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\begin{document}
\section*{Math 701 Fall 2026, Lecture 1: How symmetric polynomials appear}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}

\setcounter{section}{0}

\subsection{General}

This is a course on symmetric and quasisymmetric functions. We will start with
classical results that are well represented in textbooks and lecture notes;
then we will gradually move on to more active research topics.

The course website (where all lecture notes and homeworks will be posted) is%
\[
\text{\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fs}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fs/}}}}%
\]
But homework should be submitted via
\href{https://drexel.instructure.com/courses/13625}{Canvas}. Homework must be
done without AI and without consulting outsiders; collaboration is allowed as
long as you acknowledge your collaborators. Homework should be typeset, not handwritten.

\subsection{Some motivation}

Historically speaking, symmetric polynomials/functions
\href{https://hsm.stackexchange.com/questions/2040/how-were-the-phenomena-relating-to-symmetric-polynomials-discovered}{first
appeared} in the study of roots of polynomials, and this has given a major
impetus for their early study by Newton and Gauss. Let me show a slight
variant on this motivation.

\subsubsection{Playing with Fibonacci numbers}

Recall the \textbf{Fibonacci sequence}: This is the integer sequence $\left(
f_{0},f_{1},f_{2},\ldots\right)  $ defined recursively by $f_{0}=0$ and
$f_{1}=1$ and
\begin{equation}
f_{n}=f_{n-1}+f_{n-2}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq
2.\label{eq.mot.fib.fibrec}%
\end{equation}
The latter equality $f_{n}=f_{n-1}+f_{n-2}$ is called a \textbf{linear
recurrence}, as it expresses each term $f_{n}$ of the sequence as a linear
combination (with constant coefficients) of the previous two terms $f_{n-1}$
and $f_{n-2}$. (This is a discrete analogue of linear ODEs with constant
coefficients, a venerable subject!)

Now consider the sequence $\left(  f_{0},f_{2},f_{4},f_{6},\ldots\right)  $,
which consists of every second Fibonacci number. I claim that it also
satisfies a linear recurrence: namely,%
\[
f_{2n}=3f_{2n-2}-f_{2n-4}\ \ \ \ \ \ \ \ \ \ \text{for all }n\geq2.
\]
In fact, this is not specific to even-indexed Fibonacci numbers; more
generally, we have%
\begin{equation}
f_{m}=3f_{m-2}-f_{m-4}\ \ \ \ \ \ \ \ \ \ \text{for all }m\geq4,
\label{eq.mot.fib.fibrec2}%
\end{equation}
as we can show by applying the original Fibonacci recurrence
(\ref{eq.mot.fib.fibrec}) multiple times:%
\begin{align*}
f_{m}  &  =\underbrace{f_{m-1}}_{\substack{=f_{m-2}+f_{m-3}\\\text{(by
(\ref{eq.mot.fib.fibrec}))}}}+\,f_{m-2}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{eq.mot.fib.fibrec})}\right) \\
&  =\left(  f_{m-2}+f_{m-3}\right)  +f_{m-2}=2f_{m-2}+\underbrace{f_{m-3}%
}_{\substack{=f_{m-2}-f_{m-4}\\\text{(since (\ref{eq.mot.fib.fibrec}) yields
}f_{m-2}=f_{m-3}+f_{m-4}\text{)}}}\\
&  =2f_{m-2}+f_{m-2}-f_{m-4}=3f_{m-2}-f_{m-4}.
\end{align*}
So not only does the every-second-Fibonacci-number sequence $\left(
f_{0},f_{2},f_{4},f_{6},\ldots\right)  $ satisfy a linear recurrence, but its
offset variant $\left(  f_{1},f_{3},f_{5},f_{7},\ldots\right)  $ satisfies the
same linear recurrence.

