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\centerline{{\large \bf Solutions to May 2009 Differential Geometry Exam}} 

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

\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 $O(\lambda)$.\\


\noindent
{\bf Qu 1 iii)}
The Lie derivative is defined as,
\begin{eqnarray}
\mathcal{L}_V g = \lim_{\lambda \rightarrow 0} \frac{1}{\lambda} \left( (\sigma_V(\lambda))^* \cdot g(\sigma_V(\lambda) \cdot p) - g(p) \right)
\end{eqnarray}
Thus for small $\lambda$ we may write,
\begin{eqnarray}
g'(p) - g(p) = \lambda \, \mathcal{L}_V g + O(\lambda^2) 
\end{eqnarray}


\noindent
{\bf Qu 1 iv)}
We may use the fact that,
\begin{eqnarray}
\mathcal{L}_V dx^\mu = \frac{\partial V^\mu(x)}{\partial x^\nu} dx^\nu
\end{eqnarray}
Then writing $g = g_{\mu\nu}(x) dx^\mu \otimes dx^\nu$ we have,
\begin{eqnarray}
\mathcal{L}_V g &= &\left( \mathcal{L}_V g_{\mu\nu}(x) \right) dx^\mu \otimes dx^\nu + g_{\mu\nu}(x) \left( \mathcal{L}_V dx^\mu \right) \otimes dx^\nu + g_{\mu\nu}(x) dx^\mu \otimes \left( \mathcal{L}_V dx^\nu \right) \nonumber \\
&= & V[g_{\mu\nu}(x)]  dx^\mu \otimes dx^\nu + g_{\mu\nu}(x) \frac{\partial V^\mu(x)}{\partial x^\alpha} dx^\alpha  \otimes dx^\nu + g_{\mu\nu}(x) dx^\mu \otimes  \frac{\partial V^\nu(x)}{\partial x^\alpha} dx^\alpha \nonumber \\
& = & \left( V[g_{\mu\nu}(x)] + g_{\alpha\nu}(x) \frac{\partial V^\alpha(x)}{\partial x^\mu} + g_{\mu\alpha}(x) \frac{\partial V^\alpha(x)}{\partial x^\nu}\right) dx^\mu \otimes dx^\nu \nonumber \\
& = & \left( V^\alpha \frac{\partial}{\partial x^\alpha} g_{\mu\nu}(x) + g_{\alpha\nu}(x) \frac{\partial V^\alpha(x)}{\partial x^\mu} + g_{\mu\alpha}(x) \frac{\partial V^\alpha(x)}{\partial x^\nu}\right) dx^\mu \otimes dx^\nu
\end{eqnarray}


\newpage


\noindent
{\bf \large Question 2}
\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 $\{ \xi_i \}$, the structure constants $C_{ab}^{~~c}$ are given by $[ \xi_a, \xi_b ] = C_{ab}^{~~c} \xi_c$. 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)}\\
Take a basis $\{e_a\} \in T_eG$ and dual basis $\{e^a\} \in T_e^{*}G$ so that $\delta^a_{~b} = <e^a , e_b >$. Then take a basis for left invariant vector fields $\{ \xi_a \}$ with $\xi_a = L_{g*} (e_a)$. Now take the set of one-forms $\{ \theta^a \}$, so,
\begin{equation*}
\delta^a_{~b} = < \theta^a , \xi_b > |_g= < \theta^a |_g , L_{g*} e_b > = <  L_g^{~*} (\theta^a |_g) , e_b > 
\end{equation*}
We also have,
\begin{equation*}
\delta^a_{~b} = < \theta^a , \xi_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 iii)}\\
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 $V = v^{ij} \frac{\partial}{\partial x^{ij}}  \in T_hG$ is given as,
\begin{eqnarray}
L_{g*} \left( V|_h \right) & = & v^{ij}  \frac{\partial x^{ab}(g h)}{\partial x^{ij}(h)} \frac{\partial}{\partial x^{ab}(g h)}   \nonumber \\
& = & v^{ij}   \frac{\partial}{\partial x^{ij}(h)} \left( x^{ak}(g) x^{kb}(h)  \right) \frac{\partial}{\partial x^{ab}(g h)}   \nonumber \\
& = & v^{ij} x^{ak}(g) \delta^i_k \delta^j_b \frac{\partial}{\partial x^{ab}(g h)}   \nonumber \\
& = &  x^{ai}(g) v^{ij} \frac{\partial}{\partial x^{aj}(g h)} 
\end{eqnarray}

