Pencils, ramification, and geometric invariant theory

Proof. Fix $p=[1:0]$. Row reduction gives a basis of $U$ with vanishing orders $(0,e_p)$ at $p$. For the one-parameter subgroup $$\lambda(t)(x,y)=(t^{-1}x,ty),$$ the weight of $x^{4-i}y^i$ is $-4+2i$. Hence the least nonzero weight of a Plücker vector representing $U$ is $-8+2e_p$. Every nontrivial one-parameter subgroup is conjugate to a positive multiple of one of these. In this convention Hilbert–Mumford semistability means that each least weight is nonpositive, and stability means that it is negative. Since $1\le e_p\le4$, this proves the first two assertions.

If $e_p=4$, the pencil contains $y^4$, and a second generator has nonzero $x^4$ coefficient. The limit under $\lambda$ is $\langle x^4,y^4\rangle$. The orbit of this pencil consists of the spans of the fourth powers of two independent linear forms. Its boundary in the Grassmannian consists of tangent pencils, equivalent to $\langle x^4,x^3y\rangle$. This also follows by extending the secant map from unordered distinct pairs of points on the rational normal quartic to its diagonal. Such a tangent pencil is unstable: at $[0:1]$ its two vanishing orders are $3$ and $4$, giving least weight $-8+2(3+4)>0$. Therefore the orbit of $U_0$ is closed in the semistable locus, and every nonstable polystable pencil is equivalent to it.

Finally, two points of index four contribute all six to $\deg R_U$. Their images are distinct, and coordinates on source and target then identify the map with $z\mapsto z^4$. Thus their pencil is $U_0$. ◻

Proof. Send the unique point of index four and its image to infinity. The resulting map is a polynomial of degree four. Multiplication by a constant, translation on the source, and translation on the target put it in the form $z^4+az^2+bz$, giving the displayed pencil. If this polynomial had another point of index four, its derivative would equal $4(z-r)^3$. Comparing the coefficient of $z^2$ gives $r=0$, and then $a=b=0$, which is excluded. Any equivalence between two normal forms must preserve the unique totally ramified point and its image. The source and target changes are therefore affine. The missing cubic term forces the source translation to vanish; comparison of leading and constant terms then leaves exactly the indicated weighted scaling. ◻

Proof. Put $L=-(K_{\mathbb P^1}+\frac14R_U)$; it has degree $1/2$. All coefficients of $\frac14R_U$ are at most $3/4$, so the pair is klt. Every divisorial valuation of the function field of a smooth curve is proportional to $\operatorname{ord}_p$ for a point $p$. Writing $n_p=e_p-1$, its log discrepancy is $1-n_p/4$, while $$S_L(\operatorname{ord}_p) =\frac{1}{\deg L}\int_0^{\deg L}(\deg L-t)\,dt =\frac14.$$ Taking the infimum of the ratio of these two quantities proves the formula. The valuative criterion also shows that all these log curves are K-semistable. The usual criterion for weighted points on $\mathbb P^1$ says that equality gives a K-polystable pair only when $R_U=3p+3q$, with $p\ne q$; this is precisely the pencil $U_0$. This is the log-curve formulation in [3, Section 7.6]. ◻

Proof. On $\mathbb P^1\times\mathbb P^1\times T$, write $A,B$ for the classes of the first and second factors. The universal centre has class $D=A+4B+h$. Its normal bundle in $\mathbb P^1\times\mathbb P^2\times T$ has Chern classes $$c_1=A+8B+h,\qquad c_2=4B(A+4B+h).$$ The ambient anticanonical divisor restricts to $L=2A+6B$ on $\mathbb P^1\times\mathbb P^1\times T$. The pushforward formulas for a codimension-two blow-up give $$\varphi_*((-K_{\mathcal X/T})^4) =\operatorname{pr}_{T*} \bigl(D(-6L^2+4Lc_1-c_1^2+c_2)\bigr).$$ Indeed, the pushforwards of the square, cube, and fourth power of the exceptional divisor are $-[\mathcal C]$, $-i_*c_1$, and $-i_*(c_1^2-c_2)$, respectively. Since $A^2=B^2=0$, only the coefficient of $ABh$ contributes. Its values in $DL^2$, $DLc_1$, and $D(c_1^2-c_2)$ are $24$, $36$, and $32$. The required coefficient is therefore $-6\cdot24+4\cdot36-32=-32$, proving the assertion. The Plücker coordinates are the $2\times2$ minors of the two coefficient vectors, hence have degree two. Its hyperplane class thus pulls back to $2h$, giving the last assertion. ◻