
\documentclass[12pt]{article}

\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{epsfig}

\usepackage{bbm}

\begin{document}


\centerline{{\large \bf Solutions to May 2011 Differential Geometry Exam}} 

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

\vspace{0.5cm}

\noindent
{\bf Qu 1 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 1 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 1 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 1 iv)}
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 1 v)} 

Consider first $H_0$ and $H^0$ which are one dimensional. Then take a representative of $H_0$ to be some point $p$ on the circle. Take a representative of $H^0$ to be some (non-zero) constant function $f$. Then,
\begin{eqnarray}
\Lambda( [ p] , [ f ] ) = \int_p f = f
\end{eqnarray}
This is non-zero (since $f$ is a non-zero constant), and hence $\Lambda$ is non-degenerate on $H_0$ and $H^0$. \\

Now consider  $H_1$ and $H^1$ which again are one dimensional. Then take a representative 1-cycle for the non-trivial class of $H_1$ to be the circle itself, $c = S^1$. Take a representative of $H^1$ to be $\omega = d\theta$. Then,
\begin{eqnarray}
\Lambda( [ c ] , [ \omega ] ) = \int_c \omega =  \oint_{S^1} d \theta = 2 \pi
\end{eqnarray}
Again this is non-zero so $\Lambda$ is non-degenerate on $H_1$ and $H^1$.



\newpage


\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; $V |_g = L_g* (V |_e)$.\\



{\bf Qu 2 ii)} 

Consider the General Linear matrix group GL(n, R). An element $g \in GL(n, R)$ is 
an $n \times n$ matrix with non-zero determinant and we may take the components of this 
matrix $\{ x_{ij}(g) \}$ to give coordinates on the group manifold.\\

Take two vector fields. In the coordinate basis we may write these as $A = a_{ij}(g)  \frac{\partial}{\partial x^{ij}(g)}$ and $B = b_{ij}(g)  \frac{\partial}{\partial x^{ij}(g)}$. Then the Lie bracket is give as,
\begin{eqnarray}
[ A, B ]  & = &  \left( a_{ij} \frac{\partial}{\partial x^{ij}} b_{ab}(x) - b_{ij} \frac{\partial}{\partial x^{ij}} a_{ab}(x) \right) \frac{\partial}{\partial x^{ab}(g)}  \end{eqnarray}


{\bf Qu 2 iii)} 

\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 $x\left( L_g h\right) = x(g) \cdot x(h)$.  \\



{\bf Qu 2 iv)} 

\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}
Hence the push forward of a vector can be calculated very simply using matrix multiplication. \\

\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 $V = v^{ij} \left. \frac{\partial}{\partial x^{ij}} \right|_e \in T_eG$, again using the matrix notation. \\



{\bf Qu 2 v)} 


Finally consider the Lie bracket for,
\begin{eqnarray}
[ X_V, X_W ] |_g & = & \left. \left(  \left( x^{ik}(g) v^{kj} \right) \frac{\partial}{\partial x^{ij}(g)} \left( x^{ac}(g) w^{cb} \right) - \mathrm{sym}_{v \leftrightarrow w} \right) \frac{\partial}{\partial x^{ab}(g)} \right|_g \nonumber \\
& = & \left. \left(  x^{ik}(g) v^{kj} w^{cb} \delta^a_i \delta^c_j  - \mathrm{sym}_{v \leftrightarrow w} \right) \frac{\partial}{\partial x^{ab}(g)} \right|_g \nonumber \\
& = & \left.  x^{ak}(g) \left( v^{kc} w^{cb}  - w^{kc} v^{cb} \right) \frac{\partial}{\partial x^{ab}(g)} \right|_g \nonumber \\
& = &  \left.  \left(  g \cdot [ v, w ] \right)^{ab} \frac{\partial}{\partial x^{ab}(g)} \right|_g = X_{Y} |_g
\label{eq:comm}
\end{eqnarray}
where in the last line we have used matrix multiplication notation, with $y = [ v, w ] \equiv v \cdot w - w \cdot v$, and $Y = y^{ij} \left. \frac{\partial}{\partial x^{ij}} \right|_e \in T_eG$. \\



\newpage


\noindent
{\bf Qu 3 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 3 ii)} \\

Think of $S^1$ as embedded in $\mathbb{R}^2$ so $x^2 + y^2 = 1$. Take angular coordinate $\theta$ on circle,  $x = \cos \theta, y = \sin \theta$. 

