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%%% Characters %%%------------------------------------------------------------
%%%%% Alphabets %%%%% 
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\def\AA{\mathbb{A}} \def\BB{\mathbb{B}} \def\CC{\mathbb{C}} \def\DD{\mathbb{D}} \def\EE{\mathbb{E}} \def\FF{\mathbb{F}} \def\GG{\mathbb{G}} \def\HH{\mathbb{H}} \def\II{\mathbb{I}} \def\JJ{\mathbb{J}} \def\KK{\mathbb{K}} \def\LL{\mathbb{L}} \def\MM{\mathbb{M}} \def\NN{\mathbb{N}} \def\OO{\mathbb{O}} \def\PP{\mathbb{P}} \def\QQ{\mathbb{Q}} \def\RR{\mathbb{R}} \def\SS{\mathbb{S}} \def\TT{\mathbb{T}} \def\UU{\mathbb{U}} \def\VV{\mathbb{V}} \def\WW{\mathbb{W}} \def\XX{\mathbb{X}} \def\YY{\mathbb{Y}} \def\ZZ{\mathbb{Z}}  

\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\half{\hbox{$\frac12$}}
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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}
\def\SCALE{1}
\tikzstyle{v}=[draw, fill =black, circle, inner sep=0pt, minimum size=1.5pt]
\tikzstyle{V}=[draw, fill =black, circle, inner sep=0pt, minimum size=1.5pt]
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\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}
}


%This is the lattice behind all the examples below. 
\def\BLattice{
%Sets the coordinates for the partitions of 0 to 5:
	\coordinate (0) at (0,.4);
	\coordinate (11) at (0,1.2);
	\coordinate (20) at (-2,2);\coordinate (21) at (0,2);\coordinate (22) at (2,2);
	\foreach \x in {1, ..., 3}{\coordinate (3\x) at (-2+2*\x,3);}
	\coordinate (30) at (-2,3);
	\foreach \x in {1, ..., 5}{\coordinate (4\x) at (-1+2*\x,5);}
	\foreach \x in {0,1,2}{\coordinate (-4\x) at (-6+2*\x,5);}
%	\foreach \x in {1, ..., 7}{\coordinate (5\x) at (-8+2*\x,6);}
%edges in lattice:
	\draw (0)--(11)--(21) (20)--(11)--(22);
	\draw (20) to (30) (30)--(21)--(31) (21)--(32) (30)--(22)--(32) (22)--(33);
	\draw (31)--(41) (31)--(42) (32)--(42)  (32)--(43) (32)--(44) (33)--(44)  (33)--(45)
		(30) to (-40) (30) to (-41) (30) to (-42) (31) to (-41) (32) to (-41)  (32) to (-42) (33) to (-42) ;
%	\draw (41)--(51) (41)--(52) (42)--(52)  (42)--(53)  (42)--(54)  (43)--(53)
%		(43)--(55)  (44)--(54)  (44)--(55) (44)--(56) (45)--(56)  (45)--(57);
%partitions in lattice:
\begin{scope}[every node/.style={fill=white}]
	\node at (0) {$\emptyset$};
	\node at (11) {\PART{1}};
	\node at (22) {\PART{1,1}};
	\node at (21) {\PART{2}};
	\node at (20) {$\emptyset$};
	\node at (33) {\PART{1,1,1}};
	\node at (32) {\PART{2,1}};
	\node at (31) {\PART{3}};
	\node at (30) {\PART{1}};
	\node at (45) {\PART{1,1,1,1}};
	\node at (44) {\PART{2,1,1}};
	\node at (43) {\PART{2,2}};
	\node at (42) {\PART{3,1}};
	\node at (41) {\PART{4}};
	\node at (-42) {\PART{1,1}};
	\node at (-41) {\PART{2}};
	\node at (-40) {$\emptyset$};
%	\node at (57) {\PART{1,1,1,1,1}};
%	\node at (56) {\PART{2,1,1,1}};
%	\node at (55) {\PART{2,2,1}};
%	\node at (54) {\PART{3,1,1}};
%	\node at (53) {\PART{3,2}};
%	\node at (52) {\PART{4,1}};
%	\node at (51) {\PART{5}};
\end{scope}
}

