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

\centerline{{\large \bf Solutions to 2013 Differential Geometry Exam}} 

\vspace{0.5cm}

\noindent
{\bf Qu 1 i)} \\

Consider a vector field $V$. This generates \emph{integral curves} whose tangents at any point $p$ are given by $V_p$.

\noindent
Consider a curve,
\begin{eqnarray}
\mathcal{C}: \quad \mathbb{R} & \rightarrow & \mathcal{M} \nonumber \\
\lambda & \rightarrow & p(\lambda)
\end{eqnarray}

\noindent
The tangent vector at $p(\lambda)$ is $\frac{d}{d\lambda} |_p$. Hence for the curve to be generated by a vector field $V$ we require $V_p = \frac{d}{d\lambda} |_p$. \\

\noindent
Take a chart containing $p$ with coordinates $x^\mu$. Then we may write the curve as $x^\mu(\lambda)$ and our vector field in the chart as,
\begin{equation}
V =  V^\mu(x) \frac{\partial}{\partial x^\mu}
\end{equation}
and the tangent to the curve through $p(\lambda)$ as,
\begin{equation}
\frac{d}{d\lambda} = \left.  \frac{d x^\mu(\lambda)}{d \lambda} \frac{\partial}{\partial x^\mu} \right|_{p(\lambda)}
\end{equation}
Then if $\mathcal{C}$ is an integral curve of $V$ we have,
\begin{equation}
\frac{d x^\mu(\lambda)}{d \lambda} = V^\mu( x(\lambda) )
\label{eq:ode}
\end{equation}

\noindent
We can use this to find the integral curve explicitly. Assume that at $\lambda = 0$ the curve to passes through the point $p_0$ with coordinates $x^\mu_{(0)} = x^\mu(p_0)$ so that $x^\mu(0) = x_{(0)}^\mu$. Then given we know $V^\mu(x)$ the above equation provides a ODE to integrate to find $x^\mu(\lambda)$ and hence $\mathcal{C}$. \\

\noindent
Consider this solution which we denote as $\sigma_V(\lambda, p_0)$. In coordinates we write,
\begin{equation}
x^\mu(\lambda) = \sigma_V^\mu(\lambda, x_{(0)})
\end{equation}
so that $\sigma_V(0,x_{(0)}) = x_{(0)}$. \\

\noindent
Then taking integral curves starting at all points in $\mathcal{M}$ we may think of $\sigma_V$ as the map,
\begin{eqnarray}
\sigma_V : \quad \mathbb{R} \times \mathcal{M} & \rightarrow \mathcal{M} \nonumber \\
( \lambda, p ) & \rightarrow & p' = \sigma_V(\lambda,p)
\end{eqnarray}
This map gives the \emph{flow} defined by the vector field $V$. \\

\noindent
For a fixed $\lambda$, the flow $\sigma_V$ gives a diffeomorphism $\sigma_V(\lambda) : \quad \mathcal{M} \rightarrow \mathcal{M}$ (inducing an `active' coordinate transform). \\



\noindent
{\bf Qu 1 ii)}
From the above discussion we see that the position of a point $p$, with coordinates $x^\mu(p)$ acted on by the diffeomorphism $\sigma_V(\lambda)$ is given by the solution to,
\begin{equation}
\frac{d x^\mu(\lambda)}{d \lambda} = V^\mu( x(\lambda) )
\end{equation}
where $V = V^\mu(x) \partial / \partial x^\mu$ and we use the notation $x^\mu(\lambda) = x^\mu( \sigma_V(\lambda) \cdot p)$ so that $x^\mu(0) = x^\mu(p)$. For small $\lambda$ we may approximate the solution to this equation as,
\begin{eqnarray}
x^\mu(\lambda) = x^\mu(p) + \lambda v^\mu(p) + \frac{1}{2} \lambda^2 h^\mu(p) + O(\lambda^3)
\end{eqnarray}
and then substituting into the  equation we have,
\begin{eqnarray}
\frac{d x^\mu(\lambda)}{d \lambda} = v^\mu(p) + \lambda h^\mu(p) + O(\lambda^2)
\end{eqnarray}
from the LHS and,
\begin{eqnarray}
V^\mu( x(\lambda) ) & = & V^\mu\left( x(p) + \lambda v(p) + \frac{1}{2} \lambda^2 h(p) + O(\lambda^3) \right) \nonumber \\
& = & V^\mu( p ) + \lambda v^\nu(p) \partial_\nu V^\mu( p ) + O(\lambda^2)
\end{eqnarray}
and so we see $v^\mu(p) = V^\mu( p )$ and $h^\mu(p) = v^\nu(p) \partial_\nu V^\mu( p )$ etc... giving,
\begin{eqnarray}
x^\mu( \sigma_V(\lambda) \cdot p) = x^\mu(\lambda) = x^\mu(p) + \lambda V^\mu(x(p)) + \frac{1}{2} \lambda^2 V^\nu \partial_\nu V^\mu( p ) + O(\lambda^3)
\end{eqnarray}
which gives the result required to order $O(\lambda^2)$.\\



