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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 $f$ homogeneous of degree $d$ if for all $v\in V$ and all $\lambda\in\Bbbk^\times$, we have $f(\lambda.v)=\lambda^d f(v)$.

My question is: Does this define a grading on $\Bbbk[V]$?

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 $\Bbbk[V]$.

If it is true, I would really like to see a proof - it should use as little machinery as possible.

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    $\begingroup$ If $k$ is algebraically closed then the action of $k^*$ identifies with an action of the torus $G_{m,k}$. This is a diagonalisable group scheme and therefore any action of $G_{m,k}$ on a $k$-algebra corresponds to a $\bf Z$-grading (it does not have to be finite dimensional over $k$). See for instance SGA 3, I, 4.7.3. $\endgroup$
    – Damian Rössler
    Commented Aug 15, 2012 at 11:00
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    $\begingroup$ @Jesko: Please see my final comment to your earlier question related to this matter; Damian's SGA3 reference is precisely Ben's computation, carried out over any ring. Note also that if you already believed that "the linear span of $k^{\times}f$ decomposes since $k^{\times}$ is reductive" (as you say above) then you are done, since it would imply that every element lies in a finite-dimensional $k^{\times}$-stable subspace and so the span of any two such would be similarly exhausted in this way (hence graded, etc.). So that case done rigorously contains all of the difficulties. $\endgroup$
    – user22479
    Commented Aug 15, 2012 at 14:19
  • $\begingroup$ If you imagine that K∗ is like a circle, it will have $π_1=Z$. Then you have have Z "act on" V. This "action" could be grading. I'm am not sure if this is non-sense from a coincidence or their is something to this. I am not even sure it this can be made precise. $\endgroup$
    – Spice the Bird
    Commented Aug 16, 2012 at 8:01

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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 $V\times k^*\times k^*$ (letting $t,u$ be the two coordinates)

$$\sum_{i\in \mathbb{Z}}\phi(f_i)u^i=\sum_{i\in \mathbb{Z}} f_i t^i u^i$$

since the pull-back of the coordinate by multiplication is just the product of the coordinates . Thus, $\phi(f_i)=f_it^i$.

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.

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.

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    $\begingroup$ Wonderful! This is just perfect. $\endgroup$
    – Jesko Hüttenhain
    Commented Aug 15, 2012 at 11:47

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