\subsection{Chapter 5}

\begin{p}{Show that any 2-form $F$ on $\R\times S$ can be uniquely expressed as $B+E\wedge dt$ in
such a way that for any local coordinates $x^i$ on $S$ we have
$E=E_i dx^i$ and $B=\frac{1}{2}B_{ij} dx^i\wedge dx^j$.}
\end{p}

For some patch $U\subset S$, we can use the local coordinates $x^i$ along with $t\in\R$ to define an
element of $\R^{n+1}$. Any 2-form $F$ can be written as $F=F_{\alpha\beta} dx^\alpha\wedge dx^\beta$ where the summation
runs over $0\dots n$ and $x^0=t$. We can also express this as $F=F_{0i}dt\wedge dx^i+F_{i0}dx^i\wedge dt+F_{ij}dx^i\wedge dx^j$.
But then by
antisymmetry of the wedge product, we might as well have $F_{i0}=-F_{0i}$. Then defining $E_i=2F_{0i}$ we have
$F=E_i dx^i\wedge dt+F_{ij}dx^i\wedge dx^j$. With $B_{ij}=2F_{ij}$ we obtain $F=E\wedge dt+B$. This decomposition is unique
because if it were true using different functions $E'_i$ and $B'_{ij}$, then their difference would be zero.

\begin{p}{Show that for any form $\om$ on $\R\times S$ there is a unique way to write $d\om=dt\wedge\partial_t\om+d_S\om$
such that for any local coordinates $x^i$ on $S$, writing $t=x^0$, we have $d_S\om=\partial_i \om_I dx^i\wedge dx^I$ and
$dt\wedge \partial_t\om=\partial_0\om_I dx^0\wedge dx^I$.}
\end{p}

First write $\om=\om_I dx^I$ for $I$ a multi-index of length $p$ (i.e. a $p$-tuple whose entries range over $0\dots n$, where
$n$ is the dimension of $S$). Then $d_S\om=\partial_i \om_I dx^i\wedge dx^I$ and $\partial_t\om=\partial_0\om_I$. Again, uniqueness
follows from linearity: if there were two possibilities, their difference would be zero by the linearity of $d$, implying their equality.

\begin{p}{Use the nondegeneracy of the metric to show that the map from $V$ to $V^*$ given by $v\mapsto g(v,\cdot)$ is
an isomorphism, that is, 1-to-1 and onto.}\end{p}

Nondegeneracy of the metric means that if $g(v,w)=0$ for all $w$, then $v=0$. Now, for the map to be an isomorphism, we have to
check the two conditions. First 1-to-1: different inputs result in different outputs. So we check if the difference of the
outputs is zero: $g(v,\cdot)-g(w,\cdot)=g(v-w,\cdot)\neq 0$ unless $v=w$. Next is the onto condition: is every element of $V^*$ the image
of some $v\in V$? Yes, consider $\om\in V^*$ and consider the action on a basis: $\om_\mu=\om(e_\mu)$. Setting $\om_\mu=g(v,e_\mu)$,
we have to solve for $v$. Expanding $v=v^\mu e_\mu$ we obtain $\om_\mu=v^\nu g(e_\nu,e_\mu)=v^\nu g_{\nu,\mu}$. But
the nondegeneracy of $g$ implies that $g_{\nu,\mu}$ is invertible. Suppose we have $g(v,w)=g_{\mu,\nu}v^\mu w^\nu=0$ for all $w$. Then
$g_{\mu,\nu}v^\mu=0$ for all $\nu$. But this only occurs when $v^\mu=0$ for all $\mu$, meaning that $g_{\mu,\nu}$ never annihilates
an input vector; it has full rank. Thus, it's invertible.  So we can solve the equation for $v$.\\

There's a more direct proof for the vector spaces we are considering here: by exercise 28, we know the dual space
has the same dimension as the original space. Thus, nothing in the dual can be outside the image of the original space, since if
were, it would consititute a new basis element.

\begin{p}{Let $v=v^\mu e_\mu$ be a vector field on a chart. Show that the corresponding 1-form $g(v,\cdot)$ is equal to $v_\nu f^\nu$, where $f^\nu$ is the dual basis of 1-forms and $v_\nu=g_{\mu,\nu}v^\mu$.}\end{p}

We can follow the lines of the previous exercise to derive the form, or simply show that the equation is correct. Choosing an
arbitrary vector field $w$, $g(v,w)=g_{\mu,\nu}v^\mu w^\nu$ which is equal to $v_\nu f^\nu(w)=v_\nu w^\nu$.

\begin{p}{Let $\om=\om_\mu f^\mu$ be a 1-form on a chart. Show that the corresponding vector field is equal to
$\om^\nu e_\nu$ where $\om^\nu=g^{\mu,\nu}\om_\mu$.}\end{p}

For arbitrary $v$, $\om(v)=\om_\mu f^\mu(v)=\om_\mu v^\mu$. But there should be a $w$ such that $g(w,v)=\om_\mu v^\mu$.
Using the series expression for $v$ and $w$, we obtain $\om_\nu v^\nu=g_{\mu,\nu}w^\mu v^\nu$. This should hold for arbitrary $v$,
so $\om_\nu=g_{\mu,\nu}w^\mu$, or $w^\mu=g^{\mu,\nu}\om_\nu$.

\begin{p}{Let $\eta$ be the Minkowski metric on $\R^4$. Show that its components in the standard basis
are $\eta_{\mu,\nu}=\delta_{\mu,\nu}(1-2\delta_{\mu,0})$.}\end{p}

The Minkowski metric is defined for two vectors $v$ and $w$ as $\eta(v,w)=-v^0w^0+v^1w^1+v^2w^2+v^3w^3$. The
form of $\eta_{\mu,\nu}$ follows immediately.

\begin{p}{Show that $g^\mu_\nu=\delta^\mu_\nu$.}\end{p}

If we start with $g_{\lambda,\nu}$ we can raise the first index with the metric, obtaining $g^\mu_\nu=g^{\mu,\lambda}g_{\lambda,\nu}$.
But $g^{\lambda,\nu}$ is the matrix inverse of $g_{\lambda,\nu}$, so the result of the product is the Kronecker delta.

\begin{p}{Show that the inner product of $p$-forms is nondegenerate by supposing that $(e^1,\dots,e^n)$ is any
orthonormal basis of 1-forms in some chart, with $g(e^i,e^i)=\epsilon(i)$ for $\epsilon(i)=\pm 1$. Show the $p$-fold
wedge products $e^{i_1}\wedge\dots\wedge e^{i_p}$ form an orthonormal basis of $p$-forms with
$\langle e^{i_1} \wedge\dots\wedge e^{i_p},e^{i_1}\wedge\dots\wedge e^{i_p}\rangle=\epsilon(i_1)\cdots\epsilon(i_p)$.}\end{p}

From exercise 52 we already know that the nondegeneracy of $g$ is equivalent to the invertibility of the matrix of $g$ applied to basis
elements. The same will be true here.  First recall that the terms in
the $p$-fold wedge products must be distinct so that the $p$-form doesn't vanish.
Suppose we have two basis $p$-forms in which $m$ basis 1-forms
appear in both. Reordering so that these come first, the matrix of $g(e^{i_j},e^{k_\ell})$ will be block diagonal with the first
$m\times m$ block equal to the identity matrix, and zero everywhere else. This matrix obviously has zero determinant, thus
the basis $p$-forms with any distinct elements are orthogonal. Otherwise, the determinant yields the product of the individual
normalizations, establishing the last equation posed in the problem. This in turn means that the metric matrix
for the basis $p$-forms is diagonal with entries $\pm 1$, whence it has full rank and is therefore nondegenerate.

\begin{p}{Let $E=E_j dx^j$ be a 1-form on $\R^3$ with its Euclidean metric. Show that $\langle E,E\rangle=E_x^2+E_y^2+E_z^2$. Similarly, let $B=B_x dy\wedge dz+B_y dz\wedge dx+B_z dx\wedge dy$ be a 2-form.
Show that $\langle B,B\rangle = B_x^2+B_y^2+B_z^2$.}\end{p}

$\langle E,E\rangle=\delta^{i,j}E_i E_j=E_x^2+E_y^2+E_z^2$.  $\langle B,B\rangle=\langle B_x dy\wedge dz+B_y dz\wedge dx+B_z dx\wedge dy,
B_x dy\wedge dz+B_y dz\wedge dx+B_z dx\wedge dy\rangle= B_x^2+B_y^2+B_z^2$ since the different wedge products are orthonormal.\\

\begin{p}{In $\R^4$ let $F$ be the 2-form given by $F=B+E\wedge dt$, where $E$ and $B$ are given by the formulas
in the previous exercise. Using the Minkowski metric, calculate $-\frac{1}{2}\langle F,F\rangle$.}\end{p}

$\langle F,F\rangle=\langle B+E\wedge dt,B+E\wedge dt\rangle=\langle B,B\rangle+\langle B,E\wedge dt\rangle+\langle E\wedge dt,B\rangle
+\langle E\wedge dt,E\wedge dt\rangle$. The middle two terms are zero by orthogonality of different wedge products, leaving only the
last term to deal with. $\langle E\wedge dt,E\wedge dt\rangle={\rm det}\left(\begin{array}{cc}\langle E,E\rangle & \langle E,dt\rangle\\
\langle dt,E\rangle & \langle dt,dt\rangle\end{array}\right)={\rm det}\left(\begin{array}{cc}\langle E,E\rangle & 0\\
0 &-1\end{array}\right)=-\langle E,E\rangle$. Hence $-\frac{1}{2}\langle F,F\rangle=\frac{1}{2}(\langle E,E\rangle-\langle B,B\rangle)$,
which is the Lagrangian of the source(charge)-free field.\\

\begin{p}{Show that any even permutation of a given basis has the same orientation, while any odd permutation has the
opposite orientation.}
\end{p}

To check this we must examine the determinant. Since the determinant of a product is the product of determinants, we can
break up the permutation of basis elements into a product of pairwise swaps, whose individual determinant is minus one. Thus,
if the permutation has an even number of swaps, the new basis has the same orientation, while an odd number of swaps
implies the determinant will equal minus one.

\begin{p}{Let $M$ be an oriented manifold. Show that we can cover $M$ with oriented charts $\phi_\alpha:U_\alpha\rightarrow\rn$, that is,
charts such that the basis $dx^\mu$ of cotangent vectors on $\rn$, pulled back to $U^\alpha$ by $\phi_\alpha$, is positively
oriented.}
\end{p}

The point here is that the volume form on $\rn$ can be pulled back to $U_\alpha$ by $\phi_\alpha$.
Since $M$ is orientable, we can define a volume form $\om$ on all of $M$ to specify the standard orientation. In the chart
$U_\alpha$ we can also define a volume form by pulling back the standard volume form $\bigwedge_\mu dx^\mu$ from $\rn$.
If this specifies a different orientation, just reorder the basis of $\rn$ so as to have the same orientation as given by $\om$.
Since $\om$ exists in all charts $U_\alpha$, we can always orient these appropriately.

\begin{p}{Given a diffeomorphism $\phi:M\rightarrow N$ from one oriented manifold to another, we say that $\phi$ is
orientation-preserving if the pullback of any right-handed basis (standard orientation) of a cotangent space in $N$ is a
right-handed basis of a cotangent space in $M$. Show that if we can cover $M$ with charts
such that the transition functions $\varphi_\alpha\circ \varphi_\beta^{-1}$ are orientation-preserving, we
can make $M$ into an oriented manifold by using the charts to transfer the
standard orientation on $\rn$ to an orientation on $M$.}
\end{p}

In each chart pull the standard volume form on $\rn$ back to $M$. Since
any two charts are connected by an orientation-preserving transition function, the orientation is
the same in any chart and the manifold is oriented everywhere.\\

\begin{p}{Let $M$ be an oriented $n$-dimensional semi-Riemannian manifold and let $\{e_\mu\}$ be an oriented orthonormal
basis of cotangent vectors at some point $p\in M$. Show that $e_1\wedge\dots\wedge e_n={\rm vol}_p$ where \emph{vol} is
the volume form associated to the metric on $M$, and \emph{vol}$_p$ is its value at $p$.}
\end{p}

The oriented basis in question can be generated from the standard basis by applying a transformation $T$. Since
the input and output are orthonormal bases, $T$ is an orthogonal matrix, implying det$T=\pm 1$.
Because it's orientation preserving, det$T=1$. Since in transforming $e_1\wedge\dots\wedge e_n$ into the standard
basis at $p$, we pick up a factor equal to the determinant of $T$,  $e_1\wedge\dots\wedge e_n={\rm vol}_p$.

