{q}{r^2} \star (dt \wedge dr) = \frac{q}{r^2} \frac{\sqrt{|g|}}{2!} \epsilon^{tr}_{~~\mu\nu} dx^\mu \wedge dx^\nu = \frac{q}{r^2} \sqrt{|g|} g^{tt} g^{rr} d\theta \wedge d\phi
\end{equation*}
and $\sqrt{|g|} = r^2 \sin{\theta}$, $g^{tt} = -1$ and $g^{rr}=+1$, so we have,
\begin{equation*}
\star F = - q \sin{\theta} d\theta \wedge d\phi = + q \; d(\cos{\theta}) \wedge d\phi
\end{equation*}
Hence $d\star F = q \; d^2(\cos{\theta}) \wedge d\phi - q \; d(\cos{\theta}) \wedge d^2\phi = 0$.\\

\noindent
{\bf Qu 3 v)} Suppose we have a solution to the vacuum equations on $\tilde{\mathcal{M}}$, $d F_{vac} = d^\dagger F_{vac} = 0$. Now, these equations imply the pair,
\begin{equation*}
d F_{vac} = d \star F_{vac} = 0
\end{equation*}
Now inserting $\star \star = -(-1)^{r(m-r)}$ we can rewrite the first as,
\begin{equation*}
d F_{vac} = -(-1)^{2(4-2)} d \star \star F_{vac} = - d \star \star F_{vac} = 0
\end{equation*}
Hence we have $d \star \star F_{vac} = d \star F_{vac} = 0$. If we write $F_{dual} = \star F_{vac}$ then we obtain $d \star F_{dual} = d F_{dual} = 0$ which implies,
\begin{equation*}
d^\dagger F_{dual} = d F_{dual} = 0
\end{equation*}
Hence $F_{dual} = \star F_{vac}$ is indeed another solution to the vacuum EM equations. We have computed the dual of the point electric charge above,
\begin{equation*}
F_{mag} = \star F_{elec} = - q \sin{\theta} d\theta \wedge d\phi 
\end{equation*}
which is the field strength of the magnetic monopole.
\\


\noindent
{\bf Qu 3 vi)} If we integrate an exact 2-form $\omega = d a$ over a 2-sphere 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. Consider the integral of $F_{mag}$ over a 2-sphere at $t = 0, r = 1$. Then,
\begin{equation*}
\int_{S^2} F_{mag} = - 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*}
Hence $F_{mag}$ cannot be an exact 2-form as its integral over the closed $S^2$ does not vanish.\\


\newpage

\centerline{{\large \bf Solutions to 2007 Differential Geometry Exam }} 

\vspace{0.5cm}

\noindent
{\bf Qu 4 i)} Given an $r$-chain $c \in C_r$, it is an r-cycle (ie. $c \in Z_r$) iff $\partial c = 0$, and it is an r-boundary (ie. $c \in B_r$) iff $c = \partial b$ for some $b \in C_{r+1}$. Note that since $\partial^2 = 0$, then an $r$-boundary, $c$, is also an $r$-cycle as $\partial c = \partial^2 a = 0$. \\

\noindent
Given an $r$-form $\omega \in \Omega_r$, it is an r-cocycle (ie.  $\omega$ is a closed $r$-form, $\omega \in Z^r$) iff $d \omega = 0$, and it is an r-coboundary (ie. $\omega$ is an exact $r$-form, $\omega \in B^r$) iff $\omega = d \beta$ for some $\beta \in \Omega^{r-1}$. Note that since $d^2 = 0$, then an $r$-coboundary, $\omega$, is also an $r$-cocycle as $d \omega = d^2 \beta = 0$ (ie. any exact form is also closed). \\