\noindent
{\bf Qu 1 iii)}
The integral curve equation implies;
\begin{eqnarray}
\frac{ d f(p(\lambda) )}{d \lambda} =  \frac{ d f( \sigma_V(\lambda) \cdot p )}{d \lambda} =   \left. \frac{ \partial f( x )}{\partial x^\mu} \right|_{x(\lambda)} \cdot \frac{ d x^\mu(\lambda)}{d \lambda} = \left. V^\mu( x(\lambda) ) \frac{ \partial f( x )}{\partial x^\mu} \right|_{x(\lambda)} = \left. V[ f ] \right|_{ p(\lambda) }
\end{eqnarray}
So consider, $f(p(\lambda)) = \left. \left( e^{\lambda V} f \right) \right|_{p}$. Then differentiating,
\begin{eqnarray*}
\frac{d f(p(\lambda))}{d \lambda} &=& \frac{d }{d \lambda} \left( \left. e^{\lambda V} f  \right|_{p} \right) \nonumber \\
& = & \frac{d }{d \lambda} \left( \sum_{n=0}^\infty \left. \frac{\lambda^n}{n !}  \left( V^\mu \frac{\partial}{\partial x^\mu} \right)^n f(x)  \right|_{x = x(p)} \right) \nonumber \\
& = & \left( \sum_{n=1}^\infty \left. \frac{\lambda^{n-1}}{(n-1) !}  \left( V^\mu \frac{\partial}{\partial x^\mu} \right)^{n} f(x)  \right|_{x = x(p)} \right) \nonumber \\
& = & \left( \sum_{n=1}^\infty \left. \frac{\lambda^{n-1}}{(n-1) !}  \left( V^\mu \frac{\partial}{\partial x^\mu} \right)^{n-1} \left( V^\mu \frac{\partial}{\partial x^\mu} f(x) \right) \right|_{x = x(p)} \right) \nonumber \\
& = & \left( \sum_{n=0}^\infty \left. \frac{\lambda^{n}}{(n) !}  \left( V^\mu \frac{\partial}{\partial x^\mu} \right)^{n} \left( V^\mu \frac{\partial}{\partial x^\mu} f(x) \right) \right|_{x = x(p)} \right) \nonumber \\
& = &  \left( \left. e^{\lambda V} V[f]  \right|_{p} \right) = \left. V[f]  \right|_{p(\lambda)}
\end{eqnarray*}
as required, where we use the fact that $V[f]$ is also a function, so that $V[f](p(\lambda)) = \left. \left( e^{\lambda V} V[f] \right) \right|_{p}$.\\




\noindent
{\bf Qu 1 iv)}
So we may write,
\begin{eqnarray}
f_{(1)} &=& f|_{\sigma_U(\lambda) \cdot \sigma_W(\lambda) \cdot p} \nonumber \\
&=& \left.  \left(e^{\lambda U} f \right)\right|_{ \sigma_W(\lambda) \cdot p }
 \nonumber \\
&=& \left.  \left(e^{\lambda W}e^{\lambda U} f \right)\right|_{ p }
\end{eqnarray}
and likewise,
\begin{eqnarray}
f_{(2)} &=& f|_{\sigma_W(\lambda) \cdot \sigma_U(\lambda) \cdot p} \nonumber \\
&=& \left.  \left(e^{\lambda W} f \right)\right|_{ \sigma_U(\lambda) \cdot p }
 \nonumber \\
&=& \left.  \left(e^{\lambda U}e^{\lambda W} f \right)\right|_{ p }
\end{eqnarray}
Then,
\begin{eqnarray}
f_{(2)} - f_{(1)} &=& \left.  \left(e^{\lambda U}e^{\lambda W} f \right)\right|_{ p } - \left.  \left(e^{\lambda W}e^{\lambda U} f \right)\right|_{ p } \nonumber \\
& = &  \left.  \left( \left( e^{\lambda U}e^{\lambda W} - e^{\lambda W}e^{\lambda U} \right) f \right)\right|_{ p }  \nonumber \\
& = &   \left. \left(  1 + \lambda U + \frac{\lambda^2}{2} U^2 + \ldots \right) \left( 1 + \lambda W + \frac{\lambda^2}{2} W^2 + \ldots\right) f \right|_{ p } \nonumber \\
& & \qquad -  \left.  \left( 1 + \lambda W + \frac{\lambda^2}{2} W^2 + \ldots\right)   \left( 1 + \lambda U + \frac{\lambda^2}{2} U^2 + \ldots \right)  f \right|_{ p } 
 \nonumber \\
& = &  \left.  \left( \left( 1 + \lambda U+ \lambda W + \lambda^2 U W+ \frac{\lambda^2}{2} U^2  + \frac{\lambda^2}{2} W^2 + \ldots\right) f \right)\right|_{ p } \nonumber \\
&& \qquad - \left.  \left( \left( 1 + \lambda U+ \lambda W + \lambda^2 W U + \frac{\lambda^2}{2} U^2  + \frac{\lambda^2}{2} W^2 + \ldots\right) f \right)\right|_{ p } \nonumber \\
& = &  \left.  \lambda^2 [ U , W ]  f  \right|_{ p } + O(\lambda^3)
\end{eqnarray}

\noindent
Now if $[ U, W ] = 0$ (everywhere), then $e^{\lambda W}e^{\lambda U} = e^{\lambda U}e^{\lambda W}$ so $f_{(1)} = f_{(2)}$.