\usepackage{graphicx}

\title{Signed Brauer algebras and their translations}
\author[Zajj Daugherty]{Zajj Daugherty\\
\small The City College of New York\\ \& The CUNY Graduate Center\\~\\
Joint with M.\ Balagovic, I.\ Entova-Aizenbud,  I.\ Halacheva, \\
J.\ Hennig, M.\ S.\ Im, 
G.\ Letzter, 
E.\ Norton,\\ 
V.\ Serganova, and  C.\ Stroppel}
\date{October 2, 2019}





\begin{document}
\begin{frame}\maketitle\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,
\\ \smallskip
\centerline{$\displaystyle
dd' = \TikZ{[diagram]
\draw[|-|] (.5, 1.5) to node[left]{$d$} (.5, .25);
\draw[|-|] (.5, -1.5) to node[left]{$d'$} (.5, -.25);
\foreach\x in {1,2,...,6}{
\node[V] (t\x) at (\x,1.5){};\node[V] (M\x) at (\x,.25){};\node[V] (m\x) at (\x,-.25){};\node[V] (b\x) at (\x,-1.5){};}
\draw (t1) to (M2) (t4) to (M6)
	(t2) to [bend right] (t6) (t3) to [bend right] (t5)
	(M1) to [bend left] (M4) (M3) to [bend left] (M5);
\draw (m3) to (b5) (m5) to (b4)
	(m1) to [bend right] (m4)
	(b1) to [bend left] (b3)
	(m2) to (b2) (m6) to (b6);
\foreach\x in {1,2,...,6}{\draw[densely dotted, alert] (M\x) to (m\x);}
\foreach\x in {1,2,...,6}{\node[above, alert] at (t\x) {\tiny \x};\node[below, alert] at (b\x) {\tiny \x};}
}
=
\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);
\foreach\x in {1,2,...,6}{\node[above, alert] at (t\x) {\tiny \x};\node[below, alert] at (bb\x) {\tiny \x};}}\ .
$} 
\end{frame}




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

{\def\SCALE{.3}
\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 expected relations. \pause 

Let $V$ be a finite dimensional 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
\\ \smallskip
\centerline{$\displaystyle
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},
$} \smallskip
where $s(u \otimes v) = v \otimes u$, is a map
\\ \smallskip
\centerline{$\displaystyle
B_k(\delta) \xrightarrow{~} \End_{\fg}(V^{\otimes k})
$} \smallskip
when  $\fg = \fso(V)$ (resp.\ $\fsp(V)$), $\delta = \dim V$ (resp.\ $-\dim V$). 

\pause \medskip

\alert{Consequence:} 
Schur-Weyl duality between modules for $B_k(\delta)$ and $\fg$,
\\ \smallskip
\centerline{$\displaystyle
V^{\otimes k} = \bigoplus_{\lambda \vdash k, k-2, \dots \atop} B_k^\lambda \otimes L(\lambda),
$} 
where $B_k^\lambda$ are distinct simple Brauer modules, and $ L(\lambda)$ are distinct simple (highest weight) $\fg$-modules.
\end{frame}

 \begin{frame}{Decompositions of $V^{\otimes k}$}

\centerline{\begin{tikzpicture}[xscale=3*\UNIT, yscale=-4.5*\UNIT]
\BLattice  %Defined in preamble. 
\foreach \x in {-4,0,4} {\node at (\x , 5.75) {$\vdots$}; }
\end{tikzpicture}}


 \end{frame}

\begin{frame}{Jucys-Murphy elements}
\only<1-2>{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$}; 
}
\quad \text{ and } \quad 
e_{i,j} = \TikZ{[diagram]
\foreach\x/\y in {1/1, 2/2, 2.5/3, 4.5/4, 5/5, 6/6}{\node[V] (t\y) at (\x,1){};\node[V] (b\y) at (\x,0){};}
\draw (t1) to (b1)  (t3) to (b3) (t4) to (b4)(t6) to (b6)
		(t2) to [bend right=20]  (t5) (b5) to [bend right=20] (b2) ;
\node[above] at (t2) {\tiny\alert{$\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 


