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\newtheorem*{defn}{Definition}
\newtheorem*{ex}{Example}
\newtheorem*{fex}{For example:}

\def\be{\mathbf{e}}\def\bh{\mathbf{h}}\def\bph{\mathbf{\phi}}\def\bps{\mathbf\psi}
\def\bE{\mathbf{E}}\def\bH{\mathbf{H}}\def\bPh{\mathbf{\Phi}}\def\bPs{\mathbf\Psi}


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\begin{document}
\title{Quasisymmetric power sums}
\author{{\large Zajj Daugherty}\\~\\\includegraphics[height=.4in]{ccny}\\~\\Joint work with\\Cristina Ballantine, Angela Hicks,\\ Sarah Mason, and Elizabeth Niese\\\includegraphics[height=1in]{cballantine}\includegraphics[height=1in]{msri}}
\date{}




\begin{frame}
\maketitle
\end{frame}

\begin{frame}
\centerline{\includegraphics[width=5in]{banff}}
\end{frame}


\begin{frame}{Some combinatorics}



{\begin{minipage}{2in}
\alert{Partitions}: 
$$\PART{5,4,4,2} = (5,4,4,2) = \lambda$$
\end{minipage}
\begin{minipage}{2in}\alert{Compositions}:
$$\COMP{4,2,5,4} = (4,2,5,4) = \alpha$$
\end{minipage}}


\medskip


\pause 

For a composition $\alpha$,  
\begin{enumerate}[~]
\item let $|\alpha|$ be the size ($\#$ boxes) of $\alpha$;
\item let $\ell(\alpha)$ be the length ($\#$ parts) of $\alpha$; and 
\item let $\tilde{\alpha}$ be the rearrangement of the parts of $\alpha$ into decreasing order. 
\end{enumerate}
For example, $|\alpha| =15$, $\ell(\alpha) =4$, and $\tilde{\alpha} = \lambda$. 


\medskip

\pause For compositions $\alpha$ and $\beta$, we say $\alpha$ \alert{refines} $\beta$, written $\alpha  \cleq \beta$, if $\beta$ can be built by combining adjacent parts of $\alpha$. For example, 

\smallskip
\centerline{$
\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} \cleq \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}, \qquad\qquad \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (3,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[green!50] (0,3) rectangle (5,4);
	\filldraw[blue!50] (0,4) rectangle (4,5);
	\Comp{3,2,1,5,4}
	} \not\cleq 
	\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
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	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}
$}


\end{frame}




\begin{frame}
Let $\Sym$ be the ring of symmetric functions.

\medskip

\alert{\large Many favorite bases indexed by  integer partitions $\lambda \vdash n$:} 


\alert{$\bullet$} Monomial symmetric functions
\\ \smallskip
\centerline{$\displaystyle
m_\lambda = 
	\sum_{\tilde{\alpha} = \lambda \atop i_1 < i_2 < \cdots < i_\ell} 
	x_{i_1}^{\alpha_1} x_{i_2}^{\alpha_2} \cdots x_{i_\ell}^{\alpha_\ell}
$} \smallskip	
 

\alert{$\bullet$} Complete homogeneous symmetric functions
\\ \smallskip
\centerline{$\displaystyle
h_r = 
	\sum_{|\alpha| = r \atop i_1 < i_2 < \cdots < i_\ell} 
	x_{i_1}^{\alpha_1} x_{i_2}^{\alpha_2} \cdots x_{i_\ell}^{\alpha_\ell}
	= \sum_{|\lambda| = r} m_\lambda, \qquad h_\lambda = h_{\lambda_1} h_{\lambda_2} \cdots.
$} \smallskip 

\alert{$\bullet$} Elementary symmetric functions \\ \smallskip
\centerline{$\displaystyle
e_r = \sum_{1\leq i_1 < i_2< \cdots < i_r} x_{i_1} \cdots x_{i_r} = m_{(1,1,\dots,1)}  \qquad e_\lambda = e_{\lambda_1} e_{\lambda_2} \cdots.
$} \smallskip  





\alert{$\star$} Power sum symmetric functions
\\ \smallskip
\centerline{$\displaystyle
p_r = x_1^r + x_2^r + \cdots = m_{(r)}, \qquad p_\lambda = p_{\lambda_1} p_{\lambda_2} \cdots
$} \smallskip


