MAT 469: Advanced Topology

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1 MAT 469: Advanced Topology Notes from course given by András Némethi, Spring July 22, 2014 Dates are approximate guides. Because Némethi would often go over the same material twice in slightly different ways across consecutive classes, material is not always filed under the date it is actually given, but usually under the date in which related material is first presented. 1 Introductory Notions 1.1 February 4: Knots, invariants R 3 { } = S 3 = {x R 4 : x = 1}. Definition 1.1. A knot is a C map f : S 1 R 3 which is an injective embedding (i.e. f (x) 0 x S 1.) We usually identify f and its image. Some basic examples include the unknot, which we may write as U = {z = 0, x 2 + y 2 = 1}, and the trefoil. Definition 1.2. A link is a knot with more components S 1 S 1 R 3 S 3. e.g. the Hopf link. (a) (b) (c) Figure 1: Knot (link) diagrams of (a) the unknot; (b) the trefoil; (c) the Hopf link. Remark. Recall the Hopf fibration S 1 S 3 which we may also write as C C 2 0 S 2 CP 1 = C 2 0/ Definition 1.3. K 1 K 2 are isotopic knots (links) if there exists diffeomorphisms f t : R 3 R 3 such that t f t is smooth on t [0, 1], f 1 = id R 3, and f 2 (K 1 ) = K 2. f t is called an ambient isotopy. Question: how can we distinguish knots / links? One way is to use the invariants of algebraic topology: e.g. if K 1 K 2, then the knot complements S 3 K 1 S 3 K 2 are homeomorphic, and have equal fundamental groups π 1 (S 3 K i ) and first homology groups H 1 (S 3 K i ) = Z r, where r is the number of components in the link K 1 K 2. 1

2 K 1 # K 2 = K 1 K 2 Figure 2: Schematic diagram of a connect-sum K 1 #K 2 of two knots K 1 and K 2. We may define connect-sums of knots as follows: A non-trivial knot K (i.e. K is not the unknot) is prime if K K 1 #K 2 for non-trivial knots K 1, K 2. The set of knots is a (free) semigroup under #. We can represent knots by knot diagrams, which are projections onto the plane with transversal (generic) intersections. Remark. Link diagrams are not unique. Figure 3: Various different knot diagrams for the same knot (the trefoil.) However we can relate any two diagrams for the same link using a finite number of combinatorial moves known as Reidemeister moves: Figure 4: Type Ω 1, Ω 2, and Ω 3 Reidemeister moves (from left to right.) Theorem 1.4 (Reidemeister). The links associated with two link diagrams are isotopic iff there exists a finite sequence of moves of types Ω 1, Ω 2, Ω 3 connecting the two diagrams. Some knot invariants The Conway polynomial (z) = L (z) of a link L is defined by ( ) ( ) ( ) (a) = z (the skein relation) 2

