A NULLSTELLENSATZ FOR AMOEBAS

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416 KEVIN PURBHOO Remark 2.1 Note that we have actually determined which term is the special term m k .Thisis done in (2.2). If f n is a polynomial, then n(d − k) is the number of roots (counted with multiplicity) of f n inside the disc {|z| ≤β 0 }.Iff n is a Laurent polynomial, n(d − k) − mindeg(f n ) is the number of roots inside {0 < |z| ≤β 0 }. LEMMA 2.2 As before, let A ={z | β 0 < |z|
A NULLSTELLENSATZ FOR AMOEBAS 417 γ → 1. The best general answer for the question is of the same form; that is, n log γ −1 ≤ (D 0 + D 1 ) log n + log (c 0 c 1 ). One can see this by performing the requisite analysis on the polynomials h n (z) = (z n + 1) c 0n D 0 . As a general heuristic, the more closely the roots of f n are packed, the larger n has to be; thus the family of polynomials h n (z), where every root has as high a multiplicity as possible, is where we expect our worst-case behaviour to occur. Suppose we want n large enough to guarantee that h n {log |z 0 |} is (c 1 n D 1 )-superlopsided for |z 0 | < γ < 1. Write h n (z) = 1 + c 0 n D 0 zn + ···. We know that 1 is the dominant term as n gets large (since 1 = lim n→∞ h n (z 0 )); thus we need (c 1 n D 1 )(c 0n D 0 zn 0 ) < 1 or, equivalently, n log γ −1 ≥ (D 0 + D 1 ) log n + log(c 0 c 1 ). If we only want to guarantee that f n {a} is lopsided for a ∈ K, we need n large enough so that ∑ |m l (z 0 )|≤(1 + γ n ) d − 1 < 2 − (1 + γ n ) d ≤|m k (z 0 )| l≠k or, equivalently, (1 + γ n ) c 0n D 0 ≤ 3/2. This holds if we have ( n log γ −1 c ) 0 ≥ D 0 log n + log . log 3/2 So n needs to be only about half as big to guarantee that f n {a} is lopsided as it does to guarantee that f n {a} is superlopsided. Again, we can see that this is fairly close to the best answer by considering (z n + 1) c 0n D 0 . 3. The hypersurface case 3.1. Preliminaries In this section, we prove our main theorem characterising the amoeba of a hypersurface. If f (z) ∈ C[z 1 ,z −1 1 ,...,z r,z −1 ], we consider the Laurent polynomials ˜f n (z) = ∏n−1 k 1 =0 r ··· ∏n−1 k r =0 f (e 2πi k 1/n z 1 ,...,e 2πi k r /n z r ). Theorem 1 states that for ε>0 and for a point a ∈ R r in the complement of the amoeba A f whose distance from A f is at least ε, we can choose n large enough so that ˜f n {a} is superlopsided. Moreover, the theorem gives an upper bound on how large n needs to be, based only on ε and the Newton polytope of f .
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Page 41 and 42: 448 DAVID GINZBURG The method we us
Page 43 and 44: 450 DAVID GINZBURG Let ν denote a
Page 45 and 46: 452 DAVID GINZBURG Proof The proof
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Page 55 and 56: 462 DAVID GINZBURG where L τ (¯s)
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468 DAVID GINZBURG immediately foll
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470 DAVID GINZBURG expansion. Here
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472 DAVID GINZBURG Notice that S l
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474 DAVID GINZBURG values of m ′
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476 DAVID GINZBURG For 1 ≤ i ≤
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478 DAVID GINZBURG 2m rows, we have
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480 DAVID GINZBURG left to right. C
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482 DAVID GINZBURG unipotent orbit
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484 DAVID GINZBURG for every choice
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486 DAVID GINZBURG is equal to L(σ
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488 DAVID GINZBURG Here f π (h) is
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490 DAVID GINZBURG character ψ Um
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492 DAVID GINZBURG Proof The proof
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494 DAVID GINZBURG in the above ref
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496 DAVID GINZBURG THEOREM 7 The ir
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498 DAVID GINZBURG In the above, th
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500 DAVID GINZBURG is GL 2m+1 × SO
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502 DAVID GINZBURG [GP] S. GELBART
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A CHARACTERIZATION OF SUBSPACES AND
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DEGREE GROWTH OF MEROMORPHIC SURFAC
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DISTORTION OF HAUSDORFF MEASURES AN
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574 MARCHÉ and NARIMANNEJAD them w
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576 MARCHÉ and NARIMANNEJAD 1.1. P
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578 MARCHÉ and NARIMANNEJAD 2.1. T
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580 MARCHÉ and NARIMANNEJAD Figure
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582 MARCHÉ and NARIMANNEJAD Figure
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584 MARCHÉ and NARIMANNEJAD Defini
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586 MARCHÉ and NARIMANNEJAD [A2] [
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A NULLSTELLENSATZ FOR AMOEBAS

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