$\DeclareMathOperator\diag{diag}\DeclareMathOperator\Gr{Gr}\DeclareMathOperator\SL{SL}$I'm having some trouble computing affine Springer fibers, even in simple cases. For example, consider the group $G=\SL\_2$ over $\mathbb{C}$ and let $\mathcal{K}=\mathbb{C}((z))$ and $\mathcal{O}=\mathbb{C}[[z]]$. Then the affine Grassmannian $\Gr\_G$ of $G$ parametrizes $\mathcal{O}$ lattices in $\mathcal{K}^2$. We then define the space $S=\{(\gamma, L) \in \mathfrak{g}\_{\mathcal{O}} \times \Gr\_G \mathrel\vert \gamma L \subseteq L\}$ together with the obvious projection $\pi: S \to \mathfrak{g}\_{\mathcal{O}}$. The *affine Springer fiber* of $\gamma \in \mathfrak{g}\_{\mathcal{O}}$ is then the fiber of $\pi$ over $\gamma$. Here $\mathfrak{g}\_{\mathcal{O}}$ is the Lie algebra $\mathfrak{g} \otimes \mathcal{O}$.


For example, the reduced Springer fiber of $\gamma= \diag(x,-x)$, for $x \in \mathbb{C}-\{0\}$ is an infinite discrete space parametrized by $\mathbb{Z}$. Letting $x=z$, we get an infinite chain of $\mathbb{P}^1$'s instead, where each one intersects the next in exactly one point.


In principle, to compute these things one takes an arbitrary matrix $g$ in $\SL\_2(\mathcal{K})$, conjugates $\gamma$ by $g$, and then works out conditions on the entries for this matrix so that the conjugate matrix $g^{-1} \gamma g$ is an element of $\mathfrak{g}\_{\mathcal{O}}$, but this method is extremely tedious and difficult. Is there an elegant way to do this? If there isn't, could someone show me at least an efficient way to compute $\pi^{-1}(\gamma)$, say when $\gamma=\diag(z,-z)$?


Lastly, assuming we have found that $\pi^{-1}(\gamma)$ is parametrized by the points of $\mathbb{P}^1$, say, how does one rigorously show that the scheme structure on the fiber agrees with that of $\mathbb{P}^1$? Thanks in advance.



Computing affine Springer fibers
I came across this question while studying affine Springer fibers myself, and I hope this answer can help future learners.


Let us fix a Borel subgroup $B\subset G$ and a maximal torus $T\subset G$. Let $X\_\*(T):=\hom(\mathbb{C}^\times,T)$ denote the cocharacter lattice of $T$. Let $N:=[B,B]$ denote the maximal unipotent subgroup of $B$. The Iwasawa decomposition says $G(\mathcal{K})=\bigsqcup\_{\lambda\in X\_\*(T)}N(\mathcal{K})t^\lambda G(\mathcal{O})$. In the case $G=\operatorname{SL}\_2$, we can take $B$ to be the group of upper triangular matrices and $T$ to be the group of diagonal matrices. Then the Iwasawa decomposition in this case says that any element in $G(\mathcal{K})/G(\mathcal{O})$ can be represented as a left $G(\mathcal{O})$-coset of $\begin{pmatrix} 1 & f(t) \\ 0 & 1\end{pmatrix}\begin{pmatrix} t^n & 0 \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}$ for some $f(t)\in \mathcal{K}$ and $n\in \mathbb{Z}$. Since any term in $f(t)$ with degree $\geq 2n$ can be moved through $\begin{pmatrix} t^n & 0 \\ 0 & t^{-n}\end{pmatrix}$ and become an $N(\mathcal{O})$-element, we may assume without loss of generality that $f$ is a Laurent polynomial with degree less than $2n$.


Recall that the affine Springer fiber is defined to be $\mathfrak{X}\_\gamma:=\{x\in G(\mathcal{K})\mid x^{-1}\gamma x\in \mathfrak{g}(\mathcal{O})\}/G(\mathcal{O})$. Therefore we need those representatives $x=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}$ such that
$$
\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}^{-1}\begin{pmatrix} t & 0 \\ 0 & -t\end{pmatrix}\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix}t & 2t^{1-2n}f(t) \\ 0 & -t \end{pmatrix} \in \mathfrak{g}(\mathcal{O}).
$$
As a result, this implies that $f(t)\in t^{2n-1}\mathcal{O}$. Combining this with $\deg f<2n$ we can conclude that $f(t)=at^{2n-1}$ for some $a\in \mathbb{C}$. In other words, the affine Springer fiber contains a copy of $\mathbb{A}^1$ for each integer $n$.


To understand how these copies of $\mathbb{A}^1$'s are glued together, we need a little bit more geometry. If we think of points in the affine Grassmannian $G(\mathcal{K})/G(\mathcal{O})$ in terms of lattices, the column vectors of $x=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix} t^n & at^{n-1}\\0 & t^{-n}\end{pmatrix}$ form a basis of the corresponding lattice $\Lambda$. We claim that the closure of the $\mathbb{A}^1$ copy associated with the integer $n$ coincides with the set
$$
C\_n:=\{\text{lattices } \Lambda\mid t^{n}\mathcal{O}\oplus t^{-n+1}\mathcal{O}\subset \Lambda\subset t^{n-1}\mathcal{O}\oplus t^{-n}\mathcal{O}\}.
$$
It is not hard to verify that the $\mathbb{A}^1$ copy associated with $n$ is contained in $C\_n$. On the other hand, note that by doing elementary column operations with $\mathcal{O}$-coefficients (i.e., up to a right multiple of $G(\mathcal{O})$), $x=\begin{pmatrix} t^n & at^{n-1}\\0 & t^{-n}\end{pmatrix}$ can be turned into a lower triangular matrix $\begin{pmatrix} t^{n-1} & 0 \\ a^{-1}t^{-n} & t^{-n+1}\end{pmatrix}$. Thus, as $a\rightarrow \infty$, the lattice $\Lambda$ converges to the lattice $\Lambda\_{n-1}:=t^{n-1}\mathcal{O}\oplus t^{-n+1}\mathcal{O}$, which is also contained in the set $C\_n$. Thus, we can conclude that the closure of the $\mathbb{A}^1$ copy associated with $n$ is a copy of $\mathbb{P}^1$ and is contained in the set $C\_n$. Moreover, by using Iwasawa decomposition again one can show that there is nothing else other than this $\mathbb{P}^1$ inside $C\_n$. This shows that the affine Springer fiber $\mathfrak{X}\_\gamma$ is the union $\bigcup\_{n\in \mathbb{Z}}C\_n$ with $C\_n\cong \mathbb{P}^1$ being its irreducible components and
$$
C\_n\cap C\_m=\begin{cases}\Lambda\_{\min\{n,m\}} & \text{if $|n-m|=1$},\\
C\_n & \text{if $n=m$}, \\
\emptyset & \text{otherwise}.\end{cases}
$$
In other words, the affine Springer fiber $\mathfrak{X}\_\gamma$ is "an infinite chain of $\mathbb{P}^1$'s".


