Linearity and conjugation invariance of the finite potent trace

(Setup; reduction for (T4)) Let $F\subseteq {\mathrm{End}}_k(V)$ be finite potent with exponent $n$, and let $F_0\subseteq F$ be an arbitrary finite-dimensional $k$-subspace, with a finite basis ${\phi }_1,\dots ,{\phi }_m$; since a map out of $F$ is $k$-linear as soon as it is additive and $k$-homogeneous on every such $F_0$, it suffices to prove that ${\mathrm{Tr}}_V{\mathrm{\mid }}_{F_0}$ is linear for this arbitrary $F_0$.

(A common finite-dimensional space) If $n=0$, the empty-product condition says $V$ is finite-dimensional; set $W:=V$, which is stable under each ${\phi }_j$. If $n\ge 1$, set $W:={\sum }_{i_1,\dots ,i_n}{\phi }_{i_1}\cdots {\phi }_{i_n}(V)$. This is a finite sum of finite-dimensional spaces by finite potency of $F$, so $W$ is finite-dimensional, and it is stable under each ${\phi }_j$ because ${\phi }_j{\phi }_{i_1}\cdots {\phi }_{i_n}(V)\subseteq {\phi }_j{\phi }_{i_1}\cdots {\phi }_{i_{n-1}}(V)\subseteq W$.

(Traces are computed on $W$) For $\theta ={\sum }_j{\lambda }_j{\phi }_j\in F_0$ one has ${\theta }^n(V)\subseteq W$ and $W$ is $\theta$-stable, while $W$ is finite-dimensional and $\theta$ is finite potent, so [F5] gives ${\mathrm{Tr}}_V(\theta )={\mathrm{tr}}_W(\theta {\mathrm{\mid }}_W)$; in particular ${\mathrm{Tr}}_V({\phi }_j)={\mathrm{tr}}_W({\phi }_j{\mathrm{\mid }}_W)$ for every $j$.

((T4)) By $k$-linearity of the ordinary trace in the endomorphism, ${\mathrm{Tr}}_V(\theta )={\mathrm{tr}}_W(\theta {\mathrm{\mid }}_W)={\sum }_j{\lambda }_j{\mathrm{tr}}_W({\phi }_j{\mathrm{\mid }}_W)={\sum }_j{\lambda }_j{\mathrm{Tr}}_V({\phi }_j)$, so ${\mathrm{Tr}}_V$ is linear on the arbitrary finite-dimensional subspace $F_0\subseteq F$, and (T4) follows for $F$.

(Rectangular trace identity) If $W,W^{\mathrm{\prime }}$ are finite-dimensional and $A\text{\hspace{0.17em}⁣}:W^{\mathrm{\prime }}\to W$, $B\text{\hspace{0.17em}⁣}:W\to W^{\mathrm{\prime }}$ are $k$-linear, then ${\mathrm{tr}}_{W^{\mathrm{\prime }}}(B\circ A)={\mathrm{tr}}_W(A\circ B)$: choose ordered bases and let $(a_{ij})$, $(b_{ij})$ be the matrices of $A$ and $B$; the diagonal entries of the two products are the finite sums ${\sum }_j{b}_{ij}{a}_{ji}$ and ${\sum }_i{a}_{ij}{b}_{ji}$, which are rearrangements of one another in the commutative ring $k$.

((T5), stabilisation) Put $\alpha :=\psi \phi \text{\hspace{0.17em}⁣}:V^{\mathrm{\prime }}\to V^{\mathrm{\prime }}$ and $\beta :=\phi \psi \text{\hspace{0.17em}⁣}:V\to V$, and fix $n_0$ with ${\alpha }^{n_0}(V^{\mathrm{\prime }})$ finite-dimensional; the descending chains ${\alpha }^k(V^{\mathrm{\prime }})$ for $k\ge n_0$ and, using ${\beta }^k(V)\subseteq \phi ({\alpha }^{k-1}(V^{\mathrm{\prime }}))$, the chains ${\beta }^k(V)$ for $k\ge n_0+1$ lie inside the finite-dimensional spaces ${\alpha }^{n_0}(V^{\mathrm{\prime }})$ and $\phi ({\alpha }^{n_0}(V^{\mathrm{\prime }}))$, so both stabilise: choose $n$ large enough that $n\ge n_0+1$ and $W^{\mathrm{\prime }}:={\alpha }^n(V^{\mathrm{\prime }})={\alpha }^{n+1}(V^{\mathrm{\prime }})$ and $W:={\beta }^n(V)={\beta }^{n+1}(V)$, both finite-dimensional; then $\beta$ is finite potent.

