If an associative algebra `A`

is $\mathbb{Z}$-graded, then it is automatically $\mathbb{Z}\_2$ (aka $\mathbb{Z}/2\mathbb{Z}$) graded by defining $A\_{\bar{0}}$ to be the direct sum over the even graded elements of `A`

, and $A\_{\bar{1}}$ to be the direct sum over the odds. Conversely, what may be said to distinguish those algebras `A`

for which a $\mathbb{Z}\_2$-grading exists, but no compatible $\mathbb{Z}$-grading exists? (Of course, compatible in the sense that the induced grading just described matches the given one.)

My motivation is the study of the `B(0,n)`

(aka `osp(1|2n)`

) series of Lie superalgebras, which I have been told cannot be $\mathbb{Z}$-graded, thus making their study a bit different from several of the other classes of Lie superalgebras.

The question can also be generalized greatly from above, and I think this is the right generalization. Say $\pi:R\to S$ is a surjection of abelian groups, so that an associative algebra `A`

graded over `R`

is automatically graded over `S`

. What properties does an algebra have if it has an `S`

grading, but no compatible `R`

grading?

For a boring example, the associative algebra $\mathbb{Z}\_2$ (with itself as base field) is graded over itself as an abelian group, but clearly cannot be $\mathbb{Z}$-graded. In fact, any finite associative algebra with nontrivial grading over $\mathbb{Z}\_2$ cannot be given a compatible $\mathbb{Z}$ grading.