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I am just learning about action of $G$, algebraic group over an algebraically closed field $k$, that is locally algebraic. It states that if $G$ acts linearly on a vector space $W$, we say the action is locally algebraic if it is locally fintie and for any finite dimensional $G$-stable subspace $V$, the action $\theta: G \times V \rightarrow V$ is a morphism.

1) My first question is what is meant by " $G$ acts linearly on $W$"?

2) What is meant by " the action $\theta: G \times V \rightarrow V$ is a morphism"? Does this mean a morphism as algebraic groups or varieties?

3) Could someone explain why if $G\times V \rightarrow V$ is given by $$ (g, \sum_{i=1}^n \lambda_i e_i) \rightarrow \sum_{i} \lambda_i h_i(g^{-1})e_i $$ where $\lambda_i \in k$, $\{e_i\}$ is a basis of $V$ and $h_i \in k[G]$ then this defines a morphism?

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  • $\begingroup$ Can you explain the exact definition of this "locally finite"? $\endgroup$ – Amira Lo Mar 15 at 13:27
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1) $G$ acts linearly on $W$ means that each element of $G$ acts by a linear transformation of $W$. Equivalently the action determines a homomorphism of $G$ into $GL(W)$.

2) This means a morphism of varieties, the algebraic group structure of $V$ isn't relevant here.

3) At the level of (closed) points, a morphism of varieties is a map that's locally given by polynomials in affine charts. In this case, and $h_i(g^{-1})$ are polynomial in $g$, since inversion is a morphism on $G$, and $h_i\in k[G]$, so this map is polynomial in any affine chart of $G\times V$.

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  • $\begingroup$ Is an inversion from $G$ to $G$ a polynomial map? $\endgroup$ – Johnny T. Jun 2 '19 at 21:20
  • $\begingroup$ Yes, this is part of the definition of an algebraic group (you need the inverse map to be a morphism of varieties, hence its locally given by polynomials). $\endgroup$ – A Nonny Mouse Jun 3 '19 at 7:19
  • $\begingroup$ When you say it's locally given by polynomials, I guess you mean like $f/g$ where $f$ and $g$ are polynomials? (just thinking of $2$ by $2$ matrices the inversion involves $1/$determinant..) $\endgroup$ – Johnny T. Jun 3 '19 at 7:52
  • $\begingroup$ When I say locally polynomial, I mean that it's a morphism of varieties, equivalently that it's polynomial on affine charts. For $gl(n)$ this ends up being the same thing as being globally given by rational functions whose denominator doesn't vanish, since $gl(n)$ is quasi-affine. If you want to make more sense of all these notions (and properly learn about algebraic groups), you really ought to learn what a variety is, and what morphisms between them are. $\endgroup$ – A Nonny Mouse Jun 3 '19 at 21:26
  • $\begingroup$ Could you possibly give me an example of what you mean for the case of $GL_2$ by any chance? (where the inverse is a polynomial map) $\endgroup$ – Johnny T. Jun 3 '19 at 21:50

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