Monodromy and spectrum of quasi-ordinary surface singularities. Mirel Caibar, Gary Kennedy, Lee McEwan
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1 Monodromy and spectrum of quasi-ordinary surface singularities 2009
2 An irreducible germ of algebraic surface is quasi-ordinary if one can find a finite map to a nonsingular surface germ, whose discriminant locus is a normal crossings divisor. Thus we can find local coordinates so that the surface is the zero locus of f (x, y, z) = z d + g 1 (x, y)z d g d (x, y), and so that the discriminant locus of projection is the pair of coordinate lines in the x-y plane.
3 An irreducible germ of algebraic surface is quasi-ordinary if one can find a finite map to a nonsingular surface germ, whose discriminant locus is a normal crossings divisor. Thus we can find local coordinates so that the surface is the zero locus of f (x, y, z) = z d + g 1 (x, y)z d g d (x, y), and so that the discriminant locus of projection is the pair of coordinate lines in the x-y plane.
4 z = x 1/2 y + xy 3/2
5 We want to calculate certain invariants of a quasi-ordinary surface, and to understand the relationships among them. The invariants include: (1) The monodromy of the Milnor fibration, as recorded in the graded characteristic function det(ti m 0 ) det(ti m 2 ) det(ti m 1 ) (where m i is monodromy on H i of Milnor fiber). (2) The spectrum of the same fibration.
6 We want to calculate certain invariants of a quasi-ordinary surface, and to understand the relationships among them. The invariants include: (1) The monodromy of the Milnor fibration, as recorded in the graded characteristic function det(ti m 0 ) det(ti m 2 ) det(ti m 1 ) (where m i is monodromy on H i of Milnor fiber). (2) The spectrum of the same fibration.
7 We want to calculate certain invariants of a quasi-ordinary surface, and to understand the relationships among them. The invariants include: (1) The monodromy of the Milnor fibration, as recorded in the graded characteristic function det(ti m 0 ) det(ti m 2 ) det(ti m 1 ) (where m i is monodromy on H i of Milnor fiber). (2) The spectrum of the same fibration.
8 Consider a transverse slice of the surface by x = C, where C is sufficiently small. This is a plane curve germ, with its own Milnor fibration, which we call the horizontal fibration. Continuing our list of invariants: (3) The monodromy of the horizontal fibration: H(t) = det(ti h 0) det(ti h 1 ). (4) The spectrum of the same fibration. (5,6) Same things for transverse slice y = C.
9 Consider a transverse slice of the surface by x = C, where C is sufficiently small. This is a plane curve germ, with its own Milnor fibration, which we call the horizontal fibration. Continuing our list of invariants: (3) The monodromy of the horizontal fibration: H(t) = det(ti h 0) det(ti h 1 ). (4) The spectrum of the same fibration. (5,6) Same things for transverse slice y = C.
10 Consider a transverse slice of the surface by x = C, where C is sufficiently small. This is a plane curve germ, with its own Milnor fibration, which we call the horizontal fibration. Continuing our list of invariants: (3) The monodromy of the horizontal fibration: H(t) = det(ti h 0) det(ti h 1 ). (4) The spectrum of the same fibration. (5,6) Same things for transverse slice y = C.
11 Consider a transverse slice of the surface by x = C, where C is sufficiently small. This is a plane curve germ, with its own Milnor fibration, which we call the horizontal fibration. Continuing our list of invariants: (3) The monodromy of the horizontal fibration: H(t) = det(ti h 0) det(ti h 1 ). (4) The spectrum of the same fibration. (5,6) Same things for transverse slice y = C.
12 For the horizontal fibration we fix the value of x while varying the parameter in f (x, y, z) = ɛ. Alternatively we can fix ɛ and let x move around a small circle, obtaining the vertical fibration. We consider: (7) The monodromy of the vertical fibration: V(t) = det(ti v 0) det(ti v 1 ). (8) Is there a natural way to define a vertical spectrum? (9,10) Same things for transverse slice y = C.