\end{frame}



\begin{frame}

We have a \alert{scalar product} $\<,\>: \Sym \otimes \Sym \to \CC$ defined by \\ \smallskip
\centerline{$\displaystyle
\<h_\lambda, m_\mu\> = \delta_{\lambda, \mu},
$} \smallskip
so that the homogeneous and monomial functions are dual. \pause 
Under this pairing, we have  
\\ \smallskip
\centerline{$\displaystyle
\<p_\lambda, p_\mu\> = z_\lambda \delta_{\lambda \mu},
$} \smallskip
where $z_\lambda$ is the size of the stabilizer of a permutation of cycle type $\lambda$:
\\ \medskip
\centerline{$\displaystyle
z_\lambda =  \prod_k a_k! k^{a_k}, \quad a_k = \#\{\text{pts of length } k\}
\qquad  
\text{\alert{Ex: }} z_{\PART{3,3}} = 2! \ 3^2.
$}  

\pause 
\alert{Generating functions:}

 \smallskip
\centerline{$\displaystyle
H(t) = \sum_{k \geq 0} h_k t^k = \prod_{i \geq 1}(1-x_i t)^{-1}
$} \smallskip
\centerline{$\displaystyle
E(t) = \sum_{k \geq 0} e_k t^k = \prod_{i \geq 1}(1+x_i t)$$
$} \smallskip
Note $H(t) = 1/E(-t)$. 

 
 \smallskip
\centerline{$\displaystyle
P(t) = \sum_{k \geq 0} p_k t^k = \frac{d}{dt} \ln(H(t)) = \frac{d}{dt} \ln(1/E(-t))
$} 



\end{frame}



%\begin{frame}
%The ring of \alert{noncommutative symmetric functions} $\NSym$ is the $\CC$-algebra generated by the free group on $\be_1, \be_2, \dots$. 
%
%
%\pause \bigskip
%
%
%\alert{Think:} The elementary symmetric functions $e_1, e_2, \dots$ generate $\Sym$, and, aside from commuting, are algebraically independent. Now, we're lifting to an algebra where the elementary functions no longer commute. So the abelianization
%$$\Ab: \NSym \to \Sym$$
%is surjective (with kernel generated by commutators). 
%\end{frame}
%


\begin{frame}

The ring of \alert{noncommutative symmetric functions} $\NSym$ is the $\CC$-algebra generated freely by $\be_1, \be_2, \dots$. \pause 

 \medskip


Analogous bases indexed by compositions $\alpha$. 
\begin{enumerate}[$\bullet$]
\item Noncommutative elementary: \alert{$\be_\alpha = \be_{\alpha_1} \cdots \be_{\alpha_\ell}$}.
\hfill \pause
	$\Ab(\be_\alpha) = e_{\tilde{\alpha}}$
	\pause
\item  Noncom.\ homog.: \alert{$\bh_\alpha= \bh_{\alpha_1} \cdots \bh_{\alpha_\ell}$}, where $\bh_i$ is defined by\dots
$$\text{if } \quad \bE(t) = \sum_{k \geq 0} \be_k t^k \quad \text{ and } \quad \bH(t) = \sum_{k \geq 0} \bh_k t^k,$$
then $\bH(t) = 1/\bE(-t)$. \hfill {(Recall: $H(t) = 1/E(-t)$ in $\Sym$)}.\\
		\centerline{$\Ab(\bh_\alpha) = h_{\tilde{\alpha}}$}\pause
\item[$\star$] Noncommutative power sums: two choices, $\bps$ and $\bph$! \pause
{\def\arraystretch{1.2}
\centerline{\begin{tabular}{lc@{\quad}c}
&\alert{In $\Sym$:} & \alert{In $\NSym$:}\\
\alert{Type 1:} & $P(t) = \frac{d}{dt} \ln(H(t))$  & $\frac{d}{dt} \bH(t) = \bH(t) \bPs(t)$ \\
{\alert{Type 2:}} 
	& {$H(t) = \exp\left(\int P(t) dt\right)$} 
	&  {$\bH(t) = \exp\left(\int \bPh(t)dt\right)$}
\end{tabular}}}



Not the same! (No unique notion of log derivative for power series with noncommutative coefficients.) But \\
\centerline{$\Ab(\bps_\alpha) = p_{\tilde{\alpha}} = \Ab(\bph_\alpha)$}



\end{enumerate}







\end{frame}



\begin{frame}

The ring of \alert{quasisymmetric functions} $\QSym$ is a subring of $\CC[\![x_1, x_2, \dots]\!]$ consisting of series where the coefficients on the monomials
$$x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_\ell^{\alpha_\ell} \quad \text{ and } \quad 
	x_{i_1}^{\alpha_1} x_{i_2}^{\alpha_2} \cdots x_{i_\ell}^{\alpha_\ell}
$$
are the same, for all $i_1 < i_2 < \cdots < i_\ell$.  In particular, $\Sym \subset \QSym$. \pause 

