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Let $L$ be a Lie algebra with $L \subseteq \mathfrak{gl}(V) $ ($V$ finite dimensional over $\mathbb{C}$) and let $I$ be an abelian ideal of $L$. Given $x \in I, \lambda \in \mathbb{C} $, I am trying to show that the generalised eigenspace $\ker((x-\lambda \text{Id}_V)^{\dim V })$ is stable meaning for all $y \in L$ and all $v \in V$ we have $y(v)$ $\in \ker((x-\lambda \text{Id}_V)^{\dim V }) $.

I need to show that it $v \in \ker ((x-\lambda \text{Id}_V)^{\dim v} )$, $y \in L$ then $(x-\lambda \text{Id}_V)^{\dim V } (yv)=0 $ but I just can’t make any headway after this.

Any pointers?

Anonmath101
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  • You can prove by induction on $n$ (for $n\ge 1$) that if $x\in I, y\in L$ then $[(x-\lambda I)^n, y] = n [x,y] (x-\lambda I)^{n-1}$ with $[x, y]\in I$. Does that help? – Chad K Apr 19 '23 at 17:13
  • I can see how that would help. I actually already know that for $x \in H, y \in L$ that we have $x^n y =n[x, y]x^{n-1} +yx^n $ but I tried replacing $x$ by $x-\lambda \text{Id}_V $ but I realised that the problem with this is that $x-\lambda \text{Id}_V $ is not in $I$. How can I use what I did already to get what you did? – Anonmath101 Apr 19 '23 at 17:30
  • The associative multiplication and bracket are in $\mathfrak{gl}(V)$, not just any Lie algebra. $I$ commutes with everything, $[x,y]$ commutes with $x$ because both are in $I$ and $I$ is abelian. In a Lie bracket of associative algebra you have an identity $[uv,w] = [u,w]v + u[v,w]$. So the induction proof follows through. – Chad K Apr 19 '23 at 17:35
  • I’m sorry I’m really confused what you’re trying to say. Can you clarify this a bit with some details? – Anonmath101 Apr 19 '23 at 17:45

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This is just a proof of the formula in the comments. I'll use $\mathbf{1}$ for the identity in $\mathfrak{gl}(V)$ so as not to use $I$ which is the symbol for the ideal.

$$[x-\lambda \mathbf{1}, y] = [x, y] - \lambda [\mathbf{1}, y] = [x,y]$$ because $[\mathbf{1}, y] = 0$.

Assume by induction that $$[(x-\lambda \mathbf{1})^n, y] = n[x,y](x-\lambda \mathbf{1})^{n-1}$$ then $$\begin{align} [(x-\lambda \mathbf{1})^{n+1}, y] &= [(x-\lambda \mathbf{1})(x-\lambda\mathbf{1})^{n}, y]\\ &=[x-\lambda \mathbf{1}, y](x-\lambda \mathbf{1})^{n} + (x-\lambda \mathbf{1})[(x-\lambda\mathbf{1})^{n}, y]\\ &=[x,y](x-\lambda\mathbf{1})^{n}+(x-\lambda\mathbf{1})n[x,y](x-\lambda \mathbf{1})^{n-1}\\ &=[x,y]((x-\lambda\mathbf{1})^{n} + n (x-\lambda\mathbf{1})^{n})\\ &=(n+1)[x,y](x-\lambda\mathbf{1})^{n} \end{align}$$ where use was made that $[x,y], x\in I$ so they commute because $I$ is abelian (and so $[x,y], (x-\lambda\mathbf{1})$ also commute), and also the identity $[uv,w] = [u,w]v + u[v,w]$ in an associative algebra.

Chad K
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  • How do you deduce this identity from the one I have already shown: that if $x \in I $ and $y \in L$ then $x^n y =n[x, y]x^{n-1}+yx^n $? As I said this holds for $x \in I $ but $x-\lambda \text{Id}_V \in I $ (as it’s not even in $\mathfrak{sl}(V)$. – Anonmath101 Apr 19 '23 at 21:51
  • @Anonmath101: I don't see a way to deduce it. It has to be proven from scratch using a similar induction. Maybe if you expand the exponent using the binomial formula you can do it, but it just looks more complicated. I haven't tried. – Chad K Apr 20 '23 at 07:23
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    @Anonmath101: You can extend the identity you wrote by linearity to show that for any polynomial $f$, $[f(x), y] = [x,y]f'(x)$ where $f'(x)$ is the formal derivative. For $(x-\lambda)^n$ the formal derivative is $n(x-\lambda)^{n-1}$ by the chain rule for formal derivative. – Chad K Apr 20 '23 at 13:26