Definition of $\mathbb{Z}_2$-graded module homomorphism

Question

A $\mathbb{Z}_2$-graded module homomorphism $f$ of a $\mathbb{Z}_2$-graded algebra $A$ is defined to satisfy $f(am)=(-1)^{\text{deg}(f)\text{deg}(a)}af(m)$, for example, at the bottom of the second page of the paper [J]. When $f$ and $a$ have grading $1$, an extra $-1$ will appear.

In ordinary algebra, a simple module $M$ has a division algebra $D$ corresponding to the endomorphism $\text{End}(M)$. Conversely, if we have a simple module $M$ of a division algebra $D$, then the division algebra of the simple module $M$ is still $D$.

I believe the same thing should happen in the context of $\mathbb{Z}_2$-graded algebra. But if we consider the $\mathbb{Z}_2$-graded $\mathbb{R}$-algebra $Cl^{0,1}$, which is the Clifford algebra with one generator $e^2=1$. $Cl^{0,1}$ is a $\mathbb{Z}_2$-graded division algebra, and it has one simple $\mathbb{Z}_2$-graded module $M\cong Cl^{0,1}$ with an action on the left. Notice that the only possible grading $1$ $\mathbb{Z}_2$-graded $A$-module map has $1\mapsto e$ and $e\mapsto -1$. It is very strange because $\text{End}(M)$ becomes $Cl^{1,0}$ with one generator $(e')^2=-1$. On the other hand, if we don't require the extra $-1$ in the definition of $\mathbb{Z}_2$-graded $A$-map, then $\text{End}(M)$ is $Cl^{0,1}$.

I'm wondering which definition is more natural and how to fix this inconsistency.

[J]: Józefiak, Tadeusz, Semisimple superalgebras, Algebra: some current trends, Proc. 5th Natl. Sch. Algebra, Varna/Bulg. 1986, Lect. Notes Math. 1352, 96-113 (1988). ZBL0666.17001.

Answer 1

The sign is correct. Your mistake is here:

Conversely, if we have a simple module $M$ of a division algebra $D$, then the division algebra of the simple module $M$ is still $D$.

This is only true if you use right modules. If you use left modules, then $D$ as a left module over itself has endomorphism ring $D^{op}$; explicitly, endomorphisms are given by right multiplication by elements of $D$.

In the $\mathbb{Z}_2$-graded setting the definition of the opposite algebra also has to be modified by a Koszul sign; that is, if $A$ is a $\mathbb{Z}_2$-graded algebra with multiplication $a \star b$ then its graded opposite has multiplication

$$a \star^{op} b = (-1)^{\deg a \deg b} b \star a.$$

Then the graded opposite of $\text{Cl}_{p, q}$ is $\text{Cl}_{q, p}$, which is responsible for the explicit calculation you're seeing, since you are using left modules. So, we have some funny $\mathbb{Z}_2$-graded stuff going on here: the underlying algebra of the graded opposite is not the ordinary opposite (each of the Clifford algebras is its own ordinary opposite, I think), and the Clifford algebras $\text{Cl}_{0, 1}$ and $\text{Cl}_{1, 0}$ are commutative but not graded commutative.

If you drop the signs then you're computing the ordinary opposite rather than the graded opposite but in that case you're no longer working in the graded / Koszul / super world.