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.

Let $k$ be a field and $\mathfrak{g}=kx\oplus ky$ with $[x,y]=y$. Show that $\rho(x)=t\,\frac{d}{dt}$ and $\rho(y)=t\cdot$ (mult. by $t$) define a representation $\rho:\mathfrak{g}\to \mathfrak{gl}(k[t])$. Show that the ideals $(t^n)$ are the only non-zero subrepresentations.

The first part is easy: just check that $\rho$ is a Lie algebra map. I just want to check that I'm thinking about the second part correctly. Checking that each $(t^n)$ is stable under $\rho$ is also easy. Now given a general $\rho$-stable subspace $W\subseteq k[t]$ I want to show that $W=(t^n)$ for some $n$. Now $k[t]$ has basis $\{1,t,t^2,\ldots\}$, so $W$ has a basis coming from some subset of this basis; in particular, $W$ contains $t^n$ for some $n$. So $(t^n)\subseteq W$.

Is this the correct way to think about this? If so, how do I get $W\subseteq (t^n)$?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

No, you can't assume that the basis for $W$ is a subset of the basis for $k[t]$. For example $k^2$ is a vector space with basis $(0, 1)$ and $(1, 0)$ but we can take any vector, for example $(1, 1)$, and get a $1$ dimensional subspace. We are not just limited to choosing from the basis.

To prove that $W = (t^n)$ first we must show that $t^n \in W$ for some $n$. Choose $f = t^n + a_{n + 1}t^{n + 1} + \cdots + a_mt^m$ in $W$ with the property that $n$ is minimal among all possible choices (why can I assume it's monic?). Now act on $f$ to try and produce $t^n$. For example, start by acting by the element $\frac{1}{n + 1}x$ and then subtracting the result from $f$.

After you've figured that out we have $t^n \in W$ hence $(t^n) \subseteq W$. Now you need to argue that $W \subseteq t^n$. Do that by arguing the contrapositive. Assume $g \notin (t^n)$ and argue that also $g \notin W$. For this step you will use the fact that when we chose $f$ we chose it so that $n$ was minimal.

share|improve this answer
    
Thank you for your help –  Bey Feb 12 '13 at 3:08
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.