# How does a $\mathbb{Z}$-grading on a vector space $V$ translate to a map “$\mathbb{G}_m\rightarrow GL(V)$”?

Let $V$ be a finite dimensional vector space over $\mathbb{Q}_p$, let $C$ be the completion of the algebraic closure of $\mathbb{Q}_p$. Suppose $V\otimes_{\mathbb{Q}_p}C$ has a $\mathbb{Z}$-grading, then at the end of the first page of these notes on Hodge-Tate theory, the author claims that the grading gives a map $$\mu : (\mathbb{G}_m)_C\rightarrow GL(V)_C$$

How does one define the map in terms of the grading? What does $GL(V)_C$ mean?

I can see that in general if you have a $\mathbb{Z}$-graded commutative $k$-algebra $A$, then one obtains a comodule structure on $A$ given by: $$m : A\rightarrow A\otimes_k k[\mathbb{Z}] = A\otimes_k k[x,x^{-1}]\qquad a\mapsto \sum_i a_i\otimes x^i$$ where $a_i$ is the $i$th graded part of $a$. Taking Spec's, one gets a map $$\text{Spec}(m) : \text{Spec }A\times_{\text{Spec }k}\mathbb{G}_m\rightarrow\text{Spec }A$$ which gives an action of $\mathbb{G}_m$ on $\text{Spec }A$. In the setting of the question, instead of a commutative algebra $A$, we just have a finite dimensional $C$-vector space, so we can't really just "take Spec". So...what do?

• Every vector space over a field $k$ canonically determines a group scheme over $k$, namely the Spec of the symmetric algebra of its dual. That lets you talk about group schemes acting on it. – Qiaochu Yuan Sep 23 '17 at 8:07

Any $\Bbb Z$-graded module $M$ gives an action of $\Bbb G_m$ in the sense that for $g\in \Bbb G_m$ and $m\in M$ of degree $d$, one may define $g\cdot m= g^dm$. This action of $\Bbb G_m$ is the same thing as a map $\Bbb G_m\to Aut(M)$, and in this case, $M=V$ and $Aut(M)=GL(V)$.
$GL(V)_C$ probably means $Aut_{C}(V\otimes_{\Bbb Q_p} C)$- without adding the subscript, it could potentially mean $Aut_{\Bbb Q_p}(V)$, which is different.