3 (b) U (z) = 1 (a normalisation) ( Example 1.5. K ) ( K ) = z (U K) = (U K) = 0. (T ) + (U) = z (H), where H denotes the Hopf link, and so (T ) = 1 z (H). (H) + (U U) = z(u) = z. The Alexander polynomial of a link L can be defined as L (t) = Example 1.6. U (t) = 1 For L the Hopf link, L (t) = t 1 2 t 1 2. ( ) t 1 2 t 1 2. For L = U K where K is any knot or link, L (t) = 0. ( ) 2 For the trefoil T, T (t) = 1 + t 1 2 t 1 2 = t 1 + t 1 = t 1 (t 2 t + 1) = t 1 φ 6 (t), where φ 6 (t) = t 6 1 (t 3 1)(t 2 1) is the sixth cyclotomic polynomial Aside: the trefoil T is an example of a torus knot. For (p, q) coprime, the torus knot T p,q may be defined as the image of the line segment from (0, 0) to (p, q) under the quotient map R 2 R 2 /Z 2 = S 1 S 1. T = T 2,3. Exercise. Compute (T 2,m ) (or (T n,m) for any coprime n and m.) Alexander s original definition of K (t) was as follows: fix a knot diagram with n crossings and (n + 2) regions; create a n (n + 2) incidence matrix I whose (c, r) th entry I cr is zero if r is not incident to c, and is given by t t +1 1 otherwise. Next delete two columns corresponding to two neighboring regions, and compute the determinant of the remaining n n matrix. Theorem 1.7. K (t) as defined in the paragraph above is a polynomial in t independent, up to multiplication by a unit of Z[t 1, t], of the choice of knot diagram or of deleted columns. Sketch. We will just show why K (t) is stable w.r.t. Reidemeister moves of type Ω 1 : v K K [n] v [n + 2] old matrix 0 Figure 5: An Ω 1 Reidemeister move changes the matrix involved in the definition of (K) by the addition of a column (representing the newly-created region ) all of whose entries are zero except for the last one, which is ±t or ±1. Performing a cofactor expansion on the expanded matrix, we obtain the same (K). We leave it as an exercise to show that the same holds for Reidemeister moves of type Ω 2 and Ω 3. The Jones polynomial V (L) = V L (t) = V (t) of a link L is defined by ( ) t 1 V (L ) tv (L + ) = t 1 2 t 1 2 V (L 0 ) V (U) = 1. Proposition 1.8. V (L 1 #L 2 ) = V (L 1 ) V (L 2 ) 3

4 V (T ) = t + t 3 t 4 V ( T ) = t 1 + t 3 t 4, where here T denotes the mirror of the trefoil. The trefoil can be written as {z 2 1 = z 3 2} S 3 C 2 (z 1, z 2 ), where C 2 = R 4 S 3. Knots which can be written as the intersection of zero loci of algebraic equations with the 3-sphere are known as algebraic knots. The goal of this course, in short, is to show algebraic knots connect algebra and topology: what information can we recover from the knot about the equation, and what information can recover about the knot from the equation? 1.2 February 6: Braids A braid is defined as follows: fix n N and the points (0, 0, 0), (0, 1, 0),..., (0, N, 0), (0, 0, 1), (0, 1, 1),..., (0, N, 1). Fix a permutation σ : {1,..., n} {σ(1),..., σ(n)}, and define γ i : [0, 1] R 3 for 1 i N to be embeddings t (x(t), y(t), z(t)) satisfying γ i (0) = (0, i, 0), γ i (1) = (0, σ(i), 0), z(t) increasing, and Iγ i Iγ j = i j (up to isotopy.) Figure 6: A simple example of a braid. The set of braids of n strands is denoted B n. B n is a group under composition. The generators of B n are σ i for i = 1, 2,..., n 1, where σ i transposes i and i + 1. Note that B n is not isomorphic to the symmetric group S n ; indeed it is larger: there is a surjection from B n onto S n with nontrivial kernel. Figure 7: Example of a non-trivial braid which maps to the trivial permutation. There are obvious relations between the generators given by σ i σ j = σ j σ i if i j 2, and also the slightly less obvious ones σ i σ i+1 σ i = σ i+1 σ i σ i+1 (if we change this slightly to σ i σ i+1 σ 1 i = σ 1 i+1 σ iσ i+1 then we have a Ω 3 Reidemeister move.) These in fact generate all the relations in B n, though this is not easy to prove: Theorem 1.9. B n = σ 1,..., σ n 1 σ i σ j = σ j σ i i j 2, σ i σ i+1 σ i = σ i+1 σ i σ i+1 i. Given a braid b B n, we can obtain a link in R 3 from it by forming its closure This is the link associated with b B n. Example B 2 = σ = Z. The closure of 0 B 2 is a 2-component unlink U U. The closure of σ B 2 is the unknot U. The closure of σ 2 B 2 is the Hopf link. The closure of σ 3 B 2 is the trefoil (we may say that it has 3/2 twists.) In general, the closure of σ m B 2 is the torus knot T 2,m. 4

