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%%%%% Alphabets %%%%% 
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\def\fa{\mathfrak{a}} \def\fb{\mathfrak{b}} \def\fc{\mathfrak{c}} \def\fd{\mathfrak{d}} \def\fe{\mathfrak{e}} \def\ff{\mathfrak{f}} \def\fg{\mathfrak{g}} \def\fh{\mathfrak{h}} \def\fj{\mathfrak{j}} \def\fk{\mathfrak{k}} \def\fl{\mathfrak{l}} \def\fm{\mathfrak{m}} \def\fn{\mathfrak{n}} \def\fo{\mathfrak{o}} \def\fp{\mathfrak{p}} \def\fq{\mathfrak{q}} \def\fr{\mathfrak{r}} \def\fs{\mathfrak{s}} \def\ft{\mathfrak{t}} \def\fu{\mathfrak{u}} \def\fv{\mathfrak{v}} \def\fw{\mathfrak{w}} \def\fx{\mathfrak{x}} \def\fy{\mathfrak{y}} \def\fz{\mathfrak{z}}
\def\fgl{\mathfrak{gl}}  \def\fsl{\mathfrak{sl}}  \def\fso{\mathfrak{so}}  \def\fsp{\mathfrak{sp}}  
\def\GL{\mathrm{GL}} \def\SL{\mathrm{SL}}  \def\Sp{\mathrm{Sp}}

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\def\Aut{\mathrm{Aut}}
\def\dim{\mathrm{dim}} 
\def\End{\mathrm{End}} 
\def\ev{\mathrm{ev}} 
\def\half{\hbox{$\frac12$}}
\def\Hom{\mathrm{Hom}} 
\def\id{\mathrm{id}}
\def\img{\mathrm{img}}
\def\Ind{\mathrm{Ind}}
\def\ker{\mathrm{ker}}
\def\normeq{\unlhd}
\def\qtr{\mathrm{qtr}} 
\def\tr{\mathrm{tr}} 
\def\Tr{\mathrm{Tr}} 
\def\vep{\varepsilon}
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%%%%%%%%%%%%%%%%%
%\usepackage{etex} %fixes the fight that pictex has with every other drawing package
%\usepackage{pictexwd}
\usepackage{tikz}
%\input xy
%\xyoption{all}
%\xyoption{color}
\usepgflibrary{shapes.geometric}
\tikzstyle{V}=[draw, fill =black, circle, inner sep=0pt, minimum size=1.5pt]
\tikzstyle{D}=[draw, black, fill =green!70!white, diamond, inner sep=0pt, minimum size=5pt]
\tikzstyle{T}=[draw, black, fill =alert!70!white, isosceles triangle, inner sep=0pt, minimum size=5pt] 
\tikzstyle{bT}=[draw, black, fill =alert!70!white, isosceles triangle, inner sep=0pt, minimum size=5pt, rotate = 180] 
\tikzstyle{bV}=[draw, fill =black, circle, inner sep=0pt, minimum size=4pt]
\tikzstyle{over}=[draw=white,double=black,line width=2pt, double distance=.75pt]
\tikzstyle{diagram}=[line width=.75pt, scale=\SCALE]

 \newcommand{\TikZ}[1]{\begin{matrix}\begin{tikzpicture}#1\end{tikzpicture}\end{matrix}}

\def\Over[#1,#2][#3,#4]{ %1,2=start position; 3,4=end position
	\draw[over]   (#1,#2) .. controls ++(0,#4*.5-#2*.5) and ++(0,-#4*.5+#2*.5) .. (#3,#4);}
\def\Under[#1,#2][#3,#4]{ %1,2=start position; 3,4=end position
	\draw  (#1,#2) .. controls ++(0,#4*.5-#2*.5) and ++(0,-#4*.5+#2*.5) .. (#3,#4);}
\def\Cross[#1,#2][#3,#4]{%Mimic over, under follows
	\Under[#3,#2][#1,#4]\Over[#1,#2][#3,#4]}


\newcounter{r}
\newcommand\Part[1]{
        \setcounter{r}{1}
	 \foreach \x in {#1}{
 	{\ifnum\value{r}=1
		\draw (0,\value{r}-1)--(\x,\value{r}-1); 
		\fi}
	\draw (0,\value{r}) to (\x,\value{r});
   	\foreach \y in {0, ..., \x} {\draw (\y,\value{r})--(\y,\value{r}-1);}
	\addtocounter{r}{1}
 }}


%For colored partitions. Takes in a list of row lengths and a color. 
 \newcommand{\fPart}[2]{
 \setcounter{r}{1}
 \foreach \x in {#1}{
  \filldraw[#2] (0,\value{r}) -- (\x,\value{r})--(\x,\value{r}-1)--(0,\value{r}-1);
	\addtocounter{r}{1}
 }
 \setcounter{r}{1}
 \foreach \x in {#1}{
 	{\ifnum\value{r}=1
		\draw (0,\value{r}-1)--(\x,\value{r}-1); 
		\fi}
   \draw (0,\value{r}) to (\x,\value{r});
   \foreach \y in {0, ..., \x} {\draw (\y,\value{r})--(\y,\value{r}-1);}
\addtocounter{r}{1}
 }}
 

\def\UNIT{.25} %Sets the scale of your partitions 
%Whenever there's "yscale = - ...", that turns  the coordinates upside-down, to agree with how we count rows of partitions. 