\begin{p}{Show that if we define the Hodge star operator in a chart using the formula $\star(e^{i_1}\wedge \cdots \wedge
e^{i_p})=\pm e^{i_{p+1}}\wedge \cdots \wedge e^{i_n}$, where the second set of indices consists of those not contained in the
first and the $\pm$ is given by ${\rm sign}(i_1,\dots,i_n)\epsilon(i_1)\cdots \epsilon(i_p)$, it satisfies the property $\om\wedge\star\mu=
\langle\om,\mu\rangle{\rm vol}$.}
\end{p}

First expand $\om$ and $\mu$: $\om=\om_\alpha e^{\alpha_1}\wedge\cdots\wedge e^{\alpha_p}$,
$\mu=\mu_\beta e^{\beta_1}\wedge\cdots\wedge e^{\beta_p}$. Now compute their inner product:
$\langle \om,\mu\rangle=\om_\alpha\mu_\beta\langle e^{\alpha_1}\wedge\cdots\wedge e^{\alpha_p},
e^{\beta_1}\wedge\cdots\wedge e^{\beta_p}\rangle=\om_\alpha\mu_\beta\delta^{\alpha,\beta}\epsilon(\alpha_1)\cdots\epsilon(\alpha_p)$.
Now $\star\mu=\pm\mu_\alpha e^{\alpha_{p+1}}\wedge\cdots\wedge
e^{\alpha_n}$, and taking the wedge product with $\om$, one obtains $\om\wedge\star\mu=\pm\om_\alpha\mu_\beta\delta^{\alpha,\beta}
e^{\alpha_1}\wedge\cdots e^{\alpha_n}$. Finally, $e^{\alpha_1}\wedge\cdots e^{\alpha_n}={\rm sign}(\alpha_1,\dots,\alpha_n){\rm vol}$,
so we obtain the desired result.

\begin{p}{Calculate $\star d\om$ when $\om$ is a 1-form on $\R^3$.}\end{p}

$\star d\om$ is essentially the curl of $\om$.
First expand $\om$ in an orthonormal basis $e^j$: $\om=\om_je^j$. $d\om=\partial_k\om_j dx^k\wedge dx^j$. Finally, $\star d\om=
\partial_j\om_k\, {\rm sign}(j,k,\ell)dx^\ell=\partial_j\om_k \epsilon^{jk}_\ell dx^\ell.$ Note that $\epsilon(i)=1$ for all $i$ in this case.

\begin{p}{Calculate $\star d\,{\star}\om$ when $\om$ is a 1-form on $\R^3$.}\end{p}

This gives the divergence of $\om$. Expanding $\om$ in a basis as above, we obtain $\star\om=\om_j\epsilon^j_{k\ell}dx^k\wedge dx^\ell$.
Applying $d$ gives $d\star\om=\partial_m\om_j\epsilon^j_{k\ell}dx^m\wedge dx^k\wedge dx^\ell$. Another $\star$ yields the final
result: $\star d\star\om=\partial_m\om_j\epsilon^j_{k\ell}\epsilon^{mk\ell}$ and since $\epsilon^j_{k\ell}\epsilon^{mk\ell}=\delta^{jm}$, we
have $\star d\star\om=\partial_j\om_k\delta^{jk}$, the divergence.

\begin{p}{Give $\R^4$ the Minkowski metric and the orientation in which
$(dt,dx,dy,dz)$ is positively oriented. Calculate the Holdge star operator on all wedge products of
$dx^\mu$'s Show that on $p$-forms $\star^2=(-1)^{p(4{-}p)+1}$.}
\end{p}

First we have the definition vol$=dt\wedge dx\wedge dy\wedge dz$. Using this we can make a table
of the various starred wedge products.

\begin{tabular}{c|c}
$\omega$ & $\star\omega$\\\hline
$dt$ & $-dx\wedge dy\wedge dz$\\
$dx$ & $-dt\wedge dy\wedge dz$ \\
$dy$ & $dt\wedge dx\wedge dz$\\
$dz$ & $-dt\wedge dx\wedge dy$\\
$dt\wedge dx$ & $-dy\wedge dz$\\
$dt\wedge dy$ & $dx\wedge dz$ \\
$dt\wedge dz$ &$-dx\wedge dy$\\
$dx \wedge dy$ & $dt\wedge dz$ \\
$dx \wedge dz$ & $-dt\wedge dy$\\
$dy \wedge dz$ & $dt\wedge dx$\\
$dt\wedge dx\wedge dy$ & $-dz$ \\
$dt\wedge dx\wedge dz$ & $dy$\\
$dt\wedge dy\wedge dz$ & $-dx$\\
$dx\wedge dy\wedge dz$ & $-dt$
\end{tabular}\\

From the table we immediately see that the given condition is fulfilled.

\begin{p}{Let $M$ be an oriented semi-Riemannian manifold of dimension
$n$ and signature $(s,n{-}s)$. Show that on $p$-forms $\star^2=(-1)^{p(n-p)+s}$.}
\end{p}

$\om\wedge \star \om=\langle \om,\om\rangle {\rm vol}$ and $\star\om \wedge \star^2\om=\langle \star\om,
\star\om\rangle{\rm vol}$, so $\star\om \wedge \star^2\om=(-1)^{p(n{-}p)}\star^2\!\om\wedge \star\om=
\frac{\langle \star\om, \star\om\rangle}{\langle \om,\om\rangle}\,\om\wedge\star\om$. Hence
$\star^2=(-1)^{p(n{-}p)}\frac{\langle \star\om, \star\om\rangle}{\langle \om,\om\rangle}$.
To evaluate this expression,
let $\om$ be a basis $p$-form $e^{i_1}\wedge\dots\wedge e^{i_p}$.
Setting $\epsilon(\mu)=\langle e^\mu,e^\mu\rangle$, we have
$\langle\om,\om\rangle=\prod_{j=1}^p \epsilon(i_j)$. On the other hand,
$\star\om=\pm e^{i_{p+1}}\wedge\dots\wedge e^{i_n}$, so $\langle \star\om,\star\om\rangle=
\prod_{j=p+1}^n \epsilon(i_{p+j})$. Since all the terms in the denominator are $\pm 1$, we might as
well multiply by $\langle\om,\om\rangle$ instead of dividing. We then obtain
$\star^2=(-1)^{p(n{-}p)}\prod_{j=1}^n \epsilon(i_j)$. The latter term equals $(-1)^s$ as $s$ is the
number of basis elements with norm $-1$.

\begin{p}{Let $M$ be an oriented semi-Riemannian manifold of dimension $n$ and signature $(s,n{-}s)$.
Let $e^\mu$ be an orthonormal basis of 1-forms on some chart. Define the
Levi-Civita sympbol for $1\leq i_j\leq n$ by
\[ \epsilon_{i_1\dots i_n}=\left\{
\begin{array}{ll}
{\rm sign}(i_1\dots i_n) & \textrm{all $i_j$ distinct}\\0 &{\rm otherwise}
\end{array}
\right. \]}
Show that for any $p$-form $\om=\frac{1}{p!}\om_{i_1\dots i_p}e^{i_1}\wedge\dots\wedge e^{i_p}$ we
have $(\star\om)_{j_1\dots j_{n{-}p}}=\frac{1}{p!}\epsilon^{i_1\dots i_p}_{j_1\dots j_{n{-}p}}\om_{i_1\dots i_p}$.
\end{p}

Using the results of exercise 64, we have $\star(e^{i_1}\wedge \cdots \wedge
e^{i_p})=\epsilon(i_1)\cdots \epsilon(i_p)\epsilon^{i_1\dots i_p}_{i_{p+1}\dots i_{n}}e^{i_{p+1}}\wedge \cdots \wedge e^{i_n}$
(no sum). Now $(\star\om)_{j_1\dots j_{n{-}p}}=\epsilon(j_1)\cdots \epsilon(j_{n{-}p})\langle e^{j_1}\wedge\dots \wedge e^{j_{n{-}p}},\star \om\rangle$, so $(\star\om)_{j_1\dots j_{n{-}p}}=\frac{(-1)^s}{p!}\epsilon^{i_1\dots i_p}_{j_1\dots j_{n{-}p}}\om_{i_1\dots i_p}$.

\begin{p} {Show that the Maxwell equations $\nabla\cdot \vec{E}=\rho, \nabla\times \vec{B}-\frac{\partial \vec{E}}{\partial t}=\vec{j}$ can be rewritten as $\star_S\, d_S \star_S\! E=\rho$ and $-\partial_t E+\star_S \,d_S\star_S\! B=j$.}
\end{p}

Start from $E=E_j dx^j$. For the remainder of the exercise, $\star$ means $\star_S$. So $\star E=\frac{1}{2}E_j \epsilon^{j}_{k\ell}dx^k\wedge dx^\ell$. Then $d\star\!E=\frac{1}{2}\epsilon^j_{k\ell}\partial_m E_j dx^m\wedge dx^k\wedge dx^\ell$, and finally $\star d\star\! E=\frac{1}{2}\epsilon^{j}_{k\ell}\epsilon^{mk\ell}\partial_m E_j=\partial_j E_j=\nabla\cdot\vec{E}$. Meanwhile $B=\frac{1}{2}\epsilon^j_{k\ell} B_j dx^k\wedge dx^\ell$. Thus $\star B=\frac{1}{2}\epsilon^j_{k\ell}\epsilon^{k\ell}_m B_j dx^m=B_j dx^j$. Then $d\star\!B=\partial_k B_j dx^k\wedge dx^j$
and $\star d\star\!B=\epsilon^{kj}_\ell \partial_k B_j dx^\ell$, the components of which are simply
$\nabla\times \vec{B}$.\\

\begin{p}
{Show that on a  semi-Riemannian manifold $M$ which can
be decomposed into $\R\times S$, where $S$ is space, the general Maxwell
equations $dF=0$ and $\star d\star F=J$ can be transformed into their ``usual''
appearance $d_S E=0, \partial_t B+d_SE=0, \star_Sd_S\star_SE=\rho,$ and
$\star_Sd_S\star_SB-\partial_tE=j$.}
\end{p}

Given the decomposition of $M$, we write $F=B+E\wedge dt$ where $B$ is the
portion of $F$ defined only on $S$. Similarly $J=j-\rho dt$.
The first equation then reads $dB+dE\wedge dt=0$. Observe that
$dB=d_SB+\partial_t B\wedge dt$
while $dE\wedge dt=\partial_t E dt\wedge dt+d_S E\wedge dt$. Thus we have
$d_S B=0$ and $\partial_t B+d_S E=0$; the first equation deals with three forms
only on space, while the second involves three forms on space and time.
Assume further that
the metric can be decomposed into $g=-dt^2+{}^3g$
{\small (can this always be done when the underlying space is a product?
The tangent space can be decomposed, it seems clear,
from the picture of vectors as arrows pointing tangentially to the surface. In the case of
a product space, there are arrows for each manifold, and we take their formal product
to get the tangent to the total manifold. But from the point of view of vectors
as derivatives? I think it might be possible by showing that the value of the derivative
is equal to making changes on each submanifold separately and then adding them. In
other words, we have a curve $\gamma(t)$ on $M$, but since $M=\R\times S$, we
can write this as $(\gamma_1(t),\gamma_2(t))$. Now $\gamma'(t)[f]=\frac{\rm d}{{\rm d}t}f(\gamma(t))=\frac{\rm d}{{\rm d}t}f(\gamma_1(t),\gamma_2(t))=\frac{\partial f}{\partial \gamma_1}\frac{\partial\gamma_1}{\partial t}+\frac{\partial f}{\partial \gamma_2}\frac{\partial\gamma_2}{\partial t}=\gamma_1'(t)[f]|_{\gamma_2(t)}+\gamma_2'[f]|_{\gamma_1(t)}
\equiv(\gamma_1'(t),\gamma_2'(t))[f].$ That's
what we do in $\R^n$, after all. I guess it's obvious once you map open sets of
the manifold to $\R^n$. But this doesn't mean the metric has to be block diagonal and respect the split; after all that \emph{doesn't} always happen in $\R^n$.)}
Now let $\star_S$ be the Hodge dual on (differential forms on) $S$. Since $E$ is a one form
on $S$, $\star E\wedge dt=\star_S E$ and similarly since $B$ is a two form on $S$,
$\star B=-\star_S B\wedge dt$ (using the chart from 67).   So $\star F=\star_S E-\star_S B\wedge dt$ and then $d\star F=d_S\star_S E+\star_S(\partial_t E)\wedge dt-d_S\star_S B\wedge dt$. Note that the first term is a three form on space, so $\star d_S\star_S E$ must be a one form on time, $\star d_S\star_S E=-(\star_S d_S\star_S E) dt$, where
the minus sign can be determined again by the chart from 67. The second term is a three form on space and time, and due to the ordering we'll have $\star(\star_S(\partial_t E)\wedge dt)=-\partial_t E$. The last term is also a three form on space and time and
we end up with $\star(d_S\star_S B\wedge dt)=-\star_S d_S\star_S B$. Thus $\star_S d_S\star_S E=\rho$ and $-\partial_t E+\star_S d_S\star_SB=j$.