(The induced maps) The map $\phi$ sends $W^{\mathrm{\prime }}$ into $W$, since $\phi (W^{\mathrm{\prime }})=\phi ({\alpha }^n(V^{\mathrm{\prime }}))={\beta }^n(\phi (V^{\mathrm{\prime }}))\subseteq {\beta }^n(V)=W$. Conversely, stabilization gives $W={\beta }^n(V)={\beta }^{n+1}(V)=\phi {\alpha }^n\psi (V)\subseteq \phi (W^{\mathrm{\prime }})$, so $\phi (W^{\mathrm{\prime }})=W$. The map $\psi$ sends $W$ into $W^{\mathrm{\prime }}$, since $\psi (W)=\psi ({\beta }^n(V))={\alpha }^n(\psi (V))\subseteq {\alpha }^n(V^{\mathrm{\prime }})=W^{\mathrm{\prime }}$. Thus the restrictions have the stated domains and codomains, and their composites are $\alpha {\mathrm{\mid }}_{W^{\mathrm{\prime }}}=\psi {\mathrm{\mid }}_W\circ \phi {\mathrm{\mid }}_{W^{\mathrm{\prime }}}$ and $\beta {\mathrm{\mid }}_W=\phi {\mathrm{\mid }}_{W^{\mathrm{\prime }}}\circ \psi {\mathrm{\mid }}_W$.

((T5)) Applying [F5] to $\alpha$ on $V^{\mathrm{\prime }}$ and to $\beta$ on $V$, and the rectangular identity to $\phi {\mathrm{\mid }}_{W^{\mathrm{\prime }}}\text{\hspace{0.17em}⁣}:W^{\mathrm{\prime }}\to W$ and $\psi {\mathrm{\mid }}_W\text{\hspace{0.17em}⁣}:W\to W^{\mathrm{\prime }}$, gives ${\mathrm{Tr}}_{V^{\mathrm{\prime }}}(\psi \phi )={\mathrm{tr}}_{W^{\mathrm{\prime }}}(\alpha {\mathrm{\mid }}_{W^{\mathrm{\prime }}})={\mathrm{tr}}_{W^{\mathrm{\prime }}}(\psi {\mathrm{\mid }}_W\circ \phi {\mathrm{\mid }}_{W^{\mathrm{\prime }}})={\mathrm{tr}}_W(\phi {\mathrm{\mid }}_{W^{\mathrm{\prime }}}\circ \psi {\mathrm{\mid }}_W)={\mathrm{tr}}_W(\beta {\mathrm{\mid }}_W)={\mathrm{Tr}}_V(\phi \psi )$, which is (T5); note that $\psi \phi$ is finite potent by hypothesis and $\phi \psi$ is finite potent by step 6.1.

((T6)(a), factorisation) Assume $\theta \text{\hspace{0.17em}⁣}:V\to V$ has finite-dimensional image $W:=\theta (V)$, let $i\text{\hspace{0.17em}⁣}:W\hookrightarrow V$ be the inclusion and let $\bar{\theta }\text{\hspace{0.17em}⁣}:V\to W$ be $\theta$ with restricted codomain, so that $\theta =i\bar{\theta }$ and $\bar{\theta }i=\theta {\mathrm{\mid }}_W$; for any $k$-linear $\sigma \text{\hspace{0.17em}⁣}:V\to V$ the maps $\bar{\theta }\sigma \text{\hspace{0.17em}⁣}:V\to W$ and $\sigma i\text{\hspace{0.17em}⁣}:W\to V$ are $k$-linear, with $\theta \sigma =i(\bar{\theta }\sigma )$ and $\sigma \theta =(\sigma i)\bar{\theta }$.

((T6)(a), traces) Applying (T5) to the pairs $(i,\bar{\theta })$, $(i,\bar{\theta }\sigma )$ and $(\sigma i,\bar{\theta })$ is legitimate because in each case the composite in the finite-dimensional space $W$ is finite potent, and yields ${\mathrm{Tr}}_V(\theta )={\mathrm{Tr}}_W(\theta {\mathrm{\mid }}_W)$, ${\mathrm{Tr}}_V(\theta \sigma )={\mathrm{Tr}}_W(\bar{\theta }\sigma i)$ and ${\mathrm{Tr}}_V(\sigma \theta )={\mathrm{Tr}}_W(\bar{\theta }\sigma i)$.