13 For the horizontal fibration we fix the value of x while varying the parameter in f (x, y, z) = ɛ. Alternatively we can fix ɛ and let x move around a small circle, obtaining the vertical fibration. We consider: (7) The monodromy of the vertical fibration: V(t) = det(ti v 0) det(ti v 1 ). (8) Is there a natural way to define a vertical spectrum? (9,10) Same things for transverse slice y = C.
14 For the horizontal fibration we fix the value of x while varying the parameter in f (x, y, z) = ɛ. Alternatively we can fix ɛ and let x move around a small circle, obtaining the vertical fibration. We consider: (7) The monodromy of the vertical fibration: V(t) = det(ti v 0) det(ti v 1 ). (8) Is there a natural way to define a vertical spectrum? (9,10) Same things for transverse slice y = C.
15 For the horizontal fibration we fix the value of x while varying the parameter in f (x, y, z) = ɛ. Alternatively we can fix ɛ and let x move around a small circle, obtaining the vertical fibration. We consider: (7) The monodromy of the vertical fibration: V(t) = det(ti v 0) det(ti v 1 ). (8) Is there a natural way to define a vertical spectrum? (9,10) Same things for transverse slice y = C.
16 Since the singularity is not isolated, it is natural to compare the surface to a surface in its Yomdin series: f + L k = 0, where L is a general linear form and k is large. We consider: (11,12) The monodromy and spectrum of a surface in the Yomdin series.
17 Since the singularity is not isolated, it is natural to compare the surface to a surface in its Yomdin series: f + L k = 0, where L is a general linear form and k is large. We consider: (11,12) The monodromy and spectrum of a surface in the Yomdin series.
18 Since the singularity is not isolated, it is natural to compare the surface to a surface in its Yomdin series: f + L k = 0, where L is a general linear form and k is large. We consider: (11,12) The monodromy and spectrum of a surface in the Yomdin series.
19 Calculations Without changing the local topology, we may find local coordinates so that each branch is parametrized by ζ = e x λ i y µ i i=1 with λ i λ i 1, µ i µ i 1, and (λ i, µ i ) not contained in the group generated by the previous pairs. (Abhyankar-Lipman) Each (λ i, µ i ) is called a characteristic pair. How do the characteristic pairs determine the invariants?
20 Calculations Without changing the local topology, we may find local coordinates so that each branch is parametrized by ζ = e x λ i y µ i i=1 with λ i λ i 1, µ i µ i 1, and (λ i, µ i ) not contained in the group generated by the previous pairs. (Abhyankar-Lipman) Each (λ i, µ i ) is called a characteristic pair. How do the characteristic pairs determine the invariants?
21 An example: a recursion for vertical monodromy. Given ζ = e x λ i y µ i i=1... its truncation is ζ 1 = x λ 1 y µ 1 = x a mb y n m, (where a and b are relatively prime). Let r and s be smallest nonnegative integers so that ( ) m n r s has determinant 1.
22 An example: a recursion for vertical monodromy. Given ζ = e x λ i y µ i i=1... its truncation is ζ 1 = x λ 1 y µ 1 = x a mb y n m, (where a and b are relatively prime). Let r and s be smallest nonnegative integers so that ( ) m n r s has determinant 1.
23 An example: a recursion for vertical monodromy. Given ζ = e x λ i y µ i i=1... its truncation is ζ 1 = x λ 1 y µ 1 = x a mb y n m, (where a and b are relatively prime). Let r and s be smallest nonnegative integers so that ( ) m n r s has determinant 1.
24 Given ζ = e x λ i y µ i i=1... its derived singularity is e 1 ζ = x λ i y µ i, i=1 where the new exponents are computed by µ i = m(µ i+1 µ 1 + mbµ 1 ) λ i = b(λ i+1 λ 1 + mbλ 1 + rµ iλ 1 ).
25 Given ζ = e x λ i y µ i i=1... its derived singularity is e 1 ζ = x λ i y µ i, i=1 where the new exponents are computed by µ i = m(µ i+1 µ 1 + mbµ 1 ) λ i = b(λ i+1 λ 1 + mbλ 1 + rµ iλ 1 ).