%Self-contained tikz images for \Part and \fPart above. 
\newcommand{\PART}[1]{
\begin{tikzpicture}[xscale=\UNIT, yscale=-\UNIT] 
	\Part{#1}
\end{tikzpicture}
}
\newcommand{\fPART}[2]{
\begin{tikzpicture}[xscale=\UNIT, yscale=-\UNIT]
	\fPart{#1}{#2}
\end{tikzpicture}
}


\usepackage{graphicx}

\title{Centralizers of the Lie superalgebra $\fp(n)$\\
(where loops go to die)}
\author[Zajj Daugherty]{Zajj Daugherty}
\date{18 July, 2016}





\begin{document}
%\begin{frame}\maketitle\end{frame}
%
%
%
\begin{frame}
\vspace{-2pt}
\hbox{\hspace{-.51in}
\includegraphics[width=5.15in]{groupphoto}}

\end{frame}




\begin{frame}{The classical Brauer algebra}
\def\SCALE{.75}

The \alert{Brauer algebra} $B_k(\delta)$ is the space spanned by Brauer diagrams\\
\centerline{$d=\TikZ{[diagram]
\foreach\x in {1,2,...,5}{
\node[V, label=above:{\tiny \alert{$\x$}}] (t\x) at (\x,1.5){};\node[V, label=below:{\tiny \alert{$\x'$}}] (b\x) at (\x,0){};}
\node[V, label=above:{\tiny \alert{$k$}}] (t6) at (6,1.5){};\node[V, label=below:{\tiny \alert{$k'$}}] (b6) at (6,0){};
\draw (t1) to (b2) (t4) to (b6)
	(t2) to [bend right] (t6) (t3) to [bend right] (t5)
	(b1) to [bend left] (b4) (b3) to [bend left] (b5);
}
\qquad 
\alert{\begin{array}{l}
\text{perfect matchings of}\\
\{1, \dots, k, 1', \dots, k'\}
\end{array}}
$}
(equivalent under isotopy), with multiplication given by vertical concatenation, subject to the relation
$\TikZ{[line width=.75pt, scale=.4] \draw (1,1) ellipse [x radius=.75cm,y radius=.5cm];}=\delta.$ \pause For example,
$$dd' = \TikZ{[diagram]
\draw[|-|] (.5, 1.45) to node[left]{$d$} (.5, .05);
\draw[|-|] (.5, -1.45) to node[left]{$d'$} (.5, -.05);
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};\node[V] (bb\x) at (\x,-1.5){};}
\draw (t1) to (b2) (t4) to (b6)
	(t2) to [bend right] (t6) (t3) to [bend right] (t5)
	(b1) to [bend left] (b4) (b3) to [bend left] (b5);
\draw (b3) to (bb5) (b5) to (bb4)
	(b1) to [bend right] (b4)
	(bb1) to [bend left] (bb3)
	(b2) to (bb2) (b6) to (bb6);
}
=
\delta  \TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\node[V] (bb\x) at (\x,0){};}
\draw (t1) to (bb2) (t4) to (bb6)
	(t2) to [bend right] (t6) (t3) to [bend right] (t5)
	(bb1) to [bend left] (bb3) (bb4) to [bend left] (bb5);}\ .
$$
\end{frame}




%%%%%%
\begin{frame}{Action on tensor space}
The \alert{Brauer algebra} $B_k(\delta)$ is generated by

{\def\SCALE{.4}
\centerline{$s_i = \TikZ{[diagram]
\foreach\x in {1, 3, 4, 6}{
\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};}
\draw (t1) to (b1) (t3) to (b4) (t4) to (b3) (t6) to (b6);
\node[above] at (t3) {\tiny\alert{$\phantom{+}i\ \phantom{+}$}};\node[above] at (t4) {\tiny\alert{$\ i\!+\!1$}};\node[below, white] at (b4) {\tiny{$\ i\!+\!1$}};
\node at (2, .75){$\dots$}; \node at (5, .75){$\dots$}; 
}
\quad \text{ and } \quad 
e_i = \TikZ{[diagram]
\foreach\x in {1, 3, 4, 6}{
\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};}
\draw (t1) to (b1) (t3) to [bend right] (t4) (b4) to [bend right] (b3) (t6) to (b6);
\node[above] at (t3) {\tiny\alert{$\phantom{+}i\ \phantom{+}$}};\node[above] at (t4) {\tiny\alert{$\ i\!+\!1$}};\node[below, white] at (b4) {\tiny{$\ i\!+\!1$}};
\node at (2, .75){$\dots$}; \node at (5, .75){$\dots$}; 
}\ ,
\quad i=1, \dots, k-1,
$}}
with some nice relations. \pause 

Let $V$ be a f.d.\ vector space, with $\beta:V \otimes V \to \CC$ a non-degenerate symmetric (resp.\ skew symmetric)
bilinear form on $V$, and $\beta^*$ its dual. \pause Then the map $B_k(\delta) \to \End(V^{\otimes k})$ that sends
$$s_i \mapsto 1^{\otimes i-1} \otimes s \otimes 1^{k-i-1}, \qquad e_i \mapsto 1^{\otimes i-1} \otimes \beta^*\beta \otimes 1^{k-i-1},$$
where $s(u \otimes v) = v \otimes u$, is a map
$$B_k(\delta) \xrightarrow{~} \End_{\fg}(V^{\otimes k})$$
when  $\fg = \fso(V)$ (resp.\ $\fsp(V)$), $\delta = \dim V$ (resp.\ $-\dim V$). 
\end{frame}