\medskip

For example, \\ \smallskip
\centerline{$\displaystyle
\sum_{i < j} x_i x_j^2 = x_1 x_2^2 + x_1 x_3^2 + x_2 x_3^2 + \cdots \uncover<5->{= M_{\COMP{1,2}}}
$} \smallskip
is quasisymmetric but not symmetric (the coef.\ on $x_1^2 x_2$ is $0$). \pause 

\bigskip

Bases of $\QSym$ are also indexed by compositions. \pause Namely, the \alert{monomial basis} has a natural analog: \\ \smallskip
\centerline{$\displaystyle
M_\alpha = \sum_{i_1 < i_2 < \cdots < i_{\ell(\alpha)}} x_{i_1}^{\alpha_1} x_{i_2}^{\alpha_2} \cdots x_{i_\ell}^{\alpha_\ell}, \quad \text{so that} \quad m_\lambda = \sum_{\tilde{\alpha} = \lambda} M_\alpha.
$} \smallskip 
%\pause \pause 
%Extending linearly gives a natural surjective map $\QSym \to \Sym$.


\end{frame}




\begin{frame}

The rings $\NSym$ and $\QSym$ have Hopf algebra structures that are dual to each other under a pairing\\ \smallskip
\centerline{$\displaystyle
\<,\>: \NSym \otimes \QSym \to \CC
$} 
analogous to the pairing in $\Sym$. \pause Namely\\ \medskip
\centerline{$\displaystyle
\<h_\lambda, m_\mu\> = \delta_{\lambda, \mu} \quad \text{ in } \Sym \otimes \Sym,
$}\medskip
\centerline{$\displaystyle
\<\bh_\alpha, M_\beta\>= \delta_{\alpha, \beta} \quad \text{ in } \NSym \otimes \QSym.
$} \medskip

\pause 

Recall in  $\Sym$,  the power sum basis is (essentially) self-dual:
$$\<p_\lambda, p_\mu\> = z_\lambda \delta_{\lambda \mu}$$

\pause 


\alert{Question:} What is dual to $\psi$ in $\QSym$? to $\phi$? 


\end{frame}





\begin{frame}{Type 1}
In $\Sym$ the power sum basis is (essentially) self-dual:\\ \smallskip
\centerline{$\displaystyle
\<p_\lambda, p_\mu\> = z_\lambda \delta_{\lambda \mu}
$} \smallskip  

In  $\NSym$, the \alert{type 1 power sum basis} $\psi$ is defined by the generating function relation\\ \smallskip
\centerline{$\displaystyle
\frac{d}{dt} \bH(t) = \bH(t) \bPs(t).
$} \smallskip\pause 

This is equivalent to
$$\bh_\alpha = \sum_{\beta \cleq \alpha} \frac{1}{\pi(\beta,\alpha)}{\psi}_\beta,$$
where $\pi(\beta,\alpha)$ is a combinatorial statistic on the refinement $\beta \cleq \alpha$. 

\pause 


So, the dual in $\QSym$ will satisfy\\ \smallskip
\centerline{$\displaystyle
\psi^*_\alpha = \sum_{\beta \cgeq \alpha} \frac{1}{\pi(\alpha,\beta)} M_\beta.
$} \smallskip

\pause 


Define
$$\alert{\Psi_\alpha = z_{\tilde{\alpha}} \psi^*_\alpha}, \quad \text{ so that } \quad 
	\<\psi_\alpha, \Psi_\beta\> = z_{\tilde{\alpha}} \delta_{\alpha \beta}.$$

\end{frame}















\begin{frame}{Computing coefficients}

\centerline{$\displaystyle
{\Psi_\alpha = z_{\tilde{\alpha}} \sum_{\beta \cgeq \alpha} \frac{1}{\pi(\alpha,\beta)} M_\beta.}
$} \smallskip

\pause 

 \def\PartUNIT{.25}
 For example, we saw that 
\centerline{$
\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} \quad \cleq \quad \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}
$}


\pause 

First, for each block, we compute the product of the partial sums:
\centerline{$
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right) = \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\Comp{1}
	}\right| \cdot \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\Comp{1,2}
	}\right| \cdot \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right|   = 1 \cdot 3 \cdot 4$}


\pause 

Then, for $\alpha$ refining $\beta$, the coefficient of $M_{\beta}$ in $\psi_\alpha^*$ is $1/\pi(\alpha, \beta)$, where 

