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.