%%%%%%%
\begin{frame}{Lie superalgebras and action on tensor space (still)}
Let $V = V_0 \oplus V_1$ be a $\ZZ_2$-graded vector space. For $v \in V_i$, write $\bar{v}=i$ for its degree.\\\pause
Let $\beta: V \otimes V \to \CC$ be a nondeg., homog., bilinear form satisfying 
$$\beta(u,v) = (-1)^{\bar{u}\bar{v}}\beta(v,u)\quad  \text{(supersymmetric)}.$$
Then 
$$\fg = \{x \in \End(V) ~|~ \beta(xu,v) + (-1)^{\bar{x}\bar{u}}\beta(v,xu)\}$$
is a Lie superalgebra  ($\ZZ_2$-graded). \pause  For example, if $\beta$ is \alert{even}, $\fg=\mathfrak{osp}(V)$ the \alert{orthosymplectic} Lie superalgebra  (if $V_1 = 0$, $\fg = \fso(V)$; and if $V_0 = 0$, $\fg = \fsp(V)$).\pause



The map $B_k(\delta) \to \End(V^{\otimes k})$ that sends
$$s_i \mapsto 1^{\otimes i-1} \otimes s \otimes 1^{k-i-1}, \qquad e_i \mapsto 1^{\otimes i-1} \otimes \beta^*\beta \otimes 1^{k-i-1},$$
where $s(u \otimes v) = (-1)^{\bar{u}\bar{v}} v \otimes u$, gives 
$$B_k(\delta) \xrightarrow{~} \End_{\fg}(V^{\otimes k})$$
when $\delta = \dim V_0 - \dim V_1$.
\end{frame}


%
%
%\begin{frame}MARKED BRAUER\end{frame}



%%%%%%%%%%%
\begin{frame}

%FLIP ARROWS AND DIAMONDS! WE WANT $e_i s_i = e_i$ and $s_i e_i = \epsilon e_i$. CONVENTION: GO LEFT!

\def\SCALE{.75}

%odd Brauer supercategory in [BE]

(Kujawa-Tharp 2014) The \alert{marked Brauer algebra} $B_k(\delta, \epsilon)$, $\epsilon = \pm 1$, is the space spanned by \alert{marked Brauer diagrams}
$$d=\TikZ{[diagram]
\foreach\x in {1,2,...,5}{
\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};}
\node[V] (t6) at (6,1.5){};\node[V] (b6) at (6,0){};
\draw (t1) to (b2) (t4) to (b6)
	(t2) to [bend right] node[bT]{} (t6) (t3) to [bend right] node[T]{} (t5)
	(b4) to [bend right=40] node[D]{} (b1) (b3) to [bend left] node[D]{} (b5);
}
\qquad 
\alert{\begin{array}{l}
\text{caps get one $\TikZ{[diagram]\node[D] at (0,0){};}$ each,}\\
\text{cups get one $\TikZ{[diagram]\node[T] at (0,0){};}$ or $\TikZ{[diagram]\node[bT] at (0,0){};}$ each, }\\
\text{no two markings at same height.}
\end{array}}
$$
 with equivalence up to isotopy except for the local relations 

\smallskip

\centerline{$\begin{matrix}\TikZ{[diagram]\draw (0,0) to [bend right=40] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}\\
\TikZ{[diagram]\draw[rounded corners] (0,0) to (.75,-.25) to [bend left=40] (0,-.25) to (.75,0); \node[T] at (.75/2, -.4) {}; \draw[thin, densely dotted] (0,-.6) rectangle (.75,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT] {}(1,0); \draw[thin, densely dotted] (0,-.4) rectangle (1,.1);}
\end{matrix} 
\quad \text{and}\quad
\TikZ{[diagram]
\draw (0,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,.8); \draw[thin, densely dotted] (0,.5) rectangle (1,1.1);
\draw (1.25,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,0);  \draw[thin, densely dotted] (1.25,.3) rectangle (2.25,-.3);}
=
\epsilon\ 
\TikZ{[diagram]
\draw (0,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,0); \draw[thin, densely dotted] (0,.3) rectangle (1,-.3);
\draw (1.25,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,.8);  \draw[thin, densely dotted] (1.25,.5) rectangle (2.25,1.1);}
$}

\smallskip

for any \alert{adjacent} markings $\TikZ{[diagram] \node[draw, fill=white, circle, inner sep=1pt] at (0,0) {\scriptsize$x$};}$ and $\TikZ{[diagram] \node[draw, fill=white, circle, inner sep=1pt] at (0,0) {\scriptsize$y$};}$ (meaning no markings of height between these two). \pause Again, multiplication is given by vertical concatenation, with relations  
$\TikZ{[line width=.75pt, scale=.4] \draw (1,1) ellipse [x radius=.75cm,y radius=.5cm];}=\delta,$

$$\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[T]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}
\quad \text{ and } \quad 
\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \epsilon\ \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[bT]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}\ .$$