\vspace{-.25in}

\begin{align*}
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} , \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}\right) & =
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[yellow!50] (0,0) rectangle (2,1);
	\Comp{2}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[green!50] (0,0) rectangle (5,1);
	\Comp{5}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[blue!50] (0,0) rectangle (1,2);
	\filldraw[blue!50] (0,2) rectangle (2,3);
	\Comp{1,1,2}
	}\right)\\
	&= (1 \cdot 3 \cdot 4) (2)(5)(1 \cdot 2 \cdot4) 
\end{align*}
\end{frame}


\begin{frame}{Computing coefficients}
 \def\PartUNIT{.2}
\only<1-2>{
First, for each block, we compute the product of the partial sums:
\centerline{$
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right) = \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\Comp{1}
	}\right| \cdot \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\Comp{1,2}
	}\right| \cdot \left|\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right|   = 1 \cdot 3 \cdot 4$}


Then, for $\alpha$ refining $\beta$, the coefficient of $M_{\beta}$ in $\psi_\alpha^*$ is $1/\pi(\alpha, \beta)$, where 

\vspace{-.25in}

\begin{align*}
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} , \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}\right) & =
\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[yellow!50] (0,0) rectangle (2,1);
	\Comp{2}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[green!50] (0,0) rectangle (5,1);
	\Comp{5}
	}\right)\pi\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[blue!50] (0,0) rectangle (1,2);
	\filldraw[blue!50] (0,2) rectangle (2,3);
	\Comp{1,1,2}
	}\right)
\end{align*}}


As another example,  $z_{\COMP{2,1}} = 2$, 
$$\Psi_{\CCOMP{1/red,2/yellow}} = z_{\CCOMP{2/yellow,1/red}}\psi_{\CCOMP{1/red,2/yellow}}^* 
	= 2\left(\frac{1}{2} M_{\CCOMP{1/red,2/yellow}} + \frac{1}{3} M_{\CCOMP{3/orange}}\right),$$
$$\Psi_{\CCOMP{2/yellow,1/red}} = z_{\CCOMP{2/yellow,1/red}}\psi_{\CCOMP{2/yellow,1/red}}^*
	= 2\left(\frac{1}{2} M_{\CCOMP{2/yellow,1/red}} + \frac{1}{6} M_{\CCOMP{3/orange}}\right).$$

\pause

So 
\begin{align*}
\Psi_{\COMP{1,2}} + \Psi_{\COMP{2,1}} &=  \alert<3>{M_{\COMP{1,2}} + M_{\COMP{2,1}}} + M_{\COMP{3}} \\
	&\uncover<3->{= \alert<3>{m_{\COMP{2,1}}} + m_{\COMP{3}}} \uncover<3->{= m_{\COMP{2}} m_{\COMP{1}}= p_{2} p_{1} = p_{\COMP{2,1}}.}
\end{align*}

\pause\pause\pause \vspace{-.2in}

\begin{thm}[BDHMN]
Type 1 $\QSym$ power sums refine $\Sym$ power sums:
$$p_\lambda = \sum_{\tilde{\alpha} = \lambda} \Psi_\alpha.$$

\end{thm}

\end{frame}

\begin{frame}
\alert{\text{Theorem:}}  $\displaystyle p_\lambda = \sum_{\tilde{\alpha} = \lambda} \Psi_\alpha, \quad \text{where} \quad  \Psi_\alpha = z_{\tilde{\alpha}} \sum_{ \alpha \cleq \beta} \frac{1}{\pi(\alpha,\beta)} M_\beta.$

\medskip

\alert{Proof outline:} For compositions $\alpha$ and $\beta$, define $\cO_{\alpha,\beta}$ be the set of ordered set partitions $(B_1,\cdots, B_{\ell(\beta)})$ of $\{1,\cdots,\ell(\alpha)\}$ satisfying 
$$\beta_j=\sum_{i\in B_j}\alpha_i \text{ for } 1\leq j\leq \ell(\beta).$$

\pause \only<1-3>{For example, 
if 
$$\alpha = \COMP{1,2,1} \quad \text{and} \quad \beta = \COMP{2,2},$$
then
$\cO_{\alpha,\beta}$ contains $(\{1,3\}, \{2\})$ and $(\{2\}, \{1,3\})$.}


\pause \medskip

It has been shown that 
$$p_\lambda = \sum_{\text{part'n }\mu} |\cO_{\lambda,\mu}| m_\mu, \quad \text{ so that } \quad 
	p_\lambda = \sum_{\text{comp }\beta} |\cO_{\lambda,\beta}| M_\beta.$$
	