5 Figure 8: Illustration of braid relations σ i σ i+1 σ i = σ i+1 σ i σ i+1 (left) and σ i σ i+1 σ 1 i = σ 1 i+1 σ iσ i+1 (right). B Figure 9: Schematic illustration of a braid closure Remark. Note that the closure of (σ 1 σ 2 ) 2 B 3 is also a trefoil, so that the map c from n 1 B n to the set of isotopy classes of links defined here by braid closure is not injective. However, we do have (with a lot of work which we not do here) Theorem 1.11 (Alexander). c is onto, i.e. any link in R 3 is the closure of some braid B n. Braids in a solid torus Given a solid torus D 2 S 1, we have a projection p : D 2 S 1 S 1, so that for any t S 1, p 1 (t) = D t. Fix p 1 (t),..., p n (t) as n different points moving continuously in D 2 S 1 ; this forms a closed braid, i.e. a link in D 2 S 1 R 3. Remark. C 2 = R 4 B 4 R = { (z 1, z 2 ) R}, and define S 3 R = B4 R. 5

6 Figure 12: Embedding of T 2,3 torus link in a solid torus; we can see this as a braid closure using the construction described above. Also C 2 D 2 z 1 D 2 z 2 and (D 2 D 2 ) S 3. We may write (Dz 2 1 Dz 2 2 ) = Dz 2 1 Dz 2 2 Dz 2 1 Dz 2 2 Dz 2 D 2 1 z 2 and indeed this is one way of seeing the decomposition of S 3 into two solid tori. Example Define V = {z 3 1 = z 2 2} C 2. D z1 D z2 = { z 1, z 2 R < 1}. We claim that V (D 2 D 2 ) F T 1 := D 2 z 1 D 2 z 2 : if (z 1, z 2 ) V F T 2, then R 2 = z 2 2 = z 1 3 R 3, which is a contradiction (since R < 1.) 1 Figure 13: Schematic diagram of V (D 2 z 1 D 2 z 2 ). The two vertical sides of the box form F T 1, the two horizontal sides F T 2. so The intersection V { z 1 = R, z 2 R} is given by z 1 = Re 2πit (where t [0, 1]) and z 2 2 = z 3 1 = R 3 e 2πit 3, z 2 = ±R 3/2 e 3πit = R 3 2 e 2πit πik 2 where k {0, 1}; we may think of this as the braid formed by two points moving at speed 3/2. 1 Note that if we took R > 1, we would have obtained V (D 2 D 2 ) F T 2, and, at R = 1, V (D 2 D 2 ) lies precisely on the boundary torus which is the intersection of the two solid tori. 6

7 Exercise. Compute the link which we get from z n 1 = z m 2 (it should be a torus knot T n,m.) In general connecting to the comments at the end of the previous section on the general goals of this course, given f(z 1, z 2 ) C[[z 1, z 2 ]] with f(0, 0) = 0, and V = {f = 0}, we can construct a closed n-braid, which is the link of f in S 3. Is it possible to characterize this link from the equation of f? In the below we will use the algebraic properties / invariants of f to study the link of f (so defined). As a final aside, we note that the braid representation of a knot allows us to compute its fundamental group (i.e. the fundamental group of the knot complement.) For instance, using that the trefoil T is the closure of σ 3 B 2, we obtain that π 1 (S 3 T ) = g 1, g 2 g 1 g 2 g 1 = g 2 g 1 g 2. g 1 g 2 g 2 g 1 2 g 1g 2 g 1 2 g 1g 2 g 1 2 g 1 1 g 2g 1 g 2 Figure 14: After closing the braid, we obtain g 1 = g 1 2 g 1 1 g 2g 1 g 2 and hence g 1 g 2 g 1 = g 2 g 1 g 2. Some References For more details, and proofs, for the material below, see E. Brieskorn and H. Knörrer, Plane Algebraic Curves C. T. C. Wall, Singular Points of Plane Curves G. Fischer, Plane Algebraic Curves D. Eisenbud and W. Neumann, Three-dimensional link theory and plane curve singularities J. Milnor, On isolated singularities of hypersurfaces 2 Classification of Algebraic Links 2.1 February 11: Algebraic curves and their singularities Introduction: germs of holomorphic functions 1. C n 0; (C n, 0) is the germ of C n at 0, i.e. an arbitrarily small neighborhood of 0 C n. 2. f : (C n, 0) (C, 0) is the germ of a function f, i.e. f is defined on an arbitrarily small neighborhood U 0 and f(0) = 0. Suppose f : (C 2, 0) (C, 0) is the germ of a holomorphic function f : U C with U 0 and f(0) = 0. The set of all such germs of functions forms a ring (indeed, an Artinian ring) denoted O 2 or C{z 1, z 2 }. Proposition 2.1. C{z 1, z 2 } is a local ring, i.e. there exists a unique maximal ideal m z1,z 2 = {f : f(0) = 0} C{z 1, z 2 }. Note that, in comparison, the polynomial ring C[z 1, z 2 ] is not local. C{z 1, z 2 } is a UFD: any f C{z 1, z 2 } can be written uniquely (up to permutation of factors) as a product f = f α1 1 fγ αγ where the f i are irreducible. We note in passing that the notion of irreducibility in C{z 1, z 2 } is well-defined. 7