What about every third Fibonacci number? Does the sequence $\left(
f_{0},f_{3},f_{6},f_{9},\ldots\right)  $ satisfy a linear recurrence, too? We
can try to find such a recurrence by \textquotedblleft following our
nose\textquotedblright\ using (\ref{eq.mot.fib.fibrec}) again -- e.g. as
follows:
\begin{align*}
f_{m}  &  =f_{m-1}+f_{m-2}=2f_{m-2}+f_{m-3}=2\left(  f_{m-3}+f_{m-4}\right)
+f_{m-3}\\
&  =3f_{m-3}+2f_{m-4}=3f_{m-3}+2\left(  f_{m-5}+f_{m-6}\right)  =\cdots
\end{align*}
-- but it is not clear how to finish this: we would like to get rid of all
appearances of $f_{m-1},f_{m-2},f_{m-4},f_{m-5},f_{m-7},\ldots$ on the
right-hand side, but it is not obvious how to do so.

Yet it can be shown that the sequence $\left(  f_{0},f_{3},f_{6},f_{9}%
,\ldots\right)  $ satisfies a linear recurrence, which is again shared by its
two offset variants $\left(  f_{1},f_{4},f_{7},f_{10},\ldots\right)  $ and
$\left(  f_{2},f_{5},f_{8},f_{11},\ldots\right)  $. Namely, we have%
\begin{equation}
f_{m}=4f_{m-3}+f_{m-6}\ \ \ \ \ \ \ \ \ \ \text{for all }m\geq6.
\label{eq.mot.fib.fibrec3}%
\end{equation}
Proving this is straightforward (e.g., by strong induction), but how was this
recurrence found? And can it be generalized? What about every fourth or every
fifth Fibonacci number? More generally, what about every $m$-th Fibonacci
number for a given $m\in\mathbb{N}$ ?

\subsubsection{Linear recurrences in general}

Actually, what about linearly recurrent sequences of higher order? Let us
recall what this means:

\begin{definition}
\label{def.mot.fib.linrec}Let $k$ be a positive integer. Let $\left(
x_{0},x_{1},x_{2},\ldots\right)  $ be a sequence of numbers (e.g., complex
numbers). We say that this sequence is \textbf{linearly recurrent of order
}$k$ if it satisfies the recurrence%
\begin{equation}
x_{n}=a_{1}x_{n-1}+a_{2}x_{n-2}+\cdots+a_{k}x_{n-k}%
\ \ \ \ \ \ \ \ \ \ \text{for all }n\geq k,
\label{eq.def.mot.fib.linrec.linrec}%
\end{equation}
where $a_{1},a_{2},\ldots,a_{k}$ are fixed constants (not depending on $n$).
These constants $a_{1},a_{2},\ldots,a_{k}$ are called the
\textbf{coefficients} of this recurrence.
\end{definition}

Examples are easy to find:

\begin{itemize}
\item The Fibonacci sequence is linearly recurrent of order $2$, with
coefficients $1$ and $1$.

\item Any geometric progression is linearly recurrent of order $1$ (the
recurrence being $x_{n}=qx_{n-1}$).

\item Any arithmetic progression is linearly recurrent of order $2$ (the
recurrence being $x_{n}=2x_{n-1}-x_{n-2}$).
\end{itemize}

Note that a linearly recurrent sequence of order $k$ will also be linearly
recurrent of order $k+1$ (just add a \textquotedblleft$+0x_{n-k-1}%
$\textquotedblright\ term to (\ref{eq.def.mot.fib.linrec.linrec})), so the
coefficients belong to the recurrence, not to the sequence.

Now we can generalize the question we posed above for the Fibonacci sequence:

\begin{question}
\label{quest.mot.fib.linrec-k}Let $\left(  x_{0},x_{1},x_{2},\ldots\right)  $
be a linearly recurrent sequence of order $k$. Let $m\in\mathbb{N}$ be
arbitrary. Then, is the sequence $\left(  x_{0},x_{m},x_{2m},x_{3m}%
,\ldots\right)  $ again linearly recurrent of order $k$ ? If so, how can we
compute the coefficients of the recurrence?
\end{question}

\subsubsection{The matrix trick}

The main trick involved in linearly recurrent sequences is viewing them as
entries of matrix powers. Namely, consider a linearly recurrent sequence
$\left(  x_{0},x_{1},x_{2},\ldots\right)  $ of order $k$ with coefficients
$a_{1},a_{2},\ldots,a_{k}$. Define the $k\times k$-matrix%
\begin{equation}
A:=\left(
\begin{array}
[c]{ccccc}%
a_{1} & a_{2} & \cdots & a_{k-1} & a_{k}\\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \cdots & 1 & 0
\end{array}
\right)  \label{eq.mot.fib.linrec.A=}%
\end{equation}
(where each of the $2$-nd, $3$-rd, $\ldots$, $k$-th rows has just a single $1$
surrounded by zeroes, and these $1$'s are situated in columns $1,2,\ldots,k-1$
from top to bottom).

