Action of k* on a variety induces grading? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:15:56Z http://mathoverflow.net/feeds/question/104756 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104756/action-of-k-on-a-variety-induces-grading Action of k* on a variety induces grading? Jesko Hüttenhain 2012-08-15T10:21:25Z 2012-08-15T11:10:49Z <p>Let $V$ be a $\Bbbk$-variety such that $\Bbbk^\times$ (as an algebraic group) acts algebraically on $V$. Given any $f\in\Bbbk[V]$, let us call $f$ <b>homogeneous of degree $d$</b> if for all $v\in V$ and all $\lambda\in\Bbbk^\times$, we have $f(\lambda.v)=\lambda^d f(v)$. </p> <p>My question is: Does this define a grading on $\Bbbk[V]$? </p> <p>I was convinced that it is true, but I am running into difficulties. Let us first assume $\Bbbk=\mathbb{C}$, the ground field should not be an obstruction. The linear span of $\Bbbk^\times f$ decomposes since $\Bbbk^\times$ is reductive, but I don't see how to turn this into a grading on all of $\Bbbk[V]$. </p> <p>If it is true, I would really like to see a proof - it should use as little machinery as possible.</p> http://mathoverflow.net/questions/104756/action-of-k-on-a-variety-induces-grading/104758#104758 Answer by Ben Webster for Action of k* on a variety induces grading? Ben Webster 2012-08-15T11:10:49Z 2012-08-15T11:10:49Z <p>Turning the action map of varieties into a map of rings, we get a ring map $\phi$ from $k[V]$ to $k[V][t,t^{-1}]$, the coordinate ring with an extra invertible variable (the coordinate on $k^*$) adjoined. Now, for any function $\phi(f)=\sum_{i\in \mathbb{Z}}f_it^i$ for some $f_i$'s, almost all of which are 0. Note that $f=\sum f_i$, which we obtain by restricting the function to $t=1$. Using associativity, applying $\phi$ again to the $f_i$'s is the same as applying pull-back by the multiplication map to t. Thus, as functions on <code>$V\times k^*\times k^*$</code> (letting $t,u$ be the two coordinates)</p> <p>$$\sum_{i\in \mathbb{Z}}\phi(f_i)u^i=\sum_{i\in \mathbb{Z}} f_i t^i u^i$$ </p> <p>since the pull-back of the coordinate by multiplication is just the product of the coordinates . Thus, $\phi(f_i)=f_it^i$. </p> <p>We can define the grading by letting $f$ be homogeneous of degree $i$ if $\phi(f)=ft^i$. We have already seen that every element can be written uniquely as a sum of such elements (the $f_i$'s), and this is multiplicative since $\phi$ is a ring homomorphism.</p> <p>Alternatively, we can note that we have proven that the span of the $f_i$'s is an finite-dimensional invariant subspace containing $f$, so we can apply your argument. In general, essentially the same argument shows that the action of any affine algebraic group on the coordinate ring of any affine variety by pull-back is a locally finite action: any function is contained in a finite-dimensional invariant subspace.</p>