\newpage


\centerline{{\large \bf Solutions to 2013 Differential Geometry Exam}} 

\vspace{0.5cm}

\noindent
{\bf Qu 2 i)} \emph{Coset manifold:}
For a subgroup $H \subset G$ we may define an \emph{equivalence}: $ g \sim g'$ if $g' = g h$ for some $h \in H$. \\

\noindent
Define the \emph{equivalence class} as set of equivalent elements;
\begin{equation}
[ g ] = \{ g h \; \forall \; h \in H \}
\end{equation}
The element $g$ in $[g]$ is the `representative' of the equivalence class: $[g_1] = [g_2]$ if $g_1 \sim g_2$. \\

\noindent
A \emph{coset manifold}, denoted $G / H$, is the group $G$ with elements identified under the equivalence $\sim$. We may think of this as the group whose elements are the equivalence classes,
\begin{equation}
 G / H  \equiv \{ [ g ]  \}
\end{equation}

\noindent
\emph{Homogeneous space:} Consider a manifold $\mathcal{M}$ which admits a transitive group action, $G$. Consider any point $p_0 \in \mathcal{M}$, and the stabiliser subgroup $H_{p_0}$ of that point. Then if $\mathcal{M}$ is compact, and in certain other cases too, we may write $\mathcal{M}$ as he coset,
\begin{eqnarray}
\mathcal{M} = G / H_{p_0}
\end{eqnarray}
and we say that $\mathcal{M}$ is a homogeneous space.\\

\noindent
{\bf Qu 2 ii)}
Consider the transitive group action of $G$ on $\mathcal{M}$, 
\begin{eqnarray}
\Phi: G \times \mathcal{M} & \rightarrow & \mathcal{M} \nonumber \\
( g, p ) & \rightarrow & \Phi_g(p)
\end{eqnarray}
Take some point $p_0 \in \mathcal{M}$. The stabilizer subgroup at  $p_0$ is then $H_{p_0}$. Let $\chi(g) \equiv \Phi_g(p_0)$. This provides the smooth map,
\begin{eqnarray}
\chi : \quad G & \rightarrow & \mathcal{M} \nonumber \\
g & \rightarrow & \chi(g)
\end{eqnarray}
Now $\chi(g h) = \chi(g)$ if and only if $h \in H_{p_0}$. Hence we may define a map $\tilde{\chi}$ as,
\begin{eqnarray}
\tilde{\chi} : \quad G/H & \rightarrow & \mathcal{M} \nonumber \\
{[ g ]} & \rightarrow & \tilde{\chi}([g]) = \chi(g)
\end{eqnarray}
which is a one-to-one (injective). Since $\Phi$ is transitive, $\tilde{\chi}$ is also onto (surjective) ie. for any $p \in \mathcal{M}$ there exists a $[g] \in G / H$ such that $p = \chi(g)$. Hence it is invertable. 
A smooth invertable map is a diffeomorphism. Strictly speaking we haven't shown the inverse is smooth too, but for a compact $\mathcal{M}$ it can be proven to be. \\



\noindent
{\bf Qu 2 iii)}
$\mathbb{RP}^n \equiv$ set of undirected lines in $\mathbb{R}^{n+1}$. \\

\noindent
Label a line using `homogeneous coordinates' $(x_1, x_2, \ldots x_{n+1})$ in  $\mathbb{R}^{n+1} - \{0\}$ and identifying $(x_1, \ldots, x_{n+1}) \sim (\lambda x_1, \ldots, \lambda x_{n+1})$ for all $\lambda \in \mathbb{R} - \{0\}$. \\

\noindent
Construct $(n+1)$ charts:
\begin{eqnarray}
U_m : && \mathrm{all}\; p \in  \mathbb{RP}^n \; s.t. \; x_m \ne 0  \\
\psi_m : && \mathbb{RP}^n \qquad \qquad  \rightarrow   \mathbb{R}^n \\
&& p  = (x_1,\ldots,x_{n+1}) \rightarrow \left( \frac{x_1}{x_m}, \frac{x_2}{x_m}, \ldots,   \frac{x_{m-1}}{x_m} , \frac{x_{m+1}}{x_m} \ldots,  \frac{x_{n+1}}{x_m} \right) 
\end{eqnarray} \\

\noindent
Transition functions from $U_i' \rightarrow U_j'$ are smooth. \\

\noindent
{\bf Qu 2 iv)}
Take a point $x_i = \{ x_1, x_2, \ldots , x_{n+1} \}$ in $\mathbb{R}^{n+1}$. Then a group action of the matrix group $O(n+1)$ on $\mathbb{R}^{n+1}$ is given by matrix multiplication,
\begin{eqnarray}
\Phi:  O(n+1) \times \mathbb{R}^{n+1} & \rightarrow & \mathbb{R}^{n+1} \nonumber \\
( M_{ij}, x_i ) & \rightarrow & M_{ij} x_j
\end{eqnarray}

We may view $\mathbb{R}^{n+1}$ as being homogeneous coordinates for $\mathbb{RP}^n$ . Then a point in $\mathbb{RP}^n$ is given by the equivalence class of homogeneous coordinates $[x_i]$ where $\{ x_1, x_2 ,\ldots, x_{n+1} \} \sim \{ \lambda x_1, \lambda x_2 ,\ldots, \lambda x_{n+1} \}$ for any $\lambda \in \mathbb{R}^\star$. \\

Let us act on a point $x_i$ in $\mathbb{R}^{n+1}$ to obtain a new point $x'_i = M_{ij} x_j$. Since $\lambda x'_i = \lambda M_{ij} x_j = M_{ij} ( \lambda x_j )$ we see that the points $\{ x_i \}$ and $\{ \lambda x_i \}$ which are distinct in $\mathbb{R}^{n+1}$ but label the same point $[ x_i ]$ in $\mathbb{RP}^n$ transform to two new distinct points in $\mathbb{R}^{n+1}$ which again label the same point, $[ M_{ij} x_j ]$, in $\mathbb{RP}^n$. Hence the group action $O(n+1)$ on $\mathbb{R}^{n+1}$ also gives a group action on $\mathbb{RP}^n$. \\

Take any two points in $\mathbb{RP}^n$ given by the equivalences $[ x_i ]$ and $[ x'_i ]$. Recalling $x_i \rightarrow \lambda x_i$ then $x_i x_i  \rightarrow \lambda^2 x_i x_i $. Hence
we may choose representatives of these classes so that $x_i x_i = x'_i x'_i = 1$. Thus these points lie on the unit $m$-sphere in $\mathbb{R}^{n+1}$. Then there must exist an $O(n+1)$ matrix, $M_{ij}$, so that $x'_i = M_{ij} x_j$ (since $O(n+1)$ acts transitively on the unit $m$-sphere). Thus $O(n+1)$ acts transitively on $\mathbb{RP}^n$.