	\pause
 \only<4->{We combinatorially prove, for a fixed partition $\lambda$ with size $n$, and a fixed composition $\beta$, that \\ \smallskip
\centerline{$\displaystyle
\uncover<5->{|\cO_{\lambda\beta}| \cdot \alert<5>{|S_n^\lambda|} = }|\cO_{\lambda\beta}|\alert<5>{\frac{n!}{z_{\lambda}}}=\sum_{\substack{\alpha \cleq \beta \\ \tilde{\alpha}=\lambda}}\frac{n!}{\pi(\alpha, \beta)},
$} \smallskip
\pause where $S_n^\lambda = \{ \sigma \in S_n \text{ of cycle type } \lambda\}$.
}	

\end{frame}


\begin{frame}
\only<1-6>{Two ways of thinking about permutations:}
\begin{itemize}
\only<1-6>{\item In \alert{one-line notation}:\\ \smallskip
\centerline{$\displaystyle
\sigma = 571423689
$} \smallskip
is the permutation sending \\ \smallskip
\centerline{$\displaystyle
1 \mapsto 5, \ 2 \mapsto 7, \ 3 \mapsto 1, \text{ and so on\dots}
$} }\pause 
\item In \alert{cycle notation}:\\ \smallskip
\centerline{$\displaystyle
\sigma = (152763)(4)(8)(9).
$} \smallskip\pause 
Several equivalent ways to write in cycle notation. We say $\sigma$ is written in \alert{standard form} if \\
\quad  largest element of each cycle is last, and \\
\quad cycles ordered increasingly according to  largest element\\
 \hfill \alert{Ex:} $(\alert{4})(63152\alert{7})(\alert{8})(\alert{9})$

\pause 
\item Let $\alpha \cleq \beta$ of size $n$, and let $\sigma \in S_n$. We say $\sigma$ is \alert{consistent} with $\alpha \cleq \beta$ if\dots\pause 

\smallskip

\alert{Ex:} let $\alpha = (1,1,2,1,3,1)$ and $\beta = (2,2,5)$\\\pause 
Start in one-line notation: \hfill $571423689$\\\pause
Split according to $\beta$: \hfill $57|\!|14|\!|23689$\\\pause 
Add parentheses according to $\alpha$: \hfill $(\alert<9->{5})(\alert<9->{7})|\!|(1\alert<9->{4})|\!|(\alert<9->{2})(36\alert<9->{8})(\alert<9->{9})$\pause 

\smallskip

If the permutations in each partition are in standard form, then $\sigma$ is consistent. 

\pause 

\alert{Non-example:} $571428369$ \hfill  $\to$ \hfill  $(5)(7)|\!|(14)|\!|(2)(\alert{8}36)(9)$


\end{itemize} 
\end{frame}


\begin{frame}\only<1-3>{
\begin{align*}
\Cons_{(1,2,1) \cleq (1,2,1)}& = \{1234, 1243, 1342, 2134, 2143, 2341, 3124,\\
&\qquad\qquad  3142, 3241, 4123, 4132, 4231\},\\
&\uncover<3->{\pi((1,2,1), (1,2,1)) = 2 \phantom{\Big|}}\\
\Cons_{(1,2,1) \cleq (1,3)} &=  \{1234, 2134, 3124, 4123\},\\
&\uncover<3->{\pi((1,2,1), (1,3)) = 2 \cdot 3  \phantom{\Big|}}\\
\Cons_{(1,2,1) \cleq (3,1)} &= \{1234, 1243, 1342, 2134, 2143, 2341, 3142, 3241\},\\
&\uncover<3->{\pi((1,2,1), (3,1)) = 1 \cdot 3 \phantom{\Big|}}\\
\Cons_{(1,2,1) \cleq (4)}& = \{1234, 2134\}\\
&\uncover<3->{\pi((1,2,1), (4)) = 1 \cdot 3 \cdot 4 \phantom{\Big|}} \end{align*}

\vspace{-.2in}}

\pause 

\begin{lem}Fix  $\alpha \cleq \beta$ of size $n$ Then\\\smallskip
\centerline{$\displaystyle
n!=|\Cons_{\alpha \cleq \beta}|\cdot \pi(\alpha, \beta).
$} \smallskip
\end{lem}
%
%\pause\pause
%\alert{Proof:} Let \\ \smallskip
%\centerline{$\displaystyle
%A_{\alpha \cleq \beta} = \prod_{i=1}^{\ell(\beta)}\left( \prod_{j=1}^{\ell(\alpha^{(i)})} \ZZ/a_j^{(i)}\ZZ\right)
%$, \quad where $\displaystyle a_j^{(i)} = \sum_{r=1}^j \alpha_r^{(i)}$,} \smallskip
%so that $|A_{\alpha \cleq \beta}| = \pi(\alpha, \beta)$. \pause 
%Then there is a bijection\\ \smallskip
%\centerline{$\displaystyle
%S_n \to \Cons_{\alpha \cleq \beta} \times A_{\alpha \cleq \beta}.
%$} \smallskip
%
%\pause 
%
%\alert{Example:} $\alpha=(2,3,2,2)$, $\beta=(5,4)$, $\sigma=739628451$ ($\in \Cons_{\alpha \cleq \beta}$).\\
%
%\smallskip
%
%Split $\sigma$ according to $\beta$:  \qquad  $\underbrace{73962}_{\sigma^{(1)}}|\!|\underbrace{8451}_{\sigma^{(1)}}$\\
%For each $i$, ``rotate'' $\sigma^{(i)}$ into consistency with to $\alpha \cleq \beta$, and record rotations\dots
\end{frame}