Brauer algebra $B_k(\delta)$ has Jucys-Murphy elements \\ \smallskip
\centerline{$\displaystyle
x_j =\text{constant}+ \sum_{i=1}^{j-1} s_{i,j} - e_{i,j}, \quad  j=1, \dots, k,
$} \smallskip
that \hfill {\footnotesize 
(See Nazarov '96, D.-Ram-Virk '13 \& '14.)}
\begin{enumerate}[1.]
\item pairwise commute;
\item generate the center;  and
\item have eigenvalues in $\End(V^{\otimes k})$ given by combinatorial data from the partitions lattice.
\end{enumerate}

\pause Let $\Gamma$ be a basis for $\fg$, and $\Gamma = \{b^* ~|~ b \in \Gamma\}$ be the dual basis with respect to a nice bilinear form. The \alert{split Casimir} invariant $U\fg \otimes U\fg$ is 
\\ 
\centerline{$\displaystyle
\gamma = \sum_{b \in \Gamma} b \otimes b^*.
$} \smallskip \pause 
Then $\gamma$ acts on $V \otimes V$ by $s_1 - e_1$, so the action of $x_i$ is given by 
\\ 
\centerline{$\displaystyle x_i \Big|_{V^{\otimes k}} = \text{constant} + \sum_{j = 1, \dots, i-1} \gamma\Big|_{V^{(j)}, V^{(i)}}.$}
\end{frame}

\begin{frame}{Jucys-Murphy elements}
Now let $M$ be a simple $\fg$ module, and define an operator on $M \otimes V^{\otimes k}$ by 
\\ 
\centerline{\def\SCALE{.3}$\displaystyle
y_i 
= \text{constant} + \gamma\Big|_{M,V^{(i)}} + \sum_{j = 1, \dots, i-1} \gamma\Big|_{V^{(j)}, V^{(i)}}\only<2->{= \TikZ{[diagram]
	\foreach\x in {1, 3, 4, 5, 7}{\node[V] (t\x) at (\x,1.5){};\node[V] (b\x) at (\x,0){};}
	\draw (t1) to (b1) (t3) to (b3) (t4) to node[rV]{} (b4) (t5) to (b5) (t7) to (b7);
\node[above] at (t4) {\tiny\alert{$i$}};
\node at (2, .75){$\dots$}; \node at (6, .75){$\dots$}; 
}}.$}\smallskip
(When $M = L(0)$, this is the same as $x_i$ from before.)

\smallskip


\alert{Nice facts:} Still\dots \begin{enumerate}[1.]
\item $y_1, \dots, y_k$ commute;
\item $y_i \in \End_\fg(M \otimes V^{\otimes k})$;  
\item they generate the center of the action; and
\item have eigenvalues given by combinatorial data from the partitions lattice.
\end{enumerate}\pause 

The \alert{graded Brauer algebra} is the algebra generated by $s_i$ and  $e_i$  for $i = 1, \dots k-1$, and $y_1, \dots, y_k$ (modulo  relations), and is also in Schur-Weyl duality with symplectic and orthogonal $\fg$.  

{\def\SCALE{.65}
\smallskip
\centerline{Relations: \quad
$\TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
	\draw (t1) to node[pos=.2,rV]{} (b2) (t2) to (b1);
	}
	-
	\TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
	}
	\draw (t1) to node[pos=.8,rV]{} (b2) (t2) to (b1);
	}
	=  \TikZ{[diagram, xscale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\draw (t1) to [bend right=50] (t2) (b2) to [bend right=50]  (b1);}
	-
\TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\draw (t1) to (b1) (t2) to (b2);
}
$,  \qquad $
\begin{matrix}
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
\draw (t1) to [bend right=100] 
	node[pos=.75, rV]{}
	(t2);
}
=-\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
\draw (t1) to [bend right=100] 
	node[pos=.25, rV]{}
	(t2);
}\\\text{and}\\
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
\draw (b2) to [bend right=100] 
	node[pos=.25, rV]{}
	(b1);
},
=
-\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
\draw (b2) to [bend right=100]  
	node[pos=.75, rV]{}
	(b1);
}.
\end{matrix}
$}} 