So $O(n+1)$ acts transitively on $\mathbb{RP}^n$. Let us consider its stability group.
Consider a point $x^{(0)}_i = \{ 0 , \ldots , 0 , 1 \} \in \mathbb{R}^{n+1}$. Consider the most general matrix that preserves this point in $\mathbb{RP}^n$,
\begin{eqnarray}
M_{ij} = \left( \begin{array}{cc} A_{ab} & 0 \\ C_b & B \end{array} \right)
\end{eqnarray}
where $a,b = 1,\ldots,n$ and $B \ne 0$, so that $x'_i = M_{ij} x_j = \{ 0 , \ldots , 0 , B \}$ and hence $[ x'_i ] = [ x_i ]$.

For this to be an element of $O(n+1)$ we require $M^T M = 1$. Hence we require,
\begin{eqnarray}
 \left( \begin{array}{cc} A^T & C \\ 0 & B \end{array} \right) \cdot  \left( \begin{array}{cc} A & 0 \\ C & B \end{array} \right) =  \left( \begin{array}{cc} A^T  A + C \dot C & C B \\ C B & B^2 \end{array} \right)  = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)
\end{eqnarray}
and hence
\begin{equation}
A^T A = 1 \qquad B^2 = 1  \qquad C = 0
\end{equation}
Thus $A \in O(n)$ and $B = \pm 1$ (ie. $O(1)$ or $\mathbb{Z}_2$). So the stabiliser group is,
\begin{eqnarray}
H_{x^{(0)}} =  O(1) \times O(n) 
\end{eqnarray}

Thus we see $\mathbb{RP}^n = O(n+1) / \left( O(1) \times O(n) \right)$. \\




\newpage


\centerline{{\large \bf Solutions to 2013 Differential Geometry Exam}} 

\vspace{0.5cm}

\noindent
{\bf Qu 3 i)}

\begin{itemize}

\item An $r$-form $\omega \in \Omega^r$ is \emph{closed} if $d \omega = 0$.

\item An $r$-form $\omega \in \Omega^r$ is \emph{exact} if $\omega = d \sigma$ where $\sigma \in \Omega^{r-1}$ is some $r-1$ form.

\end{itemize}

\noindent
Since $d^2 = 0$, we immediately see that \emph{any exact form is also closed}. \\

\noindent
We define,

\begin{itemize}

\item the space of closed $r$-forms forms the vector space, $Z^r(\mathcal{M})$

\item the space of exact $r$-forms forms the vector space, $B^r(\mathcal{M})$

\end{itemize}

\noindent
As we saw earlier the exterior derivative is nilpotent - hence $d^2 \omega = 0$ for $\omega \in \Omega^r$. Thus, $B^r \subset Z^r$. ie. if $\omega \in B^r$ then $\omega \in Z^r$. \\


\noindent
Two closed $r$-forms $z_1, z_2 \in Z^r(\mathcal{M})$ are equivalent - `\emph{cohomologous}' - if,
\begin{eqnarray}
z_1 - z_2 \in B^r(\mathcal{M})
\end{eqnarray}
ie. the closed forms differ by an exact form. We write the equivalence $z_1 \sim z_2$. \\

\noindent
We denote the set of \emph{equivalent} closed $r$-forms - `equivalence classes' - by $[ z ]$ where $z$ is a \emph{representative} closed $r$-form in the class. Hence,
\begin{eqnarray}
[ z ] = \{ z' \in Z^r(\mathcal{M}) \; \mathrm{where} \; z' \sim z \}
\end{eqnarray}
is the equivalence class. \\

\noindent
The linear structure of $Z^r, B^r$ is simply inherited from $\Omega^r$. However the set of equivalence classes of closed $r$-forms, $\{ [z] \}$, forms a vector space $H^r(\mathcal{M})$ - the cohomology vector space. The linear structure is defined on $H^r$ by
\begin{eqnarray}
\alpha [z_1 ] + \beta [z_2 ] = [ \alpha z_1 + \beta z_2 ]
\end{eqnarray}
for $\alpha, \beta \in \mathbb{R}$ and $z_{1,2} \in Z^r$. \\


\noindent
{\bf Qu 3 ii)}
Recall that $\dim \Omega^r = \dim \Omega^{m-r}$ on an $m$-diml manifold. With a metric we have a natural map;
\begin{eqnarray}
\star : \qquad \Omega^r(\mathcal{M}) \quad & \rightarrow & \quad \Omega^{m-r}(\mathcal{M}) \nonumber \\
dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_r} & \rightarrow & \star \left( dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_r} \right) \equiv \frac{1}{(m-r)!} \sqrt{| g |} \epsilon^{\mu_1 \ldots \mu_r}_{~~~~~~~\mu_{r+1}\ldots\nu_m} dx^{\mu_{r+1}} \wedge \ldots \wedge dx^{\mu_m} \nonumber
\end{eqnarray}