\begin{p}{Show that in a Riemannian 4-dimensional manifold any 2-form $F$ can be written as a sum of self-dual and
anti-self-dual parts, $F=F_++F_-$ for $\star F_\pm=\pm F_\pm$, if we take $F_\pm=(F\pm\star F)/2$.}
\end{p}

Clearly $F=F_++F_-$. And
$\star F_\pm=(\star F\pm\star^2 F)/2=(\pm F+\star F)/2=\pm(F\pm\star F)/2=\pm F_\pm$.

\begin{p}{Show that in a Lorentizan 4-manifold any 2-form $F$ can be written as a sum of self-dual and
anti-self-dual parts, $F=F_++F_-$ for $\star F_\pm=\pm i F_\pm$.}
\end{p}

Take $F_\pm=(F\mp \star i F)/2$. Then $\star F_\pm=(\star F\mp\star^2 i F)/2=(\pm i F+\star F)/2=\pm i(F\mp \star  i F)/2=\pm i F_\pm$.

\begin{p}{Show that the equations $\star_S E=iB$ and $\star_S B=-i E$ are
equivalent and that they both hold if at every time $t$ we have $E=E_i dx^i$ and $B=-(i/2)\varepsilon^j_{\phantom{j}k\ell}E_j dx^k\wedge dx^\ell$.}
\end{p}

Applying $\star_S$ turns one equation into the other, to they are equivalent.
From exercise 69 we know that $\star_S E=(1/2)\varepsilon^j_{\phantom{j}k\ell}E_j dx^k\wedge dx^\ell$ which is clearly equal to $iB$.

\begin{p}{Show that the second Maxwell equation $\partial_t B+d_SE=0$ leads to ${}^3k\wedge E=k_0 B$.}
\end{p}

Start with $d_S E=d_S(e^{ik_\mu x^\mu}E_j dx^j)=\partial_k(e^{ik_\mu x^\mu}E_j)dx^j\wedge dx^k=i k_k E_jdx^j\wedge dx^k=-i{}^3k\wedge E$. Meanwhile
$\partial_t B=ik_0 B$, so $\partial_t B+d_SE=i(k_0 B-{}^3k\wedge E)=0$, the desired
result.

\begin{p}%76
{Show that ${}^3k\wedge E=-ik_0 \star_s E$ implies $k_\mu k^\mu=0$.}
\end{p}

Start from $\langle {}^3k\wedge E,{}^3k\wedge E\rangle=k_0^2\langle\star_s E,\star_s E\rangle$.
Note that the inner product should be antilinear in one of its inputs. Using the coordinate expressions for $E$
and ${}^3k$ we have $\langle {}^3k\wedge E,{}^3k\wedge E\rangle=k_\ell k_{\ell'}^*E_j E_{j'}^*\langle dx^\ell\wedge dx^j,dx^{\ell'}\wedge dx^{j'}\rangle=k_\ell k_{\ell'}^*E_j E_{j'}^*(\delta^{j j'}\delta^{\ell\ell'}-\delta^{j'\ell}\delta^{j\ell'})=\langle {}^3k,{}^3k\rangle\,\langle E,E\rangle-|\langle {}^3k,E\rangle|^2=\langle {}^3k,{}^3k\rangle\,\langle E,E\rangle$, since $E$ is orthogonal to ${}^3k$.
On the other hand, $k_0^2\langle\star_s E,\star_s E\rangle=\frac{k_0^2}{4}\varepsilon^j_{k\ell}\varepsilon^{j'}_{k'\ell'}E_jE_{j'}\langle dx^k\wedge dx^\ell,dx^{k'}\wedge dx^{\ell'}\rangle=\frac{k_0^2}{4}\varepsilon^j_{k\ell}\varepsilon^{j'}_{k'\ell'}E_jE_{j'}(\delta^{kk'}\delta^{\ell \ell'}-\delta^{k\ell'}\delta^{k'\ell})=k_0^2\langle E,E\rangle$.
Thus, $\langle {}^3k,{}^3k\rangle=k_0^2$. Using the metric, we have $k_0^2=-k_0k^0$, and the
other components are unchanged, so $k_\mu k^\mu=0$.

\begin{p}%77
{Show $\vec{E}=(0,e^{i(t-x)},-ie^{i(t-x)})$, $\vec{B}=i\vec{E}$ satisfy the vacuum Maxwell equations. [corrected from the book]}
\end{p}

Since $\vec{\nabla}\cdot \vec{E}=0$ (by inspection),
$\vec{\nabla}\cdot \vec{B}=0$, too. Now $\vec{\nabla}\times \vec{E}=\vec{E}$ and $\partial_t \vec{B}=i\partial_t \vec{E}=-\vec{E}$, so $\vec{\nabla}\times \vec{E}+\partial_t \vec{B}=0$. Finally,
$ \vec{\nabla}\times \vec{B}=i\vec{E}$, so $\vec{\nabla}\times \vec{B}-\partial_t \vec{E}=0$.

\begin{p}%{78}
{Prove that all self-dual and anti-self-dual plane wave solutions are left and right circularly polarized, respectively.}
\end{p}

Without loss of generality we can take ${}^3k$ to point in the $x$ direction. Then the self-dual case is worked
out in detail in the book. Left circular polarization can be recognized from the form of $E$, namely that the
polarization vector is proportional to $(1,-i)$; right circular polarization is given by $(1,i)$.
In the anti-self-dual case $\star_S E=-iB$ or, equivalently,
$\star_SB=iE$. The first Maxwell equation $d_SB=0$ reads as before, $B\wedge {}^3k=0$, so
$\langle E,{}^3k\rangle=0$ still holds. The second equation, $\partial_t B+d_SE=0$, leads to ${}^3k\wedge E=k_0B$ as before. Rewriting this in terms of $E$ we obtain ${}^3k\wedge E=ik_0\star_SE$, which
again leads to $\langle {}^3k\wedge E,{}^3k\wedge E\rangle=k_0^2\langle\star_s E,\star_s E\rangle$, as
in exercise 76. Thus we know that $k$ must be lightlike, so we can assume without loss of generality
that $k=dt-dx$. $E$ must be orthogonal to ${}^3k$, so $E=a dy+bdz$. We now use the second
Maxwell equation to determine $a$ and $b$. ${}^3k\wedge E=-(a dx\wedge dy+b dx\wedge dz)$,
while $ik_0\star_SE=i(a dz\wedge dx+b dx\wedge dy)$, so $b=ia$, which is the condition for right circular
polarization.

\begin{p}%{79}
{Let $P:\R^4\rightarrow\R^4$ be the parity transformation $P(t,x,y,z)=(t,-x,-y,-z)$. Show that if $F$ is a self-dual solution of Maxwell's equations, the pullback $P^*F$ is an anti-self-dual solution, and vice versa.}
\end{p}

We don't really need to show both, since the vice versa part follows from $P^*P^*F=F$. To see that $P^*F$ takes a self-dual solution to an anti-self-dual solution, note that $P^*E=-E$, while $P^*B=B$. (This is why
$B$ is called an axial vector sometimes.) If $F=dE\wedge dt+B$ was self-dual to begin with, meaning $\star_SE=iB$,  the new $F'=dE'\wedge dt'+B'=-dE\wedge dt+B$ is anti-self-dual: $\star_SE'=-\star_SE=-iB=-iB'$.


\subsection{Graph embedding}\label{subsec:graph_embedding}

\begin{definition}\label{def:quiver_geometric_realization}
  Let \( Q = (V, A, h, t) \) be a \hyperref[def:quiver]{quiver}. Our goal is to construct a \hyperref[def:topological_space]{topological space} that translates the connectivity properties of \( Q \) into their topological equivalents.

  Consider the \hyperref[def:topological_sum]{topological sum}
  \begin{equation*}
    \mscrS \coloneqq \parens[\Bigg]{ \coprod_{a \in A} [0, 1] } \amalg \parens[\Bigg]{ \coprod_{\deg(v) = 0} \set{ v } }.
  \end{equation*}

  The space \( \mscrS \) consists of disjoint unit intervals, one for each arc, and of disjoint points, one for each \hyperref[def:hypergraph/degree]{isolated vertex}.

  We now want to glue common endpoints of arcs in \( \Sigma \). We define the following function:
  \begin{equation}\label{eq:def:quiver_geometric_realization}
    \begin{aligned}
      &R: V \cup A \to \pow(\mscrS), \\
      &R(v) \coloneqq \begin{cases}
        \set[\Big]{ (v, v) },                                                           &\deg(v) = 0 \\
        \set[\Big]{ (0, a) \given h(a) = v } \cup \set[\Big]{ (1, a) \given t(a) = v }, &\deg(v) > 0,
      \end{cases} \\
      &R(a) \coloneqq \set[\Big]{ \set{ (x, a) } \given 0 \leq x \leq 1 }.
    \end{aligned}
  \end{equation}

  The family
  \begin{equation*}
    \mscrX \coloneqq \set[\Big]{ R(v) \given* v \in V } \cup \set[\Big]{ \Int R(a) \given* a \in A }.
  \end{equation*}
  is a \hyperref[def:set_partition]{partition} of \( \mscrS \). For each vertex, there is a single point in \( \mscrX \) (which is a set in \( \mscrS \)) and for each arc, the interior of the arc is a subset of \( \mscrX \).

  We can endow the partition \( \mscrX \) it with a \hyperref[def:topological_quotient]{quotient topology} \( \mscrT \). We will call the topological space \( (\mscrX, \mscrT, R) \) endowed with \( R \) the \term{geometric realization} of \( G \).

  \begin{thmenum}
    \thmitem{def:quiver_geometric_realization/undirected} For an \hyperref[def:undirected_multigraph]{undirected multigraph} \( G = (V, E, \mscrE) \), the geometric realization is any of the geometric realizations of its \hyperref[def:multigraph_orientation]{orientations}. This construction is dependent on a choice function, but fortunately all the geometric realizations are homeomorphic as shown in \fullref{thm:undirected_multigraph_geometric_realizations_homeomorphic}. Hence, for a lot of purposes, we can speak of \enquote{the} geometric realization of an undirected multigraph.

    \thmitem{def:quiver_geometric_realization/drawing} We will call any \hyperref[def:global_continuity]{continuous function} with domain \( (\mscrX, \mscrT) \) a \term{graph drawing}. The term \enquote{graph drawing} is not standard terminology, but unfortunately non-injective continuous images of the realization have no established name.

    \thmitem{def:quiver_geometric_realization/embedding} An injective graph drawing is called a \term{graph embedding}.

    Every graph can be embedded into \( \BbbR^3 \) as shown in \fullref{thm:quiver_can_be_embedded_into_r3}.

    \thmitem{def:quiver_geometric_realization/linear} If a graph can be embedded into \( \BbbR \), we say that it is \term{linear}.

    \thmitem{def:quiver_geometric_realization/planar} If a graph can be embedded into \( \BbbR^2 \), we say that it is \term{planar}.
  \end{thmenum}
\end{definition}

\begin{example}\label{ex:def:quiver_geometric_realization}
  We will give a few examples of \hyperref[def:quiver_geometric_realization/undirected]{quiver geometric realizations}.

  \begin{thmenum}
    \thmitem{ex:def:quiver_geometric_realization/edgeless} The \hyperref[def:quiver_geometric_realization/undirected]{geometric realization} of an edgeless quiver is the empty topological space.

    \thmitem{ex:def:quiver_geometric_realization/positive_integers} Consider the reduced positive integer graph \eqref{eq:ex:infinite_integer_graphs/positive}. We start with \( \aleph_0 \) copies of \( [0, 1] \) and glue both ends of each of them except for the first. Thus, we obtain (a space homeomorphic to)
    \begin{equation*}
      \bigcup_{k \geq 0} [k, k + 1] = [0, \infty).
    \end{equation*}

    Therefore, \eqref{eq:ex:infinite_integer_graphs/positive} is a \hyperref[def:quiver_geometric_realization/linear]{linear graph}.

    \thmitem{ex:def:quiver_geometric_realization/k3} The graph with vertices \( V = \set{ a, b, c } \) and arcs \( \set{ \overbrace{a \to b}^{e_1}, \overbrace{b \to c}^{e_2}, \overbrace{c \to a}^{e_3} } \) is more subtle.