\end{frame}



%%%%%%
\begin{frame}

\def\SCALE{.75}

$$\begin{matrix}\TikZ{[diagram]\draw (0,0) to [bend right=40] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}\\
\TikZ{[diagram]\draw[rounded corners] (0,0) to (.75,-.25) to [bend left=40] (0,-.25) to (.75,0); \node[T] at (.75/2, -.4) {}; \draw[thin, densely dotted] (0,-.6) rectangle (.75,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT] {}(1,0); \draw[thin, densely dotted] (0,-.4) rectangle (1,.1);}
\end{matrix}  
\qquad
\TikZ{[diagram]
\draw (0,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,.8); \draw[thin, densely dotted] (0,.5) rectangle (1,1.1);
\draw (1.25,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,0);  \draw[thin, densely dotted] (1.25,.3) rectangle (2.25,-.3);}
=
\epsilon\ 
\TikZ{[diagram]
\draw (0,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,0); \draw[thin, densely dotted] (0,.3) rectangle (1,-.3);
\draw (1.25,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,.8);  \draw[thin, densely dotted] (1.25,.5) rectangle (2.25,1.1);}
$$
$$\TikZ{[line width=.75pt, scale=.4] \draw (1,1) ellipse [x radius=.75cm,y radius=.5cm];}=\delta$$
$$\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[T]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}
\qquad
\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \epsilon\ \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[bT]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}$$






\pause

For example, \def\SCALE{.6}
$$\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.6){};\node[V] (b\x) at (\x,.1){};}
\draw (t1) to (b2) (t4) to (b6)
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5)
	(b4) to [bend right=40] node[D]{} (b1) (b3) to [bend left] node[D, pos=.6, sloped]{} (b5);
\foreach\x in {1,2,...,6}{
\node[V] (bt\x) at (\x,-.1){};\node[V] (bb\x) at (\x,-1){};
\draw[thin] (bt\x) to (b\x);}
\draw (bt1) to (bb1) (bt2) to (bb2) (bt5) to (bb5) (bt6) to (bb6)
	(bt3) to [bend right] node[T]{} (bt4) (bb4) to [bend right] node[D]{} (bb3)
	;}\pause
= \epsilon\ 
\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\coordinate (b\x) at (\x,0); \node[V] (bb\x) at (\x,-.9){};}
\draw 
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5) (bb4) to [bend right] node[D]{} (bb3);
\draw [rounded corners] (t1) to (b2) to (bb2)
	(t4) to (b6) to (bb6)
	 (bb1) to (b1) to [bend left=50] node[pos=.6] (a){} (b3)  to [bend right] node (b){} (b4) to [bend left] node (c){} (b5) to (bb5);
\draw[densely dotted, red] (3,-.5) rectangle (5,.5);
\node[D] at (a){}; \node[bT] at (b){}; \node[D] at (c){};
}\pause
=
\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\node[V]  (b\x) at (\x,0){};}
\draw 
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5) 
	(b4) to [bend right] node[D]{} (b3) (b5) to [bend right] node[D]{} (b1)
	(t1) to (b2) (t4) to [bend left] (b6);
}
$$
\pause Alternatively,
$$
= \epsilon\ 
\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\coordinate (b\x) at (\x,0); \node[V] (bb\x) at (\x,-.9){};}
\draw 
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5) (bb4) to [bend right] node[D]{} (bb3);
\draw [rounded corners] (t1) to (b2) to (bb2)
	(t4) to (b6) to (bb6)
	 (bb1) to (b1) to [bend left=50] node[pos=.6] (a){} (b3)  to [bend right=40] node (b){} (b4) to [bend left=40] node (c){} (b5) to (bb5);
\draw[densely dotted, red] (2,0) rectangle (4.75,.7);
\node[D] at (a){}; \node[bT] at (b){}; \node[D] at (c){};
}\pause =
\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\coordinate (b\x) at (\x,0); \node[V] (bb\x) at (\x,-.9){};}
\draw 
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5) (bb4) to [bend right] node[D]{} (bb3);
\draw [rounded corners] (t1) to (b2) to (bb2)
	(t4) to (b6) to (bb6)
	 (bb1) to (b2) to [bend left=20] node[pos=.6] (a){} (b3)  to [bend right=50] node (b){} (b4) to [bend left=50] node (c){} (b5) to (bb5);
%\draw[densely dotted, red] (2,0) rectangle (4.75,.7);
\node[D] at (a){}; \node[bT] at (b){}; \node[D] at (c){};
}\pause
=
\TikZ{[diagram]
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\node[V]  (b\x) at (\x,0){};}
\draw 
	(t2) to [bend right] node[bT, pos=.4]{} (t6) (t3) to [bend right] node[T]{} (t5) 
	(b4) to [bend right] node[D]{} (b3) (b5) to [bend right] node[D]{} (b1)
	(t1) to (b2) (t4) to [bend left] (b6);
%	 (bb1) to (b1) to [bend left=40] (b4)  to [bend left] (b3) to [bend left] (b5) to (bb5);
%\node[T] at (2.5,.7){};
} \checkmark
$$
\end{frame}






%%%%%%%%
\begin{frame}

\def\SCALE{.75}


(Kujawa-Tharp 2014) The \alert{marked Brauer algebra} $B_k(\delta, \epsilon)$, $\epsilon = \pm 1$, is the space spanned by \alert{marked Brauer diagrams}
$$d=\TikZ{[diagram]
\foreach\x in {1,2,...,5}{
\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};}
\node[V] (t6) at (6,1.5){};\node[V] (b6) at (6,0){};
\draw (t1) to (b2) (t4) to (b6)
	(t2) to [bend right] node[bT]{} (t6) (t3) to [bend right] node[T]{} (t5)
	(b4) to [bend right=40] node[D]{} (b1) (b3) to [bend left] node[D]{} (b5);
}
\qquad 
\alert{\begin{array}{l}
\text{caps get one $\TikZ{[diagram]\node[D] at (0,0){};}$ each,}\\
\text{cups get one $\TikZ{[diagram]\node[T] at (0,0){};}$ or $\TikZ{[diagram]\node[bT] at (0,0){};}$ each, }\\
\text{no two markings at same height.}
\end{array}}
$$
with equivalence up to isotopy except for the local relations 

