TITLE: Fredholm elements of a Lie algebra
QUESTION [5 upvotes]: An element $a$ of a Lie algebra $L$ is called a Fredholm element if the adjoint operator $\mathrm{ad}_a:L \to L$ is a Fredholm linear map. That is: its kernel is a finite-dimensional space and its range $\mathrm{ad}_a(L)$ is a finite-codimensional subspace.
Is there an infinite-dimensional Lie algebra with at least one Fredholm element? Is there a Lie algebra $L$ whose only non-Fredholm element is $0$?

REPLY [4 votes]: The infinite-dimensional Lie algebra with basis $(e_n)_{n\in\mathbf{N}}$ and brackets $[e_i,e_j]=(i-j)e_{i+j}$, over a field of characteristic zero, satisfies the required condition: every nonzero element is Fredholm. 
First let me mention it's immediate that $\mathrm{ad}_{e_i}$ is Fredholm for every $i$. Denote by $V_k$ the subspace generated by $(e_i)_{i\le k}$.
Let $f$ be a nonzero element, say $f=\sum_{i=0}^ka_ie_i$ with $a_k\neq 0$ and $T=\mathrm{ad}_f$ (let's call $k$ the degree of $f$). Then the kernel of $T$ is contained in $V_k$ (indeed, if $g\notin V_k$, say of degree $\ell$, then $[f,g]$ has degree exactly $k+\ell$). Next computing $[f,e_i]$ for each $i\ge k+1$, we see by induction that $V_i/V_k$ is contained in the image of $p_k\circ T$, where $p_k$ is the projection $L\to L/V_k$. Hence $p_k\circ T$ is surjective. Hence $\mathrm{Im}(T)+V_k=L$, thus $\mathrm{Im}(T)$ is finite-codimensional.