\end{frame}


%%%%%%%
\begin{frame}{Lie superalgebras}
A \alert{Lie superalgebra} is a  $\ZZ_2$-graded vector space $\mathfrak{g} =\mathfrak{g}_0 \oplus \mathfrak{g}_1$ with a \alert{super} Lie bracket \only<1-4>{
$$[,]:\fg \otimes \fg \to \fg$$
satisfying \\ \smallskip
\centerline{$\displaystyle
[x,y] = - (-1)^{\bar x\bar y}[y,x]
$} \smallskip
and\\ \smallskip
\centerline{$\displaystyle
[x,[y,z]]] = [[x,y], z] + (-1)^{\bar x\bar y}[y ,[x,z]],
$} \smallskip
where $x, y, z$ are each homogeneous, and $\bar{x}$ means \alert{degree}.
}\only<5->{
$[,]:\fg \otimes \fg \to \fg$ satisfying a super symmetry and super Jacobi identity.
}
\pause \medskip

Three types: \alert{basic}, \alert{Cartan type}, and \alert{strange} (two families: \alert{periplectic} and queer).

\medskip

\pause 
Let $V = V_0 \oplus V_1 = \CC^{m|n}$ be a $\ZZ_2$-graded vector space over $\CC$. \\
\only<5->{For (homogeneous) $v \in V_i$, write $\bar{v}=i$ for its \alert{degree}.}




\medskip\pause
The \alert{general linear Lie superalgebra is} \\ \smallskip
\centerline{$\displaystyle
\fgl(m|n) = \End(V)\pause = \fg_0 \oplus \fg_1,
$} \smallskip
where 
\\ \smallskip
\only<5->{\centerline{$\displaystyle
\fg_0 = \left\{\left. \begin{pmatrix} A & 0 \\ 0 & D\end{pmatrix} ~\right|~ A \in \End(V_0), D \in \End(V_1)\right\},
$} 
\smallskip
\centerline{$\displaystyle
\fg_1 = \left\{\left. \begin{pmatrix} 0 & B \\ C & 0\end{pmatrix} ~\right|~ B \in \Hom(V_1,V_0), C \in  \Hom(V_0,V_1)\right\}.
$}} \smallskip


 
\alert{Bracket:} $[x,y] = xy - (-1)^{\bar x\bar y}yx$.

\end{frame}

%%%%%%%
\begin{frame}{Lie superalgebras}
Let $\beta: V \otimes V \to \CC$ be a nondegenerate, homogeneous, bilinear form satisfying \\ \smallskip
\centerline{$\displaystyle
\beta(v,w) = (-1)^{\bar{v}\bar{w}}\beta(w,v)\quad  \text{(\alert{supersymmetric})}.
$} \smallskip
\pause 


\only<1-4>{Then 
\\ \smallskip
\centerline{$\displaystyle
\fg = \{x \in \End(V) ~|~ \beta(xu,v) + (-1)^{\bar{x}\bar{u}}\beta(v,xu)\}
$} \smallskip
is a Lie superalgebra  ($\ZZ_2$-graded).   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 $V_0 = 0$, $\fg = \fsp(V)$).

\bigskip
}

\pause 

If $\beta$ is odd,  then $\fg$ is the  \alert{periplectic} Lie superalgebra, \\ \smallskip
\centerline{$\displaystyle
\fp(V) = \fp(n)= \{x \in \End(V) ~|~ \beta(xv,w) + (-1)^{\bar{x}\bar{v}}\beta(v,xw)=0\}.
$}

\pause \medskip

Specifically, we have 
\\ \smallskip
\centerline{$\displaystyle
\fp(n) \cong \left\{\left. \begin{pmatrix} A & B \\ C & -A^t \end{pmatrix} \in \fgl(n|n) 
		~\right|~ B = B^t, C = -C^t \right\}.
$} \smallskip \pause 
Then, as vector spaces $\fp(n) = \fg_0 \oplus \fg_1 \oplus \fg_{-1} $, where  
\\ \smallskip
\centerline{$\displaystyle
\fg_0 = \left\{ \begin{pmatrix} A & 0 \\ 0 & -A^t \end{pmatrix}\right\} \cong \fgl(n)
$} \smallskip
\centerline{$\displaystyle
\fg_1  = \left\{ \begin{pmatrix} 0 & B \\ 0 & 0 \end{pmatrix}\right\}, \quad 
 \fg_{-1} = \left\{ \begin{pmatrix} 0 & 0 \\ C & 0 \end{pmatrix}\right\}.
$} \smallskip

\pause
\alert{Goal:} Study the representation theory of $\fp(n)$. \pause In particular,  study the category $\mathcal{F}_n$ of finite-dimensional integrable representations (a ``highest weight category").
\end{frame}



\begin{frame}{Translation functors}

\alert{Key ingredients for other cases:} a large center in  $\cU\fg$, and  translation functors given by tensoring with the natural representation followed by the projection onto a block (given by eigenvalues of $y_i$'s).\pause \medskip


Namely, you study the action of $\cU\fg$ on 
\\ \smallskip
\centerline{$\displaystyle
M \otimes V \otimes V \otimes \cdots \otimes V = M \otimes V^{\otimes d},
$} \smallskip
where $V$ is $\fg$'s favorite module, and $M$ is another simple module, by constructing operators in $\End_{\fg}(M \otimes V^{\otimes d})$ that commute with the $\fg$-action. Many commuting operators are generated by taking coproducts of central elements (again, like $y_i$'s). 

\pause  \medskip

\alert{Examples:} If $\fg = \fso(V)$ or $\fsp(V)$, then the commuting operators generate the graded Brauer algebra; when $\fg=\fsl(V)$, you get the ``graded Hecke algebra of type A''. \pause \medskip


\alert{Obstruction:} The center of $\cU\fp(V)$ is  trivial!  
But we'll figure it out anyway\dots
\end{frame}


\begin{frame}{Example: $V \otimes V$}
{ \alert{Recall:} \color{black} $V = V_0 \oplus V_1 = \CC^{m|n}$ is a $\ZZ_2$-graded vector space over $\CC$. \\
For (homogeneous) $v \in V_i$, write $\bar{v}=i$ for its \alert{degree}.}


\bigskip

The algebra $\End_{\fp(V)}(V\otimes V)$ is 3-dimensional with basis $1$, 
\\ \smallskip
\centerline{$\displaystyle
s: v\otimes w\mapsto (-1)^{\bar{v}\bar{w}} w\otimes v, \quad \text{and} \quad 
e={\beta^*}\beta: v\otimes w \mapsto \beta(v,w)c,
$} \smallskip
where $c$ spans the (super) sign module.
\medskip\pause

Draw: {\def\SCALE{.65}
\\ \smallskip
\centerline{
$\phantom{\alert{marked Brauer}}$\hfill
$\displaystyle
s  = \TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\draw (t1) to (b2) (t2) to (b1);
}
\quad \text{ and } \quad 
e = \TikZ{[diagram, xscale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\only<1-3>{\draw (t1) to [bend right=50] (t2) (b2) to [bend right=50] (b1);}
\only<4->{\draw (t1) to [bend right=50] node[bT]{} (t2) (b2) to [bend right=50] node[D]{} (b1);}
}\ .
$ \qquad \uncover<4->{(\alert{signed Brauer})}}} 

\medskip\pause

Relation: $e\circ s=e=-s\circ e$. \pause \pause Also, $e^2 = 0$. \hfill(non-semisimple case)
\end{frame}

%
%%%%%%%%
%\begin{frame}{Lie superalgebras and action on tensor space}
%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}

\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 
$\beta: V \otimes V \to \CC$
be 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 split 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 
%$$x_j =\text{constant}+ \sum_{i=1}^{j-1} s_{i,j} - e_{i,j}, \quad c\in \CC,\ j=1, \dots, k,$$
%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
%\\ \smallskip
%\centerline{$\displaystyle
%%\kappa = \sum_{b \in \Gamma} b  b^* \qquad 
%\gamma = \sum_{b \in \Gamma} b \otimes b^*,
%$} \smallskip
%where $\Gamma$ is a basis of $\fg$, and $\{b^*~|~b \in \Gamma\}$ is the dual basis w.r.t.\ a nice bilinear form. \pause
%Then $\gamma$ acts on $V \otimes V$ as as $s_1 - e_1$ \\{\footnotesize (i.e.\ the action of $x_j$ on $V^{\otimes k}$ is the same as that of 
%$(\text{const.} + \sum_{i=1}^{j-1} \gamma_{i,j})$)}.
%
%\end{frame}


