8
$\begingroup$

Let $(A,g)$ be a graded commutative ring, where $A$ denotes the commutative ring, and $g$ its grading. What can be said about the set $\mathcal{G}_A := \{ \mathrm{Proj}\Big((A,g)\Big)\ \vert\ g \}$? How do elements of $\mathcal{G}_A$ relate to each other ? Can we endow $\mathcal{G}_A$ with more structure ?

$\endgroup$
3
  • 1
    $\begingroup$ Interesting question! If you consider the special case of polynomial rings and the subset of gradings simply by changing weights with respect to a fixed system of coordinates, then this space is equivalent to a quotient of the unit sphere (if you allow real weights), or the set of rational points of the unit sphere in $R^n$. You need to take a quotient, since it is possible that different weights lead to isomorphic proj. I do not know the precise correspondence, but would think that there is a combinatorial characterizationn of this isomorphism. $\endgroup$
    – pinaki
    Commented Aug 23, 2018 at 10:56
  • 1
    $\begingroup$ For starters, there's en.wikipedia.org/wiki/Weighted_projective_space and also the fact that the Veronese embedding exhibits a number of different gradings with isomorphic scheme structures, namely that scaling and shifting the grading unaffect the result. $\endgroup$ Commented Aug 23, 2018 at 13:07
  • 1
    $\begingroup$ Some more good info is in Ottem's answer here: mathoverflow.net/questions/47682/…. $\endgroup$ Commented Aug 23, 2018 at 14:19

1 Answer 1

2
$\begingroup$

EDIT. I misread the question - below I'm trying to describe the set of all gradings and not of their ${\rm Proj}$.

Note that giving a $\mathbf{Z}$-grading on a $k$-algebra $A$ is equivalent to giving a $\mathbf{G}_m$-action on $X={\rm Spec}(A)$. For example, if $A=k[x]$ for $k$ algebraically closed, then every such action is linearizable, i.e. either is trivial (everything in degree zero), or there exists an $a\in k$ such that $x-a$ is homogeneous with some degree $n$. One can think of the set $\mathcal{G}_A$ in this case as the disjoint union of $\mathbf{Z}\setminus \{0\}$ copies of $\mathbf{A}^1$ and a point corresponding to the zero grading.

Suppose that $A = k[x_1,\ldots, x_n]/I$. Then giving a $\mathbf{Z}$-grading/$\mathbf{G}_m$-action on $A$ is equivalent to providing homogeneous decompositions $x_i = \sum_n x_i^n$ of the generators satisfying some relations. In other words, we want to specify a ring homomorphism $\eta: A\to A\otimes k[t^{\pm 1}]$ (whose $\rm Spec$ is the action of $\mathbf{G}_m$) satisfying the usual conditions, given by $\eta(x_i) = \sum_n x_i^n \otimes t^n$. I think you can use this down-to-earth description to endow $\mathcal{G}_A$ with the structure of an ind-scheme, but I suspect that already for $A= k[x,y]$ this will not be a scheme locally of finite type.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .