Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to understand a set of lecture notes which I have on representation theory, and I have come unstuck on the section about invariant theory. Suppose we have a finite dimensional vector space $W$ over $\mathbb{C}$, with coordinate ring $\mathbb{C}[W]$. Then my notes state the following:

For $f:W \to V$ ($V$, $W$ $\mathbb{C}(G)$ modules), we say $f$ is a concomitant if $f(gw) = gf(w)$ (for $w \in W,\,g \in G$). (We have not specified what G is in the notes.)

A G-invariant for $W$ is a concomitant $f:W \to \mathbb{C}$, so $f(gw)=f(w) \, \forall \, w \in W,\,g \in G$.

So, I'm probably just being stupid here and forgetting some really obvious representation theory, but why should the definition of a concomitant imply $f(gw)=f(w)$ when the image of $f$ is in $\mathbb{C}$? Or rather, I suppose the question is why should $gf(w)=f(w)$? f(w) should just be some complex number so maybe I'm being stupid but I don't see why $gf(w)$ should equal $f(w)$.

It is possible I simply copied the notes down incorrectly, but the name "G-invariant" made me think this was probably the right definition. I am also aware concomitant is some fairly ancient terminology, which is why I haven't managed to find much online to explain this! Many thanks for your help.

share|improve this question
1  
The action on $\mathbb{C}$ is the trivial action. –  M Turgeon Apr 24 '12 at 16:38
    
Is that the only possible action on $\mathbb{C}$? Or is it just implied that this is the chosen action? Sorry, my rep theory is very rusty, this is the first time I've done any in a while. –  Ben Apr 24 '12 at 16:43
2  
This is not the only possible action (for example, any cyclic group of order $n$ has $n$ distinct 1-dimensional representations, and so for that group there are $n$ distinct ways to act on $\mathbb{C}$. But usually, unless explicitly mentioned, the action of $G$ on $\mathbb{C}$ is taken to be the trivial one. –  M Turgeon Apr 24 '12 at 16:50
1  
Ok, that makes a lot of sense, I didn't think it seemed quite right for any general action! So I guess whenever we're working with G-invariants we will assume the action on $\mathbb{C}$ is trivial. Thanks again for your help. –  Ben Apr 24 '12 at 17:04
2  
This is strange terminology. I am used to the word "equivariant" rather than concomitant. Also, what you call a $G$-invariant vector in $W$ is really a $G$-invariant element of the dual space $W^*$. –  Justin Campbell Apr 24 '12 at 17:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.