For each $n\geq0$, define the vector%
\[
v_{n}:=\left(
\begin{array}
[c]{c}%
x_{n+k-1}\\
x_{n+k-2}\\
\vdots\\
x_{n}%
\end{array}
\right)  ;
\]
this consists of $k$ consecutive entries of our sequence, ending with $x_{n}$.
Then, our recurrence equation (\ref{eq.def.mot.fib.linrec.linrec}) yields%
\begin{equation}
v_{n}=Av_{n-1}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq1,
\label{eq.mot.fib.linrec.vecrec}%
\end{equation}
since%
\begin{align*}
v_{n}  &  =\left(
\begin{array}
[c]{c}%
x_{n+k-1}\\
x_{n+k-2}\\
\vdots\\
x_{n}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
a_{1}x_{n+k-2}+a_{2}x_{n+k-3}+\cdots+a_{k}x_{n-1}\\
x_{n+k-2}\\
\vdots\\
x_{n}%
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we rewrote the
topmost entry using (\ref{eq.def.mot.fib.linrec.linrec})}\right) \\
&  =\underbrace{\left(
\begin{array}
[c]{ccccc}%
a_{1} & a_{2} & \cdots & a_{k-1} & a_{k}\\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & \cdots & 1 & 0
\end{array}
\right)  }_{=A}\underbrace{\left(
\begin{array}
[c]{c}%
x_{n+k-2}\\
x_{n+k-3}\\
\vdots\\
x_{n-1}%
\end{array}
\right)  }_{=v_{n-1}}=Av_{n-1}.
\end{align*}


Applying this equality repeatedly, we find%
\[
v_{n}=A\underbrace{v_{n-1}}_{=Av_{n-2}}=AA\underbrace{v_{n-2}}_{=Av_{n-3}%
}=AAAv_{n-3}=\cdots=A^{n}v_{0}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq0.
\]
Thus,
\begin{equation}
v_{n}=A^{n}v_{0}\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq0.
\label{eq.mot.fib.linrec.indrec}%
\end{equation}
And this makes it easy to compute $v_{n}$ (and therefore $x_{n}$, which is the
last entry of $v_{n}$) by taking the matrix $A$ to the $n$-th power. This is
actually a useful observation in computing high values of linearly recurrent
sequences, since powers of matrices can be computed using fairly efficient
algorithms
(\href{https://en.wikipedia.org/wiki/Exponentiation_by_squaring}{exponentiation
by squaring}).

But we are interested in the sequence $\left(  x_{0},x_{m},x_{2m}%
,x_{3m},\ldots\right)  $. This sequence consists of the last entries of the
vectors $v_{0},v_{m},v_{2m},v_{3m},\ldots$ (since each $x_{n}$ is the last
entry of $v_{n}$). We want to know whether it is linearly recurrent of order
$k$. Thus we want to find linear relations between the vectors $v_{0}%
,v_{m},v_{2m},v_{3m},\ldots$. Using (\ref{eq.mot.fib.linrec.indrec}), we can
rewrite these vectors as $A^{0}v_{0},\ A^{m}v_{0},\ A^{2m}v_{0},\ A^{3m}%
v_{0},\ \ldots$. Therefore, if we have a linear relation between the matrices
$A^{0},A^{m},A^{2m},A^{3m},\ldots$, then (by multiplying with the vector
$v_{0}$) we will obtain a linear relation between the vectors $v_{0}%
,v_{m},v_{2m},v_{3m},\ldots$, and then (by taking the last entries) also
between the numbers $x_{0},x_{m},x_{2m},x_{3m},\ldots$.