For a circle we actually need 2 charts to provide an atlas. eg. 
\begin{eqnarray}
U_1 : &&\theta = (0, 2 \pi ) \\
\psi_1 :  &&p = (x = \cos \theta,y = \sin \theta) \rightarrow \theta
\end{eqnarray}
and
\begin{eqnarray}
U_2 :&& \theta' = (-\pi, +\pi ) \\
\psi_2 : && p = (x = \cos \theta',y = \sin \theta') \rightarrow \theta'
\end{eqnarray}

\noindent
This now covers $S^1$, and on the intersections $\theta = \theta'$ ($0<\theta'<\pi$) or $\theta = \theta'+ 2 \pi$ ($-\pi<\theta'<0$) so both transition functions are clearly smooth. \\




\noindent
{\bf Qu 3 iii)} \\

If $X$ is an $m$-dim'l manifold with atlas $\{ ( U_i, \psi_i ) \}$ and $Y$ is an $n$-dim'l manifold with atlas $\{ ( V_j, \xi_j ) \}$, then $X \times Y$ is the `product manifold', an $(m+n)$ dim'l manifold with altas $\{ ( W_{ij}, \sigma_{ij} ) \}$ defined as;

\begin{itemize}

\item points: a point in $X \times Y$ is labelled as $(p,q)$ with $p \in X$ and $q \in Y$

\item patches: $(p,q) \in W_{ij}$ iff $p \in U_i$ and $q \in V_j$

\item maps: 
\begin{eqnarray}
\sigma_{ij} :  X \times Y  & \rightarrow & \mathbb{R}^{m+n} \\
(p,q) & \rightarrow & \left( \psi_i(p), \xi_j(q) \right)
\end{eqnarray}

ie. vector $(x_1, \ldots, x_m, y_1, \ldots, y_n) \in \mathbb{R}^{m+n}$.

Obviously satisfies manifold conditions.

\end{itemize}




\noindent
{\bf Qu 3 iv)} \\

Think of $T^2 = S^1 \times S^1$. Label points in first $S^1$ with angle $\theta$ and for second $S^1$ with angle $\alpha$.

Then need 4 charts: eg.

\begin{eqnarray}
U_1 : && \theta = (0, 2 \pi ) \quad \alpha = (0, 2 \pi ) \\
\psi_1 :  && p(\theta,\alpha)  \rightarrow  ( \theta_1, \alpha_1 ) = ( \theta, \alpha )
\end{eqnarray}
and
\begin{eqnarray}
U_2 :&& \theta =   (-\pi, +\pi ) \quad \alpha = (0, 2 \pi )  \\
\psi_2 : && p(\theta,\alpha)  \rightarrow   ( \theta_2, \alpha_2 ) = ( \theta, \alpha )
\end{eqnarray}
and
\begin{eqnarray}
U_3 :&& \theta =   (0, 2 \pi )  \quad \alpha =  (-\pi, +\pi ) \\
\psi_3 : && p(\theta,\alpha)  \rightarrow   ( \theta_3, \alpha_3 ) = ( \theta, \alpha )
\end{eqnarray}
and
\begin{eqnarray}
U_4 :&& \theta =   (-\pi, +\pi ) \quad \alpha =  (-\pi, +\pi )   \\
\psi_4 : && p(\theta,\alpha)  \rightarrow  ( \theta_4, \alpha_4 ) = ( \theta, \alpha )
\end{eqnarray}

\noindent
This now covers $S^1$, and on the intersections $\theta_i = \theta_j$ ($0<\theta_j<\pi$) or $\theta_i = \theta_j+ 2 \pi$ ($-\pi<\theta_j<0$) and likewise for $\alpha$ so both transition functions are clearly smooth. \\


\noindent
{\bf Qu 3 v)} \\

\noindent
For example, take complex coordinates, $z_i = \theta_i + i \alpha_i$ in each chart. Then the transition functions are;
\begin{eqnarray}
z_i = z_j + a + i b
\end{eqnarray}
where $a = 0$ if $0<\theta_j<\pi$ and $2 \pi$ if $-\pi<\theta_j<0$, and likewise, $b = 0$ if $0<\alpha_j<\pi$ and $2 \pi$ if $-\pi< \alpha_j<0$. Hence $z_i = z_j + c_{ij}$ where $c_{ij}$ are various complex numbers. These are clearly analytic transition functions.