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


\noindent
{\bf Qu 2 iv)}\\
The dual basis to the coordinate basis $\partial / \partial x^{ij}(g)$ is given by $d x^{ij}(g)$ so that,
\begin{eqnarray}
< d x^{ij}(g) , \frac{\partial}{\partial x^{ab}(g)} > =  \delta^{(ij)}_{(ab)} = \delta^i_a \delta^j_b
\end{eqnarray}
We require a one-form field $\theta^{ij}$ dual to our basis left invariant vector fields $\xi_{ij}$, so,
\begin{eqnarray}
 \delta^a_i \delta^b_j = <  \theta^{ab}, \xi_{ij} > |_g =  <  \theta^{ab} |_g, x^{ki}(g) \frac{\partial}{\partial x^{kj}(g)} > 
\end{eqnarray}
Let us write the one form field in components using the dual basis so, $\theta^{ij} = \omega^{(ij)}_{ab}(g) dx^{ab}(g)$. Then,
\begin{eqnarray}
<  \theta^{ab} |_g, x^{ki}(g) \frac{\partial}{\partial x^{kj}(g)} > & = & \omega^{(ab)}_{cd}(g) x^{ki}(g) <  d x^{cd}(g) , \frac{\partial}{\partial x^{kj}(g)} > \nonumber \\
& = & \omega^{(ab)}_{cd}(g) x^{ki}(g) \delta^c_k \delta^d_j \nonumber \\
& = & \omega^{(ab)}_{cj}(g) x^{ci}(g) = \delta^{ai} \delta^b_j  \nonumber \\
\end{eqnarray}
which is solved if, $\omega^{(ab)}_{cj}(g) = x^{ac}(g^{-1}) \delta^b_j$. Thus,
\begin{eqnarray}
\theta^{ab} = \omega^{(ab)}_{cj}(g)  dx^{cj}(g)  = x^{ac}(g^{-1}) dx^{cb}(g) 
\end{eqnarray}
provides a basis for the left invariant one forms.

\newpage


\noindent
{\bf \large Question 3}
\vspace{0.5cm}


\noindent
{\bf Qu 3 i)}
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
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 3 ii)}\\
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 3 iii)} 
Poincar\'e's Lemma states that a closed 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 $U_i$. 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 3 iv)}\\
Let us consider the gauge $d^\dagger A = 0$ - the `Lorentz' gauge. Under a gauge transform we see,
\begin{eqnarray}
 d^\dagger A' = d^\dagger A + d^\dagger d \lambda =  d^\dagger A + ( d^\dagger d + d d^\dagger) \lambda = d^\dagger A + \triangle \lambda
\end{eqnarray}
where we have used the fact that $d^\dagger \lambda = 0$ since $\lambda$ is a zero form.
Hence we see that by choosing $\lambda = - \triangle^{-1} (d^\dagger A)$, then $d^\dagger A' = 0$. \\

\noindent
So since $d^\dagger A =0$ we then have,
\begin{eqnarray}
d^\dagger F = d^\dagger d A = ( d^\dagger d + d d^\dagger ) A = \triangle A = j
\end{eqnarray}


\newpage


\noindent
{\bf \large Question 4}
\vspace{0.5cm}



\noindent
{\bf Qu 4 i)} \\
There is an elegant basis invariant definition given by the following map,
\begin{eqnarray}
\mathcal{T} : \quad \mathcal{J}^1_0 \times \mathcal{J}^1_0 & \rightarrow & \mathcal{J}^1_0 \nonumber \\
X, \quad Y \quad & \rightarrow & \mathcal{T}(X,Y) \equiv \nabla_X Y - \nabla_Y X - [ X, Y ] 
\end{eqnarray}
It defines a $(1,2)$-tensor with components in a basis $\{ e_\mu \}$,
\begin{eqnarray}
T_{\mu\nu}^{~~~\rho} = \;< e^\rho , \,\mathcal{T}( e_\mu , e_\nu ) >
\end{eqnarray}

For a given metric $g$, a connection $\nabla$ is a \emph{metric connection} if at any point $p \in \mathcal{M}$ it obeys,
\begin{eqnarray}
\nabla_X g = 0 \quad \forall \; X \in T_p(\mathcal{M})
\end{eqnarray}
Another way to say this is that the metric is parallel transported along any curve. \\

The \emph{Christoffel} connection is the connection defined with connection components,
\begin{eqnarray}
C_{\alpha\beta}^{~~~\rho} \equiv \frac{1}{2} g^{\rho \sigma} \left( \partial_\alpha g_{\beta \sigma} +  \partial_\beta g_{\alpha \sigma} -  \partial_\sigma g_{\alpha \beta} \right)
\end{eqnarray}
The symmetric part of any connection can simply be written in terms of the Christoffel connection and the antisymmetric part, given by the torsion tensor,
\begin{eqnarray}
\frac{1}{2} S_{\alpha\beta}^{~~~~\rho}  = C_{\alpha\beta}^{~~~\rho} + T^{\rho}_{~~(\alpha\beta)} 
\end{eqnarray}
where we note that $T^{\rho}_{~~\alpha\beta} = g^{\rho\mu} g_{\beta\nu} T_{\mu\alpha}^{~~~\nu}$. \\

\noindent
In the case of vanishing torsion, $T = 0$, we see there is a \emph{unique metric connection} $\Gamma_{\mu\nu}^{~~~\rho} = C_{\mu\nu}^{~~~\rho}$, which is symmetric $\Gamma_{\mu\nu}^{~~~\rho} = \Gamma_{\nu\mu}^{~~~\rho}$. This is called the \emph{Levi-Civita connection}. \\



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

Since both the torsion and curvature have an explicit antisymmetry, it is natural to define forms from them. We do so as follows; \\