\smallskip

\centerline{$\begin{matrix}\TikZ{[diagram]\draw (0,0) to [bend right=40] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.5) rectangle (1.5,.1);}\\
\TikZ{[diagram]\draw[rounded corners] (0,0) to (.75,-.25) to [bend left=40] (0,-.25) to (.75,0); \node[T] at (.75/2, -.4) {}; \draw[thin, densely dotted] (0,-.6) rectangle (.75,.1);}
=
\epsilon\ \TikZ{[diagram]\draw (0,0) to [bend right=40] node[bT] {}(1,0); \draw[thin, densely dotted] (0,-.4) rectangle (1,.1);}
\end{matrix} 
\quad \text{and}\quad
\TikZ{[diagram]
\draw (0,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,.8); \draw[thin, densely dotted] (0,.5) rectangle (1,1.1);
\draw (1.25,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,0);  \draw[thin, densely dotted] (1.25,.3) rectangle (2.25,-.3);}
=
\epsilon\ 
\TikZ{[diagram]
\draw (0,0) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$x$} (1,0); \draw[thin, densely dotted] (0,.3) rectangle (1,-.3);
\draw (1.25,.8) to node[draw, fill=white, circle, inner sep=1pt]{\scriptsize$y$} (2.25,.8);  \draw[thin, densely dotted] (1.25,.5) rectangle (2.25,1.1);}
$}

\smallskip

for any \alert{adjacent} markings $\TikZ{[diagram] \node[draw, fill=white, circle, inner sep=1pt] at (0,0) {\scriptsize$x$};}$ and $\TikZ{[diagram] \node[draw, fill=white, circle, inner sep=1pt] at (0,0) {\scriptsize$y$};}$ (meaning no markings of height between these two). Again, multiplication is given by vertical concatenation, with relations  
$\TikZ{[line width=.75pt, scale=.4] \draw (1,1) ellipse [x radius=.75cm,y radius=.5cm];}=\delta,$

\smallskip

\centerline{$
\begin{matrix}
\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[bT]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[T]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}\\
\text{and}\\
\TikZ{[diagram]\draw (0,0) to [bend left] node[D]{} (.75,0) to [bend right] node[T]{} (1.5,0); \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);} 
= \epsilon\ \TikZ{[diagram]\draw (0,0) to (.75,0); \draw[thin, densely dotted] (0,-.3) rectangle (.75,.3);}
=\TikZ{[diagram]\draw (0,0) to [bend right] node[bT]{} (.75,0) to [bend left] node[D]{} (1.5,0);  \draw[thin, densely dotted] (0,-.3) rectangle (1.5,.3);}\ .
\end{matrix}$
\quad
\fbox{\begin{minipage}{5.5cm}
\alert{Note:}\\
\alert{(1)} $B_k(\delta, 1) = B_k(\delta)$.\\
\alert{(2)} If $\epsilon = -1$, then multiplication is well-defined exactly when $\delta =0$.
\end{minipage}}
}



\end{frame}


\begin{frame}
\def\SCALE{.5}

The marked Brauer algebra $B_k(\delta, \epsilon)$ is generated by 
{\def\SCALE{.65}
\centerline{$s_i = \TikZ{[diagram]
\foreach\x in {1, 3, 4, 6}{
\node[V] (t\x) at (.75*\x,1){};\node[V] (b\x) at (.75*\x,0){};}
\draw (t1) to (b1) (t3) to (b4) (t4) to (b3) (t6) to (b6);
\node[above] at (t3) {\tiny\alert{$\phantom{+}i\ \phantom{+}$}};\node[above] at (t4) {\tiny\alert{$\ i\!+\!1$}};\node[below, white] at (b4) {\tiny{$\ i\!+\!1$}};
\node at (2*.75, .5){$\dots$}; \node at (5*.75, .5){$\dots$}; 
}
\quad \text{ and } \quad 
e_i = \TikZ{[diagram]
\foreach\x/\y in {1.5/1, 3/3, 4/4, 5.5/6}{\node[V] (t\y) at (\x,1){};\node[V] (b\y) at (\x,0){};}
\draw (t1) to (b1) (t3) to [bend right=40] node[bT]{} (t4) (b4) to [bend right] node[D]{} (b3) (t6) to (b6);
\node[above] at (t3) {\tiny\alert{$\phantom{+}i\ \phantom{+}$}};\node[above] at (t4) {\tiny\alert{$\ i\!+\!1$}};\node[below, white] at (b4) {\tiny{$\ i\!+\!1$}};
\node at (2+.25, .5){$\dots$}; \node at (5-.25, .5){$\dots$}; 
}\ ,
$}}
for $i=1, \dots, k-1$,
with relations exactly analogous to those for the Brauer algebra, with some $\epsilon$'s. 

