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.