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Background:

Let $A$ be a finite-dimensional (associative and unital) algebra over $\mathbb{C}$. Assume there is a quiver $Q=(Q_0,Q_1)$, where $Q_0$ is the set of vertices and $Q_1$ is the set of arrows, associated to $A$, i.e., $A\cong\mathbb{C}Q/I$, with $0 \neq I\subset \mathbb{C}Q$ an admissible ideal. Further assume that the algebra $A$ is such that the quiver $Q$ is acyclic.

For a dimension vector $\beta$, the representation space $$\DeclareMathOperator\rep{rep}\DeclareMathOperator\GL{GL} \rep_{\beta}(Q,I):=\biggl\lbrace M\in\prod_{a\in Q_1}\operatorname{Mat}_{\beta(ha)\times\beta(ta)}(\mathbb{C}) \;\bigg|\; \text{$M(r) = 0$ for all $r \in I$}\biggr\rbrace $$ parametrizes $\beta$-dimensional representations of $(Q,I)$. The linear algebraic group $$ \GL_{\beta}:=\prod_{x\in Q_0}\GL_{\beta(x)}(\mathbb{C}) $$ acts on $\rep_{\beta}(Q,I)$ by change of basis.

My setting:

Now, I take an irreducible component $\mathcal{C}\subseteq\rep_{\beta}(Q,I)$ and restrict the action of $\GL_{\beta}$ to $\mathcal{C}$. Suppose I have a character $\chi$ of $\GL_{\beta}$ defined by the element $\theta\in\mathbb{Z}^{Q_0}$. Assume that there are two $\theta$-semi-stable representations $M,N\in\mathcal{C}$ (following the definition of A. D. King see Moduli of representations of finite dimensional algebras) and $\theta$-polystable representation $S\in\mathcal{C}$ such that we have one-parameter subgroups $\lambda_1,\lambda_2:\mathbb{G}_m\rightarrow \GL_{\beta}$ with $$\lim_{t\rightarrow0}\lambda_1(t)\cdot M=S\text{ and }\lim_{t\rightarrow0}\lambda_2(t)\cdot N=g\cdot S,\text{ for some }g\in\GL_{\beta}.$$ Also, assume that there exists a $G$-invariant rational function $h\in\mathbb{C}(\mathcal{C})$ such that $h\rvert_{\mathcal{O}(M)}=0$ and $h\rvert_{\mathcal{O}(N)}=1$.

My question:

Is $h$ defined on the orbit of $S$? If not, is there any result which says when the invariant rational function will be defined on the limit, or is there any condition on $S$ or $\theta$ that guarantees that $h$ will be defined on the orbit of $S$?

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  • $\begingroup$ Could you give a reference for "the definition of A. D. King"? \\ TeX note: \vert (or, even better, \rvert) for restriction works better than \mid; compare $h\mid_{\mathcal O(M)} = 0$ h\mid_{\mathcal O(M)} = 0 to $h\rvert_{\mathcal O(M)} = 0$ h\rvert_{\mathcal O(M)} = 0. (I think that \mid is essentially just \mathbin\vert.) I have edited accordingly. $\endgroup$
    – LSpice
    Jan 7, 2023 at 0:40
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    $\begingroup$ @LSpice Thanks for the suggestion. I have added the link. $\endgroup$
    – It'sMe
    Jan 7, 2023 at 1:15
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    $\begingroup$ Cross posted at math stack exchange: math.stackexchange.com/q/4615913/884739 $\endgroup$
    – It'sMe
    Jan 10, 2023 at 23:29

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