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The following is just a basic definition in Invariant Theory, which I copied from wikipedia

"Let $G$ be a group, and ${\displaystyle V}$ a finite-dimensional vector space over a field ${\displaystyle k}$ (which in classical invariant theory was usually assumed to be the complex numbers). A representation of ${\displaystyle G}$ in ${\displaystyle V}$ is a group homomorphism ${\displaystyle \pi :G\to GL(V)}$, which induces a group action of ${\displaystyle G}$ on ${\displaystyle V}$. If ${\displaystyle k[V]}$ is the space of polynomial functions on ${\displaystyle V}$, then the group action of ${\displaystyle G}$ on ${\displaystyle V}$ produces an action on ${\displaystyle k[V]}$ by the following formula:

${\displaystyle (g\cdot f)(x):=f(g^{-1}(x))\qquad \forall x\in V,g\in G,f\in k[V].} $"

I did not quite understand the group action at the end. I mean, we need to prove $((gh)\cdot f)(x)=g\cdot (h\cdot f)(x)\qquad \forall x\in V,g,h\in G,f\in k[V]$

I have two ways to understand the right hand side:

1) $g\cdot (h\cdot f)(x)=g\cdot f(h^{-1}(x))=f(g^{-1}h^{-1}x)$

2) $g\cdot (h\cdot f)(x)= (h\cdot f)(g^{-1}x)=f(h^{-1}g^{-1}x)$

Of course, to make this a group action, the second way is correct. However, I want to ask how can one just look at $g\cdot (h\cdot f)(x)$ and tell which way is correct? In fact, I think the first way is more rational as we need to compute $h\cdot f$ first.

Thank you in advance

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$ g \cdot (h \cdot f) (x) $ is by definition the evaluation of $ g \cdot (h \cdot f) $ on $ x $ .

This function is of the form $ g \cdot m $ for $ m = h \cdot f $, thus by definition it evaluation on $ x $ is $ m(g^{-1}x) $ .

This is the evaluation of $ h \cdot f $ on $ g^{-1} x $ which is by definition $ f(h^{-1} (g^{-1}x)) =f (h^{-1}g^{-1}x) $ .

So there is only one way to evaluate this, the other one is a common mistake.

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If you denote $f':=h.f$ , then your asking what is $g.f'$. So lets see what it does to $x$, for convenience denote $x'=g^{-1}.x$

Now: $$ g.f'(x)=f'(g^{-1}.x)=f'(x')=f(h^{-1}.x')=f(h^{-1}g^{-1}.x)$$

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