    We start with three copies of the interval \( [0, 1] \), depicted in \eqref{eq:ex:def:quiver_geometric_realization/k3/relization} as upward-pointing arrows, and use dashed lines to connect the endpoints that we want to glue together.
    \begin{equation}\label{eq:ex:def:quiver_geometric_realization/k3/relization}
      \begin{aligned}
        \includegraphics[page=1]{output/ex__def__graph_geometric_realization.pdf}
      \end{aligned}
    \end{equation}

    After contracting the dashed lines, we obtain a topological space that can easily be \hyperref[def:quiver_geometric_realization/embedding]{embedded} into \( \BbbR^2 \). An obvious embedding corresponds to \enquote{pulling up} \( e_2 \) and \( e_3 \):
    \begin{equation}\label{eq:ex:def:quiver_geometric_realization/k3/embedding}
      \begin{aligned}
        \includegraphics[page=2]{output/ex__def__graph_geometric_realization.pdf}
      \end{aligned}
    \end{equation}

    This is only one possible embedding of the geometric realization. It is sufficient, however, for proving that the graph is \hyperref[def:quiver_geometric_realization/planar]{planar}. The underlying undirected graph is the \hyperref[ex:complete_graph]{complete graph} \( K_3 \), hence we have shown that \( K_3 \) is also planar.

    \thmitem{ex:def:quiver_geometric_realization/k4} \Cref{fig:ex:complete_graph} shows that the complete graph \( K_4 \) is planar.

    This is not-at-all obvious from its geometric realization, however.
    \begin{equation}\label{eq:ex:def:quiver_geometric_realization/k4/realization}
      \begin{aligned}
        \includegraphics[page=3]{output/ex__def__graph_geometric_realization.pdf}
      \end{aligned}
    \end{equation}

    This example shows that constructing embeddings can be a tedious task.
  \end{thmenum}
\end{example}

\begin{proposition}\label{thm:linear_quiver_equivalence}
  If finite quiver is \hyperref[def:quiver_geometric_realization/linear]{linear}, it has degree at most \( 2 \).
\end{proposition}
\begin{proof}
  Let \( Q = (V, A, t, h) \) be a quiver and let \( (\mscrX, \mscrT, R) \) be its geometric realization. Let \( f: \mscrX \to \BbbR \) be an injective continuous function, i.e. a topological embedding.

  Suppose that the vertex \( v \) has degree larger than \( 2 \). It is sufficient to consider the case where \( a \), \( b \) and \( c \) are distinct arcs incident to \( v \).

  We have
  \begin{equation*}
    R(v) \in R(a) \cap R(b) \cap R(c)
  \end{equation*}
  thus
  \begin{equation*}
    f(R(v)) \in f(R(a)) \cap f(R(b)) \cap f(R(c))
  \end{equation*}

  Since \( R(a) \), \( R(b) \) and \( R(c) \) are \hyperref[def:connected_space]{connected}, so are their images under \( f \). If \( f(R(a)) \) has a point to the right of \( f(R(v)) \), then \( f(R(b)) \) must be left of \( R(v) \) and there remains nowhere to place \( R(c) \).

  Therefore, \( \deg(v) \leq 2 \) and, since \( v \) was arbitrary, \( \deg(Q) \leq 2 \).
\end{proof}

\begin{proposition}\label{thm:quiver_can_be_embedded_into_r3}
  Every finite quiver can be embedded into \( \BbbR^3 \).
\end{proposition}
\begin{proof}
  Let \( Q = (V, A, t, h) \) be a finite quiver of order \( n \). By definition of cardinality, there exists a bijection from \( n \) to \( V \).

  Place the vertices of \( Q \) along the \hyperref[thm:moment_curve]{moment curve} by using \( \gamma(k) \) as the position for the \( k \)-th vertex of \( V \). Then by \fullref{thm:moment_curve}, no four of these points are \hyperref[def:collinear_complanar]{complanar}. Hence, if we connect their vertices using a straight line where there is an arc, no two lines would intersect.

  Therefore, this is an embedding.
\end{proof}

\begin{proposition}\label{thm:def:quiver_geometric_realization/properties}
  Let \( Q = (V, A, h, t) \) be a \hyperref[def:quiver]{quiver} and let \( (\mscrX, \mscrT, R) \) be its \hyperref[def:quiver_geometric_realization]{geometric realization}.

  \begin{thmenum}
    \thmitem{thm:def:quiver_geometric_realization/properties/isolated} A vertex \( v \in V \) is isolated in \( Q \), i.e. has degree zero, if and only if \( R(v) \) is an isolated point of \( \mscrX \).

    \thmitem{thm:def:quiver_geometric_realization/properties/t1} If \( Q \) is \hyperref[def:hypergraph/degree]{locally finite}, then the space \( (\mscrX, \mscrT) \) satisfies the \ref{def:separation_axioms/T1} separation axiom.
  \end{thmenum}
\end{proposition}
\begin{proof}
  \SubProofOf{thm:def:quiver_geometric_realization/properties/isolated} For every isolated vertex \( v \), the point \( R(v) = \set{ (v, v) } \) is isolated by definition.

  Now suppose that \( v \) is not an isolated vertex. Then \( R(v) \) is defined differently in \eqref{eq:def:quiver_geometric_realization}. If there exists an arc \( a \) such that \( v = h(a) \), then \( (a, 0) \in R(v) \) and hence there exists no neighborhood of \( R(v) \) disjoin from \( R(a) \), hence \( R(v) \) is not a disjoint point of \( \mscrX \). The case \( v = t(a) \) is handled analogously.

  \SubProofOf{thm:def:quiver_geometric_realization/properties/t1} Let \( x \in \mscrX \).

  If \( x = \set{ (t, a) } \) for some arc \( a \) and \( 0 < t < 1 \), then \( \set{ x } \) is closed because \( [0, 1] \) satisfies \hyperref[def:separation_axioms/T1]{T1} and hence \( \set{ t } \) is closed in \( [0, 1] \).

  If \( x = R(v) = \set{ (v, v) } \) for some isolated vertex \( v \), then \( R(v) \) is clopen by definition.

  If \( x = R(v) \) for some vertex \( v \) of positive degree, then
  \begin{equation*}
    R(v) = \set[\Big]{ (0, a) \given h(a) = v } \cup \set[\Big]{ (1, a) \given t(a) = v }.
  \end{equation*}

  Both \( \set{ 0 } \) and \( \set{ 1 } \) are closed in \( [0, 1] \), hence \( \set{ 0, 1 } \) is also closed in \( [0, 1] \). Since \( Q \) is locally finite, \( R(v) \) is the union of finitely many closed sets and is thus itself closed.
\end{proof}

\begin{proposition}\label{thm:undirected_multigraph_geometric_realizations_homeomorphic}
  Let \( G = (V, E, \mscrE) \) be an \hyperref[def:undirected_multigraph]{undirected multigraph}.

  Let \( (\mscrX_c, \mscrT_c, R_c) \) and \( (\mscrX_d, \mscrT_d, R_d) \) be geometric realizations corresponding to the \hyperref[def:multigraph_orientation]{orientations} \( O_c(G) \) and \( O_d(G) \) of \( G \).

  Then \( (\mscrX_c, \mscrT_c) \) and \( (\mscrX_c, \mscrT_d) \) are homeomorphic.
\end{proposition}
\begin{proof}
  Let \( h_c \) and \( h_d \) be the head functions from the \hyperref[def:quiver]{quivers} \( O_c(G) \) and \( O_d(G) \).

  Define the function
  \begin{equation*}
    \begin{aligned}
      &f: \mscrX_c \to \mscrX_d \\
      &f(x) \coloneqq \begin{cases}
        \set[\Big]{ (0, e) \given h_d(e) = v } \cup \set[\Big]{ (1, e) \given t_d(e) = v }, &x = R_c(v) \T{and} \deg(v) > 0 \\
        \set[\Big]{ (e, 1 - t) },                                                           &x = \set{ (e, t) } \T{and} h_c(e) \neq h_d(e) \\
        x,                                                                                &\T{otherwise.}
      \end{cases}
    \end{aligned}
  \end{equation*}

  This function \enquote{reverses} the direction of some of the intervals in the construction of the realizations and fixes everything else in place. It is clearly bijective. It is also continuous because it satisfies \fullref{def:global_continuity/closure}. Finally, it is a homeomorphism because the inverse function is defined in the same way by interchanging \( c \) and \( d \).
\end{proof}

\begin{proposition}\label{thm:quiver_geometric_realization_paths}
  Let \( Q = (V, A, h, t) \) be a \hyperref[def:quiver]{quiver} and let \( (\mscrX, \mscrT, R) \) be its \hyperref[def:quiver_geometric_realization]{geometric realization}.

  \begin{thmenum}
    \thmitem{thm:quiver_geometric_realization_paths/quiver_to_realization} If there exists an \hyperref[def:quiver_path/undirected]{undirected path} \( p = (v, e_1, \ldots, e_n) \) from \( s \) to \( f \), then there exists a \hyperref[def:parametric_curve]{continuous path} \( \gamma: [0, 1] \to \mscrX \) from \( R(s)  \) to \( R(f) \).

    \thmitem{thm:quiver_geometric_realization_paths/realization_to_quiver} If \( Q \) is finite and if there exists a simple continuous path \( \gamma: [0, 1] \to \mscrX \) from \( R(s)  \) to \( R(f) \), then there exists a simple undirected path from \( s \) to \( f \).
  \end{thmenum}
\end{proposition}
\begin{proof}
  The case \( s = f \) is trivial, hence we assume that \( s \neq f \).

  \SubProofOf{thm:quiver_geometric_realization_paths/quiver_to_realization} Suppose that \( p \) is an undirected path from \( s \) to \( f \). We will use \hyperref[rem:induction/well_founded]{strong induction} on the length of \( p \) to show that there exists a continuous path between the points \( R(s) \) and \( R(f)  \) of \( \mscrX \).

  Suppose that the statement holds for paths of length smaller than \( n \) and let
  \begin{equation*}
    p = (v, e_1, \ldots, e_{n-1}, e_n)
  \end{equation*}
  be a path of length \( n \) from some vertex \( s \) to \( f \). The inductive hypothesis holds for the initial segment \( (e_1, \ldots, e_{n-1}) \) of \( p \), hence there exists a continuous path \( \gamma: [0, 1] \to \mscrX \) from \( s \) to an endpoint of \( e_{n-1} \).
  \begin{itemize}
    \item If both \( e_{n-1} \) and \( e_n \) are positively oriented, then \( t(e_{n-1}) = h(e_n) \) and thus \( \gamma \) is a continuous path from \( R(s)  \) to \( R(h(e_n)) \). We can then append to \( \gamma \) the continuous path from \( R(h(e_n)) \) to \( R(t(e_n)) = R(f)  \) to obtain a path from \( R(s)  \) to \( R(f)  \).

    \item If \( e_{n-1} \) is positively oriented but \( e_n \) is not, then \( \gamma \) is a continuous path from \( R(s)  \) to \( R(h(e_{n-1})) \). We can then append to \( \gamma \) the paths from \( R(h(e_{n-1})) \) to \( R(t(e_{n-1})) = R(t(e_n)) \) and from \( R(t(e_n)) \) to \( R(h(e_n)) = R(f)  \).

    \item Similarly, if \( e_{n-1} \) is negatively oriented but \( e_n \) is not, then \( \gamma \) is a continuous path from \( R(s)  \) to \( R(t(e_{n-1})) \), and we can append to it the paths from \( R(t(e_{n-1})) \) to \( R(h(e_{n-1})) = R(h(e_n)) \) and from \( R(h(e_n)) \) to \( R(t(e_n)) \).

    \item Finally, if both \( e_{n-1} \) and \( e_n \) are negatively oriented, then \( \gamma \) is a continuous path from \( R(s)  \) to \( R(t(e_{n-1})) \), and we can append to it the paths from \( R(t(e_{n-1})) \) to \( R(h(e_{n-1})) = R(t(e_n)) \) and from \( R(t(e_n)) \) to \( R(h(e_n)) = R(f)  \).
  \end{itemize}

  We have shown that there exists a continuous path from \( R(s) \) to \( R(f)  \).

  \SubProofOf{thm:quiver_geometric_realization_paths/realization_to_quiver} Suppose that \( Q \) is finite and let \( \gamma: [0, 1] \to \mscrX \) be a continuous path from \( R(s) \) to \( R(f) \). We will show that there is an undirected path from \( s \) to \( f \).

  \SubProof*{\( \gamma \) contains no isolated vertices} We have
  \begin{equation}\label{eq:thm:quiver_geometric_realization_paths/full_preimage}
    \gamma^{-1}(\mscrX) = \bigcup_{a \in A} \gamma^{-1}(R(a)) \cup \bigcup_{\mathclap{\deg(v) = 0}} \gamma^{-1}(R(v)).
  \end{equation}

  For each arc \( a \), the set \( R(a) \) is closed as a homeomorphic image of the unit interval \( [0, 1] \). From \fullref{thm:def:quiver_geometric_realization/properties/t1} it follows that \( \set{ R(v) } \) is a closed set for every vertex \( v \in V \). Therefore, \( \gamma^{-1}(\mscrX) \) is a union of disjoint closed sets.