\noindent
Take $\alpha, \beta \in \Omega^r(\mathcal{M})$, so that, 
\begin{eqnarray}
\alpha = \frac{1}{r!} \alpha_{\mu_1 \ldots \mu_r} dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_r} \qquad \beta = \frac{1}{r!} \beta_{\mu_1 \ldots \mu_r} dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_r} 
\end{eqnarray}
and then,
\begin{eqnarray}
\alpha \wedge (\star \beta) = \beta \wedge (\star \alpha) = \frac{1}{r!} \left( \alpha_{\mu_1 \ldots \mu_r} \beta^{\mu_1 \ldots \mu_r} \right) \Omega_g
\end{eqnarray}
which is natural to integrate over $\mathcal{M}$. 
We use this to define the following inner product. 
\begin{eqnarray}
( \cdot , \cdot ) : \quad \Omega^r(\mathcal{M}) \times  \Omega^r(\mathcal{M}) & \rightarrow & \mathbb{R} \nonumber \\
\alpha \quad , \quad \beta \quad & \rightarrow & ( \alpha, \beta ) \equiv \int_{\mathcal{M}} \alpha \wedge ( \star \beta ) 
\end{eqnarray}

\noindent
We see this product is {\bf symmetric}: $( \alpha , \beta ) = ( \beta , \alpha ) =  \frac{1}{r!} \int_\mathcal{M} \left( \alpha_{\mu_1 \ldots \mu_r} \beta^{\mu_1 \ldots \mu_r} \right) \Omega_g$ \\

\noindent
We note that for the case of a Riemannian metric, signature $(0,m)$, then using Riemann normal coordinates at a point the metric is the Euclidean metric, so, 
\begin{eqnarray}
\alpha \wedge (\star \alpha)  = \frac{1}{r!} \left(  \alpha_{\mu_1 \ldots \mu_r} \alpha^{\mu_1 \ldots \mu_r} \right) \Omega_g  = \frac{1}{r!} \left( \alpha_{\mu_1 \ldots \mu_r} \alpha_{\nu_1 \ldots \nu_r} \delta^{\mu_1\nu_1} \ldots  \delta^{\mu_r \nu_r} \right) \Omega_g = \frac{1}{r!} | \alpha_{\mu_1 \ldots \mu_r} |^2 \Omega_g \ge 0
\end{eqnarray}
with equality only if $\alpha$ vanishes at the point.
This is true for any point, and hence, $\alpha \wedge (\star \alpha) \ge 0$.
Hence for a Riemannian metric the product is {\bf positive definite} so that $( \omega , \omega ) \ge 0$ with equality only if $\omega = 0$. \\

\noindent
So \emph{up to signs}, and using Stokes' (on a compact manifold with no boundary)
\begin{eqnarray}
( d^\dagger \alpha, \beta ) &=& ( \star d \star \alpha, \beta ) = ( \beta,  \star d \star \alpha ) = \int \beta \wedge \star ( \star d \star \alpha ) \nonumber \\
& = & \int \beta \wedge ( \star  \star ) d \star \alpha =   \int \beta \wedge d \star \alpha \nonumber \\
&  =&  \int d \beta \wedge \star \alpha  = ( d \beta , \alpha ) = ( \alpha, d \beta )
\end{eqnarray}

\noindent
{\bf Qu 3 iii)}
We now define `harmonic' , `co-closed' and `co-exact' forms; An $r$-form $\omega \in \Omega^r(\mathcal{M})$ is;
\begin{eqnarray}
\mathrm{co-closed} \quad \quad d^\dagger \omega & = & 0 \nonumber \\
\mathrm{co-exact} \quad \qquad  \omega & = & d^\dagger \beta \qquad \beta \in \Omega^{r+1}(\mathcal{M}) \nonumber \\
\mathrm{harmonic} \quad \triangle \omega \equiv (d^\dagger d + d d^\dagger) \omega & = & 0 \nonumber 
\end{eqnarray}
\\

\noindent
Consider a Riemannian manifold $(\mathcal{M},g)$. Then since the inner product $(\cdot, \cdot)$ is positive, so that $(\alpha,\alpha) \ge 0$ with equality on if $\alpha = 0$, then we have,
\begin{eqnarray}
(\omega , \triangle \omega) = ( \omega , d^\dagger d \omega ) +  ( \omega , d d^\dagger \omega ) = ( d \omega , d \omega ) + ( d^\dagger \omega , d^\dagger \omega ) \ge 0
\end{eqnarray}
We see then that $\omega$ is harmonic, so $\triangle \omega = 0$, if and only if $d \omega = 0$ and $d^\dagger \omega = 0$ ie. a form is harmonic iff it is closed and co-closed. \\


%\noindent
%{\bf Qu 3 iv)}
% \emph{Hodge decomposition theorem:}
%On a compact Riemannian manifold the space of $r$-forms admits a unique decomposition into exact, co-exact and harmonic forms; ie.
%\begin{eqnarray}
%\Omega^r = B^r(\mathcal{M}) \oplus B^{\dagger r}(\mathcal{M}) \oplus \mathrm{Harm}^r(\mathcal{M})
%\end{eqnarray}
%This implies any $r$-form $\omega \in \Omega^r$ can be written \emph{uniquely} as;
%\begin{eqnarray}
%\omega = d \alpha + d^\dagger \beta + \gamma
%\end{eqnarray}
%for some $\alpha \in \Omega^{r-1}$, $\beta \in \Omega^{r+1}$ and $\gamma \in \mathrm{Harm}^r$. \\
%
%
%\noindent
%Consider a closed $r$-form $\omega \in Z^r$ on a compact Riemannian manifold. Hodge's theorem tells us we may write any $r$-form as $\omega = d \alpha + d^\dagger \beta + \gamma$ for harmonic $\gamma$. However we wish this to be closed. Consider,
%\begin{eqnarray}
%( d \omega , \beta ) = ( d d^\dagger \beta , \beta ) = ( d^\dagger \beta, d^\dagger \beta )
%\end{eqnarray}
%Since we want $d \omega = 0$ this must vanish. However the RHS can only vanish if $d^\dagger \beta = 0$. Hence we see $Z^r = B^r \oplus \mathrm{Harm}^r$, so we may write a closed form as,
%\begin{eqnarray}
%\omega = d \alpha + \gamma
%\end{eqnarray}
%for $\gamma \in \mathrm{Harm}^r$. \\
%
%\noindent
%Therefore an important consequence of Hodge's theorem is that every cohomology class $[ \omega ] \in H^r(\mathcal{M})$ has a \emph{unique harmonic} representative on a compact Riemannian manifold. Take a given representative, $\omega$, which then has a unique decomposition, $\omega = d \alpha + \gamma$ for $\gamma \in \mathrm{Harm}^r$. Then $\gamma$ is the harmonic representative, so $[ \omega ] = [ \gamma ]$. \\

