Describe invariants using coaction.

Question

Let $X$ be an algebraic variety and $G$ be an algebraic group which acts on $X$. We know that an invariant function $f$ in the coordinate ring $\mathbb{C}[X]$ is a function such that $g(f) = f$ for all $g \in G$. For the action of $G$ on $X$, we have a corresponding coaction $\delta: \mathbb{C}[X] \to \mathbb{C}[G] \otimes \mathbb{C}[X]$. Are the invariants functions just the functions $f$ such that $\delta(f) = 1 \otimes f$? Here $1 \in \mathbb{C}[G]$ is the constant function on $G$. Thank you very much.

Answer 1

Yes, you are completely correct. Let's assume $\delta(f)=\sum_{i=1}^n \alpha_i \otimes \beta_i$ with all $\beta_i$ linearly independent. If $f$ is invariant, then $$\sum_{i=1}^n \alpha_i(1)\cdot\beta_i(x) = f(1.x) = f(g.x)=\sum_{i=1}^n \alpha_i(g)\cdot\beta_i(x) $$ for all $x\in X$ and $g\in G$. This means $\sum_{i=1}^n \beta_i\cdot(\alpha_i(g)-\alpha_i(1))=0$ for any $g\in G$ (I am using that we are over an infinite field) and because the $\beta_i$ are linearly independent, this means $\alpha_i(g)=\alpha_i(1)$ for all $g\in G$, whence $\alpha_i\in\mathbb{C}$ for all $i$. Thus, $$\delta(f)=\sum_{i=1}^n \alpha_i\otimes \beta_i =\sum_{i=1}^n 1\otimes\alpha_i\beta_i =1\otimes \left(\sum_{i=1}^n \alpha_i\beta_i\right)$$ as required.

Edit As pointed out in the comments, we can conclude that $\delta(f)=1\otimes f$ in the following way: We already know $\delta(f)=1\otimes f'$ for some $f'\in\mathbb C[X]$ and we have $f(x)=f(1.x)=1\cdot f'(x)=f'(x)$ for all $x\in X$, so again because $\mathbb{C}$ is an infinite field, $f'=f$.

The other direction is easy.