  If we assume that \( \gamma \) passes through an isolated vertex \( v \), then \( \gamma^{-1}(v) \) would be a nonempty closed set. Then \( \img(\gamma) = \set{ v } \) because otherwise \( [0, 1] \) would be the union of finitely many nonempty disjoint closed sets, which would contradict \fullref{def:connected_space/closed_union} because \( [0, 1] \) is \hyperref[def:connected_space]{connected}.

  But \( \gamma \) passes through at least two vertices because \( s \neq t \), and hence it doesn't pass through any isolated vertex.

  \SubProof*{\( \gamma \) contains the entirety of each arc it intersects} Suppose that \( R(a) \cap \img(\gamma) = \varnothing \) for some arc \( a \).

  Let \( l \coloneqq \inf\set{ 0 < t < 1 \given \gamma(t) \in R(a) } \) and \( r \coloneqq \sup\set{ 0 < t < 1 \given \gamma(t) \in R(a) } \). The closed interval \( [l, r] \) is compact, hence \( \gamma([l, r]) \) is a continuous image of a compact set and hence is itself compact.

  Since \( \Int R(a) \) is a subset of \( \gamma([l, r]) \) and since \( \gamma([l, r]) \) is closed, its closure \( R(a) \) is also a subset of \( \gamma([l, r]) \).

  Therefore, if an internal point of an arc belongs to \( \img(\gamma) \), so does the entire arc.

  \SubProof*{\( \gamma \) contains an arc} From \eqref{eq:thm:quiver_geometric_realization_paths/full_preimage} if follows that
  \begin{equation*}
    \gamma^{-1}(\mscrX)
    =
    \bigcup_{a \in A} \gamma^{-1}(R(a))
    \reloset{\ref{thm:topological_closure_operator_axioms/CO3}} =
    \cl\parens*{ \bigcup_{a \in A} \gamma^{-1}(\Int R(a)) }.
  \end{equation*}

  Hence, \( \img(\gamma) \) contains at least one internal point of some arc and thus the entire arc.

  \SubProof*{\( \gamma \) induces an undirected path from \( s \) to \( f \)} We have that \( \gamma(0) = R(s) \). By what we have already shown, \( \img(\gamma) \) contains no isolated points, hence \( \gamma \) contains the set \( R(a) \), where \( a \) is some arc incident to \( s \).

  Suppose that \( h(a) = s \). Then \( p = (s, a) \) is an undirected path from \( s \) to \( t(a) \). If \( t(a) = f \), the proof is finished. Otherwise, define
  \begin{equation*}
    x_0 \coloneqq \sup\set{ x \in [0, 1] \given \gamma(x) = R(t(a)) }
  \end{equation*}
  and
  \begin{equation*}
    \begin{aligned}
      &\delta: [0, 1] \to \mscrX, \\
      &\delta(x) \coloneqq \gamma(x_0 + (1 - x_0) x).
    \end{aligned}
  \end{equation*}

  Then \( \delta \) is a continuous path from \( R(t(a)) \) to \( R(f) \).

  We now proceed by \fullref{thm:bounded_transfinite_induction} bounded by the number of arcs to define an undirected path \( p \) from \( s \) to \( f \). Since the path \( \gamma \) is simple, it does not intersect itself and hence the image of the arc \( a \) at each step will not be in \( \delta \).
\end{proof}

\begin{corollary}\label{thm:quiver_geometric_realization_connectedness}
  A finite \hyperref[def:quiver]{quiver} is \hyperref[def:quiver_connectedness/weak]{weakly connected} if and only if its \hyperref[def:quiver_geometric_realization/undirected]{geometric realization} is \hyperref[def:path_connected]{path connected}.
\end{corollary}
\begin{proof}
  Follows from \fullref{thm:quiver_geometric_realization_paths}.
\end{proof}

\begin{corollary}\label{thm:undirected_multigraph_geometric_realization_connectedness}
  A finite \hyperref[def:undirected_multigraph]{undirected multigraph} is \hyperref[def:undirected_multigraph_connectedness]{connected} if and only if its \hyperref[def:quiver_geometric_realization/undirected]{geometric realization} is a \hyperref[def:path_connected]{path connected}.
\end{corollary}
\begin{proof}
  Follows from \fullref{thm:quiver_geometric_realization_connectedness}.
\end{proof}


\subsection{The Lanczos Tridiagonalization Step}

Golub explains in \cite{golub13} that one of the problems with the
Power Method, is that it does not take advantage of the previously
calculated information. During the iterations of the Power Method, say
until step $k$, we have calculated already the set of vectors
$K(A,q_0,k) = \{A\vec{q_0},A\vec{q_1},\dots,A\vec{q_k}\}$; still, they
are not used
at all when looking for an estimate of the eigenvector. Such
limitation is addressed by the Lanczos Process, named after its
creator in 1950 (\cite{lanczos1950}). The subspace spanned by the
$K(A,q_0,k)$ is called Krylov Subspace of order $k$ \footnote{The
  concept itself of Krylov Spaces is thanks for
  Krylov and dates back to 1931 (see \cite{krylov1931})}, which is why
the Lanczos Process is usually cataloged as a Krylov Subspace
method.  \\

Going back to Lanczos, this subsection will only explain the iterative
step (called Lanczos Tridiagonalization Step). It works as follows; let us
suppose that we have an square symmetric matrix $A^{n \times n}$, and
that we want a few of its biggest eigenvalues (as it is the case in LSI
applications) \footnote{The Lanczos Process can also calculate a few
  of the smallest eigenvalues, but we are not interested in such case
  for LSI applications.}. Each iteration $k$ of the algorithm generates a
tridiagonal matrix $T_k \in \R{k \times
  k}$ \footnote{Matrices with the middle, upper and lower
  diagonals.}, and the whole sequence ${T_k}$ is  progressively
approximating the biggest eigenvalues of the original matrix
$A$. \\

There are several ways of stating the algorithm for the Lanczos
Tridiagonalization Step, the following is taken from Golub \cite{golub13};
though it is not the the most numerically stable. That honor
corresponds to the ones created by Paige
(\cite{paige71},\cite{paige76}); we preferred Golub's one for our
exposition, aiming to have an easier introduction to the procedure:

\begin{algorithm}
  \label{alg:lanczos-step}
  \caption{The Lanczos Tridiagonalization Step}
%
  \setstretch{1.5}
  \SetKwInOut{Input}{Input}
  \SetKwInOut{Output}{Output}
  \DontPrintSemicolon
%
    \Input{A unit vector $\vec{q_1} \in \R{n}$ and a symmetric matrix $A^{n
        \times n}$}
%
    \Output{The sequences $\{\alpha_i\}$, $\{\beta_i\}$ and matrix $Q
      = [ \vec{q_1} | \vec{q_2} | \cdots ]$ }
%
    $k \gets 0, \beta_0 \gets 1, \vec{q_0} \gets 0, r_0 \gets \vec{q_1}$ \;
    \While {$k = 0 \lor \beta_k \ne 0$}
    {
      $\vec{q_{k+1}} \gets \dfrac{\vec{r_k}}{B_k}$ \;
      $k \gets k + 1$ \;
      $\alpha_k \gets \trans{\vec{q_k}}A\vec{q_k}$ \;
      $\vec{r_k} \gets A\vec{q_k} - \alpha_kq_k - \beta_{k-1}\vec{q_{k-1}}$ \;
      $\beta_k \gets \norm{\vec{r_k}}_2$ \;
    }
%
    return $(\{\alpha_i\}, \{\beta_i\},
             Q = [ \vec{q_1} | \vec{q_2} | \cdots ])$ \;
\end{algorithm}
\hfill

The \cref{alg:lanczos-step} is essentially applying Gram-Schmidt process,
but only against the last two vectors. Golub derives the algorithm
from a relation between tridiagonalization, and the QR factorization of
the matrix formed by vectors $K(A,q_0,k)$; see \cite{golub13} for
further details. \\

Golub goes even further in the cited book, and proves the following
properties about \cref{alg:lanczos-step}. We will omit the theorem
statement, and just comment directly its results: \\

\begin{itemize}
  \item The algorithm runs until $k = m = rank(K(A,q_0,k))$. This contrasts
    with the unknown number of steps of the Power Method
    (\cref{alg:power-method}). \\

  \item For $k = 1:m$ we have $AQ_k = Q_kT_k + \vec{r_k}\trans{e_k}$,
    where $Q = [\vec{q_1} | \cdots | \vec{q_k} ]$ has orthonormal
    columns that span the Krylov subspace $K(A,\vec{q_1},k)$, and $e_k
    = I_n(:,k)$ (the $k$ column of the identity matrix). This
    justifies the orthogonalization step of the algorithm (line 6),
    which only considers the last two vectors; whether that is enough to
    guarantee that all the $\vec{q}$\apos{s} will be orthogonal is
    certainly not evident, and gets proved on the same theorem. \\

  \item The matrix $T_k$ has tridiagonal shape, that is: \\
    \[
    \begin{bmatrix}
      \alpha_1 & \beta_1 & \cdots      & 0          \\
      \beta_1  & \ddots  & \ddots      & \vdots     \\
      \vdots   & \ddots  & \ddots      & \beta_{k-1} \\
      0        & \cdots  & \beta_{k-1} & \alpha_k
    \end{bmatrix}
    \]
    \hfill

    This shape allows us to calculate its eigenvalues with much less
    effort than for original matrix $A$ (which was the whole
    motivation on the beginning). There are several options for such
    calculation, but we will consider only the (implicit) QL Algorithm
    (see \cite{dubrulle71}), as that is the one used by Berry for his
    famous routines in the context of LSI (see further
    subsections).
\end{itemize}
\hfill

In addition to the above properties, which kind of guarantee the
``correctness'' of the \cref{alg:lanczos-step} (to some extent); Golub
also cites in \cite{golub13} another theorem that establishes the
approximation quality of matrix $T_k$ as a function of $k$. This is
the result that justifies our original claim that the sequence of
matrices $\{T_k\}$, approximates better the eigenvalues of $A$ as $k$
increases. \\

Finally, Golub adds in \cite{golub13} that not everything is flakes
and honey with this
algorithm; the orthogonality that we expect on vectors \vec{q}\apos{s}
is at jeopardy as $\tilde{\beta_k}$, the numerical approximation of
$\beta_k$, becomes really small; this is because that implies the
cancellation of $\vec{r_k}$). Main credit of this result goes again to
Paige (\cite{paige71},\cite{paige76}), and we will come back to
it on next subsection, when we show the full Lanczos Algorithm.


\section*{Exercises}

\begin{ex}
Let $V=\R^{3}$ and let
\begin{equation*}
W=\sspan (S),  \mbox{ where } S=\set{\begin{mymatrix}{r}
1 \\
-1 \\
1
\end{mymatrix} ,\begin{mymatrix}{r}
-2 \\
2 \\
-2
\end{mymatrix},\begin{mymatrix}{r}
-1 \\
1 \\
1
\end{mymatrix},\begin{mymatrix}{r}
1 \\
-1 \\
3
\end{mymatrix} }
\end{equation*}
Find a basis of $W$ consisting of vectors in $S$.

\begin{sol}
In this case $\dim (W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (non-zero) vector from $S$.
\end{sol}
\end{ex}

\begin{ex}
 Let $T$ be a linear transformation given by
\[
T \begin{mymatrix}{r}
x\\
y
\end{mymatrix} = \begin{mymatrix}{rrr}
1 &1  \\
1 & 1
\end{mymatrix}
\begin{mymatrix}{r}
x\\
y
\end{mymatrix}
\]
Find a basis for $\ker (T)$ and $\func{im} (T)$.

\begin{sol}
A basis for $\ker (T)$ is
$\set{\begin{mymatrix}{r}
1 \\
-1
\end{mymatrix} }$
and a basis for $\func{im} (T)$ is
$\set{\begin{mymatrix}{r}
1 \\
1
\end{mymatrix} }$. \\
There are many other possibilities for the specific bases, but in this case
$\dim (\ker (T))=1 $ and $\dim (\func{im} (T))=1$.
\end{sol}

\end{ex}

\begin{ex}
 Let $T$ be a linear transformation given by
\[
T \begin{mymatrix}{r}
x\\
y
\end{mymatrix} = \begin{mymatrix}{rrr}
1 & 0  \\
1 & 1
\end{mymatrix}
\begin{mymatrix}{r}
x\\
y
\end{mymatrix}
\]
Find a basis for $\ker (T)$ and $\func{im}
(T)$.

\begin{sol}
In this case $\ker (T) =\set{0}$
and $\func{im} (T) = \R^2$ (pick any basis of $\R^2$).
\end{sol}

\end{ex}

\begin{ex}
Let $V=\R^{3}$ and let
\begin{equation*}
W=\sspan\set{\begin{mymatrix}{r}
1 \\
1 \\
1
\end{mymatrix} ,\begin{mymatrix}{r}
-1 \\
2 \\
-1
\end{mymatrix} }
\end{equation*}
Extend this basis of $W$ to a basis of $V$.

\begin{sol}
There are many possible such extensions, one is (how do we know?):
\begin{equation*}
\set{\begin{mymatrix}{r}
1 \\
1 \\
1
\end{mymatrix} ,\begin{mymatrix}{r}
-1 \\
2 \\
-1
\end{mymatrix} ,\begin{mymatrix}{r}
0  \\
0\\
1
\end{mymatrix}
}
\end{equation*}
\end{sol}
\end{ex}

\begin{ex}
 Let $T$ be a linear transformation given by
\[
T \begin{mymatrix}{r}
x\\
y \\
z
\end{mymatrix} = \begin{mymatrix}{rrr}
1 & 1 & 1 \\
1 & 1 & 1
\end{mymatrix}
\begin{mymatrix}{r}
x\\
y \\
z
\end{mymatrix}
\]
What is $\dim  ( \ker (T) )$?

\begin{sol}
We can easily see that $\dim  ( \func{im} (T) ) =1$, and thus
$\dim  ( \ker (T) ) = 3 - \dim  ( \func{im} (T) ) = 3- 1 = 2$.
\end{sol}
\end{ex}


The description of \textit{Fuzzy Logic} has been used by Lotfi A. Zadeh for the very first time within his famous published paper "Fuzzy Sets" .