%\begin{frame}
%
%Then there is a bijection\\ \smallskip
%\centerline{$\displaystyle
%S_n \to \Cons_{\alpha \cleq \beta} \times A_{\alpha \cleq \beta}:
%$} \smallskip 
%
%\alert{Example:} $\alpha=
%		\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[red!50] (0,0) rectangle (2,1);
%		\filldraw[red!50] (0,1) rectangle (3,2);
%		\filldraw[yellow!50] (0,2) rectangle (2,3);
%		\filldraw[yellow!50] (0,3) rectangle (2,4);
%		\Comp{2,3,2,2}
%		}
%		$, $\beta=\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[red!50] (0,0) rectangle (5,1);
%		\filldraw[yellow!50] (0,1) rectangle (4,2);
%		\Comp{5,4}
%		}$, $\sigma=739628451$ ($\in \Cons_{\alpha \cleq \beta}$).\\
%
%\smallskip
%
%Split $\sigma$ according to $\beta$:  \quad  $\underbrace{73962}_{\sigma^{(1)}}|\!|\underbrace{8451}_{\sigma^{(2)}}$\\
%For each $i$, ``rotate'' $\sigma^{(i)}$ into consistency with to $\alpha \cleq \beta$, and record rotations\dots
%\\ \smallskip
%\centerline{$i=1$: $\sigma^{(1)} = 73962$, \quad $\beta_1$ parts of $\alpha$: $\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[red!50] (0,0) rectangle (2,1);
%		\filldraw[red!50] (0,1) rectangle (3,2);
%		\Comp{2,3}
%		}$} \smallskip\pause
%\centerline{
%$\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[red!50] (0,0) rectangle (2,1);
%		\filldraw[red!50] (0,1) rectangle (3,2);
%		\Comp{2,3}
%		}$ block: \pause\quad $73\alert{9}62\pause \xrightarrow{\text{rotate left by 3}} 6273\alert{9}, \qquad s_2^{(1)} = 3
%$} \smallskip\pause
%\centerline{
%$\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[red!50] (0,0) rectangle (2,1);
%		\Comp{2}
%		}$ block:\pause \quad $\alert{6}2|739 \pause\xrightarrow{\text{rotate left by 2}} 2\alert{6}|739, \qquad s_1^{(1)} = 2
%$} \smallskip\pause
%\centerline{$i=2$: $\sigma^{(2)} = 8451$, \quad $\beta_2$ parts of $\alpha$: $\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[yellow!50] (0,0) rectangle (2,1);
%		\filldraw[yellow!50] (0,1) rectangle (2,2);
%		\Comp{2,2}
%		}$} \smallskip\pause
%\centerline{
%$\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[yellow!50] (0,0) rectangle (2,1);
%		\filldraw[yellow!50] (0,1) rectangle (2,2);
%		\Comp{2,2}
%		}$ block: \pause\quad $\alert{8}451\pause \xrightarrow{\text{rotate left by 1}} 451\alert{8}, \qquad s_2^{(2)} = 1
%$} \smallskip\pause
%\centerline{
%$\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
%		\filldraw[yellow!50] (0,0) rectangle (2,1);
%		\Comp{2}
%		}$ block:\pause \quad $4\alert{5}|18 \pause\xrightarrow{\text{rotate left by 0}} 4\alert{5}|18 \pause, \qquad s_1^{(2)} = 0
%$} \smallskip\pause 
%
%So $739628451 \mapsto (267394518, ((2,3),(0,1)))$. \pause \hfill Invertible!
%
%\end{frame}