8 Example 2.2. f(z 1, z 2 ) = z 1 is irreducible, whereas f(z 1, z 2 ) = z 1 z 2 is not irreducible (it has two irreducible components); indeed we can also see this geometrically. Example 2.3. f(z 1, z 2 ) = z 1 z 2 +z 3 1 +z 3 2 is irreducible in C[z 1, z 2 ], but not in C{z 1, z 2 }: this curve is globally irreducible, but not locally irreducible near the origin. This can be seen geometrically, or by writing (z 1 z 2 + z z 3 2 = (z 1 + a 2 + a )(z 2 + b 2 + b ) where a i, b i are the homogenous parts of degree i of the repsective factors, and iteratively solving: z 1 z 2 = z 1 z 2, z1 3 + z2 3 = z 1 b 2 + a 2 z 2 = (a 2, b 2 ) = (z2, 2 z1), 2 etc. locally reducible globally irreducible Figure 15: The joys of working in C[z 1, z 2 ] instead of C{z 1, z 2 }. Remark. This decomposition is unique up to multiplication by units. Remark. The invertibles (units) of O 2 = C{z1, z 2 } are the convergent power series with non-vanishing constant term; they form the complement of the maximal ideal, whose elements are precisely the convergent power series which have a vanishing constant term. Singular and non-singular points Given f : (C n, 0) (C, 0), let V (f) deote the zero set of f, i.e. V (f) = {f = 0} C n. X V (f) is singular if f z j (x) = 0 for all 1 j n. In other words, X is non-singular iff f z (x) 0. Remark. Let f = f 0 + f f d +... be the Taylor expansion of f at 0, where f i is the homogenous term of degree i. f(0) = 0 f 0 = 0, and f 1 = f i z i (0)z i, so f singular at 0 iff f 1 = 0 iff the normal vector vanishes iff T 0 V (f) is not well-defined as a tangent plane. If 0 V (f) is non-singular, then (by the implicit function theorem) there exists a diffeomorphism ϕ : (C n, 0) (C n, 0) such that f = (z z 1 ) ϕ, i.e. in suitable local coordinates f is a straight line. V (f) (C n, 0) ϕ f (C, 0) z z 1 Figure 16: Using the implicit function theorem to simplify the local description of our variety: a schematic illustration. Moreover there exists a diffeomorphism ψ : (C n 1, 0) (V (f), 0). e.g. consider f : (C 2, 0) (C, 0) given by f(z 1, z 2 ) = z 1. Then V (f) = {z 1 = 0}. 8