%
%\begin{frame}{Action on $M \otimes V^{\otimes k}$ and cyclotomic quotients}
%Again, define the graded Brauer algebra $\cB_k(\delta)$ 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 $B_k(\delta)$: 
%Again, draw Jucys-Murphy elements as 
% {\def\SCALE{.65}
%\\ \smallskip
%\centerline{
%$y_i  = \TikZ{[diagram]
%\foreach\x in {1, 3, 4, 5, 7}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
%	\draw (t\x) to (b\x);}
%	\node[rV] at (.75*4, .5){};
%	\foreach \x in {2, 6}{\node at (.75*\x,.5) {$\cdots$};}
%	\node[above] at (t4) {\small $i$};
%	}
%$}}  
%and define the \alert{affine signed Brauer algebra} $\cB_k(0, -1)$ as the algebra generated by $B_k(0, -1)$ and $y_1, \dots, y_d$, together with relations 
%{\def\SCALE{.65}
%\\ \smallskip
%\centerline{
%$\TikZ{[diagram]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
%	\draw (t\x) to (b\x);}
%	\node[rV] at (.75*2, .5){};
%	}
%	=
%	\TikZ{[diagram]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
%	}
%	\draw[rounded corners] (t2) to (.75*1,.5) to (b2);
%	\draw[rounded corners] (t1) to (.75*2,.5) to (b1);
%	\node[rV] at (.75*1+.1, .5){};
%	}
%	+ \TikZ{[diagram]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
%\draw (t1) to (b2) (t2) to (b1);
%}
%	+ \TikZ{[diagram, xscale=1.5]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
%\draw (t1) to [bend right=50] node[bT]{} (t2) (b2) to [bend right=50] node[D]{} (b1);
%}
%$,  \qquad $
%\begin{matrix}
%\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
%\draw (t1) to [bend right=100] node[bT]{} 
%	node[pos=.75, rV]{}
%	(t2);
%}
%=\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
%\draw (t1) to [bend right=100] node[bT]{} 
%	node[pos=.25, rV]{}
%	(t2);
%}-
%\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
%\draw (t1) to [bend right=100] node[bT]{} 
%%	node[pos=.25, rV]{}
%	(t2);
%},\\\text{and}\\
%\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
%\draw (b2) to [bend right=100] node[D]{} 
%	node[pos=.25, rV]{}
%	(b1);
%}
%=
%\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
%\draw (b2) to [bend right=100] node[D]{} 
%	node[pos=.75, rV]{}
%	(b1);
%}+
%\TikZ{[diagram, scale=1.5]
%\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
%\draw (b2) to [bend right=100] node[D]{} 
%%	node[pos=.75, rV]{}
%	(b1);
%}.
%\end{matrix}
%$}} 
%\end{frame}
%
%\begin{frame}{Action on $M \otimes V^{\otimes k}$ and cyclotomic quotients}
%Again, define the graded Brauer algebra $\cB_k(\delta)$ 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 $B_k(\delta)$.
%
%\smallskip
%
%Let $M = L(\lambda)$ be the f.d.\ module indexed by partition $\lambda$, and let 
%\\ \smallskip
%\centerline{$\displaystyle
%y_j \text{ act on } M \otimes V^{\otimes k} \text{ by } \text{const.} + \sum_{i = 0}^{j-1} \gamma_{i,j},
%$} \smallskip
%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 $B_k(\delta, \epsilon)$}
\def\SCALE{.65}