\begin{frame}

\begin{lem}Fix  $\alpha \cleq \beta$ of size $n$ Then\\\smallskip
\centerline{$\displaystyle
n!=|\Cons_{\alpha \cleq \beta}|\cdot \pi(\alpha, \beta).
$} \smallskip
\end{lem}
\only<1-2>{
\alert{Proof:} Let \\ \smallskip
\centerline{$\displaystyle
A_{\alpha \cleq \beta} = \prod_{i=1}^{\ell(\beta)}\left( \prod_{j=1}^{\ell(\alpha^{(i)})} \ZZ/a_j^{(i)}\ZZ\right)
$, \quad where $\displaystyle a_j^{(i)} = \sum_{r=1}^j \alpha_r^{(i)}$,} \smallskip
so that $|A_{\alpha \cleq \beta}| = \pi(\alpha, \beta)$. 
Then there is a bijection\\ \smallskip
\centerline{$\displaystyle
S_n \to \Cons_{\alpha \cleq \beta} \times A_{\alpha \cleq \beta}.
$} \smallskip
{\small (For each permutation written in one-line notation, ``cycle'' into consistency with $\alpha \cleq \beta$, and record cycles with elements of $A_{\alpha \cleq \beta}$. See slides from ``Discrete Math Day'' October 2017.)}


}

\pause 


\begin{lem}Fix  $\alpha \cleq \beta$ of size $n$ Then\\\smallskip
\centerline{$\displaystyle
|\cO_{\alpha \cleq \beta}| \cdot |S_n^\lambda| = \sum_{\alpha \cleq \beta \atop \tilde{\alpha} = \lambda} |\Cons_{\alpha \cleq \beta}|.
$} 
\end{lem}
(Similar proof.)
\pause 

\medskip

Therefore \\ \smallskip
\centerline{$\displaystyle
|\cO_{\lambda\beta}| \cdot |S_n^\lambda| =\sum_{\substack{\alpha \cleq \beta \\ \tilde{\alpha}=\lambda}}\frac{n!}{\pi(\alpha, \beta)},
$} \smallskip
so that 
\centerline{$\displaystyle p_\lambda =\sum_{\text{comp }\beta} |\cO_{\lambda,\beta}| M_\beta= \sum_{\tilde{\alpha} = \lambda} \Psi_\alpha, \quad \text{where} \quad \Psi_\alpha = z_{\tilde{\alpha}} \sum_{ \alpha \cleq \beta} \frac{1}{\pi(\alpha,\beta)} M_\beta,$}
as desired.
\end{frame}









\begin{frame}{Type 2}
In $\Sym$ the power sum basis is (essentially) self-dual:
$$\<p_\lambda, p_\mu\> = z_\lambda \delta_{\lambda \mu}.$$

In  $\NSym$, the \alert{type 2 power sum basis} is defined by the generating function relation\\ \smallskip
\centerline{$\displaystyle
\bH(t) = \exp\left(\int \bPh(t)dt\right)
$} \smallskip\pause 

This is equivalent to 
$$\bh_\alpha = \sum_{\beta \cleq \alpha} \frac{1}{\sp(\beta,\alpha)} \phi_\beta,$$
where $\sp(\beta,\alpha)$ is a combinatorial statistic on the refinement $\beta \cleq \alpha$.

\pause 


So, the dual in $\QSym$ will satisfy\\ \smallskip
\centerline{$\displaystyle
\phi^*_\alpha = \sum_{\beta \cgeq \alpha}  \frac{1}{\sp(\alpha, \beta)} M_\beta.
$} \smallskip


\pause 
Define
$$\Phi_\alpha = z_{\tilde{\alpha}} \phi^*_\alpha, \quad \text{ so that } \quad 
	\<\phi_\alpha, \Phi_\beta\> = z_\alpha \delta_{\alpha \beta}.$$

\end{frame}


\begin{frame}{Computing coefficients}

\centerline{$\displaystyle
\Phi_\alpha= z_{\tilde{\alpha}} \sum_{\beta \cgeq \alpha}  \frac{1}{\sp(\alpha, \beta)} M_\beta.
$} \smallskip

 \def\PartUNIT{.2}
 \only<1-2>{ \def\PartUNIT{.25}}
 For example, we saw that 
\centerline{$
\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} \quad \cleq \quad\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}
$} \pause 

First, for each block, we compute $\sp(\gamma) = \ell(\gamma)! \prod_k \gamma_j$:
\centerline{$
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right) =  3! (1 \cdot 2 \cdot 1)$}