9 Now S 3 = {z : (z 1, z 2 ) = 1}, and S 3 S 3 V (f) = {z 1 = 0, z 2 = 1} = S 1, and [it is an exercise to show that] this S 1 S 3 is in fact an unknot. If f : (C n, 0) (C, 0) is in singular, take the Taylor expansion f = f 0 +f where, recall, f 0 = f 1 = 1. We define the multiplicity of f at 0 to be min{m : f m 0}; in other words, if we write f = α a αz α, then m is the degree of the smallest-degree monomial. Example 2.4. f(z 1, z 2 ) = z z 3 2 has multiplicity 2 at the origin. Exercise. Let f(z 1, z 2 ) C{z 1, z 2 } and f(0) = 0. (A) If f is a generic linear function az 1 +bz 2 (i.e. a, b are generic coefficients), show that # (V (f) {l = ɛ}) = m in a small neighbourhood of radius ɛ > 0 around the origin (i.e. we fix beforehand B η = { z η} with η 1, and then choose ɛ η.) Figure 17: Example of generic (dashed red) vs. exceptional (solid red) lines with the plane curve z 2 1 +z 3 2 = 0. (B) We can project (V (f), 0) down to (C, 0) along the direction l; this projection p is locally ramified at 0. Then the restriction p : V (f) 0 C 0 is a singular covering of degree m. Example 2.5. {z m 1 = z2 m } consists of m lines, which may be exhibited as z1 m z2 m = (z 1 z 2 ζ) ζ m =1 and so has multiplicity equal to its number of irreducible components m. {z m 1 = z n 2 } consists of min(m, n) lines, but its multiplicty is given by (m, n), so that the multiplicty is the number of irreducible components. In general, we can write V (f) = V (f 1 ) V (f r ), where the f i are the irreducible components of f, and the number of irreducible components is equal to the number of orbits of the monodromy. Tangent cone The { tangent cone of a non-singular plane curve f = f 1 + f (where, recall, f 1 0) is {f 1 = 0} = } f i z i (0)z i = 0. We extend this notion to singular curves by defining the tangent cone of a singular plane curve f = f m + f m (where m > 1 and f m 0) to be {f m = 0}. Example 2.6. For f(z 1, z 2 ) = z 1 z 2, m = 2, and the tangent cone is given by f 2 = z 1 z 2 = 0. In this case the tangent cone is again V (f). Example 2.7. For f(z 1, z 2 ) = z 1 z 2 + z z 3 2, the tangent cone is still given by f 2 = z 1 z 2 = 0. Globally our variety is not quite the same as that from the previous example, but locally (near the origin) we do not see this. Example 2.8. For f(z 1, z 2 ) = z z 3 2, the tangent cone is given by f 2 = z 2 1 = 0. 9

10 double line multiplicity 2 (a) (b) (c) Figure 18: Tangent cones from Examples (a)??,??; (b)??; (c)??. We may view the tangent cones as a union of non-degenerate lines or limits of tangent lines: given f : (C 2, 0) (C, 0), write f = f f i +... where f m is homogeneous of degree m in z 1, z 2. Then write f m = a i z1z i j 2 = z m 2 i+j=m = a i z m 2 m ( z1 a i i=0 m k=1 z 2 ) i ( ) z1 α k z 2 m = a i (z 1 α k z 2 ) k=1 where we may think of this as dehomogenizing f m. Example 2.9. For f(z 1, z 2 ) = (z z 3 2)(z z 2 2), the tangent cone is given by f 4 = z 2 1z 2 2 = 0. Remark. If f is irreducible, then f m = l m is given by the union of m lines. Isolated singularities Definition Define {x V (f) : f(x) = 0} =: Sing V (f) to be the singular set of V (f). Note that Sing V (f) 0. f has an isolated singularity at 0 if 0 is not a limit point of Sing V (f). Example If f(z 1, z 2 ) = z1 a + z2, b then f z = (aza 1, bz b 1 ), and so we see that f has an isolated singularity at zero. Example The same is true for the hypersurface f(z 1,..., z n ) = n i=1 zαi i. Example Given f(z 1, z 2 ) = z1, 2 we have f z = (2z, 0) and Sing V (f) = V (f) as a set; in particular, f does not have an isolated singularity at 0. Example Suppose f = f α1 1 fr αr : (C 2, 0) (C, 0) where the f i are irreducible. If α i > 1 for any i {1,..., r}, then V (f i ) Sing V (f) implies that f does not have an isolated singularity at 0. Example Suppose f(z 1, z 2, z 3 ) = z 2 1z 2 z 2 3. (V (f) in this case is the complex Whitney umbrella.) Then Sing V (f) = {z 1 = z 3 = 0} does not have an isolated singularity at zero. 10