\pause\medskip



Back to Lie superalgebras: $V = V_0 \oplus V_1$, let 
$\beta: V \otimes V \to \CC$
is a non-degenerate, homogeneous, bilinear form on $V$, and let $\fg$ be the corresponding $\beta$-invariant Lie superalgebra. \pause Then with 
$$\beta^* : \CC \to V \otimes V
 \quad \text{ and } \quad 
	\begin{array}{rl} s: V \otimes V &\to V \otimes V \\
				u \otimes v &\mapsto (-1)^{\bar{u}\bar{v}} v \otimes u,\end{array}
	$$
the map	
$$e_i \mapsto 1^{\otimes i-1} \otimes \beta^* \beta \otimes 1^{k-i-1}, \quad 
	s_i \mapsto 1^{\otimes i-1} \otimes s \otimes 1^{k-i-1},$$
	for $ i=1, \dots, k-1$,
 gives 
 $$B_k(\delta, \epsilon) \xrightarrow{~} \End_{\fg}(V^{\otimes k})$$
 when $\delta = \dim V_0 - \dim V_1$ and $\epsilon = (-1)^{\bar{\beta}}$ [KT14].




\end{frame}


\begin{frame}{The peculiar Lie superalgebra $\fp(V)$}
As we saw, when $\beta$ is even, $\fg$ is $\mathfrak{osp}(V)$. But what about when $\beta$ is odd? \pause 

\medskip


Let $V = V_0 \oplus V_1$, and let 
$\beta: V \otimes V \to \CC$
be a non-degenerate, homogeneous, \alert{odd} bilinear form on $V$, and let $\fg$ be the corresponding $\beta$-invariant Lie superalgebra. \pause 

\medskip

First, this requires that $\dim V_0 = \dim V_1$ (so we will have $\delta = \dim V_0 - \dim V_1 = 0$). \pause
The corresponding Lie superalgebra is  $\fg = \fp(V)$, one of the ``strange'', not contragredient  (determined by their Cartan) Lie superalgebras.  \pause 

Specifically, with $n = \dim V_0 = \dim V_1$, 

$$\fp(V) \cong \left\{ \begin{pmatrix} A & B \\ C & -A^t \end{pmatrix} \in \fgl(n|n) 
		~|~ B = B^t, C = -C^t \right\}.$$


\end{frame}


\begin{frame}{The peculiar Lie superalgebra $\fp(V)$}
$$\fp(V) \cong \left\{ \begin{pmatrix} A & B \\ C & -A^t \end{pmatrix} \in \fgl(n|n) 
		~|~ B = B^t, C = -C^t \right\}.$$


The representation theory of $\fp(V)$ is still mysterious. In particular, $B_k(0,-1)$ was first defined by Moon in 2003 to help study $\fp(V)$; Kujawa and Tharp aimed to push further, getting that $V^{\otimes k}$ decomposes into the sum of indecomposables indexed by partitions of $k, k-2, k-4, \dots >0$. \pause Moon calculated the highest weight vectors for $\fp(V)$ in $V \otimes V$ and $V \otimes V \otimes V$ in detail. \pause \medskip




Specifically
$$V \otimes V = \mathrm{Sym}^2 V \oplus \Wedge^2 V,$$
where $\mathrm{Sym}^2 V$ and $\Wedge^2 V$ are both indecomposible, but not simple:
{\def\UNIT{.15} 
$$0 \to L(\PART{2}) \to  \mathrm{Sym}^2 V \xrightarrow{\beta} \CC \to 0$$
$$0 \to \CC \xrightarrow{\beta^*}   \Wedge^2 V\to L(\begin{tikzpicture}[xscale=\UNIT, yscale=-\UNIT] 
	\Part{1,1}
\end{tikzpicture})\to0.$$}




\end{frame}





\begin{frame}{Jucys-Murphy elements and the Casimir}
For $i<j$, let 
{\def\SCALE{.65}
\centerline{$s_{i,j} = \TikZ{[diagram]
\foreach\x/\y in {1/1, 2/2, 2.5/3, 4.5/4, 5/5, 6/6}{\node[V] (t\y) at (\x,1){};\node[V] (b\y) at (\x,0){};}
\draw (t1) to (b1) (t2) to (b5) (t3) to (b3) (t4) to (b4)  (t5) to (b2) (t6) to (b6);
\node[above] at (t2) {\tiny\alert{$\phantom{j}i\ \phantom{j}$}};\node[above] at (t5) {\tiny\alert{$j$}};\node[below, white] at (b4) {\tiny{$j$}};
\node at (1.5, .5){\tiny$\dots$}; \node at (5.5, .5){\tiny$\dots$};  \node at (3.5, .75){\tiny$\dots$}; 
}
\quad \text{ and } \quad 
e_{i,j} = \TikZ{[diagram]
\foreach\x/\y in {1/1, 2/2, 2.5/3, 4.5/4, 5/5, 6/6}{\node[V] (t\y) at (\x,1){};\node[V] (b\y) at (\x,0){};}
\draw (t1) to (b1)  (t3) to (b3) (t4) to (b4)(t6) to (b6)
		(t2) to [bend right=20] node[bT]{} (t5) (b5) to [bend right=20] node[D]{} (b2) ;
\node[above] at (t2) {\tiny\alert{$\phantom{j}i\ \phantom{j}$}};\node[above] at (t5) {\tiny\alert{$j$}};\node[below, white] at (b4) {\tiny{$j$}};
\node at (1.5, .5){\tiny$\dots$}; \node at (5.5, .5){\tiny$\dots$};  \node at (3.5, .5){\tiny$\dots$}; 
}\ .
$}}\pause 