\pause

Then, for $\alpha$ refining $\beta$, the coefficient of $M_{\beta}$ in $\psi_\alpha^*$ is $1/\sp(\alpha, \beta)$, where 
\begin{align*}
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} , \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}\right) & =
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[yellow!50] (0,0) rectangle (2,1);
	\Comp{2}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[green!50] (0,0) rectangle (5,1);
	\Comp{5}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[blue!50] (0,0) rectangle (1,2);
	\filldraw[blue!50] (0,2) rectangle (2,3);
	\Comp{1,1,2}
	}\right)\\
	&= 3!(1 \cdot 2 \cdot 1) \cdot 1!(2)\cdot 1!(5)\cdot 3!(1 \cdot 1 \cdot2) 
\end{align*}




\end{frame}



\begin{frame}{Computing coefficients}
 \def\PartUNIT{.15}
\only<1-4>{
\centerline{$
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right) =  \ell(\gamma)! \prod_k \gamma_j = 3! (1 \cdot 2 \cdot 1)$}
	
\begin{align*}
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\filldraw[yellow!50] (0,3) rectangle (2,4);
	\filldraw[green!50] (0,4) rectangle (5,5);
	\filldraw[blue!50] (0,5) rectangle (1,7);
	\filldraw[blue!50] (0,7) rectangle (2,8);
	\Comp{1,2,1,2,5,1,1,2}
	} , \TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (4,1);
	\filldraw[yellow!50] (0,1) rectangle (2,2);
	\filldraw[green!50] (0,2) rectangle (5,3);
	\filldraw[blue!50] (0,3) rectangle (4,4);
	\Comp{4,2,5,4}
	}\right) & =
\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[red!50] (0,0) rectangle (1,1);
	\filldraw[red!50] (0,1) rectangle (2,2);
	\filldraw[red!50] (0,2) rectangle (1,3);
	\Comp{1,2,1}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[yellow!50] (0,0) rectangle (2,1);
	\Comp{2}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[green!50] (0,0) rectangle (5,1);
	\Comp{5}
	}\right)\sp\left(\TikZ{[xscale=\PartUNIT, yscale=-\PartUNIT]
	\filldraw[blue!50] (0,0) rectangle (1,2);
	\filldraw[blue!50] (0,2) rectangle (2,3);
	\Comp{1,1,2}
	}\right)
\end{align*}
}

 \def\PartUNIT{.2}
As another example, $z_{\COMP{2,1}} = 2$, 
$$\Phi_{\CCOMP{1/red,2/yellow}} 
	= z_{\CCOMP{2/yellow,1/red}}\phi_{\CCOMP{1/red,2/yellow}}^* 
	= 2\left(\frac{1}{2} M_{\CCOMP{1/red,2/yellow}} + \frac{1}{4} M_{\CCOMP{3/orange}}\right),$$
$$\Phi_{\CCOMP{2/yellow,1/red}} = z_{\CCOMP{2/yellow,1/red}}\phi_{\CCOMP{2/yellow,1/red}}^*
		= 2\left(\frac{1}{2} M_{\CCOMP{2/yellow,1/red}} + \frac{1}{4} M_{\CCOMP{3/orange}}\right).$$

\pause

So 
\begin{align*}
\Phi_{\COMP{1,2}} + \Phi_{\COMP{2,1}} 
	&=  M_{\COMP{1,2}} + M_{\COMP{2,1}} + M_{\COMP{3}} \\
	&\uncover<3->{= m_{\COMP{2,1}} + m_{\COMP{3}}} 
	\uncover<3->{= m_{\COMP{2}} m_{\COMP{1}}= p_2 p_1 = p_{\COMP{2,1}}.}
\end{align*}

\pause\pause\pause

\begin{thm}[BDHMN]
Type 2 $\QSym$ power sums refine $\Sym$ power sums:
$$p_\lambda = \sum_{\tilde{\alpha} = \lambda} \Phi_\alpha.$$

\end{thm}

\end{frame}


\begin{frame}{Other results}
We also give 
\begin{enumerate}[$\bullet$]
\item Transition matrices between type 1 and type 2, and to fundamentals. 
\item Combinatorial proofs of product formulas for type 1 and 2: 
\\ \smallskip
\centerline{$\displaystyle
\Psi_\alpha \Psi_\beta = \frac{z_\alpha z_\beta}{z_{\alpha\cdot\beta}}\sum_{\gamma \in \alpha\shuffle\beta} \Psi_{\wrd(\gamma)},
\quad
\Phi_\alpha\Phi_\beta = \frac{z_\alpha z_\beta}{z_{\alpha\cdot\beta}}\sum_{\gamma\in \alpha\shuffle\beta}\Phi_{\wrd(\gamma)}.
$} \smallskip



\item Some comments about plethysm, and other algebraic connections. 
\end{enumerate}

~\\~\\~

See arXiv:1710.11613 for more details. 


~\\~\\~

\centerline{Thanks!}
\end{frame}



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