I'm having some trouble understanding how to extend representation theory from Lie algebras to super Lie algebras, in particular with $psl(2|2)$. Ultimately I'm interested in 2D quantum sigma models with a supergroup target space.

Despite reading several review papers, I have not found a good worked example. Let me first set up some notation. You may prefer to jump to the questions first.

The Set-up

Consider $psl(2|2)$.

A super matrix $$ {t_A}^B = \left [ \begin{array}{c|c} {t_\alpha}^\beta & {t_\alpha}^{\beta'} \\ \hline {t_{\alpha'}}^{\beta} & {t_{\alpha'}}^{\beta'} \end{array} \right ] $$

where $\alpha,\beta,... = 1,2$ and $\alpha',\beta',... = 1',2'.$

Redefining for example ${t_1}^2$ now to be a matrix with all zero entries except for ${t_1}^2$ which is one. i.e. $$ {t_1}^2 = \left [ \begin{array}{c c|c c} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array} \right ] $$

A Borel decomposition is given by $psl(2|2) = \mathbf{h}\oplus\mathbf{n^+}\oplus\mathbf{n^-}$ where

The even Cartan generators $\mathbf{h}$ are:

$h = 1/2({t_1}^1 - {t_2}^2)$ and $h'=1/2({t_{1'}}^{1'} - {t_{2'}}^{2'})$

The raising operators $\mathbf{n^+}$ are:

even: ${t_1}^2$, ${t_{1'}}^{2'}$ expressed as $t_{(1, 0)}^+$ and $t_{(0, 1)}^+$

odd: ${t_\alpha}^{\beta'}$ expressed as $t_{(1/2, 1/2)}^+$, $t_{(1/2, -1/2)}^+$, $t_{(-1/2, 1/2)}^+$, $t_{(-1/2, -1/2)}^+$

The lowering operators $\mathbf{n^-}$ are:

even: ${t_2}^1$, ${t_{2'}}^{1'}$ expressed as $t_{(-1, 0)}^-$ and $t_{(0, -1)}^-$

and odd: ${t_{\alpha'}}^{\beta}$ expressed as $t_{(-1/2, -1/2)}^-$, $t_{(-1/2, 1/2)}^-$, $t_{(1/2, -1/2)}^-$, $t_{(1/2, 1/2)}^-$

For example, $[h, {t_1}^2] = {t_1}^2$ and $[h' , {t_1}^2]=0$ we see ${t_1}^2$ has root (1,0) so we write it as $t_{(1, 0)}^+$.

The questions:

1a. Assuming the above choice of $\mathbf{n^+}$, we should think of these has having positive roots. What is a good definition of positive in terms of the roots?

1b. What do we take for a basis of simple “positive root” operators for $psl(2|2)$?

  1. Highest weight state $|(m,n')>_+$ is annihilated by operators in $\mathbf{n^+}$. We can produce all the states in the usual way, by acting with monomials of operators in $\mathbf{n^-}$ on $|(m,n')>_+$. Besides that for an odd operator $t^- \in \mathbf{n^-}$, $(t^-)^2=0$, how do we know which operators in $\mathbf{n^-}$ are redundant? See the example below:

Adjoint Representation Example

Consider the adjoint representation since we can see explicitly how operators act on states.

The highest weight state annihilated by all of $\mathbf{n^+}$ is ${t_1}^{2’}$=$|(1/2, 1/2)>_+$. It is the unique state $t$, such that $ [\mathbf{n^+}, t] = 0$.

By acting with the odd generators $t_{(1/2, -1/2)}^- , t_{(-1/2, 1/2)}^- \in \mathbf{n^-}$ on it, we form new highest weight states with respect to the even subalgebra $sl(2)\oplus sl(2)$ with weights (1,0), (0,1) and (1/2,1/2) respectively.

The remaining states of the representation are found in the usual way as representations of $sl(2)\oplus sl(2)$ by acting on the four states with the even generators $t_{(-1, 0)}^-$ and $t_{(0, -1)}^-$.

My issue is that the remaining odd generators $t_{(1/2, 1/2)}^-, t_{(-1/2, -1/2)}^- \in \mathbf{n^-}$ are not needed:

$t_{(1/2, 1/2)}^-={t_{1’}}^{2}$ annihilates $|(1/2, 1/2)>_+$ since $[{t_{1}’}^{2}, {t_{1}}^{2'}] = 0$.

and $t_{(-1/2, -1/2)}={t_{2}’}^{1}$ produces a state which is already made.

How are we to know not to use these operators? What’s the rule for highest weight representation (3/2,3/2) for example and why?

Apologies for the extra long text! Suggestions for shortening or clarifying it are gratefully received.

  • 1
    $\begingroup$ For $1.a$ and $1.b$, there is a discussion pages $23, 24$ (document numerotation) (and more generally chapter $2.2.1$) in this paper $\endgroup$ – Trimok Sep 15 '14 at 10:15

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