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\usepackage{quantumgravity}
\title{Quantum Gravity Homework 5}
\author{Erin Pearse}

\pagestyle{myheadings}
    \markboth{\hfill \textsc{Quantum Gravity Homework 5} \hfill }
             {\hfill \textsc{Quantum Gravity Homework 5} \hfill }
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\begin{document}
\maketitle

\begin{Prob}
  \item To put a $P(k)$-structure on the $n$-element set $S$, we
    first have to totally order the elements. There are $n!$ ways
    to do this. Then, each of the numbers $j \in \{1,2,\ldots,k\}$ must
    be assigned to one of the $n$ elements of $S$. Since repeats
    are allowed (more than one $j$ may be assigned to a given
    point of $S$), there are $n$ possibilities for each $j$.
    Hence,
    \[|P(k)_n| = n!n^k.\]

  \item \(\displaystyle |P(k)|(z) = \sum_{n \geq 0} \frac{|P(k)_n|}{n!} z^n = \sum_{n \geq 0} n^k z^n.\)

  \item \(\displaystyle |P(0)|(z) = \sum_{n \geq 0} n^0 z^n = \sum_{n \geq 0} z^n = \frac{1}{1-z}.\)

  \item Define the number operator by
    \(N\gY = A^*A\gY.\)
    Then
    \begin{align*}
      \text{put a $N\gY$}&\text{-structure on a set $S$} \\
      & = \text{put an $A^* A \gY$-structure on $S$} \\
      &\cong \text{choose $x \in S$ and put an $A\gY$-structure on 
      $S\less\{x\}$}  \\
      &\cong \text{choose $x \in S$ and put $\gY$-structure on 
      $(S\less\{x\})+1$} \\
    \end{align*}
    This corresponds to `pointing' $S$ (i.e., giving it one
    distinguished point) and putting a $\gY$-structure on it.

  \item Putting an $NP(k)$ structure on $S$ amounts to pointing
    $S$, and then totally ordering $S$ and giving it a
    $k$-pointing. But this is the same as totally ordering $S$,
    giving it a distinguished point (say we give one point a star),
    and then $k$-pointing $S$, because surely the order doesn't
    matter. Finally, by `reindexing'
    \[\{*\}\cup\{,1,2,\ldots,k\} = \{*,1,2,\ldots,k\} \iso 
    \{1,2,\ldots,k+1\}\]
    so that we've really just put a total ordering on $S$ and then
    given it a $(k+1)$-pointing, i.e., put a $P(k+1)$-structure on
    it. Hence,
    \[NP(k) \cong P(k+1).\]
  \begin{comment}
    To see that $NP(k) \cong P(k+1)$, first note that a
    $AP(k)$-structure on the $n$-element set $S$ is the property of being a
    $k$-pointed totally ordered $(n+1)$-element set. Thus,
    \[|AP(k)_n| = (n+1! (n+1)^k.\]
    Then an $A^*AP(k)$-structure on the $n$-element set $S$ is:
    pick $x \in S+1$ and put the structure of being a
    $k$-pointed totally ordered $n$-element set on $S\less \{x\}$.
  \end{comment}

  \item Decategorifying $P(k+1) \cong NP(k)$,
    \begin{align*}
      |P(k+1)|(z) &= |NP(k)|(z) \\
        &= |A^*AP(k)|(z)
          &\text{def of } N \\
        &= z\tfrac{d}{dz}|P(k)|(z)
          &|A^*|=z, |A|=\tfrac{d}{dz}
    \end{align*}

  \item By the previous problem we know
    \begin{align*}
      |P(1)|(z) &= z\tfrac{d}{dz}|P(0)|(z) \\
        &= z\tfrac{d}{dz}\left(\tfrac{1}{1-z}\right) \\
        &= \tfrac{z}{(1-z)^2}
    \end{align*}

  \item By the previous formula,
    \[|P(1)|(-1) = \tfrac{-1}{(1+1)^2} = -\tfrac{1}{4}.\]
    But since $|P(1)|(z) = \sum_{n \geq 0} n z^n$ by the
    definition of $|P(k)|$ found in 3, so
    \[|P(1)|(-1) = 1-2+3-4+ \ldots = -\tfrac{1}{4}.\]

  \item Using the definition of Abel summation:
    \[A \sum_n an := \lim_{t\uparrow1} \sum_n t^n a_n,\]
    we see that
    \begin{align*}
      A \sum_{n =1}^\iy (-1)^{n+1}n
      &= \lim_{t\uparrow1} \sum_{n=1}^\iy t^n (-1)^{n+1}n
        &\text{by def} \\
      &= -\lim_{t\uparrow1} \sum_{n=1}^\iy (-t)^nn
        &\text{collecting} \\
      &= -\lim_{t\uparrow1} |P(1)|(-t)
        &\text{by 2} \\
      &= -\lim_{t\uparrow1} \tfrac{-t}{(1+t)^2}
        &\text{by 7} \\
      &= - \left(\tfrac{-1}{2^2}\right)
        &\text{continuity} \\
      &= \tfrac{1}{4}
    \end{align*}

  \item First we compute:
    \begin{align*}
      |P(2)|(z) &= z\tfrac{d}{dz}|P(1)|(z) \\
      &= z\tfrac{d}{dz}\tfrac{z}{(1-z)^2} \\
      &= z\left(\tfrac{1}{(1-z)^2} + \tfrac{2z}{(1-z)^3}\right) \\
      &= \tfrac{z}{(1-z)^2} + \tfrac{2z^2}{(1-z)^3} \\
    \end{align*}
    Now, to compute the Abel sum of $1^2-2^2+3^2-4^2+\ldots$.
    \begin{align*}
      |P(2)|(-1) &= A \sum_{n=1}^\iy n^2(-1)^{n} \\
      &= \lim_{t \uparrow 1} \sum_{n=1}^\iy t^n n^2(-1)^{n} \\
      &= \lim_{t \uparrow 1} \sum_{n=1}^\iy n^2(-t)^{n} \\
      &= \lim_{t \uparrow 1} |P(2)|(-t) \\
      &= \lim_{t \uparrow 1} \tfrac{-t}{(1+t)^2} + \tfrac{2t^2}{(1+t)^3} \\
      &= \tfrac{-1}{4} + \tfrac{2}{8} \\
      &= 0
    \end{align*}
    Lastly, we wish to find $\gz(-2)=1^2 + 2^2 + 3^2 + 4^2 + \ldots$.
    From
    \[\gz(-2) = \sum_{n=1}^\iy n^2 = 1^2 + 2^2 + 3^2 + 4^2 + \ldots,\]
    we get
    \[2^2 \gz(-2) = 2^2\sum_{n=1}^\iy n^2 %= \sum_{n=1}^\iy 2^2 n^2
      = \sum_{n=1}^\iy (2n)^2
      = 2^2 + 4^2 + 6^2 + 8^2 + \ldots.\]
    Subtracting,
    \[\gz(-2) - 2 \cdot 2^2 \gz(-2)
      = 1^2 - 2^2 + 3^2 - 4^2 + \ldots = 0.\]
    Thus,
    \[\gz(-2) - 2 \cdot 2^2 \gz(-2)
      = (1-2^3)\gz(-2)
      = -7\gz(-2)
      =0
      \implies \gz(-2)=0,\]
    as expected.
\end{Prob}

\end{document}