\noindent
{\bf Qu 4 ii)} Define an equivalence of 2 closed forms $\omega,\omega' \in Z^r$ by $\omega \sim \omega'$ iff $w' = w + b$ with $b\in B^r$, ie. equivalent if they differ by an exact form. \\
Define an equivalence class of closed forms by; $[ \omega ] = \{ \omega' \in Z^r | \omega \sim \omega' \}$, where $\omega \in Z^r$ is the representative of the class. The class $[\omega]$ is the set of closed forms equivalent to $\omega$. If we have two closed $r$-forms $\alpha,\beta$ which are equivalent, $\alpha \sim \beta$, then the equivalence classes defined by them are the same ie. $[\alpha] = [\beta]$. \\

\noindent 
The $r^{th}$ cohomology vector space $H^r = \{ [\omega] \; \forall \; \omega \in Z^r \}$, ie. it is the set of equivalence classes of closed $r$-forms. 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 4 iii)}
We have,
\begin{eqnarray*}
\Lambda : \quad H_r \times H^r & \rightarrow & \mathbb{R} \\
 \; [c] \; , \quad [\omega] & \rightarrow & \Lambda([c],[\omega]) = \int_c \omega
\end{eqnarray*}
and we wish to show that if we have $c \sim c'$ and $\omega \sim \omega'$, then $\Lambda([c],[\omega]) = \Lambda([c'],[\omega'])$, ie. the inner product $\Lambda$ defined above is independent of the representatives of the equivalence classes chosen in the definition. So, let us take $c' = c + \partial b$ and $\omega' = \omega + d \beta$ with $b \in C_{r+1}$ and $\beta \in \Omega^{r-1}$. Then,
\begin{eqnarray*}
\Lambda([c'],[\omega']) & = & \int_{c'} {\omega'} = \int_{c + \partial b} {\omega + d \beta} \\
& = & \int_{c} {\omega} + \int_{\partial b} {\omega} + \int_{c} {d \beta} + \int_{\partial b} {d \beta}  = \int_{c} {\omega} = \Lambda([c],[\omega])
\end{eqnarray*}
since,
\begin{eqnarray*}
\int_{\partial b} {\omega} & = & \int_{b} {d \omega} = 0 \\
\int_{c} {d \beta} & = & \int_{\partial c} {\beta} = 0 \\
\int_{\partial b} {d \beta} & = & \int_{b} {d^2 \beta} = 0 \\
\end{eqnarray*}
using Stoke's thm, and the fact that $d\omega = 0$ since $\omega$ is closed, and $\partial c = 0$ since $c$ is an $r$-cycle, and using $d^2 = 0$. \\

\noindent
{\bf Qu 4 iv)}
A basis for $H_1(T^3)$ is given by taking the $1$-cycles $\{ c_i \}$ with $i = 1,2,3$ and $c_i \in Z_1$, defined by,
\begin{eqnarray*}
c_1 & = & \{ \theta_2 = \theta_3 = 0 \; , \; \forall \; 0 \le \theta_1 < 2 \pi \} \\
c_2 & = & \{ \theta_3 = \theta_1 = 0 \; , \; \forall \; 0 \le \theta_2 < 2 \pi \} \\
c_3 & = & \{ \theta_1 = \theta_2 = 0 \; , \; \forall \; 0 \le \theta_3 < 2 \pi \} 
\end{eqnarray*}
Note that the cycles $c_i$ indeed have no boundary, but are not themselves the boundary of any 2-cycle. Then calling $A_i = [c_i]$ we can use these to give the basis $\{ A_i \}$ with $i=1,2,3$ for $H_1(T^3)$. \\

A basis for $H^1(T^3)$ is given by taking the closed $1$-forms $\{ \omega_i \}$ defined by;
\begin{equation*}
\omega_1 = \frac{1}{2 \pi} d \theta_1 \; , \quad \omega_2 = \frac{1}{2 \pi} d \theta_2 \; , \quad \omega_3 = \frac{1}{2 \pi} d \theta_3
\end{equation*}
where we note that whilst $\omega_i$ are closed, they are not exact as $\theta_i$ are not well defined functions on the $T^3$. Then calling $B^i = [ \omega_i ]$ we have a basis $\{ B^i \}$ with $i=1,2,3$ for $H^1(T^3)$. \\