Zadeh defined Fuzzy Logic ' a kind of logic using graded or qualified statements rather than ones that are strictly true or false . The results of fuzzy reasoning are not as definite as those derived by strict logic but they cover a larger field of discourse ' \cite{lotfi}  .

Fuzzy logic doesn't represents the logic that is fuzzy but the idea that all things can be a member of any cluster at least partly member. The idea behind fuzzy logic describes that all transmissions between clusters (memberhsips) are not so strict that you can change memberships just between 0 and 1. There are also soft and mid transitions between 0-1 .

To understand fuzzy logic we ought get into crisp sets too. In short comparision of \textit{Crispt Sets}  and \textit{Fuzzy Sets} both sets are questioning the classical "tall men" problem's memberships according to height of men (See Fig. \ref{fig:comparison})   .

\subsection{Crisp Sets and Fuzzy Sets}

In crisp sets theory a member can be in only one set. For example, for the relationship with the set “A”, the characteristic function gives us 1 (true) if an element named “u” is the member of the set “A” or 0 (false) if an element named “u” is not the member of the set “A” in the crisp logic \cite{Kasabov:1996:FNN:525657}. Hence, this logic cannot represent vague and ambiguous solutions of any paradoxes.

In fuzzy set theory all elements can be definitely true or false and partly true or false. Fuzzy Logic uses the continuum of logical values between 0 and 1 . When you put your height(cm) in Fig.\ref{fig:comparison} you can see that your degree of membership can be fractional and ambiguous .

\begin{figure}[ht]
    \centering
    \includegraphics[scale=.3]{Images/crisp_and_fuzzy.png}
    \caption{Crisp and Fuzzy Sets of "\textit{tall men}"}
    \label{fig:comparison}
\end{figure}

\subsection{Fundamentals of Fuzzy Logic}

Fundamentals of fuzzy decision making can be listed as : Fuzzification , inference and defuzzification.

\textit{1) Fuzzification}

\begin{figure}[ht]
    \centering
    \includegraphics[scale=.55]{Images/fuzzification.png}
    \caption{Example of nonlinear scaling of an input measurement. \cite{jantzen} }
    \label{fig:fig_fuzzification_jantzen}
\end{figure}

Fuzzification explains the process of converting every data to degrees of membership in one or more membership functions. The fuzzificaiton method thus matches the input data with the conditions of the rules . Every data has its own degree of memberships within a fuzzy set which can be explained with a linguistic term .

\textit{2) Inference}

Afterwards of fuzzification an inference is made based on rules . The process which called 'inference' basicly combines all results that evaluated from each rule to obtain a final result. This output given after inference step is also called \textit{fuzzy output set} .

\textit{3) Defuzzification}

After the inference there will be fuzzy output sets. But , its not recognizable in real life . In order to make it usable and recognizable in real life  those values should be translated into real values. This translation named \textit{defuzzification} \cite{defuzzification} .


\documentclass[12pt]{article}
\usepackage[fleqn]{amsmath}
\usepackage{textcomp}
\usepackage{gensymb}
\usepackage{amsfonts}
\usepackage{enumitem}
%\usepackage{tikz}  % Include for figures.
%\usepackage{subfiles}  % Include for subfiles.

\newcommand{\HOMEWORKNUM}{5}
\newcommand{\NAME}{D. Choi}
\newcommand{\DATE}{2020-05-28}

\title{\vspace{-4\baselineskip}MATH 225 - Homework \#\HOMEWORKNUM}
\author{\NAME}
\date{\DATE}

%\pagenumbering{gobble}  % Include for single-page document.

\begin{document}
\maketitle

\section*{1.}
\textit{Let $A$ be the projection matrix that projects onto the plane
$z = \sqrt{3}y$ in $\mathbb{R}^3$. \\ Find $A$.}
\\[\baselineskip]
Note that $z = \sqrt{3}y = \tan(60 \degree)y$.
\begin{gather*}
	\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
	\overset{A}{\longrightarrow}
	\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
	\\
	\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
	\overset{A}{\longrightarrow}
	\begin{pmatrix}
		0 \\
		\cos(60 \degree) \cos(60 \degree) \\
		\cos(60 \degree) \sin(60 \degree)
	\end{pmatrix}
	=
	\begin{pmatrix}
		0 \\
		\frac{1}{4} \\
		\frac{\sqrt{3}}{4}
	\end{pmatrix}
	\\
	\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
	\overset{A}{\longrightarrow}
	\begin{pmatrix}
		0 \\
		\cos(-30 \degree) \cos(60 \degree) \\
		\cos(-30 \degree) \sin(60 \degree)
	\end{pmatrix}
	=
	\begin{pmatrix}
		0 \\
		\frac{\sqrt{3}}{4} \\
		\frac{3}{4}
	\end{pmatrix}
\end{gather*}
\begin{equation*}
	A =
	\boxed{
		\begin{pmatrix}
			1 & 0 & 0 \\
			0 & \frac{1}{4} & \frac{\sqrt{3}}{4} \\
			0 & \frac{\sqrt{3}}{4} & \frac{3}{4}
		\end{pmatrix}
	}
\end{equation*}
\newpage

\section*{2.}
\textit{Let $B$ be the reflection matrix over the plane
$z = \sqrt{3}y$ in $\mathbb{R}^3$. \\ Find $B$.}
\\[\baselineskip]
Note (still) that $z = \sqrt{3}y = \tan(60 \degree)y$.
\begin{gather*}
	\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
	\overset{B}{\longrightarrow}
	\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
	\\
	\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
	\overset{B}{\longrightarrow}
	\begin{pmatrix}
		0 \\
		\cos(2 \cdot 60 \degree) \\
		\sin(2 \cdot 60 \degree)
	\end{pmatrix}
	=
	\begin{pmatrix}
		0 \\
		-\frac{1}{2} \\
		\frac{\sqrt{3}}{2}
	\end{pmatrix}
	\\
	\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
	\overset{B}{\longrightarrow}
	\begin{pmatrix}
		0 \\
		\cos(90 - 2 \cdot 30 \degree) \\
		\sin(90 - 2 \cdot 30 \degree)
	\end{pmatrix}
	=
	\begin{pmatrix}
		0 \\
		\frac{\sqrt{3}}{2} \\
		\frac{1}{2}
	\end{pmatrix}
\end{gather*}
\begin{equation*}
	B =
	\boxed{
		\begin{pmatrix}
			1 & 0 & 0 \\
			0 & -\frac{1}{2} & \frac{\sqrt{3}}{2} \\
			0 & \frac{\sqrt{3}}{2} & \frac{1}{2}
		\end{pmatrix}
	}
\end{equation*}

\section*{3.}
\textit{With $A$ and $B$ from \textbf{(1)} and \textbf{(2)},
find the following.}
\begin{enumerate}[label=(\alph*)]
	\item $A^{100} \begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix} =
	A \begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix} =
	\boxed{
		\begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix}
	}$
	\item $B \begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix}  =
	\boxed{
		\begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix}
	}$
	\item $B^3 \begin{pmatrix} 10 \\ -\sqrt{3} \\ 1 \end{pmatrix} =
	B^2 B \begin{pmatrix} 10 \\ -\sqrt{3} \\ 1 \end{pmatrix} =
	B \begin{pmatrix} 10 \\ -\sqrt{3} \\ 1 \end{pmatrix} =
	\boxed{
		\begin{pmatrix} 10 \\ \sqrt{3} \\ -1 \end{pmatrix}
	}$
\end{enumerate}

\section*{4.}
\textit{Find the matrix that rotates space $30 \degree$ about the y-axis in
$\mathbb{R}^3$.}
\\[\baselineskip]
Let $\text{Rot}_y(\theta)$ be the rotation matrix where
\begin{equation*}
	\text{Rot}_y(\theta) =
	\begin{pmatrix}
		\cos \theta & 0 & \sin \theta \\
		0 & 1 & 0 \\
		-\sin \theta & 0 & \cos \theta
	\end{pmatrix}
	.
\end{equation*}
\begin{equation*}
	\text{Rot}_y(30 \degree)
	=
	\begin{pmatrix}
		\cos(30 \degree) & 0 & \sin(30 \degree) \\
		0 & 1 & 0 \\
		-\sin(30 \degree) & 0 & \cos(30 \degree)
	\end{pmatrix}
	=
	\boxed{
		\begin{pmatrix}
			\frac{\sqrt{3}}{2} & 0 & \frac{1}{2} \\
			0 & 1 & 0 \\
			-\frac{1}{2} & 0 & \frac{\sqrt{3}}{2}
		\end{pmatrix}
	}
\end{equation*}

\end{document}


\subsection{Trees}\label{subsec:trees}

In this section, we will regard the edges of simple undirected graphs as sets of unordered tuples and edges of simple directed graphs as sets of ordered tuples.

\begin{definition}\label{def:tree}
  The possibly infinite \hyperref[def:undirected_multigraph]{simple undirected graph} \( T= (E, V, \mscrE) \) is called a \term{tree} if any of the following equivalent conditions hold:
  \begin{thmenum}
    \thmitem{def:tree/connected_acyclic} \( T \) is \hyperref[def:undirected_multigraph_connectedness]{connected} and \hyperref[def:undirected_multigraph_path/cycle]{acyclic}.
    \thmitem{def:tree/maximally_acyclic} \( T \) is \term{maximally acyclic}, meaning that adding an edge between existing vertices would create a \hyperref[def:undirected_multigraph_path/cycle]{cycle}.
    \thmitem{def:tree/minimally_connected} \( T \) is \term{minimally connected}, meaning that removing an edge will make the graph \hyperref[def:quiver_connectedness]{disconnected}.
    \thmitem{def:tree/single_path} For every pair of vertices in \( T \), there exists a unique \hyperref[def:undirected_multigraph_path]{path} connecting them.

    This condition motivates the definition of \hyperref[def:arborescence]{arborescences}.
  \end{thmenum}

  It is conventional to use the term \term{node} for the vertices of a tree.
\end{definition}
\begin{proof}
  \ImplicationSubProof{def:tree/connected_acyclic}{def:tree/maximally_acyclic} Suppose that \( T \) is connected and acyclic. There is always an edge not in \( T \) because the \hyperref[ex:complete_graph]{complete graph} \( K_V \) contains cycles and \( T \) does not. Let \( \set{ u, v } \) be an edge not in \( T \) and let \( T' \) be the supergraph that adjoins only this new edge.

  Since \( T \) is connected, there exists a path \( p \) connecting \( u \) and \( v \). Appending the edge \( \set{ u, v } \) to this path creates a cycle in \( T' \).

  Therefore, \( T' \) is not acyclic.

  \ImplicationSubProof{def:tree/maximally_acyclic}{def:tree/connected_acyclic} Suppose that \( T \) is maximally acyclic. We will show that it is connected.

  Suppose that \( T \) is not connected. Then there exist vertices \( u \) and \( v \) with no path between them. Let \( T' \) be the supergraph that adjoins only the edge \( \set{ u, v } \).

  Since there is no path between \( u \) and \( v \) in \( T \), \( T' \) is also acyclic. But this contradicts the maximality of \( T \).

  Therefore, \( T \) is connected.

  \ImplicationSubProof{def:tree/connected_acyclic}{def:tree/minimally_connected} Suppose that \( T \) is connected and acyclic. Let \( \set{ u, v } \) be any edge of \( T \) and let \( T' \) be the subgraph that does not contain this edge.