A well-known source of linear relations between powers of a matrix is the
\textbf{Cayley--Hamilton theorem}. This theorem (which is proved in any good
linear algebra text) says that if $B$ is any $k\times k$-matrix, and if
$\chi_{B}\left(  t\right)  :=\det\left(  tI_{k}-B\right)  $ is its
characteristic polynomial\footnote{Some authors define the characteristic
polynomial to be $\det\left(  B-tI_{k}\right)  $ instead. This differs from
our definition only by a sign.} (in the indeterminate $t$), then
\begin{equation}
\chi_{B}\left(  B\right)  =0. \label{eq.mot.fib.CH}%
\end{equation}
Recall that $\chi_{B}$ is always a monic polynomial of degree $k$, that is, a
polynomial of the form $\chi_{B}\left(  t\right)  =t^{k}-c_{1}t^{k-1}%
-c_{2}t^{k-2}-\cdots-c_{k}t^{0}$ for some numbers $c_{1},c_{2},\ldots,c_{k}$.
Writing $\chi_{B}$ in this form, we can rewrite (\ref{eq.mot.fib.CH}) as%
\begin{equation}
B^{k}-c_{1}B^{k-1}-c_{2}B^{k-2}-\cdots-c_{k}B^{0}=0, \label{eq.mot.fib.CH-1}%
\end{equation}
that is, as%
\begin{equation}
B^{k}=c_{1}B^{k-1}+c_{2}B^{k-2}+\cdots+c_{k}B^{0}. \label{eq.mot.fib.CH-2}%
\end{equation}


We can apply this to $B=A^{m}$, and obtain%
\begin{equation}
A^{mk}=c_{1}A^{m\left(  k-1\right)  }+c_{2}A^{m\left(  k-2\right)  }%
+\cdots+c_{k}A^{0},\label{eq.mot.fib.CHAm-1}%
\end{equation}
where $c_{1},c_{2},\ldots,c_{k}$ are constant numbers such that $\chi_{A^{m}%
}\left(  t\right)  =t^{k}-c_{1}t^{k-1}-c_{2}t^{k-2}-\cdots-c_{k}t^{0}$ (that
is, $c_{1},c_{2},\ldots,c_{k}$ are the first $k$ coefficients\footnote{By
\textquotedblleft the first $k$ coefficients\textquotedblright, we mean the
coefficients of the monomials $t^{0},t^{1},\ldots,t^{k-1}$.} of $\chi_{A^{m}%
}\left(  t\right)  $, multiplied by $-1$). Hence, for each $n\geq mk$, we have%
\begin{equation}
A^{n}=c_{1}A^{n-m}+c_{2}A^{n-2m}+\cdots+c_{k}A^{n-km}\label{eq.mot.fib.CHAm-2}%
\end{equation}
(this follows from (\ref{eq.mot.fib.CHAm-1}) by multiplying both sides by
$A^{n-mk}$). Multiplying both sides of (\ref{eq.mot.fib.CHAm-2}) by $v_{0}$,
we obtain%
\[
A^{n}v_{0}=c_{1}A^{n-m}v_{0}+c_{2}A^{n-2m}v_{0}+\cdots+c_{k}A^{n-km}v_{0}.
\]
In view of (\ref{eq.mot.fib.linrec.indrec}), we can rewrite this as%
\[
v_{n}=c_{1}v_{n-m}+c_{2}v_{n-2m}+\cdots+c_{k}v_{n-km}.
\]
Taking the last entries of all the vectors in this equality, we find%
\begin{equation}
x_{n}=c_{1}x_{n-m}+c_{2}x_{n-2m}+\cdots+c_{k}x_{n-km}\label{eq.mot.fib.CHAm-6}%
\end{equation}
(since the last entry of each vector $v_{i}$ is $x_{i}$).

