I am working through the Mirror Symmetry book, available here.

I already had a question about an earlier part of the same Exercise 9.2.1 on page 157:

We are given the following action with one boson and two fermions ($h$ is some function of $X$): $$\begin{align*} S(X, \Psi_1, \Psi_2) = \frac{1}{2} (\partial h)^2 - \partial^2 h \Psi_1 \Psi_2. \end{align*}\tag{9.29}$$

We are told to consider the following transformations:

$$\begin{align*} X' &= X +\epsilon^1 \Psi_1 + \epsilon^2 \Psi_2 \\ \Psi_1' &= \Psi_1 + \epsilon^2 \partial h \\ \Psi_2' &= \Psi_2 - \epsilon^1 \partial h. \end{align*}\tag{9.30}$$

$\epsilon^i$ and $\Psi_i$ are Grassmann odd variables, and so anticommute.

We are asked to show the integration measure is invariant. It says in doing so, we will develop a concept of superdeterminant and supertrace. I'm not sure if we are really expected to develop these notions ourselves here, but I resorted to referencing Wikipedia: Berezin Integral

From that wiki page:

\begin{align*} dX' d\Psi_1'd\Psi_2' = \frac{det(A - B D^{-1}C)}{det(D)}dX d\Psi_1 d\Psi_2 \end{align*}

where $A = \frac{\partial X'}{\partial X}$, $B = \frac{\partial X'}{\partial \Psi_j}$, $C = \frac{\partial \Psi_i'}{\partial X}$, and $D = \frac{\partial \Psi_i'}{\partial \Psi_j'}$.

I got

\begin{align*} dX' d\Psi_1'd\Psi_2' = (1 - 2 (\partial^2 h) \epsilon_1 \epsilon_2)dX d\Psi_1 d\Psi_2 \end{align*}

This comes from $A = 1$, $B = (\epsilon_1, \epsilon_2)$, $C = (\epsilon_2 \partial^2 h, - \epsilon_1 \partial^2 h)^{T}$, and $D = I_2)$.

In case it matters, I am assuming derivatives act from the right.

Have I made a calculation error? Or is there some reason why the $\epsilon_1 \epsilon_2$ term should be zero?


1 Answer 1



  1. It is enough to check for an infinitesimal transformation $$\begin{align} \delta x ~=~&\epsilon^i\psi_i,\cr \delta\psi_i ~=~& \epsilon_{ij}\epsilon^j h^{\prime}(x), \cr i,j ~\in~&\{1,2\}. \end{align}\tag{9.30}$$

  2. In that case the Jacobian factor [minus 1] is given by the supertrace of [the change in] the Jacobian matrix $$ \frac{\partial_R(\delta x,\delta\psi_1,\delta\psi_2)}{\partial (x,\psi_1,\psi_2)}.\tag{1}$$

  3. If the Grassmann-odd infinitesimal parameters $$\epsilon^i~=~ a^i(x)\psi_1\psi_2+ b^{ij}(x)\psi_i +c^i(x)\tag{2}$$ are non-constant functions of the variables $(x,\psi_1,\psi_2)$ [as they are on the bottom of the very same page!], then clearly the supertrace does not need to vanish. [That is e.g. the main point of eq. (9.34).]

  4. So Ref. 1 is presumably implicitly assuming that $\epsilon^i$ are constants in Exercise 9.2.1. Then the whole diagonal of the matrix (1) [and hence the supertrace] indeed vanish.


  1. K. Hori, S. Katz, A. Klemm, R. Pandharipande, R. Thomas, C. Vafa, R. Vakil, and E. Zaslow, Mirror Symmetry, 2003; Sections 9.2-9.3. The pdf file is available here.

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