((T6)(a), conclusion) Both $\theta \sigma$ and $\sigma \theta$ have finite-dimensional image, respectively contained in $W$ and $\sigma (W)$. The finite-image endomorphisms form a $k$-subspace: sums have image in the sum of the two finite-dimensional images, and scalar multiples still have finite-dimensional image. This subspace is finite potent with exponent $1$, so (T4) gives ${\mathrm{Tr}}_V([\theta ,\sigma ])={\mathrm{Tr}}_V(\theta \sigma )-{\mathrm{Tr}}_V(\sigma \theta )$. Step 10.1 computes both terms as ${\mathrm{Tr}}_W(\bar{\theta }\sigma i)$, so their difference is zero.

((T6)(b), first case) Assume $\gamma \in E_0$ and $\psi \in E$. Choose a finite-dimensional subspace $U\subseteq V$ with $\psi A\subseteq A+U$, using $\psi A<A$. Then $(\gamma \psi )(V)\subseteq \gamma (V)<A$, and $(\gamma \psi )(A)\subseteq \gamma (A)+\gamma (U)$ is finite-dimensional because $\gamma A$ is finite-dimensional and $\gamma (U)$ is the image of a finite-dimensional space. Thus $\gamma \psi \in E_0$. Separately choose a finite-dimensional subspace $W\subseteq V$ with $\gamma V\subseteq A+W$, using $\gamma V<A$. Then $(\psi \gamma )(V)\subseteq \psi (A)+\psi (W)\subseteq A+U+\psi (W)$, so $\psi \gamma V<A$; also $(\psi \gamma )(A)\subseteq \psi (\gamma A)$ is finite-dimensional because $\gamma A$ is finite-dimensional. Thus $\psi \gamma \in E_0$. The witnesses $U$ and $W$ serve different containments and need not be equal.

(The common finite-potent subspace $E_0$) For any ${\rho }_1,{\rho }_2\in E_0$, choose finite-dimensional $W$ with ${\rho }_{2}V\subseteq A+W$. Then ${\rho }_1{\rho }_{2}V\subseteq {\rho }_{1}A+{\rho }_{1}W$, which is finite-dimensional because ${\rho }_{1}A$ is finite-dimensional and ${\rho }_{1}W$ is the image of a finite-dimensional space. Thus every product of two elements of the subspace $E_0$ has finite-dimensional image, so $E_0$ is finite potent with exponent $2$. In particular every element of $E_0$, including $\gamma \psi$ and $\psi \gamma$ from step 12.1, is finite potent.

((T6)(b), first case concluded) For $\gamma \in E_0$ and $\psi \in E$, step 12.1 puts $\gamma \psi$, $\psi \gamma$ and their difference $[\gamma ,\psi ]$ in the common finite-potent subspace $E_0$. By (T4) on $E_0$, ${\mathrm{Tr}}_V([\gamma ,\psi ])={\mathrm{Tr}}_V(\gamma \psi )-{\mathrm{Tr}}_V(\psi \gamma )$. Apply the domain-correct (T5) with $\phi =\gamma$ and $\psi =\psi$; its hypothesis $\psi \phi =\psi \gamma$ is finite potent by step 13.1, and it gives ${\mathrm{Tr}}_V(\gamma \psi )={\mathrm{Tr}}_V(\psi \gamma )$. Hence the commutator trace is zero.

((T6)(b), second case) Assume $\gamma \in E_1$ and $\psi \in E_2$. Then $\gamma \psi \in E_0$ since $(\gamma \psi )(A)=\gamma (\psi A)$ is finite-dimensional and $(\gamma \psi )(V)\subseteq \gamma V<A$. Also $\psi \gamma \in E_0$: choose finite-dimensional $W$ with $\gamma V\subseteq A+W$; then $(\psi \gamma )(V)\subseteq \psi (A)+\psi (W)$ is finite-dimensional because $\psi A$ is finite-dimensional, and $(\psi \gamma )(A)\subseteq (\psi \gamma )(V)$ is finite-dimensional. The difference $[\gamma ,\psi ]$ lies in $E_0$ as the difference of two elements of that subspace. By (T4) on the common finite-potent subspace $E_0$ from step 13.1, ${\mathrm{Tr}}_V([\gamma ,\psi ])={\mathrm{Tr}}_V(\gamma \psi )-{\mathrm{Tr}}_V(\psi \gamma )$. The domain-correct (T5), with $\phi =\gamma$ and $\psi =\psi$, applies because $\psi \gamma$ is finite potent and gives equality of these two traces. Therefore ${\mathrm{Tr}}_V([\gamma ,\psi ])=0$.

Combining the steps: (T4) is step 4.1, (T5) is step 8.1, (T6)(a) is step 11.1, and both cases of (T6)(b) are steps 14.1 and 14.2; the only use of the Axiom of Choice is the selection of bases through [F2], in the rectangular identity of step 5.1.