\noindent
{\bf Torsion 2-form} \\
We define the vector valued 2-form;
\begin{eqnarray}
\hat{T}^{a} \equiv \frac{1}{2} T_{bc}^{~~a} \, \hat{\theta}^b \wedge \hat{\theta}^c
\end{eqnarray}

\noindent
{\bf Connection 1-form} \\
We define the matrix valued 1-form;
\begin{eqnarray}
\hat{\omega}_c^{~a} \equiv \Gamma_{bc}^{~~a} \, \hat{\theta}^b
\end{eqnarray}




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

\noindent
Take our non-coordinate bases, and then we may write the connection in terms of components,
\begin{eqnarray}
\nabla_a e_b & = & \nabla_{e^a} e_b = \Gamma_{ab}^{~c} e_c
\end{eqnarray}
and then on the dual basis we have,
\begin{eqnarray}
\nabla_a \theta^b & = & - \Gamma_{ac}^{~b} \theta^c
\end{eqnarray}
For a metric connection, $\nabla_a g = 0$, which implies,
\begin{eqnarray}
0 = \nabla_a \left( \delta_{bc} \hat{\theta}^b \otimes \hat{\theta}^c \right) = \delta_{bc}  \nabla_a \left( \hat{\theta}^b \otimes \hat{\theta}^c \right) = - \delta_{bc} \left( \Gamma_{ad}^{~~b} \hat{\theta}^d \otimes \hat{\theta}^c + \Gamma_{ad}^{~~c} \hat{\theta}^b \otimes \hat{\theta}^d \right)
\end{eqnarray}
and hence,
\begin{eqnarray}
\Gamma_{abc} = - \Gamma_{acb}
\end{eqnarray}
where $\Gamma_{abc} = \Gamma_{ab}^{~~d} \, \delta_{dc}$. Note we used Riemannian signature, but the same applies in Lorentzian signature, and in the above steps one just replaces  $\delta_{ab} \rightarrow \eta_{ab}$. Then from the definition of $\hat{\omega}_{ab}$ we see,
\begin{eqnarray}
\hat{\omega}_{ab} &= & \hat{\omega}_a^{~k} \delta_{bk} = \Gamma_{ca}^{~~k} \, \hat{\theta}^c \delta_{bk} = \Gamma_{cab} \, \hat{\theta}^c \nonumber \\
&= & - \Gamma_{cba} \, \hat{\theta}^c 
= - \hat{\omega}_{ba} 
\end{eqnarray}




\noindent
{\bf Qu 4 iv)} \\
We have,
\begin{eqnarray}
\tilde{g} & = & \frac{1}{(X^2 + Y^2 + 1)^2} ( dX\otimes dX + dY \otimes dY )
\end{eqnarray}
Take the non-coordinate basis,
\begin{eqnarray}
\theta^X = \frac{1}{X^2 + Y^2 + 1} dX \, , \quad  \theta^Y = \frac{1}{X^2 + Y^2 + 1} dY
\end{eqnarray}
so that,
\begin{eqnarray}
\tilde{g} & = & \theta^X \otimes \theta^X + \theta^Y \otimes \theta^Y = \delta_{ij} \theta^i \otimes \theta^j
\end{eqnarray}
Since the connection 1-form is antisymmetric, then $\omega_{XX} = \omega_{YY} = 0$ and $\omega \equiv \omega_{XY} = - \omega_{YX}$.

The first Cartan structure equation then reads,
\begin{eqnarray}
d \theta^X + \omega_{i}^{~X} \wedge \theta^i = d \theta^X + \omega_{YX} \wedge \theta^Y = d \theta^X - \omega \wedge \theta^Y = 0 \nonumber \\
d \theta^Y + \omega_{i}^{~Y} \wedge \theta^i = d \theta^Y + \omega_{XY} \wedge \theta^X = d \theta^Y + \omega \wedge \theta^X = 0 
\end{eqnarray}
Now we compute,
\begin{eqnarray}
d \theta^X & = & \frac{\partial}{\partial Y} \left( \frac{1}{X^2 +Y^2 + 1} \right) dY \wedge dX \nonumber \\
& = & +\frac{2 Y}{(X^2 + Y^2 + 1)^2} dX \wedge dY
\end{eqnarray}
and likewise find, $d \theta^Y = - (2 Y)/(X^2 + Y^2 + 1)^2 dX \wedge dY$. Then the equations above, $d \theta^X - \omega \wedge \theta^Y = 0$ and $\theta^Y + \omega \wedge \theta^X = 0$ imply,
\begin{eqnarray}
\omega & = & \frac{2 Y}{X^2 + Y^2 + 1} dX + h(X,Y) dY \nonumber \\
\omega & = & - \frac{2 X}{X^2 + Y^2 + 1} dY + h'(X,Y) dX
\end{eqnarray}
for some functions $h, h'$. Compatiblity of these requires,
\begin{eqnarray}
\omega & = & \frac{2}{X^2 + Y^2 + 1} \left( Y dX - X dY \right)
\end{eqnarray}


\end{document}

