Fixed mistake in the definiton of the supermatrix of a superoperator, Appendix A

1 files changed, 22 insertions, 8 deletions

diff --git a/marginal.tex b/marginal.tex
index 508b674..8efa79f 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -1772,21 +1772,30 @@ Integrals involving superfields contracted into such operators result in schemat
 \end{equation}
 where the usual role of the determinant is replaced by the superdeterminant.
 The superdeterminant can be defined using the ordinary determinant by writing a
-block version of the matrix $M$: if $\mathbf e(1)=\{1,\bar\theta_1\theta_1\}$ is
+block version of the matrix $M$. If $\mathbf e(1)=\{1,i\bar\theta_1\theta_1\}$ is
 the basis vector of the even subspace of the superspace and $\mathbf
-f(1)=\{\bar\theta_1,\theta_1\}$ is that of the odd subspace, then we can form a
+f(1)=\{i\bar\theta_1,i\theta_1\}$ is that of the odd subspace, dual bases $\mathbf e^\dagger(1)=\{i\bar\theta_1\theta_1,1\}$ and $\mathbf f^\dagger(1)=\{\theta_1,-\bar\theta_1\}$ can be defined by the requirement that
+\begin{align}
+  \int d1\,\mathbf e(1)\mathbf e^\dagger(1)=iI
+  &&
+  \int d1\,\mathbf f(1)\mathbf f^\dagger(1)=iI \\
+  \int d1\,\mathbf e(1)\mathbf f^\dagger(1)=0
+  &&
+  \int d1\,\mathbf f(1)\mathbf e^\dagger(1)=0
+\end{align}
+With such bases and dual bases defined, we can form a
 block representation of $M$ in analogy to the matrix form of an operator in quantum mechanics by
 \begin{equation}
   \int d1\,d2\,\begin{bmatrix}
-    \mathbf e(1)M(1,2)\mathbf e(2)^T
+    \mathbf e(1)M(1,2)\mathbf e^\dagger(2)
     &
-    \mathbf e(1)M(1,2)\mathbf f(2)^T
+    \mathbf e(1)M(1,2)\mathbf f^\dagger(2)
     \\
-    \mathbf f(1)M(1,2)\mathbf e(2)^T
+    \mathbf f(1)M(1,2)\mathbf e^\dagger(2)
     &
-    \mathbf f(1)M(1,2)\mathbf f(2)^T
+    \mathbf f(1)M(1,2)\mathbf f^\dagger(2)
   \end{bmatrix}
-  =\begin{bmatrix}
+  =i\begin{bmatrix}
     A & B \\ C & D
   \end{bmatrix}
 \end{equation}
@@ -1802,7 +1811,12 @@ save for the inverse of $\det D$. Likewise, the supertrace of $M$ is is given by
 \end{equation}
 The same method can be used to calculate the
 superdeterminant and supertrace in arbitrary superspaces, where for $\mathbb R^{N|2D}$ each
-basis has $2^{2D-1}$ elements. For instance, for $\mathbb R^{N|4}$ we have $\mathbf e(1,2)=\{1,\bar\theta_1\theta_1,\bar\theta_2\theta_2,\bar\theta_1\theta_2,\bar\theta_2\theta_1,\bar\theta_1\bar\theta_2,\theta_1\theta_2,\bar\theta_1\theta_1\bar\theta_2\theta_2\}$ and $\mathbf f(1,2)=\{\bar\theta_1,\theta_1,\bar\theta_2,\theta_2,\bar\theta_1\theta_1\bar\theta_2,\bar\theta_2\theta_2\theta_1,\bar\theta_1\theta_1\theta_2,\bar\theta_2\theta_2\theta_1\}$.
+basis has $2^{2D-1}$ elements. For instance, for $\mathbb R^{N|4}$ we have
+\begin{align}
+  &\mathbf e(1,2)=\{1,i\bar\theta_1\theta_1,i\bar\theta_2\theta_2,i\bar\theta_1\theta_2,i\bar\theta_2\theta_1,i\bar\theta_1\bar\theta_2,i\theta_1\theta_2,\bar\theta_1\theta_1\bar\theta_2\theta_2\}\notag \\
+  &\mathbf f(1,2)=\{i\bar\theta_1,i\theta_1,i\bar\theta_2,i\theta_2,\bar\theta_1\theta_1\bar\theta_2,\bar\theta_2\theta_2\theta_1,\bar\theta_1\theta_1\theta_2,\bar\theta_2\theta_2\theta_1\}
+\end{align}
+with the dual bases defined analogously to those above.
 
 \section{BRST symmetry}
 \label{sec:brst}