The Brauer algebra $B_k(\delta) = B_k(\delta, 1)$ has Jucys-Murphy elements 
$$x_j =c+ \sum_{i=1}^{j-1} s_{i,j} - e_{i,j}, \quad c\in \CC,\ j=1, \dots, k,$$
that pairwise commute (Nazarov 1996). \pause 

\alert{Action on tensor space:} 
Let %$\kappa \in Z(U\fg)$ and 
$\gamma \in U\fg \otimes U \fg$ be the 
%Casimir and 
split Casimir invariant, given by
$$%\kappa = \sum_{b \in \Omega} b  b^* \qquad 
\gamma = \sum_{b \in \Omega} b \otimes b^*,$$
where $\Omega$ is a basis of $\fg$, and $\{b*~|~b \in \Omega\}$ is the dual basis w.r.t.\ $\beta$. \pause
Then $\gamma$ acts on $V \otimes V$ as as $s_1 - e_1$. \pause So the action of $x_j$ on $V^{\otimes k}$ is the same as that of 
$\sum_{i=1}^{j-1} \gamma_{i,j}$.

\end{frame}



\begin{frame}{Action on $M \otimes V^{\otimes k}$ and cyclotomic quotients}
Define the degenerate affine version $\cB_k(\delta)$ by 
$$\cB_k(\delta) = \CC[y_1, \dots, y_k] \otimes B_k(\delta)/\<y_i\text{-relations}\>,$$
where relations for the $y_i$'s are those satisfied between the $x_i$'s in $B_k(\delta)$. \pause Let $M = L(\lambda)$ be the f.d.\ module indexed by partition $\lambda$, and let 
$$y_j \text{ act on } M \otimes V^{\otimes k} \text{ by } \sum_{i = 0}^{j-1} \gamma_{i,j},$$
where $\gamma_{0,j}$ is $\gamma$ applied to $M$ and the $j$th factor of $V$, and $\gamma_{i,j}$ for $i>0$ is as before. \pause  Then letting the finite part act on $V^{\otimes k}$ as before, and as the identity on $M$, we have a surjection 
$$B_k(\delta) \twoheadrightarrow \End_\fg (M \otimes V^{\otimes k}).$$
\pause Further, let $(y_1 - a_1) (y_1 - a_2) \cdots (y_1 - a_d)$ be the minimal polynomial for the action of $y_1$ on $M \otimes V$. Then for nice $M$ and $k$, 
$$\cB_k(\delta)/\<(y_1 - a_1) (y_1 - a_2) \cdots (y_1 - a_d)\> \xrightarrow{\sim} \End_{\fg}(M \otimes V^{\otimes k}).$$



\end{frame}



\begin{frame}{Jucys-Murphy elements for $B_k(\delta, \epsilon)$ and the sneaky Casimir}
For the marked Brauer algebra, 
$$x_j =c+ \sum_{i=1}^{j-1} s_{i,j} - e_{i,j}, \quad c\in \CC,\ j=1, \dots, k,$$
are still the Jucys-Murphy elements. So we define the degenerate affine version similarly, with $\epsilon$'s where needed,
$$\cB_k(\delta, \epsilon) = \CC[y_1, \dots, y_k] \otimes B_k(\delta, \epsilon)/\<y_i\text{-relations}\>.$$

\pause 

\alert{Questions:} For $B_k(0, -1)$,   \\
\alert{(1)} what tensor space do we want analogous to $M \otimes V^{\otimes k}$? \\
\alert{(2)} what's the action of the $y_i$'s? 

\pause 

\medskip

Start with \alert{(2)}: $\fp(V)$ has trivial center! Namely, if $\Omega$ is a basis of $\fp(V)$, then $\fp(V)$ does not contain a dual basis with respect to $\beta$.


\end{frame}



\begin{frame}{Jucys-Murphy elements and the sneaky Casimir}

$$\cB_k(\delta, \epsilon) = \CC[y_1, \dots, y_k] \otimes B_k(\delta, \epsilon)/\<y_i\text{-relations}\>.$$


\alert{Questions:} For $B_k(0, -1)$,   \\
\alert{(1)} what tensor space do we want analogous to $M \otimes V^{\otimes k}$? \\
\alert{(2)} what's the action of the $y_i$'s? 

\medskip


Start with \alert{(2)}: $\fp(V)$ has trivial center! Namely, if $\Omega$ is a basis of $\fp(V)$, then $\fp(V)$ does not contain a dual basis with respect to $\beta$.