\only<1>{For the marked Brauer algebra, 
\\ \smallskip
\centerline{$\displaystyle
x_j =\text{constant}+ \sum_{i=1}^{j-1} s_{i,j} - e_{i,j}, \quad c\in \CC,\ j=1, \dots, k,
$} \smallskip
are still the Jucys-Murphy elements. So w}\only<2->{W}e define the graded version similarly, with $\epsilon$'s where needed,
\\ \smallskip
\centerline{$\displaystyle
\cB_k(\delta, \epsilon) = \CC[y_1, \dots, y_k] \otimes B_k(\delta, \epsilon)/\<y_i\text{-relations}\>$} \only<1>{\bigskip}\only<2->{\smallskip}
Namely, if we draw
\smallskip
\centerline{
$y_i  = \TikZ{[diagram]
\foreach\x in {1, 3, 4, 5, 7}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
	\draw (t\x) to (b\x);}
	\node[rV] at (.75*4, .5){};
	\foreach \x in {2, 6}{\node at (.75*\x,.5) {$\cdots$};}
	\node[above] at (t4) {\small $i$};
	}
$} 
then
\\ \smallskip
\centerline{
$\TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
	\draw (t\x) to (b\x);}
	\node[rV] at (.75*2, .5){};
	}
	-
	\TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};
	}
	\draw[rounded corners] (t2) to (.75*1,.5) to (b2);
	\draw[rounded corners] (t1) to (.75*2,.5) to (b1);
	\node[rV] at (.75*1+.1, .5){};
	}
	= \TikZ{[diagram]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\draw (t1) to (b2) (t2) to (b1);
}
	+ \TikZ{[diagram, xscale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};\node[v] (b\x) at (.75*\x,0){};}
\draw (t1) to [bend right=50] node[bT]{} (t2) (b2) to [bend right=50] node[D]{} (b1);
}
$,  \qquad $
\begin{matrix}
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
\draw (t1) to [bend right=100] node[bT]{} 
	node[pos=.75, rV]{}
	(t2);
}
=\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
\draw (t1) to [bend right=100] node[bT]{} 
	node[pos=.25, rV]{}
	(t2);
}-
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (t\x) at (.75*\x,1){};}
\draw (t1) to [bend right=100] node[bT]{} 
%	node[pos=.25, rV]{}
	(t2);
},\\\text{and}\\
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
\draw (b2) to [bend right=100] node[D]{} 
	node[pos=.25, rV]{}
	(b1);
}
=
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
\draw (b2) to [bend right=100] node[D]{} 
	node[pos=.75, rV]{}
	(b1);
}+
\TikZ{[diagram, scale=1.5]
\foreach\x in {1, 2}{\node[v] (b\x) at (.75*\x,0){};}
\draw (b2) to [bend right=100] node[D]{} 
%	node[pos=.75, rV]{}
	(b1);
}.
\end{matrix}
$}


\pause 

\alert{Questions:} For $\cB_k(0, -1)$,   \\
\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 $\Gamma$ is a basis of $\fp(V)$, then $\fp(V)$ does not contain a dual basis with respect to $\beta$.


\end{frame}



\begin{frame}{Sneaky split 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 $\Gamma$ 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 $\{ b^* ~|~ b \in \Gamma\}$ is a basis for $\fp(V)^\perp \subseteq \fgl(V)$.  So 
$$\gamma = \sum_{b \in \Gamma} b \otimes b^* \in U\fp(V) \otimes U\fp(V)^\perp.$$

\pause 

However, $\gamma$ \emph{does} have a natural action on $M \otimes V$, since $V$ is also a  $\fgl(V)$-module. And since it commutes with the action of $\fgl(V)$, it commutes with $\fp(V)$. 
 In particular, as before,
$$\gamma_{i,j} \text{ acts on } V^{\otimes k} \text{ as } s_{i,j} - e_{i,j}.$$


\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 $V^{\otimes \ell}$.  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) Not big enough.} {In $V \otimes V$, the minimal polynomial for $\gamma$ is $(\gamma -1)(\gamma+1)$.  So the image 
 of $\cB_1(0,-1)$ in $\End(V \otimes V)$ (think $M = V$, $k=1$) is at most  \\
 \centerline{$\cB_1(0,-1)/\<(y_1 - 1)(y_1+1)\>$ \quad (dim $ = 2$).}
But $\End_{\fp(V)}(V \otimes V) \cong B_2(0,-1)$ (dim $ = 3$).}\pause 

\medskip

\alert{(b)} \alert{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<1-4>{\medskip\alert{Try 1:} $M(\lambda) \otimes V^{\otimes k} \subseteq V^{\otimes |\lambda| + k}$ (nope)} 