Thus, we have proved that each $n\geq km$ satisfies the equality
(\ref{eq.mot.fib.CHAm-6}). That is, we have proved the following:

\begin{theorem}
\label{thm.mot.fib.linrec.xnm}Let $a_{1},a_{2},\ldots,a_{k}$ be $k$ numbers.
Let $\left(  x_{0},x_{1},x_{2},\ldots\right)  $ be a linearly recurrent
sequence of order $k$ with coefficients $a_{1},a_{2},\ldots,a_{k}$. Let
$m\in\mathbb{N}$. Define a $k\times k$-matrix $A$ by
(\ref{eq.mot.fib.linrec.A=}). Let $c_{1},c_{2},\ldots,c_{k}$ be the first $k$
coefficients of the characteristic polynomial $\chi_{A^{m}}\left(  t\right)
$, multiplied by $-1$. Then, we have%
\begin{equation}
x_{n}=c_{1}x_{n-m}+c_{2}x_{n-2m}+\cdots+c_{k}x_{n-km}%
\ \ \ \ \ \ \ \ \ \ \text{for each }n\geq km.
\label{eq.thm.mot.fib.linrec.xnm.rec}%
\end{equation}
In particular, the sequence $\left(  x_{0},x_{m},x_{2m},x_{3m},\ldots\right)
$ is linearly recurrent of order $k$ with coefficients $c_{1},c_{2}%
,\ldots,c_{k}$; more generally, for each $i\in\mathbb{N}$, the sequence
$\left(  x_{i},x_{i+m},x_{i+2m},x_{i+3m},\ldots\right)  $ is linearly
recurrent with the same coefficients.
\end{theorem}

For example, let us apply this theorem to the Fibonacci sequence $\left(
f_{0},f_{1},f_{2},\ldots\right)  $ and to $m=2$. Here, we have $k=2$ and
$a_{1}=1$ and $a_{2}=1$, so that $A=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 0
\end{array}
\right)  $. Thus, $A^{2}=\left(
\begin{array}
[c]{cc}%
2 & 1\\
1 & 1
\end{array}
\right)  $, which has characteristic polynomial%
\[
\chi_{A^{2}}\left(  t\right)  =\det\left(  tI_{2}-\left(
\begin{array}
[c]{cc}%
2 & 1\\
1 & 1
\end{array}
\right)  \right)  =\det\left(
\begin{array}
[c]{cc}%
t-2 & -1\\
-1 & t-1
\end{array}
\right)  =t^{2}-3t+1.
\]
Thus, when we apply Theorem \ref{thm.mot.fib.linrec.xnm}, we must set
$c_{1}=3$ and $c_{2}=-1$ (corresponding to the coefficients $-3$ and $1$ of
$\chi_{A^{2}}\left(  t\right)  $). Hence, (\ref{eq.thm.mot.fib.linrec.xnm.rec}%
) says that%
\[
f_{n}=3f_{n-2}+\left(  -1\right)  f_{n-4}\ \ \ \ \ \ \ \ \ \ \text{for each
}n\geq2\cdot2.
\]
This is precisely the equality (\ref{eq.mot.fib.fibrec2}) that we found above.
Likewise, we can find (\ref{eq.mot.fib.fibrec3}) and similar recurrences for
higher values of $m$. More generally, this shows that the answer to Question
\ref{quest.mot.fib.linrec-k} is: Yes, all these sequences are linearly
recurrent -- even better: linearly recurrent, all with the same coefficients
--, and yes, these coefficients can be computed.

\subsubsection{Computing $\chi_{A^{m}}$ from $\chi_{A}$ ?}

Now let us try to improve on the above. Theorem \ref{thm.mot.fib.linrec.xnm}
gives \textbf{some} algorithm for computing the coefficients $c_{1}%
,c_{2},\ldots,c_{k}$ in terms of $a_{1},a_{2},\ldots,a_{k}$, but this
algorithm is rather cumbersome: first, form the matrix $A$; then take its
power $A^{m}$; then expand its characteristic polynomial $\chi_{A^{m}}$. Isn't
there an easier way?

