2.5. Algebraic geometric perspective [040D]

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2.5. Algebraic geometric perspective

We now take a closer examination of the complex geometry on $\mathbb{C}^{3}$ . We first raise two conceptual puzzles, and then we propose two conceptual explanations which suggest different directions of future investigations.

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A priori speaking $M\simeq\mathbb{C}^{3}$ is only equipped with a complex structure, but the assignment of holomorphic coordinates $z_{0},z_{1},z_{2}$ canonically induces an algebraic structure. What is the origin of this algebraicity?
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It is well known that $(\mathbb{C}^{3},\Omega)$ viewed as a complex manifold or an algebraic variety has a huge automorphism group preserving the holomorphic volume form. But our construction of coordinate functions are canonical up to multiplying by constants. What is the conceptual explanation?

The first explanation is that $\mathbb{C}^{3}$ has a toric structure. This comes from the holomorphic isometric action of $T^{3}$ , acting diagonally on $z_{0},z_{1},z_{2}$ . This induces a $(\mathbb{C}^{*})^{3}$ -action with an open dense orbit in $\mathbb{C}^{3}$ , making $\mathbb{C}^{3}$ a toric manifold and in particular algebraic. The canonical coordinates come from the eigenfunctions of this algebraic torus action, and $z_{0},z_{1},z_{2}$ up to constant scale factors are special because they have minimal vanishing orders on the toric boundary.

This explanation is simpler, but there are two possible criticisms. First, the $T^{3}$ has a preferred subgroup $T^{2}$ whose action has very different nature from the additional $U(1)$ -action, so it seems unnatural to put them on the same conceptual footing. Second, the additional $U(1)$ -symmetry is accidental to this particular example, which may not survive for other examples generalising our construction. A conjectural example without this $U(1)$ -symmetry is described in subsection 2.11.2.

The second and deeper explanation is based on the principle that algebraic structures arise from the ring of holomorphic functions with controlled growth (cf. [5]).

Lemma 2.12.

Any algebraic function $f$ on $\mathbb{C}^{3}$ satisfies the growth estimate

(2.18)

|f|\leq K_{1}e^{K_{2}(|\mu_{1}|+|\mu_{2}|)}(|\eta|+1)^{K_{3}}.

for some constants $K_{1},K_{2},K_{3}$ depending on $f$ .

Proof.

It suffices to prove the growth estimate for $z_{1},z_{2},z_{0}$ . By elementary calculation $|\frac{\partial\alpha_{1}}{\partial\eta}|\leq\frac{Ca_{22}|\eta|}{(\mu_{1}^{2}+a_{22}|\eta|^{2})^{3/2}}$ , so upon integration

\int_{\mu_{1}}^{\infty}|\frac{\partial\alpha_{1}}{\partial\eta}|d\mu_{1}\leq\frac{C}{|\eta|}(\frac{\mu_{1}}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}-1)\leq\frac{C}{|\eta|},

hence $|\beta_{i}|\leq\frac{C}{|\eta|}$ by the integral definition of $\beta_{i}$ . From

d\log|z_{1}|=d(a_{11}\mu_{1}+a_{12}\mu_{2})+\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})+\text{Re}(\beta_{1}d\eta),

we integrate to obtain the growth bound on $z_{1}$ for $A^{1/4}|\vec{\mu}|_{a}\geq 1$ . But $|z_{1}|$ is a continuous function, so the bound holds also near the origin. Similarly we can bound $|z_{0}|$ and $|z_{2}|$ . ∎

Proposition 2.13.

The ring of algebraic functions on $\mathbb{C}^{3}$ coincides with the holomorphic functions satisfying the growth estimate (2.18) for some $K_{1},K_{2},K_{3}$ .

Proof.

We need to prove the converse to Lemma 2.12. The $T^{2}$ -symmetry acts on functions via

(e^{i\theta_{1}},e^{i\theta_{2}})\cdot f=f(e^{i(\theta_{1}+\theta_{2})}z_{0},e^{-i\theta_{1}}z_{1},e^{-i\theta_{2}}z_{2}).

