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edited Nov 14, 2021 at 1:50

Degree of a regular function on a Algebraic Groupan algebraic group

Let $k$ be a field and $G$ be an affine agebraicalgebraic group. We assume that $G$ is a $k$-subvariety of $GL(V) \subset End(V)$$\operatorname{GL}(V) \subset \operatorname{End}(V)$ for appropriate vector space $V$ of dimension $d= \dim{V}$. Let $k[G]$ be the coordinate ring of $G$.

Why and how is it possible to associate to every element $P \in k[G]$ a degree $deg(P)$$\operatorname{deg}(P)$?

Although the assomption $G \subset End(V)$$G \subset \operatorname{End}(V)$ implies that $k[G]= k[X^iY^j \ \vert \ 1 \le i,j \le d]/I$$k[G]= k[X^iY^j \mathrel\vert 1 \le i,j \le d]/I$ is a quotient of a polynomial ring with ideal $I=I(G)$, there is no reason why it should be possible to associate a degree to an element from $k[G]$.

But seemingly that's exactly what is done in the proof of Proposition 4.6 (Chevalley) from the notes these notesRéseaux des groupes de Lie by Yves Benoist on Lie groups (page 38).

Note the script is writen in frenchFrench and seemingly the notation for degree of a $P \in k[G]$ is $d^{\circ} P$. But as I explained above this definition make no sense (at least) to me as long as $k[G]$ is not isomorphic to a polynomial ring.

Does somebody know what Benoist has there in mind there?

Degree of a regular function on a Algebraic Group

Let $k$ be a field and $G$ be an affine agebraic group. We assume that $G$ is a $k$-subvariety of $GL(V) \subset End(V)$ for appropriate vector space $V$ of dimension $d= \dim{V}$. Let $k[G]$ the coordinate ring of $G$.

Why and how is it possible to associate to every element $P \in k[G]$ a degree $deg(P)$?

Although the assomption $G \subset End(V)$ implies that $k[G]= k[X^iY^j \ \vert \ 1 \le i,j \le d]/I$ is a quotient of a polynomial ring with ideal $I=I(G)$, there is no reason why it should be possible to associate a degree to an element from $k[G]$.

But seemingly that's exactly what is done in the proof of Proposition 4.6 (Chevalley) from these notes by Yves Benoist on Lie groups (page 38)

Note the script is writen in french and seemingly the notation for degree of a $P \in k[G]$ is $d^{\circ} P$. But as I explained above this definition make no sense (at least) to me as long as $k[G]$ is not isomorphic to a polynomial ring.

Does somebody know what Benoist has there in mind?

Degree of a regular function on an algebraic group

Let $k$ be a field and $G$ be an affine algebraic group. We assume that $G$ is a $k$-subvariety of $\operatorname{GL}(V) \subset \operatorname{End}(V)$ for appropriate vector space $V$ of dimension $d= \dim{V}$. Let $k[G]$ be the coordinate ring of $G$.

Why and how is it possible to associate to every element $P \in k[G]$ a degree $\operatorname{deg}(P)$?

Although the assomption $G \subset \operatorname{End}(V)$ implies that $k[G]= k[X^iY^j \mathrel\vert 1 \le i,j \le d]/I$ is a quotient of a polynomial ring with ideal $I=I(G)$, there is no reason why it should be possible to associate a degree to an element from $k[G]$.

But seemingly that's exactly what is done in the proof of Proposition 4.6 (Chevalley) from the notes Réseaux des groupes de Lie by Yves Benoist (page 38).

Note the script is writen in French and seemingly the notation for degree of a $P \in k[G]$ is $d^{\circ} P$. But as I explained above this definition make no sense (at least) to me as long as $k[G]$ is not isomorphic to a polynomial ring.

Does somebody know what Benoist has in mind there?