# (in-)compatible gradings of an associative algebra tell us…?

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.

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I can't seem to get that first line to render correctly. –  alekzander Nov 9 '09 at 20:08
Fixed that for you. (A rare un-ironic use of that phrase. :) ) This is a well known bug with the underscore character. The temporary workaround is to write (backslash)(underscore) whenever you mean (underscore). –  David Speyer Nov 9 '09 at 20:11
The last statement of the post (In fact, any finite associative algebra with nontrivial grading over Z_2 cannot be given a compatible Z grading.) is clearly wrong for the Grassmann algebra? You should be a bit more careful here. –  Vladimir Dotsenko Nov 10 '09 at 11:48
Finite, not finite-dimensional. –  alekzander Nov 10 '09 at 19:35

A $\mathbb{Z}/2$ grading on a ring is equivalent to an action of $\mathbb{Z}/2$ on the ring: act by $(-1)$ on the odd part. Hence, it gives a $\mathbb{Z}/2$ action on the underlying variety. Similarly, a $\mathbb{Z}$ grading gives an action of $\mathbb{C}^\*$: the element $g \in \mathbb{C}^\*$ acts on $A\_i$ by $g^i$.
So, the question is when a $\mathbb{Z}/2$ action extends to an action of $\mathbb{C}^*$. In that generality, I can't think of anything intelligent to say beyond what I've said.