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 $0$?



Fredholm elements of a Lie algebra
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.


