Proposition 9.11.23511E6797B7CD

Proof. On the torus use $x,y,z$ as characters, so $w=yz/x$. The character lattice is $\mathbb Z^3$, with semigroup generators

The dual cone has primitive rays $(0,1,0)$, $(0,0,1)$, $(1,1,0)$ and $(1,0,1)$. It is full-dimensional with four rays. The residue form $\Omega =\mathrm dx\wedge \mathrm dy\wedge \mathrm dz/x$ generates the canonical sheaf of this hypersurface; it equals $yz$ times the logarithmic torus form. Hence $w_0=(0,1,1)$, which pairs to one with each primitive ray.

Write $d(m)=m_1+m_2+m_3$. Each semigroup generator has degree one. The logarithmic function $g$ differs by a bounded term from $\lambda $ times the maximum of the four linear forms given by those generators. Their convex hull is the section of $\sigma ^\vee $ by $d=1$, so

Since $w_0\in \operatorname {int}\sigma ^\vee $ and $d(w_0)=2$, the character condition is $d(m)+2>\lambda $. The smallest allowed integer degree is $k=\max \{ 0,\lfloor \lambda \rfloor -1\} $.

This also gives the ideal for whole holomorphic germs. Lift a germ to a convergent power series in $x,y,z,w$. On a smaller invariant neighborhood, angular Parseval and Tonelli force every nonzero character coefficient to satisfy the same test. All homogeneous terms of degree less than $k$ therefore vanish in the coordinate ring. They are multiples of the homogeneous relation $xw-yz$, and subtracting those finitely many terms leaves a lift of order at least $k$. Thus the germ belongs to $\mathfrak m_ o^ k$. Conversely, every degree-$k$ monomial is integrable by the character test, and bounded holomorphic coefficient germs preserve integrability. This proves the identity and its strict endpoint. $\square$