%\noindent
%{\bf Qu 3 vi)}
%Maxwell equations are $F$ is closed i.e.. $d F = 0$ and also $d^\dagger F = j$. Since $(d^\dagger)^2 = 0$, then $d^\dagger j = 0$.\\
%
%\noindent
%Poincare's Lemma states a closed form is exact on a contractible open set, an implies therefore that $\mathbb{R}^n$ has trivial cohomology since it is contractible. Thus any closed two form is exact. Hence we may always write the 2 form field strength $F = d A$ on $\mathbb{R}^n$. \\
%
%\noindent
%A gauge transformation is $A \to A' = A + d \lambda$ for some function $\lambda$ which leaves $F$ invariant since $d^2 = 0$.\\
%
%\noindent
%Recall $d^\dagger F = j$. Then we simply have,
%\begin{eqnarray}
%d^\dagger d A = j
%\end{eqnarray}
%Furthermore using Lorentz gauge, $d^\dagger A =0$, we can write this as,
%\begin{eqnarray}
%j = d^\dagger d A = ( d^\dagger d + d d^\dagger ) A = \triangle A
%\end{eqnarray}
%\\
%

\noindent
{\bf Qu 3 iv)}
Use the Hodge decomposition to write the source $\rho = d \alpha + d^\dagger \beta + \gamma$ for unique forms $\alpha$, $\beta$ and Harmonic $\gamma$. Then since $\triangle \Phi = \rho$ then,
\begin{eqnarray}
0 = ( \triangle \Phi - \rho, \gamma ) &=& ( d d^\dagger \Phi + d^\dagger d \Phi , \gamma ) - (  d \alpha + d^\dagger \beta + \gamma , \gamma ) \nonumber \\
& = & ( d^\dagger \Phi  , d^\dagger \gamma ) + ( d \Phi , d \gamma )  - (  d \alpha + d^\dagger \beta + \gamma , \gamma ) 
\end{eqnarray}
Now since $\gamma$ is Harmonic then $d \gamma = d^\dagger \gamma = 0$. Then,
\begin{eqnarray}
0 & = & (  d \alpha + d^\dagger \beta + \gamma , \gamma ) \nonumber \\
& = &  (  d \alpha , \gamma ) +  ( d^\dagger \beta  , \gamma ) +  (  \gamma , \gamma )  \nonumber \\
& = &   (  \alpha , d^\dagger \gamma ) +  (  \beta  , d \gamma ) +  (  \gamma , \gamma )  \nonumber \\
& = &  (  \gamma , \gamma ) 
\end{eqnarray}
again as $\gamma$ is harmonic. But $( \gamma , \gamma ) = 0$ only if $\gamma = 0$. Then we see that the harmonic component of the decomposition of the source $\rho$ must indeed vanish.


\newpage


\centerline{{\large \bf Solutions to 2013 Differential Geometry Exam}} 

\vspace{0.5cm}

\noindent
{\bf Qu 4 i)}
A Lie group $G$ is a manifold with the following smooth maps defined;
\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}

\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.\\



{\bf Qu 4 ii)} \\
\noindent
Using the canonical coordinates on $GL$ we may write a vector as $V = v^{ij} \left. \frac{\partial}{\partial x^{ij}} \right|_g \in T_gGL$. The components $v_{ij}$ form a matrix. There is no restriction on the matrix $v_{ij}$ that defines a vector. (In particular while the matrices of the $GL$ group have non-zero determinant, there is no determinant condition on $v_{ij}$). \\

\noindent
Consider the left action $L_g h = g h$. Since this is given by matrix multiplication we have,
\begin{equation}
x^{ij} \left( L_g h\right) = x^{ik}(g) x^{kj}(h) 
\end{equation}
which we might write in matrix terms as $x\left( L_g h\right) = x(g) \cdot x(h)$.  \\

\noindent
The push forward of a vector $V = v^{ij} \left. \frac{\partial}{\partial x^{ij}} \right|_h \in T_hG$ is given as,
\begin{eqnarray}
L_{g*} \left( V|_h \right) & = & v^{ij} \left.  \frac{\partial x^{ab}(g h)}{\partial x^{ij}(h)} \frac{\partial}{\partial x^{ab}(g h)} \right|_{g h}  \nonumber \\
& = & v^{ij} \left.  \frac{\partial}{\partial x^{ij}(h)} \left( x^{ak}(g) x^{kb}(h)  \right) \frac{\partial}{\partial x^{ab}(g h)} \right|_{g h}  \nonumber \\
& = & v^{ij} x^{ak}(g) \delta^i_k \delta^j_b \left. \frac{\partial}{\partial x^{ab}(g h)} \right|_{g h}  \nonumber \\
& = &  x^{ai}(g) v^{ij} \left. \frac{\partial}{\partial x^{aj}(g h)} \right|_{g h}  = v'^{ij} \left. \frac{\partial}{\partial x^{ij}(g h)} \right|_{g h}  \in T_{g h}G
\end{eqnarray}
Thinking of the vector components $v'^{ij}, v^{ij}$ as matrices, and since $\{ x^{ij}(g) \}$ is simply the matrix $g$ itself, we can write the relation between components neatly as,
\begin{equation}
v' = g \cdot v
\end{equation}

