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I was reading S. Mukai's- An Introduction to Invariants and Moduli, and I came across Proposition 6.16 on page 195 (see the screenshot below)

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It says that "every invariant rational function can be expressed as ratio of semiinvariants of weight $\chi$."

Should it be $n\chi$, for some $n\in\mathbb{N}$?

The reason I'm confused is because- say I consider the weight space determined by $\chi$, i.e., all the semiinvariants of weight $\chi$. It has the structure of a $\mathbb{K}$-vector space. Say I choose two of the basis element $p,q\in\mathfrak{B}$. Now consider the invariant rational function $\frac{p^2}{q^2}$, which is ratio of semiinvariants of weight $2\chi$.

I don't understand how can I write $\frac{p^2}{q^2}$ as ratio of semiinvariants of weight $\chi$?

Also, does the Definition 6.15 about "stable with respect to $\chi$" coincide with the King's stability condition (see here) in the context of quiver representations?

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    $\begingroup$ I think you are right. There might not even exist semiinvariants of weight $\chi$. $\endgroup$
    – Friedrich Knop
    Commented Dec 1, 2023 at 9:07
  • $\begingroup$ @FriedrichKnop But say there exists semiinvariants of weight $\chi$, then is the statement of the proposition true? Or should it still be $n\chi$? And, in this case is the $n$ fixed or does $n$ depend on the invariant rational function that I choose? $\endgroup$
    – It'sMe
    Commented Dec 1, 2023 at 19:02
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    $\begingroup$ It should still be $n\chi$ because there might not be enough semiinvariants of weight $\chi$. Possibly the simplest example is $X=\mathbf A^2$ and $G=\mathbf G_m$ acting by $t\cdot(x,y)=(tx,t^2y)$. Take $\chi(t)=t$. Then all points are $\chi$-stable except for the origin and $f=x^2/y$ is a counterexample. So there is a glitch in Mukai's book. The last sentence of the Proposition is not affected, though. $\endgroup$
    – Friedrich Knop
    Commented Dec 4, 2023 at 14:11

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