\newpage


\noindent
{\bf Qu 4 i)} \\

\noindent
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 $\mathcal{M}$. \\

\noindent
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 that this product is {\bf symmetric}: $( \alpha , \beta ) = ( \beta , \alpha )$ \\



\noindent
We define the adjoint exterior derivative as;
\begin{eqnarray}
d^\dagger : \quad \Omega^r(\mathcal{M}) & \rightarrow & \Omega^{r-1}(\mathcal{M}) \nonumber \\
\omega & \rightarrow & d^\dagger \omega \equiv - (-1)^{m (r+1 )} \star d \star \omega \qquad ( 0 , m ) \qquad \mathrm{Riem} \end{eqnarray}
We find that $d^\dagger$ is nilpotent so $(d^\dagger)^2 = 0$. This is since $d^\dagger d^\dagger = \star d \star \star d \star = \pm \star d^2 \star = 0$ since $d^2 = 0$. \\



\noindent
{\bf Qu 4 ii)} \\

\noindent
Take a compact orientable manifold with metric $(\mathcal{M}, g)$. Take $\alpha \in \Omega^r(\mathcal{M})$, $\beta \in \Omega^{r-1}(\mathcal{M})$
 relation;
\begin{eqnarray}
( d \beta , \alpha )  = ( \beta , d^\dagger \alpha )
\end{eqnarray}
Hence $d^\dagger$ is the \emph{adjoint} to $d$ with respect to the inner product $( \cdot , \cdot )$.

\noindent
Using Stokes we have,
\begin{eqnarray}
0 =  \int_{\partial \mathcal{M}} \beta \wedge ( \star \alpha ) = \int_{\mathcal{M}} d \left( \beta \wedge ( \star \alpha ) \right) = \int_{\mathcal{M}} d \beta \wedge ( \star \alpha ) + (-1)^{r+1} \int_{\mathcal{M}}  \beta \wedge d \star \alpha \nonumber
\end{eqnarray}
and so
\begin{eqnarray}
( d \beta , \alpha ) = (-1)^r (-1)^{\tilde{r} (m - \tilde{r})} \int_{\mathcal{M}} \beta \wedge ( \star \star d \star \alpha )
\end{eqnarray}
where $(d \star \alpha)$ is an $\tilde{r}$-form, so $\tilde{r} = m - r + 1$ and we have inserted the identity $(-1)^{\tilde{r} ( m - \tilde{r} ) } \star \star$ acting on this. Then we have,
\begin{eqnarray}
( d \beta , \alpha ) & = & (-1)^{(m - r +1 ) (r-1) + r} \int_{\mathcal{M}} \beta \wedge ( \star \star d \star \alpha ) \nonumber \\
& = & (-1)^{m r - r^2 + r - m - 1 } \int_{\mathcal{M}} \beta \wedge ( \star \star d \star \alpha ) \nonumber \\
& = & -(-1)^{m r + m} (-1)^{ r( r - 1) } \int_{\mathcal{M}} \beta \wedge ( \star \star d \star \alpha ) \nonumber \\
& = &  (-1)^{ r( r - 1) } \int_{\mathcal{M}} \beta \wedge ( \star d^\dagger \alpha ) \nonumber  = ( \beta, d^\dagger \alpha )
\end{eqnarray}
in gory detail.\\


\noindent
{\bf Qu 4 iii)} \\

In a chart we have $g = g_{\mu\nu}(x) dx^\mu \otimes dx^\nu$. In a new chart with coordinates $y^\mu$ we have,
\begin{eqnarray}
g =  g'_{\alpha\beta}(y)  dy^\alpha \otimes dy^\beta = g_{\mu\nu}(x) \frac{\partial x^\mu(y)}{\partial y^\alpha} \frac{\partial x^\nu(y)}{\partial y^\beta} dy^\alpha \otimes dy^\beta
\end{eqnarray}
Choose $x^\mu = a^\mu_{~\alpha} y^\alpha$ for a constant invertable matrix $a^\mu_{~\alpha}$. Then the new components $g'_{\alpha\beta}$ at a point $p$ are given as,
\begin{eqnarray}
g'_{\alpha\beta}(p) = g_{\mu\nu}(p) a^\mu_{~\alpha} a^\nu_{~\beta}
\end{eqnarray}
As a matrix equation we may write this, $g' = a^T \cdot g \cdot a$. Since at some point $p$ then the components of $g$ are a symmetric matrix with positive eigenvalues we may use the fact that there exists an $a$ such that at that point $g' = \mathbf{1}$, ie. $g'_{\mu\nu} = \delta_{\mu\nu}$ at this point $p$.\\