  Suppose that \( T' \) is connected. Let \( w \) be any vertex and let \( p \) be a path in \( T' \) from \( u \) to \( w \). Then adding \( \set{ u, v } \) creates a cycle in \( T \), which contradicts our assumption that \( T \) is acyclic.

  Therefore, \( T' \) is not connected.

  \ImplicationSubProof{def:tree/minimally_connected}{def:tree/connected_acyclic} Suppose that \( T \) is minimally connected. We will show that it is acyclic.

  Suppose that \( T \) has a cycle \( p \). We can thus remove any edge of \( p \) from \( T \) and the resulting subgraph will also be connected. This contradicts the minimality of \( T \).

  Therefore, \( T \) is acyclic.

  \ImplicationSubProof{def:tree/connected_acyclic}{def:tree/single_path} Suppose that \( T \) is connected and acyclic. Since \( T \) is connected, there exists at least one path between any two vertices. Since it is acyclic, this path must be unique because otherwise we could easily create a cycle by joining two such paths.

  \ImplicationSubProof{def:tree/single_path}{def:tree/connected_acyclic} Suppose that, for every pair of vertices in \( T \), there exists a unique path connecting them. It is clear that \( T \) is connected.

  Suppose that \( T \) has a cycle \( p \) from \( u \) to \( u \) passing through \( v \). We thus obtain two paths from \( u \) to \( v \) --- one not containing the first edge in \( p \) and one not containing the last edge. But this contradicts our assumption that any two paths from \( u \) to \( v \) are equal.

  Therefore, \( T \) is acyclic.
\end{proof}

\begin{example}\label{ex:def:tree}
  The reduced infinite integer graphs \eqref{eq:ex:infinite_integer_graphs/positive}, \eqref{eq:ex:infinite_integer_graphs/negative} and \eqref{eq:ex:infinite_integer_graphs/two_sided} from \fullref{ex:infinite_integer_graphs} are simple examples of infinite trees.

  Examples of finite trees are \hyperref[def:proof_tree]{proof trees} from \fullref{subsec:deductive_systems} and \hyperref[def:concrete_syntax_tree]{concrete} and \hyperref[def:abstract_syntax_tree]{abstract syntax trees} from \hyperref[def:formal_grammar]{formal grammar}.
\end{example}

\begin{definition}\label{def:arborescence}\mcite[ch. 4, sec. 3.1]{GondranMinoux1984Graphs}
  Let \( T = (V, A) \) be a \hyperref[def:quiver/simple]{simple directed graph}. Suppose that there exists a vertex \( r \in V \) such that for every other vertex \( v \in V \), there exists a unique \hyperref[def:quiver_path/directed]{directed path} from \( r \) to \( v \). The pair \( (G, r) \) is called a (directed) \term{arborescence}. It is often more convenient to consider the triple \( T = (V, A, r) \) instead. The vertex \( r \) is called the \term{root} of the arborescence.

  \begin{thmenum}
    \thmitem{def:arborescence/undirected} A \hyperref[def:tree]{tree} with a distinguished vertex \( r \) is called a \term{rooted tree}.

    For every tree \( T \), as a consequence of \fullref{def:tree/single_path}, every vertex \( r \) of \( T \) induces an \hyperref[def:multigraph_orientation]{orientation} \( O(T) \) of \( T \) such that \( O(T) \) is an arborescence.

    For this reason, we identify rooted trees with their induced arborescences.

    \thmitem{def:arborescence/depth} The \term{depth} of a node \( v \) is the length of the path from \( r \) to \( v \). The term \term{level} is also sometimes used but with a slightly different meaning --- the greater the depth, the lower the level.

    \thmitem{def:arborescence/ancestry} If \( v \) has a strictly greater depth than \( u \), we say that \( u \) is an \term{ancestor} of \( v \) and that \( v \) is a \term{descendant} of \( u \).

    The ancestor of \( v \) at the lowest possible level is called the \term{parent} of \( v \). If \( u \) is a parent of \( v \), \( u \) is a \term{child} of \( v \). If a node has no children, we say that it is a \term{leaf node}.

    Finally, if \( u \) and \( v \) are on the same level, we call them \term{siblings}.

    \thmitem{def:arborescence/height} The \term{height} or \term{depth} of the entire tree \( T \), if it exists, is the supremum of the depths of all nodes.

    \thmitem{def:arborescence/width} The \term{width} or \term{breadth} of the entire tree \( T \), if it exists, is the supremum of the number of siblings among all vertices. It is equal to the degree of \( T \) minus \( 1 \).

    We use terminology similar to \fullref{def:relation/arity}, e.g. \enquote{binary tree}, \enquote{ternary tree}, \ldots.

    \thmitem{def:arborescence/sub-arborescence} A \hyperref[def:quiver/submodel]{subgraph} of an arborescence that is itself an arborescence is called a \term{sub-arborescence}. The same holds for rooted subtrees.

    For every node \( v \) of \( T \), we define the \term{induced sub-arborescence} of \( T \) with root \( v \) as the subgraph \hyperref[def:quiver/submodel]{induced} by \( V \setminus \dom(p) \), where \( p \) is the unique path from \( r \) to \( v \).
  \end{thmenum}
\end{definition}

\begin{theorem}[K\"onig's lemma]\label{thm:konigs_lemma}
  Every \hyperref[def:hypergraph/degree]{locally finite} \hyperref[def:arborescence]{arborescence} of \hyperref[def:hypergraph/order]{infinite} \hyperref[def:hypergraph/order]{order} contains a \hyperref[def:undirected_multigraph_path/simple]{simple path} of infinite length.
\end{theorem}
\begin{proof}
  Let \( T = (V, A, r) \) be a locally finite infinite arborescence. We will build an infinite simple path
  \begin{equation*}
    p = (p_1, p_2, \cdots)
  \end{equation*}
  using \hyperref[rem:natural_number_recursion]{natural number recursion} starting at one. Let \( c \) be a \hyperref[def:choice_function]{choice function} on the family
  \begin{equation*}
    \pow(V) \setminus \set{ \varnothing }.
  \end{equation*}

  Such a choice function exists by the \hyperref[def:zfc/choice]{axiom of choice}.

  \begin{itemize}
    \item Since \( T \) is locally finite, there are finitely many children of \( r \). If we suppose that the \hyperref[def:arborescence/sub-arborescence]{subarborescence} induced by \( v \) is finite for every child \( v \) of \( r \), then we would obtain that \( T \) itself is finite, which is a contradiction. Therefore, for at least one child, the induced sub-arborescence is infinite. Using the choice function \( c \), pick one such child and denote it by \( v_1 \).

    Define \( p_1 \) to be the arc \( r \to v_1 \).

    \item Fix \( k > 1 \). Using the same argument on the children of \( t(e_{k-1}) \), we can pick a child \( v_k \) that has an infinite induced sub-arborescence.

    Define \( e_k \) to be the arc \( t(e_{k-1}) \to v_k \). The path
    \begin{equation*}
      (e_1, e_2, \ldots, e_{k-1}, e_k)
    \end{equation*}
    is simple by construction because \( T \) is an arborescence and thus contains no directed cycles.
  \end{itemize}

  Thus, we have constructed an infinite simple path.
\end{proof}


\chapter{Lagrange's Equations of the Second Kind}\label{c2}
\section{Generalised coordinates}\label{c2s1}
In the previous chapter we observed that in a system of $N$ particles and $r$
constraints, the $3N$ virtual displacements $\delta x^{(n)}_i$ are not
independent. Any subset of $f = 3N - r$ virtual displacements is independent.
$f$ is called the \emph{number of degrees of freedom} of the system.

Consider a system of two particles connected by a rigid rod. The two particles
each need three Cartesian coordinates. But the constraint makes only five of
them independent. It is not clear which five to choose. Although the lack of
independence of virtual displacements in all coordinates is taken into account
by the Lagrange's equations of the first kind, one still has to describe the
system in terms of $6$ coordinates. This situation gets worse in the case of
a rigid body. If it made up of $N$ particles then we need $3N$ coordinates to
describe its position when only $6$ of them are independent.

Generalised coordinates are any $f$ numbers that suffice to describe the
position of a system of $f$ degrees of freedom. For example, a rigid body
with $N$ particles can be described by three Cartesian coordinates of its
centre of mass and three Euler angles. We shall denote the $f$ generalised
coordiates by $\{q_1, \ldots, q_f\}$.

Let us therefore consider a system of $N$ particles and $f$ degrees of freedom
and let $q_1, \ldots, q_f$ be its generalised coordinates. Then the Cartesian
coordinates of its particles can be written as
\begin{eqnarray}
x_i &=& x_i(q_1, \ldots, q_f, t) \\
y_i &=& y_i(q_1, \ldots, q_f, t) \\
z_i &=& z_i(q_1, \ldots, q_f, t).
\end{eqnarray}
Note that the Cartesian coordinates of a particle are allowed to depend on
\emph{all} generalised coordinates. Unlike their Cartesian counterparts, the
generalised coordinates are not tied to a single particle. Further, note that
the Cartesian coordinates may depend on time as well. From these equations
we have
\begin{eqnarray}
dx_i &=& \pd{x_i}{q_j}dq_j + \pd{x_i}{t}dt \\
dy_i &=& \pd{y_i}{q_j}dq_j + \pd{y_i}{t}dt \\
dz_i &=& \pd{z_i}{q_j}dq_j + \pd{z_i}{t}dt
\end{eqnarray}
We have used the summation convention in these three equations. Although
the functional forms of $x_i, y_i, z_i$ are arbitrary, the differentials
$dx_i, dy_i, dz_i$ are linear functions of $dq_j$ and $dt$.

