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The power sums of the slice zeros vary holomorphically

Statement

Under the neighbourhood and radius supplied by Nearby slices of a regular germ have the same zero count, define for each integer $k \geq 0$

$p_{k} (z^{'}) : = \frac{1}{2 π i} \int_{∣ ζ ∣ = r} ζ^{k} \frac{\partial f / \partial z_{m} (z^{'}, ζ)}{f (z^{'}, ζ)} d ζ .$

Then $p_{k}$ is holomorphic in $z^{'}$ . If $λ_{1} (z^{'}), \dots, λ_{d} (z^{'})$ are the zeros of $ζ \mapsto f (z^{'}, ζ)$ in $∣ ζ ∣ < r$ , counted with multiplicity, then

$p_{k} (z^{'}) = λ_{1} (z^{'})^{k} + \dots + λ_{d} (z^{'})^{k} .$

Facts & Assumptions

Given: A representative of $f$ on a neighbourhood of the closed cylinder $V \times {∣ ζ ∣ \leq r}$ and the radius $r$ and neighbourhood $V$ from Nearby slices of a regular germ have the same zero count.

[L1]

Every slice $ζ \mapsto f (z^{'}, ζ)$ has no zero on $∣ ζ ∣ = r$ and has exactly $d$ interior zeros counted with multiplicity (Nearby slices of a regular germ have the same zero count).

[L2]

The derivative $\partial f / \partial z_{m}$ is holomorphic, holomorphic functions are separately holomorphic and continuous, and quotients by nonvanishing holomorphic functions stay holomorphic (Holomorphic functions of several variables are smooth and their complex derivatives are holomorphic, A holomorphic function of several variables is continuous and separately holomorphic, Sums, products and nonvanishing quotients of holomorphic functions are holomorphic).

[L3]

A contour integral with continuous integrand that is holomorphic in one complex parameter is holomorphic in that parameter (A contour integral of a jointly continuous, parameter-holomorphic integrand is holomorphic).

[L4]

A locally bounded separately holomorphic function is holomorphic (Locally bounded and separately holomorphic implies holomorphic).

[L5]

The weighted argument principle gives $\frac{1}{2 π i} \int_{Γ} g (ζ) \frac{h^{'} (ζ)}{h (ζ)} d ζ = \sum {ord}_{a} (h) g (a)$ for a holomorphic test function $g$ and a zero-free boundary (The weighted argument principle).

Proof

technique · direct

1.1

By [L1], the denominator $f (z^{'}, ζ)$ is nonzero on $V \times {∣ ζ ∣ = r}$ . Hence [L2] makes $Φ_{k} (z^{'}, ζ) : = ζ^{k} \frac{\partial f / \partial z_{m} (z^{'}, ζ)}{f (z^{'}, ζ)}$ continuous on that compact cylinder. Fixing all coordinates of $z^{'}$ except one, [L2] makes $Φ_{k}$ holomorphic in the remaining coordinate and [L3] makes the corresponding slice of $p_{k}$ holomorphic. The same compact continuity gives a uniform bound on $Φ_{k}$ , so the ML estimate makes $p_{k}$ locally bounded on $V$ . Therefore [L4] makes $p_{k}$ holomorphic on $V$ .

L1L2L3L4

2.1

Fix $z^{'} \in V$ and apply [L5] to the one-variable holomorphic function $h (ζ) : = f (z^{'}, ζ)$ on the disc $∣ ζ ∣ < r$ with test function $g (ζ) = ζ^{k}$ . By [L1], the boundary circle is zero-free and the only singularities of $h^{'} / h$ inside are the zeros $λ_{j} (z^{'})$ , counted with their multiplicities. Thus $p_{k} (z^{'}) = \sum_{j = 1}^{d} λ_{j} (z^{'})^{k} .$

givenL1L5∎

Depends on

Used by

Finite Newton recurrences for the slice zeros Lemma
Weierstrass preparation theorem Theorem

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jiří Lebl, Tasty Bits of Several Complex Variables, Section 6.3 (standard reference, not scraped)
Jaap Korevaar and Jan Wiegerinck, Several Complex Variables, Section 4.4 (standard reference, not scraped)