\noindent
A left invariant vector field, $X_V$, is then given as,
\begin{eqnarray}
X_V = x^{ai}(g) v^{ij} \left. \frac{\partial}{\partial x^{aj}(g)} \right|_{g}  = \left( g \cdot v \right)^{ij} \left. \frac{\partial}{\partial x^{ij}(g)} \right|_{g}  
\end{eqnarray}
for $V = v^{ij} \left. \frac{\partial}{\partial x^{ij}} \right|_e \in T_eGL$, again using the matrix notation. \\

\noindent
Hence indeed $x^{ai}(g) v^{ij}  \frac{\partial}{\partial x^{aj}(g)}$ is left invariant, and there is no restriction on the matrix $v^{ij}$.\\


{\bf Qu 4 iii)} \\
\noindent
Now, for $SU(2)$ we have matrices of the form,
\begin{eqnarray*}
M = \left( \begin{array}{cc} a & b \\ -\bar{b} & \bar{a} \end{array} \right) 
\end{eqnarray*}
where $a,b \in \mathbb{C}$ and $| a |^2 + | b |^2 = 1$. Writing $a = t + i x$ and $b = y + i z$ where $(x,y,z,t)\in \mathbb{R}^4$ then,
\begin{eqnarray*}
1 = | a |^2 + | b |^2 = x^2 + y^2 + z^2 + t^2
\end{eqnarray*}
and hence we see that we have embedded $SU(2)$ as a unit 3-sphere in $\mathbb{R}^4$.\\

\noindent
Now using this embedding into $\mathbb{R}^4$ we can use stereographic projection. We wish the North pole to be the identity. Now the identity corresponds to,
\begin{eqnarray*}
M = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) 
\end{eqnarray*}
and hence $(x, y, z, t) = (0, 0, 0, 1)$.\\

\noindent
{\bf Chart N} Define `North' chart $(U_N,\phi_N)$ as follows. We take the open set $U_N = \{ (x, y, z, t) \in S^3 | -1 \le t < 1\}$ ie. we exclude only the North pole point of the sphere. The homeomorphism $\phi_N$, from $U_N$ to $(X,Y,Z) \in \mathbb{R}^3$ is given by associating to a point in $S^3$ the coordinates of the intersection of the line from the North pole, $t = 1$, through the point, with the plane $t = 0$. Label the intersection point $(X, Y, Z, 0) \in \mathbb{R}^{n+1}$. Simple geometry gives,
\begin{eqnarray*}
\phi_N : \quad U_N & \rightarrow & \mathbb{R}^n \nonumber \\
a = t + i x \, , b = y + i z & \rightarrow & (X, Y, Z) = \left(  \frac{x}{1-t},  \frac{y}{1-t},  \frac{z}{1-t}\right)
\end{eqnarray*}
Recall that $x^2 + y^2 + z^2 + t^2 = 1$.\\

\noindent
{\bf Chart S} Define `Sorth' chart $(U_S,\phi_S)$ with  $U_S = \{ (x, y, z, t) \in S^3 | -1 < t \le 1\}$ in an analogous manner projecting from the South pole, so,
\begin{eqnarray*}
\phi_S : \quad U_S & \rightarrow & \mathbb{R}^n \nonumber \\
a = t + i x \, , b = y + i z & \rightarrow & (X', Y', Z') = \left(  \frac{x}{1+t},  \frac{y}{1+t},  \frac{z}{1+t} \right)
\end{eqnarray*}

\noindent
The two pairs $(U_N,\phi_N)$ and $(U_S,\phi_S)$ form the Altas. Note the transition functions are smooth.\\


\noindent
{\bf Qu 4 iv)} \\
\noindent
The chart $(U_S,\phi_S)$ contains the identity $t = +1$ where $(X', Y',Z') = (0,0,0)$. Then an embedding of $SU(2)$ into $GL(2, \mathbb{C})$ which preserves the group structure is simply given by,
\begin{eqnarray}
f : \quad SU(2) & \to & GL(2, \mathbb{C}) \nonumber \\
(X', Y', Z') & \to & x^{ij}(X', Y', Z') =  \left( \begin{array}{cc} x^{11} & x^{12} \\ x^{21} & x^{22} \end{array} \right) = \left( \begin{array}{cc} a & b \\ -\bar{b} & \bar{a} \end{array} \right) 
\end{eqnarray}
Now since,
\begin{eqnarray*}
| \vec{X'} |^2 = X'^2 + Y'^2 + Z'^2 = \frac{x^2 + y^2 + z^2}{( 1 + t)^2} = \frac{1 - t^2}{( 1 + t)^2} = \frac{1 - t}{1 + t}
\end{eqnarray*}
so that,
\begin{eqnarray*}
t = \frac{1 -  | \vec{X'} |^2}{1 + | \vec{X'} |^2 }
\end{eqnarray*}
and so,
\begin{eqnarray*}
x &=& X' (1 + t) = \frac{2 X'}{1 + | \vec{X'} |^2 } \; , \quad y = \frac{2 Y'}{1 + | \vec{X'} |^2 } \; , \quad z = \frac{2 Z'}{1 + | \vec{X'} |^2 }
\end{eqnarray*}
we then find the embedding explicitly given as,
\begin{eqnarray}
x^{11} &=& t + i x =  \frac{1 -  | \vec{X'} |^2 +  2 i X'}{1 + | \vec{X'} |^2 } \nonumber \\
x^{12} &=& y+ i z =  2 \frac{Y' + i Z'}{1 + | \vec{X'} |^2 } \nonumber \\
x^{22} &=& t - i x =  \frac{1 -  | \vec{X'} |^2 - 2 i  X'}{1 + | \vec{X'} |^2 } \nonumber \\
x^{21} &=& - y + i z =  - 2 \frac{Y' - i Z'}{1 + | \vec{X'} |^2 }
\end{eqnarray}