The three Cartesian velocities are
\begin{eqnarray}
\dot{x}_i &=& \pd{x_i}{q_j}\dot{q}_j + \pd{x_i}{t} \label{c2s1e7} \\
\dot{y}_i &=& \pd{y_i}{q_j}\dot{q}_j + \pd{y_i}{t} \label{c2s1e8} \\
\dot{z}_i &=& \pd{z_i}{q_j}\dot{q}_j + \pd{z_i}{t} \label{c2s1e9}
\end{eqnarray}
Note that the symbol $\dot{x}_i$ denotes the total time derivative of $x_i$ and
it differs from $\partial x_i/\partial t$. The generalised coordinates, however,
do not depend explicitly on time. The kinetic energy of the system is
\begin{eqnarray}
T &=& \frac{1}{2}m_i(\dot{x}_i^2 + \dot{y}_i^2 + \dot{z}_i^2) \nonumber \\
 &=& \frac{1}{2}m_i\left(\pd{x_i}{q_j}\dot{q}_j + \pd{x_i}{t}\right)
                   \left(\pd{x_i}{q_k}\dot{q}_k + \pd{x_i}{t}\right) +
     \nonumber \\
 & & \frac{1}{2}m_i\left(\pd{y_i}{q_j}\dot{q}_j + \pd{y_i}{t}\right)
                   \left(\pd{y_i}{q_k}\dot{q}_k + \pd{y_i}{t}\right) +
	 \nonumber \\
 & & \frac{1}{2}m_i\left(\pd{z_i}{q_j}\dot{q}_j + \pd{z_i}{t}\right)
                   \left(\pd{z_i}{q_k}\dot{q}_k + \pd{z_i}{t}\right)
	 \label{c1s1e10}
\end{eqnarray}
We now compute the derivative of $T$ with respect to the generalised
velocity $\dot{q}_l$.
\begin{eqnarray}
\pd{T}{\dot{q}_l} &=&
  m_i\left(\pd{x_i}{q_j}\dot{q}_j + \pd{x_i}{t}\right)\pd{x_i}{q_l} +
 m_i\left(\pd{y_i}{q_j}\dot{q}_j + \pd{y_i}{t}\right)\pd{y_i}{q_l} +
  \nonumber \\
& &  m_i\left(\pd{z_i}{q_j}\dot{q}_j + \pd{z_i}{t}\right)\pd{z_i}{q_l}
  \label{c2s1e11}
\end{eqnarray}
Using equations \eqref{c2s1e7}, \eqref{c2s1e8} and \eqref{c2s1e9} we get
\begin{equation}\label{c2s1e12}
\pd{T}{\dot{q}_l} = m_i\left(\dot{x}_i\pd{x_i}{q_l} + \dot{y}_i\pd{y_i}{q_l} +
 \dot{z}_i\pd{z_i}{q_l}\right).
\end{equation}
We now take the total time derivative of this equation.
\begin{eqnarray}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) &=&
  m_i\left(\ddot{x}_i\pd{x_i}{q_l} + \dot{x}_i\frac{d}{dt}
  \left(\pd{x_i}{q_l}\right)\right)  +
  m_i\left(\ddot{y}_i\pd{y_i}{q_l} + \dot{y}_i\frac{d}{dt}
  \left(\pd{y_i}{q_l}\right)\right) + \nonumber \\
& & m_i\left(\ddot{z}_i\pd{z_i}{q_l} + \dot{z}_i\frac{d}{dt}
  \left(\pd{z_i}{q_l}\right)\right)
\end{eqnarray}
Evaluating the total time derivatives on the right hand side,
\begin{eqnarray}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) &=&
 m_i\left(\ddot{x}_i\pd{x_i}{q_l} +
 \dot{x}_i\dot{q}_k\frac{\partial^2 x_i}{\partial q_k\partial q_l} +
 \dot{x_i}\frac{\partial}{\partial q_l}\pd{x_i}{t}\right) \nonumber \\
 & & m_i\left(\ddot{y}_i\pd{y_i}{q_l} +
 \dot{y}_i\dot{q}_k\frac{\partial^2 y_i}{\partial q_k\partial q_l} +
 \dot{y_i}\frac{\partial}{\partial q_l}\pd{y_i}{t}\right) \nonumber \\
 & & m_i\left(\ddot{z}_i\pd{z_i}{q_l} +
 \dot{z}_i\dot{q}_k\frac{\partial^2 z_i}{\partial q_k\partial q_l} +
 \dot{z_i}\frac{\partial}{\partial q_l}\pd{z_i}{t}\right) \label{c2s1e14}
\end{eqnarray}
We now observe that
\[
\pd{\dot{q_1}}{q_k} = \frac{\partial}{\partial t}\delta_{lk} = 0
\]
so that we can write
\begin{eqnarray}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) &=&
 m_i\left(\ddot{x}_i\pd{x_i}{q_l} +
 \dot{x}_i\frac{\partial}{\partial q_l}
 \left(\dot{q}_k\pd{x_i}{q_k} + \pd{x_i}{t}\right)\right) \nonumber \\
 & &  m_i\left(\ddot{y}_i\pd{y_i}{q_l} +
 \dot{y}_i\frac{\partial}{\partial q_l}
 \left(\dot{q}_k\pd{y_i}{q_k} + \pd{y_i}{t}\right)\right) \nonumber \\
 & &  m_i\left(\ddot{z}_i\pd{z_i}{q_l} +
 \dot{z}_i\frac{\partial}{\partial q_l}
 \left(\dot{q}_k\pd{z_i}{q_k} + \pd{z_i}{t}\right)\right) \label{c2s1e15}
\end{eqnarray}
Using equations \eqref{c2s1e7}, \eqref{c2s1e8} and \eqref{c2s1e9} we get
\[
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) =
m_i\left[\left(\ddot{x}_i\pd{x_i}{q_l} + \dot{x}_i\pd{\dot{x}_i}{{q}_l}\right)
 + \nonumber \\
\left(\ddot{y}_i\pd{y_i}{q_l} + \dot{y}_i\pd{\dot{y}_i}{{q}_l}\right)
 + \nonumber \\
\left(\ddot{z}_i\pd{z_i}{q_l} + \dot{z}_i\pd{\dot{z}_i}{{q}_l}\right)\right].
\]
Rearranging it
\[
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) =
\pd{T}{q_l} + m_i\left(\ddot{x}_i\pd{x_i}{q_l} +
\ddot{y}_i\pd{y_i}{q_l} +\ddot{z}_i\pd{z_i}{q_l}\right)
\]
or
\begin{equation}\label{c2s1e16}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) - \pd{T}{q_l} =
 m_i\left(\ddot{x}_i\pd{x_i}{q_l} +
 \ddot{y}_i\pd{y_i}{q_l} +\ddot{z}_i\pd{z_i}{q_l}\right)
\end{equation}
Multiplying both sides by $\delta q_l$,
\begin{equation}\label{c2s1e17}
\left(\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) - \pd{T}{q_l}\right)\delta q_l
= m_i\left(\ddot{x}_i\delta x_i + \ddot{y}_i\delta y_i +
 \ddot{z}_i\delta z_i\right)
\end{equation}
Comparing with equation \eqref{c1s2e13} we surmise that the left hand side
of the above equation is
\begin{equation}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) - \pd{T}{q_l}\delta q_l =
X_i\delta x_i + Y_i\delta y_i + Z_i \delta z_i,
\end{equation}
where $(X_i, Y_i, Z_i)$ are components of the external force on the $i$th
particle. If this force is conservative, we can write it in terms of a
potential $V$ as
\begin{eqnarray}
\frac{d}{dt}\left(\pd{T}{\dot{q}_l}\right) - \pd{T}{q_l}\delta q_l &=&
\pd{V}{x_i}\delta x_i + \pd{V}{z_i}\delta y_i + \pd{V}{z_i} \delta z_i \nonumber \\
&=&
\pd{V}{x_i}\pd{x_i}{q_l}\delta q_l + \pd{V}{z_i}\pd{y_i}{q_l}\delta q_l +
\pd{V}{z_i} \pd{z_i}{q_l}\delta q_l \nonumber \\
&=& \pd{V}{q_1}\delta q_l \label{c1s1e20}
\end{eqnarray}
Note that the last term here is not $3\partial V/\partial q_l$ because $V$
depends on all $x_1, \ldots, z_N$ and each of these is a function of all
generalised coordinates. If $V$ \emph{does not} depend on generalised
velocities then we can write \eqref{c1s1e20} as
\begin{equation}\label{c2s1e21}
\frac{d}{dt}\pd{L}{\dot{q}_l} - \pd{L}{q_l} = 0,
\end{equation}
where the function
\begin{equation}\label{c1s1e22}
L(q_1, \ldots, q_f, \dot{q}_1, \ldots, \dot{q}_f) = T - V
\end{equation}
is the Lagrangian of the system and equations \eqref{c2s1e21} are called
Lagrange's equation of the second kind.

A few remarks about \eqref{c2s1e21}:
\begin{itemize}
\item Often times the phrase `second kind' is omitted and equations
are called Lagrange's equations.
\item They equations are presently derived only for conservative forces.
\item They do not involve the forces of constraint. Neither do they insist on
the Cartesian coordinates.
\item They are valid in all frames of references. However $T$ depends on
generalised velocities and will change across frames of reference. Therefore,
$T$ should always be calculated with respect to an inertial frame of
reference \cite[p. 31]{akr}.
\end{itemize}

Problems.
\begin{enumerate}
\item A uniform rod leans against a wall \cite[Problem 1, chapter 4]{akr}.
Let $l$ be its length. Its motion is confined to a plane. Its unconstrained
motion has three degrees of freedom, the position of its centre of mass and
its orientation. In the present problem, the $x$ coordinate of its end
touching the wall is constrained to be $0$ and the $y$ coordinate of its
end touching the floor is constrained to be $0$. Therefore, one generalised
coordinate suffices to describe the motion.

We choose $\theta$ the (smaller) angle made by the rod with the $x$ axis. If
the rod is in the first quadrant then it makes and angle $\pi -\theta$ with
the positive $x$ axis.

The centre of the rod is at point $(0, l/2)$ when $\theta = \pi/2$ and it is
at $(0, l/2)$ when $\theta = 0$. It traces the arc of a circle of radius
$l/2$ as the rod slides down the path.

The moment of inertia of the rod abouts its centre of mass is $ml^2/12$.
Therefore, the rotational kinetic energy is
\[
T_r = \frac{I}{2}\dot{\theta}^2 = \frac{m}{24}l^2\dot{\theta}^2
\]
The translational kinetic energy of its centre of mass is
\[
T_c = \frac{m}{2}v^2 = \frac{m}{8}l^2\dot{\theta}^2
\]
The total kinetic energy is
\[
T = \frac{ml^2}{12}\dot{\theta}^2.
\]

We will now find the potential energy. If $dm$ is the mass element of the
rod at a height $h$ then
\[
dV = gy dm
\]
If $\lambda$ be the linear mass density of the rod, $dm = \lambda ds = mds/l$.
Thus,
\[
dV = \frac{mg}{l} yds = \frac{mg}{l}y \sqrt{(dx)^2 + (dy)^2} =
\frac{mg}{l}ydy\sqrt{1 + \left(\frac{dx}{dy}\right)^2}.
\]
Now the equation of the straight line representing the rod is $y = y_0 -
\tan\theta x$ so that
\[
\frac{dx}{dy} = \cot\theta
\]
so that
\[
dV = \frac{mg}{l}\csc\theta ydy
\]
or
\[
V = \int_{lsin\theta}^0 dV = -\frac{mg}{2}l\sin\theta.
\]
Therefore, the Lagrangian is
\[
L = \frac{ml^2}{12}\dot{\theta}^2 + \frac{mg}{2}l\sin\theta
\]
and the equation of motion is
\[
\frac{ml^2}{6}\ddot{\theta} - \frac{mgl}{2}\cos\theta = 0 \Rightarrow
\frac{l}{3}\ddot{\theta} - g\cos\theta = 0 \Rightarrow \ddot{x} =
\frac{3g}{l}\cos\theta.
\]

\item An old model of the He atom consisted of a fixed nucleus and two
electrons on the opposite ends of the diameter of the circle centred at
the nucleus \cite[Problem 3, chapter 4]{akr}. Without the constraints, the
motion of the two electrons in a plane would have required four coordinates.
Their constraints are:
\begin{itemize}
\item They are always at the opposite ends of a diameter. This means that
we can consider them to be end of a rigid, weightless rod.
\item Their centre of mass is fixed at the nucleus. This fixes the two
coordinates of their centre of mass.
\end{itemize}
This leaves the system with just one degree of freedom. If $\theta$ is the
angle made by one electron with the positive $x$ axis, the other electron
makes an angle $\theta + \pi/2$. If $m$ is the mass of the electrons, their
kinetic energy is
\[
T = \frac{m}{2}r^2\dot{\theta}^2 + \frac{m}{2}r^2\dot{\theta}^2 =
mr^2\dot{\theta}^2.
\]
Their potential energy is
\[
V = -\frac{2e^2}{r} - \frac{2e^2}{r} = -\frac{4e^2}{r}
\]
so that the Lagrangian is
\[
L = mr^2\dot{\theta}^2 + \frac{4e^2}{r^2}.
\]
The equation of motion is $\ddot{\theta} = 0$ or that $\theta = \alpha t +
\beta$, where $\alpha$ and $\beta$ are the initial conditions.
\end{enumerate}


\section{Proposed method}

\begin{frame}{Proposed method}
    Some mathematics formula
    \begin{equation}
		f(n) =
            \begin{cases}
                n/2       & \quad \text{if } n \text{ is even}\\
                -(n+1)/2  & \quad \text{if } n \text{ is odd}
            \end{cases}
	\end{equation}
\end{frame}


%\columnbreak
\section{Model Selection for Clustering}

% ===
What is the appropriate \#clusters $k$ for my data?

\emph{General approach:} Measure quality (neg. log-likelihood) for different $k$ \enspace $\to$ \textbf{elbow}.

% ===
\subsection{Complexity-based Model Selection}

\textbf{Strategy:} add a complexity term to neg. log-likelihood

\textbf{Attention:} MDL/BIC rely on likelihood optimisation $\to$ \underline{not generally applicable}

\emph{Ocam's razor:}\enspace
Choose the model that provides the shortest description of the data.

% ---
\subsubsection{Min. Description Length\quad (MDL)}

Minimise \textbf{descr. length}:\enspace
$-\log p(\bm X\mid \theta) - \log p(\theta)$

Approx.:\enspace
$\hat k \in \arg\min_k \highlight*{-\log p(\bm X\mid \hat\theta) + \frac{k'}{2} \log n}$

% ---
\subsubsection{Bayesian Information Crit.\quad (BIC)}

Parametrise likelihood $p(\bm X\mid M)$ by $\theta$:\\
\enspace $p(\bm X\mid M) = \int_{\Theta_M} \! \exp(\log p(\bm X\mid M, \theta)) \cdot p(\theta\mid M) \diff\theta$

Assume flat prior $p(\theta\vert M) \approx \mathit{const}$ and\\
expand log-likelihood by ML estimator $\hat\theta$:\\
$\overline\ell (\theta) = \frac{\ell(\theta)}{n} = \frac1n \log p(\bm X\vert M,\theta) \overset{\text{i.i.d.}}{=} \frac1n \sum_i \ell(\theta, X_i) \overset{\textrm{Taylor}}{\approx} \ldots$

$\implies p(\bm X\mid M) = \mathit{const}_2 \cdot \exp( \highlight*{\ell (\hat\theta) - \frac{k'}{2} \log n} )$\\
\quad where $k'$ : dimension of (trainable) parameters

% ===


\section{Delta Method}%
\label{sec:delta_method}

\begin{thm}
    If $\sqrt{n}(T_n - \theta) \convl N(0, \tau^2)$ and $f'(\theta) \ne 0$, then
    \begin{equation*}
        \sqrt{n}(f(T_n) - f(\theta)) \convl N(0, \tau^2 [f'(\theta)]^2).
    \end{equation*}
\end{thm}