26 Given ζ = e x λ i y µ i i=1... its derived singularity is e 1 ζ = x λ i y µ i, i=1 where the new exponents are computed by µ i = m(µ i+1 µ 1 + mbµ 1 ) λ i = b(λ i+1 λ 1 + mbλ 1 + rµ iλ 1 ).
27 Example: ζ = x 1/2 y 3/2 + x 1/2 y 7/4 + x 2/3 y 11/6 ζ = x 17/4 y 13/2 + x 9/2 y 20/3 ζ = x 1438/3 y 157/3
28 Example: ζ = x 1/2 y 3/2 + x 1/2 y 7/4 + x 2/3 y 11/6 ζ = x 17/4 y 13/2 + x 9/2 y 20/3 ζ = x 1438/3 y 157/3
29 Example: ζ = x 1/2 y 3/2 + x 1/2 y 7/4 + x 2/3 y 11/6 ζ = x 17/4 y 13/2 + x 9/2 y 20/3 ζ = x 1438/3 y 157/3
30 Let V 1 and V denote the vertical monodromy of the truncation and the derived singularity. Let d denote the number of sheets of the derived singularity. Then V(t) = (V 1(t)) d V (t b ) (t b 1) d.
31 Let V 1 and V denote the vertical monodromy of the truncation and the derived singularity. Let d denote the number of sheets of the derived singularity. Then V(t) = (V 1(t)) d V (t b ) (t b 1) d.
32 Relationships among invariants An example: the formula of Steenbrink and Saito, worked out for z n = x a y b by McEwan. The horizontal and vertical monodromies commute; thus there is a common eigenbasis, and thus a pairing {horizontal eigenvalues} {vertical eigenvalues}.
33 Relationships among invariants An example: the formula of Steenbrink and Saito, worked out for z n = x a y b by McEwan. The horizontal and vertical monodromies commute; thus there is a common eigenbasis, and thus a pairing {horizontal eigenvalues} {vertical eigenvalues}.
34 McEwan s pairing (writing eigenvalues as elements of Q/Z): i b + j n ai b (for 1 i b 1 and 1 j n 1). The horizontal spectral numbers are h ij = i b + j n 1. Let v ij be the fractional part of ai b (but use 1 if it s an integer). By Steenbrink and Saito, Sp(f +L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice].
35 McEwan s pairing (writing eigenvalues as elements of Q/Z): i b + j n ai b (for 1 i b 1 and 1 j n 1). The horizontal spectral numbers are h ij = i b + j n 1. Let v ij be the fractional part of ai b (but use 1 if it s an integer). By Steenbrink and Saito, Sp(f +L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice].
36 McEwan s pairing (writing eigenvalues as elements of Q/Z): i b + j n ai b (for 1 i b 1 and 1 j n 1). The horizontal spectral numbers are h ij = i b + j n 1. Let v ij be the fractional part of ai b (but use 1 if it s an integer). By Steenbrink and Saito, Sp(f +L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice].
37 Sp(f + L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice]. Note that all the ingredients in the formula are spectra, with the possible exception of the v ij. It s natural to want to call these vertical spectral numbers, but what does that really mean? The Steenbrink-Saito formula applies to all quasi-ordinary surfaces. Can we find compatible recursions for computing all the ingredients?
38 Sp(f + L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice]. Note that all the ingredients in the formula are spectra, with the possible exception of the v ij. It s natural to want to call these vertical spectral numbers, but what does that really mean? The Steenbrink-Saito formula applies to all quasi-ordinary surfaces. Can we find compatible recursions for computing all the ingredients?
39 Sp(f + L k ) Sp(f ) = 1 t [ 1 t 1/k t h ij +v ij /k + same for other slice]. Note that all the ingredients in the formula are spectra, with the possible exception of the v ij. It s natural to want to call these vertical spectral numbers, but what does that really mean? The Steenbrink-Saito formula applies to all quasi-ordinary surfaces. Can we find compatible recursions for computing all the ingredients?
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