\only<1-4>{\medskip

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

\medskip

\alert{Try 3:} Kac modules of two types: $K(\lambda)$ (thin) and $\tilde{K}(\lambda)$ (thick) . \pause \\
Let \alert{$\displaystyle \phi = \sum_{\text{neg. roots } \alpha} \alpha$} {\footnotesize(like the staircase partition)} and let $V(\lambda)$ be the simple $\fg_0$-module of highest weight $\lambda$. Define \\ \smallskip
\centerline{$\displaystyle
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).
$} \smallskip
\pause 
Then 
\alert{$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 (add or remove a box to $\lambda$), or replace $K(*)$ with 0 whenever they're not (similar statement for $\tilde{K}$). {\small (Proof uses eigenvalues of $\gamma$ on $K(\lambda) \otimes V$ and $\tilde{K}(\lambda) \otimes V$, which are combinatorial data in terms of boxes added/removed.)}


\end{frame}

 \begin{frame}
\small
 \alert{$K(\lambda) \otimes V \cong M_1 \oplus \cdots \oplus M_n$}
where \\ \smallskip
\centerline{$\displaystyle
0 \to K(\lambda+\vep_i) \to M_i \to K(\lambda-\vep_i) \to 0,
$} \smallskip
whenever $\lambda\pm\vep_i$ are dominant (add or remove a box to $\lambda$), or replace $K(*)$ with 0 whenever they're not (similar statement for $\tilde{K}$).
 

\normalsize

\centerline{\begin{tikzpicture}[xscale=3*\UNIT, yscale=-4.5*\UNIT]
\BLattice  %Defined in preamble. 
\draw[thick, alert] (20) to (11) to (21) (11) to (22);
\foreach \x in {-42, -41, 44,43,42}{\draw[thick, red] (32) to (\x);}
	\node[draw, thick, densely dotted, red, inner sep=.55cm, label={below, red}:{$M_3$}] at (44) {};
	\begin{scope}[every node/.style={fill=white}]
	\node at (11) {\fPART{1}{alert!50}};
	\node at (21) {\fPART{2}{alert!50}};
	\node at (22) {\fPART{1,1}{alert!50}};
	\node[alert, fill=white] at (20) {$\emptyset$};
	\draw[alert,thick,densely dotted] (-2.5,1.7) rectangle (.5,2.3);
	\node[alert, left, fill=white] at (-2.5,2) {$M_1$};
	\draw[alert,thick,densely dotted] (1.5,1.7) rectangle (2.5,2.3);
	\node[alert, right, fill=white] at (2.5,2) {$M_2$};
	%
	\node at (32) {\fPART{2,1}{red!50}};
	\node (M1a) at (-42) {\fPART{1,1}{red!50}};
	\node (M2a) at (-41) {\fPART{2}{red!50}};
	\node (M3) at (44) {\fPART{2,1,1}{red!50}};
	\node (M2b) at (43) {\fPART{2,2}{red!50}};
	\node (M1b) at (42) {\fPART{3,1}{red!50}};
	\draw[thick, red, <->, bend left=20] (M2a) to node[below] {$M_2$} (M2b);
	\draw[thick, red, <->, bend left=10] (M1a) to node[below] {$M_1$} (M1b);
	\end{scope}
\end{tikzpicture}}

\end{frame}

\begin{frame}

\alert{Some more results: } ([BDEHHILNSS-1\&2])
\begin{itemize}
\item Presentation of the graded signed Brauer algebra and related algebras/categories.
\begin{itemize}
\item Basis and spanning sets in terms of decorated diagrams.
\item Center given by a certain class of symmetric functions.
\item Filtrations and specializations similar to the classical cases.
\end{itemize}
\item Action on tensor space and translation functors.
\begin{itemize}
\item Translation functors given by actions on ``weight diagrams'' (akin to spin chain diagrams). \item  Algebraic structure: 0-Temperley-Lieb algebra.
\end{itemize}
\end{itemize}

\bigskip

\footnotesize \alert{With Martina Balagovic, Inna Entova-Aizenbud, Iva Halacheva, Johanna Hennig, Mee Seong Im, Gail Letzter, Emily Norton, Vera Serganova, and Catharina Stroppel:}
\begin{enumerate}[{[}1{]}]
\item ``Translation functors and Kazhdan-Lusztig multiplicities for the Lie superalgebra $\fp(n)$, Mathematical Research Letters Vol.26, no.3.
\item ``The affine VW supercategory'', to appear in Selecta. arXiv:1801.04178
\end{enumerate}

\centerline{https://zdaugherty.ccnysites.cuny.edu}


\end{frame}

\end{document}