\medskip

In particular, considering $\fp(V) \subseteq \fgl(V)$, then $\{ b^* ~|~ b \in \Omega\}$ is a basis for $\fp(V)^\perp \subseteq \fgl(V)$. \pause So 
$$\gamma = \sum_{b \in \Omega} b \otimes b* \in U\fp(V) \otimes U\fp(V)^\perp.$$

\pause 

Still, we can consider its action as a element of $U\fgl(V) \otimes U\fgl(V)$, and indeed, we get 
$$\gamma_{i,j} \text{ acts on } V^{\otimes k} \text{ as } s_{i,j} - e_{i,j}.$$
Good start! \pause But now for \alert{(1)}\dots

\end{frame}


\begin{frame}{What should $M$ be in  $M \otimes V^{\otimes k}$?}


\alert{Try 1:} For the partition $\lambda$ of size $\ell$, take the indecomposable $M(\lambda)$ indexed by $\lambda$ (the one paired with $B^\lambda$ by Moon, Kujawa-Tharp) in $V^{\otimes \ell}$. \pause Write the action of $\cB_k(0, -1)$ on $M(\lambda) \otimes V^{\otimes k}$ in terms of the the action of $B_k(0,-1)$  on $V^{\otimes  \ell + k}$; make an inductive argument. \pause

\medskip

\alert{Issues:} \\
\alert{(a)} In $V \otimes V$, the minimal polynomial for $\gamma$ is $(\gamma -1)(\gamma+1)$. \\ \pause
So img of $\cB_1(0,-1)$ in $\End(V \otimes V)$ (think $M = V$, $k=1$) is at most  $\cB_1(0,-1)/\<(y_1 - 1)(y_1+1)\>$ (dimension 2).\\
But $\End_{\fp(V)}(V \otimes V) \cong B_2(0,-1)$ (dimension 3).\pause 

\medskip

\alert{(b)} Non-semisimple actions! In $V \otimes V = \mathrm{Sym}^2 V \oplus \Wedge^2 V$, 
$$e_1: \mathrm{Sym}^2 V \xrightarrow{\beta}\CC \xrightarrow{\beta^*} \Wedge^2 (V)$$
has non-trivial image. So, for example, the action of $B_3(0,-1)$ on $V^{\otimes 3}$ does not restrict to a closed action on $(\mathrm{Sym}^2 V) \otimes V$. 


\end{frame}

\begin{frame}{What should $M$ be in  $M \otimes V^{\otimes k}$?}


\only<1-5>{\medskip\alert{Try 1:} $M(\lambda) \otimes V^{\otimes k} \subseteq V^{\otimes |\lambda| + k}$ (nope)} \pause

\only<1-5>{\medskip

\alert{Try 2:} Induce $\fgl(V) = \fg_0$ modules $L(\lambda)$ up to $\fp(V)$. Again, the dimensions to not match. (maybe something else)} \pause 

\medskip

\alert{Try 3:} Kac modules of two types: $K(\lambda)$ (small) and $\tilde{K}(\lambda)$ (big) . \pause \\
Let $\phi = \sum_{\alpha \in \Delta(\fg_{-1})} \alpha$ and let $V(\lambda)$ be the simple $\fg_0$-module of highest weight $\lambda$. Define 
$$K(\lambda) = \Ind_{\fg_0 \oplus \fg_1}^\fg V(\lambda - \phi) \qquad \tilde{K}(\lambda) = \Ind_{\fg_0 \oplus \fg_{-1}}^\fg V(\lambda).$$
\pause 
Then 
\alert{$K(\lambda) \otimes V \cong M_1 \oplus \cdots \oplus M_n$}
where 
$$0 \to K(\lambda+\vep_i) \to M_i \to K(\lambda-\vep_i) \to 0,$$ 
whenever $\lambda\pm\vep_i$ are dominant, or replace $K(*)$ with 0 whenever they're not (similar statement for $\tilde{K}$). \pause
Proof uses eigenvalues of $\gamma$ on $K(\lambda) \otimes V$ and $\tilde{K}(\lambda) \otimes V$, which are combinatorial in terms of boxes added/removed (good), but do not differentiate between adding or removing (not as great).\\\pause

\alert{To do:} What are the minimal polynomials for $\gamma$? What happens at the next step $K(\lambda) \otimes V\otimes V$ when $M_i$ doesn't split? What are the dimensions?

\pause 


\end{frame}




%{with relations
%$$\TikZ{[diagram]
%\coordinate (b0) at (0,0); \coordinate (t0) at (0,1.5);
%\coordinate (b1) at (.75,0);\coordinate (t1) at (.75,1.5);
%\draw (b0) .. controls (.75, .75) .. (t0) (t1) .. controls (0, .75) .. (b1);
%}
% = 
%\TikZ{[diagram]
%\coordinate (b0) at (0,0);\coordinate (t0) at (0,1.5);
%\coordinate (b1) at (.75,0);\coordinate (t1) at (.75,1.5);
%\draw(b0) .. controls (.25, .75) .. (t0) (b1) .. controls (.5, .75) .. (t1);
%}\ ,
%\qquad 
%%
%\TikZ{[diagram]
%\coordinate (b0) at (0,0);\coordinate (t0) at (0,1.5);
%\coordinate (b1) at (.75,0);\coordinate (t1) at (.75,1.5);
%\coordinate (b2) at (1.5,0);\coordinate (t2) at (1.5,1.5);
%\draw(b0) to (t2) (b2) to (t0) (b1) .. controls (0, .75) .. (t1);
%}
% = 
%\TikZ{[diagram]
%\coordinate (b0) at (0,0);\coordinate (t0) at (0,1.5);
%\coordinate (b1) at (.75,0);\coordinate (t1) at (.75,1.5);
%\coordinate (b2) at (1.5,0);\coordinate (t2) at (1.5,1.5);
%\draw(b0) to (t2) (b2) to (t0) (b1) .. controls (1.5, .75) .. (t1);
%}
%$$
%$$
%\TikZ{[diagram] \draw (1,1) ellipse [x radius=.75cm,y radius=.5cm];}=\delta
%$$}

\end{document}