Let us compute the inner product of $A_i$ and $B^j$. Consider,
\begin{eqnarray*}
\Lambda(A_1,B^1) & = & \int_{c_1} \omega_1 =  \frac{1}{2 \pi} \int_{\begin{matrix}\theta_2 = \theta_3 = 0\\ 0  \le \theta_1 < 2 \pi\end{matrix}} d \theta_1 = \frac{1}{2 \pi}  \left[ \theta_1 \right]_0^{2 \pi} = 1 \\
\Lambda(A_1,B^2) &=& \int_{c_1} \omega_2 =  \frac{1}{2 \pi} \int_{\begin{matrix}\theta_2 = \theta_3 = 0\\ 0  \le \theta_1 < 2 \pi\end{matrix}} d \theta_2 = 0 \\
\Lambda(A_2,B^1) &=& \int_{c_2} \omega_1 =  \frac{1}{2 \pi} \int_{\begin{matrix}\theta_3 = \theta_1 = 0\\ 0  \le \theta_2 < 2 \pi\end{matrix}} d \theta_1 = 0 \\
\end{eqnarray*}
and similarly for other $i,j$, giving $\Lambda(A_i,B^j) = \delta^i_{~j}$. Hence $\{ A_i \}$ and $\{ B^i \}$ are indeed dual with respect to $\Lambda$.\\


\noindent
{\bf Qu 4 v)}
We have,
\begin{eqnarray*}
<\cdot,\cdot> : \quad H^r \times H^{m-r} & \rightarrow & \mathbb{R} \\
 \; [\omega] \; , \quad [\eta] & \rightarrow & <[\omega],[\eta]> = \int_\mathcal{M} \omega \wedge \eta
\end{eqnarray*}
and we wish to show that if we have $\omega \sim \omega'$ and $\eta \sim \eta'$, then $<[\omega'],[\eta']> = <[\omega],[\eta]>$, ie. the inner product $<\cdot,\cdot>$ defined above is independent of the representatives of the equivalence classes chosen in the definition. So, let us take $\omega' = \omega + d \alpha$ and $\eta' = \eta + d \beta$ with $\alpha \in \Omega^{r-1}$ and $\beta \in \Omega^{m-r-1}$. Then,
\begin{eqnarray*}
<[\omega'],[\eta']> & = & \int_{\mathcal{M}} \omega' \wedge \eta'  \\
& = & \int_{\mathcal{M}} {\omega \wedge \eta} + \int_{\mathcal{M}} {(d\alpha) \wedge \eta} + \int_{\mathcal{M}} {\omega \wedge (d\beta)} + \int_{\mathcal{M}} {(d\alpha) \wedge (d\beta)}  \\
&=& \int_{\mathcal{M}} {\omega \wedge \eta} = <[\omega],[\eta]>
\end{eqnarray*}
Now we have $d( \alpha \wedge \eta) = (d \alpha) \wedge \eta + (-1)^{r-1} \alpha \wedge (d\eta) = (d \alpha) \wedge \eta$ (since $\eta$ is closed) and so using Stoke's thm.,
\begin{equation*}
\int_{\mathcal{M}} (d \alpha) \wedge \eta = \int_{\mathcal{M}} d( \alpha \wedge \eta) = \int_{\partial \mathcal{M}} \alpha \wedge \eta = 0
\end{equation*}
since $\partial \mathcal{M} = 0$ as $\mathcal{M}$ is compact. Likewise using, $d( \omega \wedge \beta) = (d \omega) \wedge \beta + (-1)^{r} \omega \wedge (d\beta) = (-1)^{r} \omega \wedge (d \beta)$ so,
\begin{equation*}
\int_{\mathcal{M}}  \omega \wedge (d \beta) = (-1)^{r}  \int_{\mathcal{M}} d( \omega \wedge \beta) = (-1)^{r}  \int_{\partial \mathcal{M}} \omega \wedge \beta = 0
\end{equation*}
and also using $d( (d\alpha) \wedge \beta) = (d^2 \alpha) \wedge \beta + (-1)^{r} (d\alpha) \wedge (d\beta) = (-1)^{r} (d\alpha) \wedge (d\beta)$ so,
\begin{equation*}
\int_{\mathcal{M}}  (d\alpha) \wedge (d\beta) = (-1)^{r}  \int_{\mathcal{M}} d( (d\alpha) \wedge \beta) = (-1)^{r}  \int_{\partial \mathcal{M}} (d\alpha) \wedge \beta = 0
\end{equation*}
and hence we obtain,
\begin{eqnarray*}
<[\omega'],[\eta']> & = & \int_{\mathcal{M}} {\omega \wedge \eta} + \int_{\mathcal{M}} {(d\alpha) \wedge \eta} + \int_{\mathcal{M}} {\omega \wedge (d\beta)} + \int_{\mathcal{M}} {(d\alpha) \wedge (d\beta)}  \\
&=& \int_{\mathcal{M}} {\omega \wedge \eta} = <[\omega],[\eta]>
\end{eqnarray*}
as required.\\

