I asked this question in MathStackExchange 9 days ago but get no response (not a vote nor a comment), so I'm copying it here below. The link to the original question is:


A free super module over superring $R$ of rank $p|q$ with standard basis $(e_1,\cdots,e_{p+q}) $ is defined by

$$R^{p|q}:= e_1R\oplus \cdots \oplus e_{p+q}R,$$

where $e_1,\cdots,e_p$ are even and $e_{p+1},\cdots,e_{p+q}$ are odd. It is a both sided $R$-module with scalar multiplication induced by Koszul sign rule. A morphism $F$ from for example $R^{p|q}$ to itself is an $R$-linear map that preserves the parity. Clearly $F$ is determined by its value on the basis, and we can write

$$Fe_i=\sum_j r_{ji}e_j = \sum_{j\leq p} e_j r_{ji} + \sum_{j\geq p+1} e_j (-1)^{|r_{ji}|} r_{ji}=: \sum_{j\leq p} e_j \tilde{r}_{ji} + \sum_{j\geq p+1} e_j \tilde{r}_{ji}.$$

Now, to represent $F$ by a matrix under the basis $(e_1,\cdots,e_{p+q}) $, one may put the matrix to be $(r_{ji})$ or $(\tilde{r}_{ji})$. The former translates the composition of morphisms as tranposed right-multiplication while the latter translates it as left-multiplication. In literature of supergeometry, for example the chain rule, I suppose that we prefer the latter as we are used to left-multiplication, right?

Deligne & Morgan's Notes on Supersymmetry (following Joseph Bernstein) never explains which choice the matrix representation should be but only computes the Berezinian of super matrices. Leites' Introduction to the Theory of Supermanifolds, in order to make the (modified) Jacobian (of a morphism from coordinates $(u,\xi)$ to $(v,\eta)$) behaves well in the manner of left-multiplication, puts

$$ J(v,\eta)=\left( \begin{matrix}\partial_u v & -\partial_\xi v\\ \partial_u\eta &\partial _\xi \eta \end{matrix}\right). $$

However, my equation above about $F$ tells that it seems more reasonable to put

$$ \tilde{J}(v,\eta)=\left( \begin{matrix}\partial_u v & \partial_\xi v\\ -\partial_u\eta &\partial _\xi \eta \end{matrix}\right). $$

Both choices are compatible with left-multiplication and they have identical Berezinian, but I really wonder the reason why it is not $\tilde{J}$ that is chosen. Is it totally a historical reason or is there any situation where $J$ behaves better than $\tilde{J}$?

Moreover, in purely algebra context, is the matrix representation of $F$ chosen to be $(\tilde{r}_{ji})$ as the equation implies or is it chosen to be $\left((-1)^{B(i\geq p+1 \wedge j\leq p)}r_{ji}\right)$, where $B$ is the Boolean function, to be consistent with the choice of the Jacobian $J$? This is really weird to me, because $(\tilde{r}_{ji})$ is not compatible with the Jacobian $J$ while $J$ is not compatible with the literature of algebra -- and this can be fixed by simply choosing $\tilde{J}$ instead, so why not?

Thanks in advance for any help.



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