A general fact in linear algebra says that the characteristic polynomial
$\chi_{A^{m}}$ (for an arbitrary $k\times k$-matrix $A$, not necessarily the
one given by (\ref{eq.mot.fib.linrec.A=})\footnote{For the matrix $A$ given in
(\ref{eq.mot.fib.linrec.A=}), the situation is particularly nice, since we can
show that
\[
\chi_{A}\left(  t\right)  =t^{k}-a_{1}t^{k-1}-a_{2}t^{k-2}-\cdots-a_{k}t^{0}%
\]
(exercise!), which renders the computation of $\chi_{A}$ trivial. But
everything we will be doing below applies to any $k\times k$-matrix $A$.}) is
completely determined by $m$ and $\chi_{A}$; we don't have to know the matrix
$A$ itself. The easiest way to see this is using eigenvalues: If the $k\times
k$-matrix $A$ has eigenvalues $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$
(listed with algebraic multiplicity), then $A^{m}$ has eigenvalues
$\lambda_{1}^{m},\lambda_{2}^{m},\ldots,\lambda_{k}^{m}$ (by the spectral
mapping theorem\footnote{If you don't know this fact, prove it, e.g., using
\href{https://en.wikipedia.org/wiki/Schur_decomposition}{Schur triangulation}%
.}). The eigenvalues $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ of $A$ are
precisely the roots of $\chi_{A}$, so that we have the polynomial identity%
\begin{equation}
\chi_{A}\left(  t\right)  =\left(  t-\lambda_{1}\right)  \left(  t-\lambda
_{2}\right)  \cdots\left(  t-\lambda_{k}\right)
,\label{eq.mot.fib.chiA=roots}%
\end{equation}
which shows that $\chi_{A}$ both determines and is determined by $\lambda
_{1},\lambda_{2},\ldots,\lambda_{k}$. Hence,%
\begin{equation}
\chi_{A^{m}}\left(  t\right)  =\left(  t-\lambda_{1}^{m}\right)  \left(
t-\lambda_{2}^{m}\right)  \cdots\left(  t-\lambda_{k}^{m}\right)
,\label{eq.mot.fib.chiAm=roots}%
\end{equation}
since $A^{m}$ has eigenvalues $\lambda_{1}^{m},\lambda_{2}^{m},\ldots
,\lambda_{k}^{m}$. We can theoretically use this formula to compute
$\chi_{A^{m}}$ from $\chi_{A}$, since $\lambda_{1},\lambda_{2},\ldots
,\lambda_{k}$ are the roots of $\chi_{A}$.

But this is even less of a practically useful algorithm than the one in
Theorem \ref{thm.mot.fib.linrec.xnm}, because the roots of a general
polynomial cannot be computed algebraically! (At least not in the naive
\textquotedblleft quadratic formula\textquotedblright\ sense; see
\href{https://en.wikipedia.org/wiki/Abel-Ruffini_theorem}{the Abel--Ruffini
theorem}.) Numerical algorithms exist, but we don't want to lose precision.

So what we ideally want is a way to find the coefficients of $\chi_{A^{m}}$ in
terms of the coefficients of $\chi_{A}$ without having to factor $\chi_{A}$.
To this aim, we first expand (\ref{eq.mot.fib.chiA=roots}):%
\begin{align}
\chi_{A}\left(  t\right)   &  =\left(  t-\lambda_{1}\right)  \left(
t-\lambda_{2}\right)  \cdots\left(  t-\lambda_{k}\right) \nonumber\\
&  =t^{k}-\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\right)
t^{k-1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \lambda_{1}\lambda_{2}+\lambda
_{1}\lambda_{3}+\cdots+\lambda_{k-1}\lambda_{k}\right)  }_{=\sum_{i<j}%
\lambda_{i}\lambda_{j}}t^{k-2}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\left(  \lambda_{1}\lambda_{2}\lambda
_{3}+\lambda_{1}\lambda_{2}\lambda_{4}+\cdots+\lambda_{k-2}\lambda
_{k-1}\lambda_{k}\right)  }_{=\sum_{i<j<\ell}\lambda_{i}\lambda_{j}%
\lambda_{\ell}}t^{k-3}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \pm\cdots\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  -1\right)  ^{k}\lambda_{1}\lambda_{2}%
\cdots\lambda_{k}\nonumber\\
&  =\sum_{i=0}^{k}\left(  -1\right)  ^{i}e_{i}\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{k}\right)  t^{k-i}, \label{eq.mot.fib.chiA=sumei}%
\end{align}
where we set
\begin{align*}
e_{i}\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)   &
:=\sum_{1\leq u_{1}<u_{2}<\cdots<u_{i}\leq k}\lambda_{u_{1}}\lambda_{u_{2}%
}\cdots\lambda_{u_{i}}\\
&  =\sum_{\substack{L\subseteq\left[  k\right]  ;\\\left\vert L\right\vert
=i}}\ \ \prod_{\ell\in L}\lambda_{\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{where
}\left[  k\right]  :=\left\{  1,2,\ldots,k\right\}  \right)
\end{align*}
(in particular, $e_{k}\left(  \lambda_{1},\lambda_{2},\ldots,\lambda
_{k}\right)  =\lambda_{1}\lambda_{2}\cdots\lambda_{k}$ and $e_{0}\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  =1$, because the empty
product is $1$). This latter number $e_{i}\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{k}\right)  $ is called the $i$\textbf{-th elementary
symmetric polynomial} in the $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$. So
the formula (\ref{eq.mot.fib.chiA=sumei}) says that the coefficients of
$\chi_{A}$ are precisely these $e_{i}\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  $, up to sign.

