Commensurable subspaces and the ideals E_0, E_1, E_2 of E

Assume the Axiom of Choice as inherited from the linear algebra suppliers (The Axiom of Choice). Let $k$ be a field, let $V$ be a $k$-vector space (Vector space over a field), and let $A,B$ be $k$-subspaces of $V$.

Commensurability. Write $A<B$, and say that $A$ is not much bigger than $B$, when the quotient $(A+B)\mathrm{/}B$ (The quotient vector space $V\mathrm{/}W$ and its canonical projection) is finite-dimensional (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis). Write $A\sim B$, and say that $A$ and $B$ are commensurable, when $A<B$ and $B<A$. Thus $A<B$ says that $A$ is contained in $B$ up to a finite-dimensional error: equivalently, and this reformulation is used throughout, $A<B$ holds if and only if $A\subseteq B+W$ for some finite-dimensional subspace $W\subseteq V$.

The relation $<$ has the following elementary properties, each an immediate consequence of the definition; here $A$, $B$, $C$ are $k$-subspaces of $V$ and $r\ge 0$.

1. Reflexivity and monotonicity. $A<A$ and $A<A+B$; if $A\subseteq B$ then $A<B$; if $A<B$ and $B\subseteq C$ then $A<C$; and $A<B$ if and only if $A<A\cap B$, if and only if $A+B<B$.
2. Transitivity. If $A<B$ and $B<C$, then $A<C$.
3. $k$-linear maps. If $A<B$ and $\theta \text{\hspace{0.17em}⁣}:V\to V^{\mathrm{\prime }}$ is $k$-linear, then $\theta (A)<\theta (B)$ in $V^{\mathrm{\prime }}$ (Linear map between vector spaces over the same field).
4. Finite sums. If $A_i<B_i$ for $i=1,\dots ,r$, then ${\sum }_i{A}_i<{\sum }_i{B}_i$.
5. Commensurability is an equivalence relation. The relation $\sim$ is reflexive, symmetric and transitive, and $A\sim B$ if and only if both $(A+B)\mathrm{/}B$ and $(A+B)\mathrm{/}A$ are finite-dimensional.

For instance, $A<A$ because $(A+A)\mathrm{/}A$ is the zero space; if $B\subseteq C$, then $(A+C)\mathrm{/}C\cong A\mathrm{/}(A\cap C)$ is a quotient of $(A+B)\mathrm{/}B\cong A\mathrm{/}(A\cap B)$. For transitivity, if $A<B$ and $B<C$, choose finite-dimensional subspaces $U,W\subseteq V$ with $A\subseteq B+U$ and $B\subseteq C+W$. Then $A\subseteq B+U\subseteq C+W+U$, so $A<C$. A $k$-linear $\theta$ induces a surjection $(A+B)\mathrm{/}B\twoheadrightarrow \theta (A+B)\mathrm{/}\theta (B)$, which proves 3. The reformulation $A<B\iff A\subseteq B+W$ for a finite-dimensional $W$ is obtained by lifting a finite spanning set of $(A+B)\mathrm{/}B$ to $A$: every class has a representative in $A$, since each $A+B$ element differs from an element of $A$ by an element of $B$; the span of representatives of a finite basis is such a $W$. Conversely, if $A\subseteq B+W$ for finite-dimensional $W$, then $(A+B)\mathrm{/}B$ is a quotient of the image of $W$ and is finite-dimensional.

Now let $K$ be a commutative $k$-algebra acting on $V$ (equivalently, $V$ is a $K$-module whose structure map is $k$-linear) and assume $fA<A\text{for every }f\in K\mathrm{.}$ Attached to the pair $(V,A)$ is the ideal filtration $E:=\{\theta \in {\mathrm{End}}_k(V):\theta A<A\},E_1:=\{\theta \in {\mathrm{End}}_k(V):\theta V<A\},E_2:=\{\theta \in {\mathrm{End}}_k(V):\theta A<0\},E_0:=E_1\cap E_2,$ where $\theta A<0$ is shorthand for the statement that $\theta A$ is finite-dimensional, the zero space $0$ being finite-dimensional. Thus $E_0=\{\theta \in {\mathrm{End}}_k(V):\theta V<A\text{ and }\theta A\text{ is finite-dimensional}\}\mathrm{.}$ Properties 1 and 4 show that $E$, $E_1$ and $E_2$ are closed under addition and under multiplication by scalars, hence are $k$-subspaces of ${\mathrm{End}}_k(V)$. The image of $K$ is contained in $E$ by the hypothesis $fA<A$ for every $f\in K$; it need not be contained in $E_1$ or $E_2$. Also $E_0=E_1\cap E_2$ by definition. An element of $E_0$ is finite potent in the sense of The trace of a finite potent endomorphism exists and is unique: if $\theta V<A$ with $\theta V\subseteq A+W$ for a finite-dimensional $W$, then ${\theta }^{2}V\subseteq \theta A+\theta W$, a sum of two finite-dimensional spaces, so ${\theta }^{2}V$ is finite-dimensional and the trace ${\mathrm{Tr}}_V(\theta )$ is defined.

The subspaces $E$, $E_1$, $E_2$ depend only on the commensurability class of $A$: if $A^{\mathrm{\prime }}\sim A$ is a further $k$-subspace of $V$ with $f{A}^{\mathrm{\prime }}<A^{\mathrm{\prime }}$ for all $f\in K$, then $\theta A^{\mathrm{\prime }}\sim \theta A$ for every $\theta \in {\mathrm{End}}_k(V)$ by properties 2 and 3. If $\theta \in E(A)$, then $\theta A^{\mathrm{\prime }}<\theta A<A<A^{\mathrm{\prime }}$, hence $\theta \in E(A^{\mathrm{\prime }})$; conversely, if $\theta \in E(A^{\mathrm{\prime }})$, then $\theta A<\theta A^{\mathrm{\prime }}<A^{\mathrm{\prime }}<A$, hence $\theta \in E(A)$. Also, $\theta V<A$ iff $\theta V<A^{\mathrm{\prime }}$ by transitivity and $A\sim A^{\mathrm{\prime }}$, so $E_1(A)=E_1(A^{\mathrm{\prime }})$. Finally, if $\theta A$ is finite-dimensional, the relation $\theta A^{\mathrm{\prime }}<\theta A$ places $\theta A^{\mathrm{\prime }}$ inside $\theta A$ plus a finite-dimensional subspace, so $\theta A^{\mathrm{\prime }}$ is finite-dimensional; the reverse implication follows symmetrically from $\theta A<\theta A^{\mathrm{\prime }}$. Thus $E_2(A)=E_2(A^{\mathrm{\prime }})$, and consequently $E_0(A)=E_0(A^{\mathrm{\prime }})$. These are the subspaces in which Tate's abstract residue theory states its residue map on $KdK$; they are abstract linear-algebraic objects and involve no geometry.