\noindent
{\bf Qu 4 iv)} \\
Recall that taking $\omega = \frac{1}{r!} \omega_{\mu_1 \ldots \mu_r} dx^{\mu_1} \wedge \ldots \wedge dx^{\mu_r}$ then,
\begin{eqnarray}
\star \; \omega = \frac{ \sqrt{|g|} }{ r! (m-r)! } \omega_{\mu_1 \ldots \mu_r} \epsilon^{\mu_1 \ldots \mu_r}_{~~~~~~\mu_{r+1} \ldots \mu_m} dx^{\mu_{r+1}} \wedge \ldots \wedge dx^{\mu_m}
\end{eqnarray}
so that,
\begin{eqnarray}
\omega \wedge \star \; \omega &=& \frac{ \sqrt{|g|} }{ r!^2 (m-r)! }  \omega_{\nu_1 \ldots \nu_r} \omega_{\mu_1 \ldots \mu_r} \epsilon^{\mu_1 \ldots \mu_r}_{~~~~~~\mu_{r+1} \ldots \mu_m} ( dx^{\nu_{1}} \wedge \ldots \wedge dx^{\nu_r} \wedge dx^{\mu_{r+1}} \wedge \ldots \wedge dx^{\mu_m} ) \nonumber \\
\omega \wedge \star \; \omega &=& \frac{ \sqrt{|g|} }{ r!^2 (m-r)! }  \omega_{\nu_1 \ldots \nu_r} \omega_{\mu_1 \ldots \mu_r} \epsilon^{\mu_1 \ldots \mu_r}_{~~~~~~\nu_{r+1} \ldots \nu_m} ( dx^{\nu_{1}} \wedge \ldots \wedge dx^{\nu_r} \wedge dx^{\nu_{r+1}} \wedge \ldots \wedge dx^{\nu_m} ) \nonumber \\
& =& \frac{ \sqrt{|g|} }{ r!^2 (m-r)! }  \omega_{\nu_1 \ldots \nu_r} \omega^{\mu_1 \ldots \mu_r} \epsilon_{\mu_1 \ldots \mu_r\nu_{r+1} \ldots \nu_m} ( ( \det{g} ) \epsilon^{\nu_1 \ldots \nu_m} dx^{1} \wedge \ldots \wedge dx^{m} ) \nonumber \\
& =& \frac{1 }{ r!  }  \omega_{\nu_1 \ldots \nu_r} \omega^{\nu_1 \ldots \nu_r} \sqrt{|g|} dx^{1} \wedge \ldots \wedge dx^{m} 
\end{eqnarray}

Now consider a point $p$ and take Rieman normal coordinates there. Then $g_{\mu\nu} = \delta_{\mu\nu}$ so that,
\begin{eqnarray}
\omega \wedge \star \; \omega |_p &=& \frac{1 }{ r!  }  \omega_{\nu_1 \ldots \nu_r} \omega_{\mu_1 \ldots \mu_r} \delta^{\nu_1 \mu_1} \ldots \delta^{\nu_r \mu_r}\,  dx^{1} \wedge \ldots \wedge dx^{m} \nonumber \\
& = &  \frac{1 }{ r!  }  | \omega_{\nu_1 \ldots \nu_r} |^2 dx^{1} \wedge \ldots \wedge dx^{m} \ge 0
\end{eqnarray}
with equality iff $\omega |_p = 0$.\\

Integrating $\omega \wedge \star \; \omega$ over the manifold gives $(\omega, \omega)$ so that $(\omega, \omega) \ge 0$ with equality iff $\omega = 0$ everywhere.


\noindent
{\bf Qu 4 v)} \\
The Laplacian $\triangle = d d^\dagger + d^\dagger d$. Then,
\begin{eqnarray}
( \omega , \triangle \omega ) = ( \omega , d d^\dagger \omega ) + ( \omega , d^\dagger d \omega ) =  ( d^\dagger \omega ,  d^\dagger \omega )  +  ( d \omega ,  d \omega ) 
\end{eqnarray}
Then for $ \triangle \omega = 0$ we have $( \omega , \triangle \omega )$ vanishes, and so,
\begin{eqnarray}
( d^\dagger \omega ,  d^\dagger \omega )  +  ( d \omega ,  d \omega ) = 0
\end{eqnarray}
Since $(\alpha, \alpha) = 0$ iff $\alpha = 0$ this implies,
\begin{eqnarray}
 d^\dagger \omega = d \omega  = 0
\end{eqnarray}

\end{document}