Therefore, in order to compute $\chi_{A^{m}}$ from $\chi_{A}$, we just need to
compute the $e_{i}\left(  \lambda_{1}^{m},\lambda_{2}^{m},\ldots,\lambda
_{k}^{m}\right)  $ from the $e_{i}\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  $. This is best done algebraically, without
solving for the $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$. To do so, we
forget about what the $\lambda_{i}$s are, and just study the elementary
symmetric polynomials of any $k$ inputs.

Before we do this systematically, let us do the cases $\left(  m,i\right)
=\left(  2,1\right)  $ and $\left(  m,i\right)  =\left(  2,2\right)  $ by
hand. So we want formulas for $e_{1}\left(  \lambda_{1}^{2},\lambda_{2}%
^{2},\ldots,\lambda_{k}^{2}\right)  $ and $e_{2}\left(  \lambda_{1}%
^{2},\lambda_{2}^{2},\ldots,\lambda_{k}^{2}\right)  $. The former is easy:%
\begin{align*}
&  e_{1}\left(  \lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{k}^{2}\right)
\\
&  =\lambda_{1}^{2}+\lambda_{2}^{2}+\cdots+\lambda_{k}^{2}\\
&  =\left(  \underbrace{\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}}%
_{=e_{1}\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  }\right)
^{2}-2\underbrace{\sum_{i<j}\lambda_{i}\lambda_{j}}_{=e_{2}\left(  \lambda
_{1},\lambda_{2},\ldots,\lambda_{k}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the well-known formula}\\
\left(  \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\right)  ^{2}=\lambda
_{1}^{2}+\lambda_{2}^{2}+\cdots+\lambda_{k}^{2}+2\sum_{i<j}\lambda_{i}%
\lambda_{j}%
\end{array}
\right)  \\
&  =\left(  e_{1}\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)
\right)  ^{2}-2e_{2}\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)
.
\end{align*}
Computing $e_{2}\left(  \lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{k}%
^{2}\right)  $ is harder with this method, however.

\begin{noncompile}%
\begin{align*}
e_{2}\left(  \lambda_{1}^{2},\lambda_{2}^{2},\ldots,\lambda_{k}^{2}\right)
&  =\sum_{i<j}\lambda_{i}^{2}\lambda_{j}^{2}\\
&  =\left(  \sum_{i<j}\lambda_{i}\lambda_{j}\right)  ^{2}-2\sum
_{\substack{i<j;\\u<v;\\\left(  i,j\right)  <\left(  u,v\right)  \text{
lexicographically}}}\lambda_{i}\lambda_{j}\lambda_{u}\lambda_{v}\\
&  =?
\end{align*}
(we want a polynomial in the $e_{i}\left(  \lambda_{1},\lambda_{2}%
,\ldots,\lambda_{k}\right)  $'s).
\end{noncompile}


\end{document}