Quiver Harder–Narasimhan filtrations exist, are unique, and descend

Let $M\mathrm{\ne }0$ have dimension vector $\mathbf{d}$. The nonzero dimension vectors of its subrepresentations form a nonempty subset of the finite set $\{\mathbf{e}:0\le e_i\le d_i\}$. Choose the largest attained phase and, among subobjects attaining it, the largest attained total dimension. Such a phase and dimension are maxima of finite sets of attained values, so they are attained without an infinite choice. Let $S\subseteq M$ attain both maxima. Every nonzero subobject of $S$ is a subobject of $M$, so its phase is at most ${\varphi }_Z(S)$; hence $S$ is semistable.

If $A,B$ are semistable and ${\varphi }_Z(A)>{\varphi }_Z(B)$, then $\mathrm{Hom}(A,B)=0$. Indeed, a nonzero map has a nonzero image $I$. From $0\to \ker f\to A\to I\to 0$, semistability of $A$ and [F3] force ${\varphi }_Z(I)\ge {\varphi }_Z(A)$: if the kernel is zero the phases are equal; if the kernel is nonzero and ${\varphi }_Z(I)<{\varphi }_Z(A)$, the phase of the sum lies strictly below ${\varphi }_Z(A)$, contradicting $Z(A)=Z(\ker f)+Z(I)$. But $I\subseteq B$ and semistability of $B$ give ${\varphi }_Z(I)\le {\varphi }_Z(B)$, a contradiction.

If $T\subseteq M\mathrm{/}S$ is nonzero, let $P\subseteq M$ be its inverse image. Then $0\to S\to P\to T\to 0$ is exact. If ${\varphi }_Z(T)>{\varphi }_Z(S)$, [F3] gives ${\varphi }_Z(P)>{\varphi }_Z(S)$, contrary to maximality of the phase of $S$. If ${\varphi }_Z(T)={\varphi }_Z(S)$, additivity and [F3] give ${\varphi }_Z(P)={\varphi }_Z(S)$, while $\mathrm{\mid }\mathbf{d}(P)\mathrm{\mid }>\mathrm{\mid }\mathbf{d}(S)\mathrm{\mid }$, contrary to the second maximum. Thus every nonzero subobject of $M\mathrm{/}S$ has phase strictly less than ${\varphi }_Z(S)$.

Consider two filtrations with first subobjects $S,S^{\mathrm{\prime }}$ of phases $p,p^{\mathrm{\prime }}$. If $p>p^{\mathrm{\prime }}$, every factor in the quotient filtration of $M\mathrm{/}{S}^{\mathrm{\prime }}$ has phase below $p^{\mathrm{\prime }}$. Step 1.2 therefore makes every map from $S$ to each such factor zero. Induction along the finite quotient filtration shows $\mathrm{Hom}(S,M\mathrm{/}{S}^{\mathrm{\prime }})=0$: a map whose projection to the last factor is zero factors through the preceding subobject. Thus the inclusion of $S$ in $M$ factors through $S^{\mathrm{\prime }}$, giving a nonzero map $S\to S^{\mathrm{\prime }}$, contrary to $\mathrm{Hom}(S,S^{\mathrm{\prime }})=0$ from step 1.2. Interchanging $S,S^{\mathrm{\prime }}$ similarly rules out $p^{\mathrm{\prime }}>p$; hence $p=p^{\mathrm{\prime }}$. Applying step 1.2 to all factors of both quotients now gives $S\subseteq S^{\mathrm{\prime }}$ and $S^{\mathrm{\prime }}\subseteq S$, hence $S=S^{\mathrm{\prime }}$. Induction on the quotient proves equality of the full filtrations.

Induct on $\mathrm{\mid }\mathbf{d}(M)\mathrm{\mid }$. The zero representation has the empty filtration. For $M\mathrm{\ne }0$, the quotient $M\mathrm{/}S$ has smaller total dimension, so by induction it has a filtration with semistable factors in strictly decreasing phase order. Step 2.1 places every one of those phases below ${\varphi }_Z(S)$. Pulling that filtration back along $M\to M\mathrm{/}S$ and placing $S$ first gives the required filtration of $M$.

Let $M$ be defined over $k$, set $M_{\bar{k}}=M{\otimes }_k\bar{k}$, and let $\sigma (x)=x^q$ be Frobenius. Applying $\sigma$ to a filtration preserves the arrow maps of $M_{\bar{k}}$, every dimension vector, and every phase. It therefore gives another HN filtration; uniqueness proved above fixes each term. At each vertex, a Frobenius-invariant subspace has a unique reduced row-echelon basis matrix. Applying $\sigma$ gives another such matrix for the same subspace, so uniqueness makes the matrix fixed entrywise. Its entries lie in ${\mathbb{F}}_q$, and the arrow maps of $M$ preserve these descended subspaces. Thus the filtration descends to $k$.

If an ${\mathbb{F}}_q$-object is geometrically unstable, it has a geometric subrepresentation of phase greater than its total phase. Step 1.1 chooses a subobject of maximal phase, so the first HN term has phase greater than the total phase. Step 4.1 descends that term and its phase inequality to ${\mathbb{F}}_q$, contradicting ${\mathbb{F}}_q$-semistability. Conversely, an ${\mathbb{F}}_q$-subrepresentation extends to $\bar{k}$ with the same dimension vector and phase. This proves the stated equivalence and the result.