\noindent
The coordinate basis at the identity are;
\begin{eqnarray}
e_{1} &=& \left. \frac{\partial}{\partial {X' }} \right|_{e} \nonumber \\
e_{2} &=& \left. \frac{\partial}{\partial {Y' }} \right|_{e} \nonumber \\
e_{3} &=& \left. \frac{\partial}{\partial {Z' }} \right|_{e} 
\end{eqnarray}
Then,
\begin{eqnarray}
f_\star e_{1} &=& f_\star \left. \frac{\partial}{\partial {X' }} \right|_{e} \nonumber \\
&=& \left. \frac{\partial x^{ij}}{\partial {X' }} \frac{\partial}{\partial {x^{ij} }} \right|_{e} \nonumber \\
\end{eqnarray}
Now,
\begin{eqnarray}
\left. \frac{\partial x^{11}}{\partial {X' }} \right|_{e} &=& \left.  \frac{\partial}{\partial {X' }} \left( \frac{1 - X'^2 - Y'^2 - Z'^2 +  2 i X'}{1 + X'^2 + Y'^2 + Z'^2 } \right) \right|_{e} \nonumber \\
&=& \left.  \left( \frac{ - 2 X' + 2 i}{1 + X'^2 + Y'^2 + Z'^2 } - 2 X' \frac{ 1 - X'^2 - Y'^2 - Z'^2 +  2 i X'}{(1 + X'^2 + Y'^2 + Z'^2 )^2} \right) \right|_{e} \nonumber \\
&=& 2 i
\end{eqnarray}
Likewise,
\begin{eqnarray}
\left. \frac{\partial x^{12}}{\partial {X' }} \right|_{e} &=& \left.  \left( 2 \frac{Y' + i Z'}{1 + X'^2 + Y'^2 + Z'^2 } \right) \right|_{e} \nonumber \\
&=& 0
\end{eqnarray}
and, 
\begin{eqnarray}
\left. \frac{\partial x^{22}}{\partial {X' }} \right|_{e} &=& \left. \frac{\partial (x^{11})^\star}{\partial {X' }} \right|_{e} = - 2 i \nonumber \\
\left. \frac{\partial x^{21}}{\partial {X' }} \right|_{e} &=& - \left. \frac{\partial (x^{12})^\star}{\partial {X' }} \right|_{e} = 0
\end{eqnarray}
So that,
\begin{eqnarray}
f_\star e_{1} &=& \left( \begin{array}{cc} 2 i & 0 \\ 0 & - 2 i \end{array} \right)^{ij} \left. \frac{\partial}{\partial {x^{ij} }} \right|_{e} \end{eqnarray}
Similar calculations give,
\begin{eqnarray}
f_\star e_{2} &=& \left( \begin{array}{cc}  0 & 2 \\ -2 & 0 \end{array} \right)^{ij} \left. \frac{\partial}{\partial {x^{ij} }} \right|_{e} \end{eqnarray}
and,
\begin{eqnarray}
f_\star e_{3} &=& \left( \begin{array}{cc} 0 & 2 i \\ 2 i & 0 \end{array} \right)^{ij} \left. \frac{\partial}{\partial {x^{ij} }} \right|_{e} 
\end{eqnarray}

\noindent
Thus the corresponding left invariant vector fields are,
\begin{eqnarray}
X_{f_\star e_{1}} &=& \left(  \left( \begin{array}{cc} x^{11} & x^{12} \\ x^{21} & x^{22} \end{array} \right) \cdot \left( \begin{array}{cc} 2 i & 0 \\ 0 & - 2 i \end{array} \right) \right)^{ij} \frac{\partial}{\partial {x^{ij} }} \nonumber \\
&=& 2 i \left( \begin{array}{cc} x^{11} & - x^{12} \\ x^{21} & - x^{22} \end{array} \right) ^{ij} \frac{\partial}{\partial {x^{ij} }} \nonumber \\
X_{f_\star e_{2}} &=& \left(  \left( \begin{array}{cc} x^{11} & x^{12} \\ x^{21} & x^{22} \end{array} \right) \cdot \left( \begin{array}{cc}0 & 2 \\ -2 & 0 \end{array} \right) \right)^{ij} \frac{\partial}{\partial {x^{ij} }} \nonumber \\
&=& 2 \left( \begin{array}{cc} - x^{12} & x^{11} \\  -x^{22} & x^{21} \end{array} \right) ^{ij} \frac{\partial}{\partial {x^{ij} }} \nonumber \\
X_{f_\star e_{3}} &=& \left(  \left( \begin{array}{cc} x^{11} & x^{12} \\ x^{21} & x^{22} \end{array} \right) \cdot \left( \begin{array}{cc} 0 & 2 i \\ 2 i & 0 \end{array} \right) \right)^{ij} \frac{\partial}{\partial {x^{ij} }} \nonumber \\
&=& 2 i \left( \begin{array}{cc} x^{12} & x^{11} \\ x^{22} & x^{21} \end{array} \right) ^{ij} \frac{\partial}{\partial {x^{ij} }} 
\end{eqnarray}

\end{document}



