A NULLSTELLENSATZ FOR AMOEBAS

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426 KEVIN PURBHOO this more general situation; however, the answer depends on these polynomials. It is certainly still true that SA ˜f n1 converges uniformly to A ,...,nr f , as this argument really only depends on the fact that degrees of ˜f n1 ,...,n r are growing only polynomially, while the terms are becoming suitably sparse. 4. Approximating a hypersurface amoeba by linear inequalities 4.1. Locating the dominant term Theorem 1 tells us that for n sufficiently large, one term of ˜f n dominates, but it does not specify which one. The answer depends on in which component of the amoeba complement our point a lies. Since ˜f n {a} varies continuously with a, it depends only on the component of the amoeba complement. Now, the number of components is relatively small compared to the number of terms of ˜f n . There is a natural injective map ind : components of R r \A f ↩→ (f ) ∩ Z r (cf. [FPT, Definition 2.1]). This is called the index of the component; a complete definition is given below. So only a few of the terms of ˜f n can possibly be dominant terms. Fortunately, it is relatively simple to determine which these are. The Newton polytope of ˜f n is n r (f ), and candidates for dominant term are, in fact, what one expects them to be; namely, they are the images of the integral points of (f ) under this scaling. PROPOSITION 4.1 Let a ∈ R r \A f , and let ind(a) = −→ k = (k 1 ,...,k r ) be the corresponding point in (f ). If ˜f n {a} is lopsided, then the term of ˜f n (z) which dominates has exponent vector n r−→ k (i.e., it is the (z nr k 1 1 ···z nr k r )-term). r In order to make complete sense of the statement, we need to know a definition of the index −→ k . There are a number of equivalent definitions, but the simplest for our purposes is the following. Let ζ ∈ Log −1 (a). For each i ∈{1,...,r}, consider the polynomial, f i,ζ (z) = f (ζ 1 ,...,ζ i−1 ,z,ζ i+1 ,...,ζ r ). If f is a polynomial, then k i is the number of roots (with multiplicity) of f i,ζ inside the open disc {|z|
A NULLSTELLENSATZ FOR AMOEBAS 427 ζ, this number is independent of ζ.Iff is a Laurent polynomial, then k i = #roots of f i,ζ (z) inside { 0 < |z|
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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
Page 47 and 48: 454 DAVID GINZBURG correspond to th
Page 49 and 50: 456 DAVID GINZBURG Next, consider t
Page 51 and 52: 458 DAVID GINZBURG To state the fol
Page 53 and 54: 460 DAVID GINZBURG check that up to
Page 55 and 56: 462 DAVID GINZBURG where L τ (¯s)
Page 57 and 58: 464 DAVID GINZBURG To define the li
Page 59 and 60: 466 DAVID GINZBURG cuspidality of t
Page 61 and 62: 468 DAVID GINZBURG immediately foll
Page 63 and 64: 470 DAVID GINZBURG expansion. Here
Page 65 and 66: 472 DAVID GINZBURG Notice that S l
Page 67 and 68: 474 DAVID GINZBURG values of m ′
Page 69 and 70: 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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