\noindent
{\bf Qu 4 vi)}
A basis for $H^2(T^3)$ is given by taking the $2$-cycles $\{ \eta_i \}$ with $i = 1,2,3$ and $\eta_i \in Z^2$, defined by,
\begin{equation*}
\eta_1 = \frac{1}{(2 \pi)^2} d \theta_2 \wedge d \theta_3 \; , \quad \eta_2 = \frac{1}{(2 \pi)^2} d \theta_3 \wedge d \theta_1 \; , \quad \eta_3 = \frac{1}{(2 \pi)^2} d \theta_1 \wedge d \theta_2
\end{equation*}
where we note that whilst $\eta_i$ are closed, they are not exact as $\theta_i$ are not well defined functions on the $T^3$. Then calling $C^i = [ \eta_i ]$ we have a basis $\{ C^i \}$ with $i=1,2,3$ for $H^2(T^3)$. \\

Let us compute the inner product of $B^i$ and $C^j$ with respect to $<\cdot,\cdot>$. Consider,
\begin{eqnarray*}
<B^1,C^1> & = & \int_{\mathcal{M}} \left( \frac{1}{2\pi} d\theta_1 \right) \wedge \left( \frac{1}{(2 \pi)^2} d \theta_2 \wedge d \theta_3 \right) = \frac{1}{(2 \pi)^3} \int_{\mathcal{M}} d \theta_1 \wedge d \theta_2 \wedge d \theta_3 \\
& = & \frac{1}{(2 \pi)^3} \int_0^{2\pi} d\theta_1 \int_0^{2\pi} d\theta_2 \int_0^{2\pi} d\theta_3 = 1
\end{eqnarray*}
whereas,
\begin{eqnarray*}
<B^1,C^2> & = & \int_{\mathcal{M}} \left( \frac{1}{2\pi} d\theta_1 \right) \wedge \left( \frac{1}{(2 \pi)^2} d \theta_3 \wedge d \theta_1 \right) = - \frac{1}{(2 \pi)^3} \int_{\mathcal{M}} d \theta_1 \wedge d \theta_1 \wedge d \theta_3 = 0 \\
<B^2,C^1> & = & \int_{\mathcal{M}} \left( \frac{1}{2\pi} d\theta_2 \right) \wedge \left( \frac{1}{(2 \pi)^2} d \theta_2 \wedge d \theta_3 \right) = + \frac{1}{(2 \pi)^3} \int_{\mathcal{M}} d \theta_2 \wedge d \theta_2 \wedge d \theta_3 = 0 .
\end{eqnarray*}
Likewise we can compute the other $<B^i,C^j>$ and find, $<B^i,C^j> = \delta^{ij}$. Hence $\{ B^i \}$ and $\{ C^i \}$ are indeed dual with respect to $<\cdot,\cdot>$.



\end{document}