This action allows us to expand any holomorphic function $f$ as a Fourier series on every $T^{2}$ -fibre:

f=\sum_{n,m\in\mathbb{Z}^{2}}f_{n,m},

where $f_{n,m}$ has weight $(n,m)$ with respect to the $T^{2}$ action. Since the $T^{2}$ action is holomorphic, the Fourier components $f_{n,m}$ are also holomorphic. Furthermore, these $f_{n,m}$ satisfy the same growth condition as $f$ after perhaps increasing $K_{1}$ .

We claim every $f_{n,m}$ is algebraic. To see this, we can find a suitable monomial of $z_{0},z_{1},z_{2}$ which has the same weight as $f_{n,m}$ , such that $f_{n,m}$ divided by this monomial has no pole along $\mathfrak{D}_{i}$ . But this quotient function is $T^{2}$ -invariant and holomorphic, so depends only on $\eta$ , and in fact has to be a polynomial of $\eta$ by the growth condition.

By applying the Parseval identify to every $T^{2}$ -fibre, we obtain

\frac{1}{4\pi^{2}}\int_{T^{2}}|f|^{2}\vartheta_{1}\wedge\vartheta_{2}=\sum_{n,m}|f_{n,m}|^{2}.

Both LHS and RHS are functions of $\mu_{1},\mu_{2},\eta$ , and LHS has a bound of type (2.18) by assumption. But for any given $K_{2},K_{3}$ , only finitely many monomials of $z_{0},z_{1},z_{2}$ satisfy the growth bound (2.18) globally, so only finitely many $f_{n,m}$ can appear as summands. Hence $f$ is algebraic as required. ∎

The proof in fact gives a double-index increasing filtration structure on the ring of algebraic functions:

\mathcal{F}_{K_{2},K_{3}}=\{f:\text{There exists $K_{1}$ such that }|f|\leq K_{1}e^{K_{2}(|\mu_{1}|+|\mu_{2}|)}(|\eta|+1)^{K_{3}}\},

such that every filtered piece is finite. This is the deeper mechanism why the complex automorphism group is cut down to finite size.

The insight from this discussion is that on our $\mathbb{C}^{3}$ the algebraic structure has a transcendental origin. The growth of holomorphic functions naturally involve transcendental functions such as $\exp$ and $\log$ . The ultimate reason is that torus fibrations are inherently transcendental in nature; this exponential growth behaviour already happened on the flat $\mathbb{C}^{*}$ .

At this moment we still have the freedom to normalise

(z_{0},z_{1},z_{2})\mapsto(\lambda_{0}z_{0},\lambda_{1}z_{1},\lambda_{2}z_{2}),

where $\lambda_{i}$ are constants satisfying $\lambda_{0}\lambda_{1}\lambda_{2}=1$ . Fixing a normalisation is important for keeping track of how estimates depend on the scaling parameter $A$ . We now make a choice so that the region $\{A^{1/4}|\vec{\mu}|_{a}\lesssim 1\}$ resemble a complex ball. Pick a point such that $|\mu_{1}|,|\mu_{2}|,|\mu_{1}-\mu_{2}|,A^{1/4}|\eta|$ are all comparable to $A^{-1/2}$ , so $A^{1/4}\text{dist}_{g_{a}}(\cdot,\Delta)\sim A^{1/4}|\vec{\mu}|_{a}\sim 1$ , and we demand $|z_{0}|=|z_{1}|=|z_{2}|=|\eta|^{1/3}$ at this point. This convention is compatible with both the $A$ -scaling and the functional equation $z_{0}z_{1}z_{2}=\eta$ . We did not mention the phase of $z_{i}$ because $T^{2}$ -gauge symmetry renders different phase choices equivalent. Under this convention, on the annulus region $1\lesssim A^{1/4}|\vec{\mu}|_{a}\lesssim C_{1}$ , the holomorphic functions $A^{1/4}z_{0},A^{1/4}z_{1},A^{1/4}z_{2}$ are bounded independent of scaling factor, and the metric $\omega^{(1)}$ is $C^{\infty}$ -equivalent to $\sum_{i}\sqrt{-1}dz_{i}\wedge d\bar{z}_{i}$ .