Why does wikipedia define $G$ invariant polynomials this way?

Question

Why does wikipedia(https://en.wikipedia.org/wiki/Invariant_theory) define it as:

$$(g \cdot f)(x) := f(g^{-1}(x))?$$ Instead of, $$ (g \cdot f)(x) := f(g(x)).$$

Answer 1

It has to do with making sure that compositions work in the correct order, so that $$ (g_1 g_2) \cdot f = g_1 \cdot ( g_2 \cdot f). \label{1}\tag{1} $$ These conditions are equivalent to the fact that the representation $\pi: G \to GL(k[V])$ associated to the group action is in fact a group homomorphism: $$ \pi(g_1g_2) = \pi(g_1) \, \pi(g_2). $$ Let's verify (\ref{1}), using the definition $$ (g \cdot f) (x) = f( g^{-1} \cdot x) $$ Calculate for all $x \in V$: \begin{align} (g_1g_2 \cdot f) (x) &= f \bigl( (g_1g_2)^{-1} \cdot x \bigr) \\ &= f \bigl( (g_2^{-1}g_1^{-1}) \cdot x \bigr) \\ &= f \bigl( g_2^{-1} \cdot (g_1^{-1} \cdot x) \bigr) \\ &= (g_2 \cdot f)\, ( g_1^{-1} \cdot x) \\ &= \bigl( g_1 \cdot (g_2 \cdot f) \bigr) \, (x) \end{align}

It works! The left action on $V$ (the source of the functions in $k[V]$) naturally becomes a right action on the space $k[V]$, so if we want it to be a left action, then we act by inverses.

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Another way to see this it to draw everything in terms of compositions of functions. An element $g \in G$ acts by precomposition on functions $f \in k[V]$: $$ V \xrightarrow{f} k \qquad \longmapsto \qquad V \xrightarrow{g^{-1}} V \xrightarrow{f} k $$ so the element $g_1g_2 \in G$ acts by $$ V \xrightarrow{f} k \qquad \longmapsto \qquad V \xrightarrow{g_1^{-1}} V \xrightarrow{g_2^{-1}} V \xrightarrow{f} k $$ which is the same as $$ V \xrightarrow{(g_1g_2)^{-1}} V \xrightarrow{f} k $$

Answer 2

If you define it as the latter one, then you may got $$ ((g_{1}g_{2})\cdot f)(x) = f((g_{1}g_{2})x) = (g_{1}\cdot f)(g_{2}x) = (g_{2}\cdot (g_{1}\cdot f))(x), $$ which gives $(g_{1}g_{2})\cdot f = g_{2}\cdot (g_{1}\cdot f)$. One needs to include inverse to make $(g_{1}g_{2})\cdot f = g_{1}\cdot (g_{2}\cdot f)$.