11 2.2 February 13: Local intersection multiplicity Let f, g : (C 2, 0) (C, 0) be two complex plane curves; or, more precisely, take f, g C{z 1, z 2 } (or C[[z 1, z 2 ]], the space of formal power series.) Assume (f, g) = 1, i.e. f and g have no common component, so that V (f) V (g) = {0} (in some small neighborhood of the origin.) Definition We define the local intersection multiplicity of f and g as i 0 (f, g) = dim C C{x, y} I(f, g) where the I(f, g) in the denominator denotes the ideal of C{x, y} generated by f and g. Example Given f(z 1, z 2 ) = z 1 and g(z 1, z 2 ) = z 2, I(f, g) = I(z 1, z 2 ), and then and so i 0 (f, g) = 1. C{z 1, z 2 } I(z 1, z 2 ) = C 1 Example Given f(z 1, z 2 ) = z 1 and g(z 1, z 2 ) = z 1 + z 2 2, I(f, g) = I(z 1, z 2 2), and so i 0 (f, g) = 2. C{z 1, z 2 } I(z 1, z 2 2 ) = C 1, z 2 1 z 1 z 2 z 2 1 z 1 z 2 z Example Given f(z 1, z 2 ) = z z 3 2 and g(z 1, z 2 ) = z z 2 2, (z z 3 2) z 2 (z z 2 2 = z 2 1(1 z 1 z 2 ) I(f, g) =: I and since (1 z 1 z 2 ) is invertible in C{z 1, z 2 }, we may conclude that z 2 1 I. Symmetrically, z 2 2 I; however (it can be shown that) z 1, z 2 / I, and so I = I(z 2 1, z 2 2); hence i 0 (f, g) = 4 in this case. 1 z 1 z 2 z 2 1 z 1 z 2 z 2 2 z 3 1 z 2 1z 2 z 1 z 2 2 z We recall (with one slight change) one bit of notation: write C{z 1, z 2 } =: O. Properties of the local intersection multiplicity Remark. {f = t} {g = s} is a transversal intersection for 0 < t, s 1, and so i 0 (f, g) counts the number of intersection points between f and g when f, g have no common components. Indeed, we may generalize this: 11

12 Proposition Given f 1, f 2, g O such that f 1 f 2 and g have no common components, i 0 (f 1 f 2, g) = i 0 (f 1, g) + i 0 (f 2, g). Proof. We have the short exact sequence I(f1,g) 0 I(f 1, g) I(f 1 f 2, g) O I(f 1 f 2, g) O I(f 1, g) 0 and the map α : O I(f given by α(h) = h f 1f 2,g) 1. Note that α is onto; we now show that ker α = (f 2, g), O so that the first term in the exact sequence is isomorphic to I(f 2,g) ; then, since the dimensions of the first and third terms sum to the dimension of the middle term in the long exact sequence, we are done. Suppose h ker α; then h f 1 I(f 1 f 2, g), i.e. hf 1 = af 1 f 2 + bg, which implies f 1 bg, which implies f 1 b, and so h (g, f 2 ) (since h = af 2 + b g where b f 1 = b.) Hence ker α (f 2, g); the opposite inclusion clearly holds. Example i 0 (z1 n + z2 n, z1 m 2z2 m ) = nm (by the remarks above, the zero set of the former consists of n lines, and the zero set of the latter of m lines.) Exercise. Given f O and l a generic line, i 0 (f, l) = m f, the multiplicty of f (hint: use suitable local coordinates see the comments above about the implicit function theorem and the most recent remarks on additivity of i 0. Proposition Given f O and 0 f a nonsingular point (i.e. f 1 0), (a) i 0 (f, l) = 1 where l is any generic line (b) i 0 (f, T 0 f) 2. Definition is an inflection point of f if i 0 (f, T 0 f) 3. Proposition i 0 (f, g) m f m g, where m f denotes the multiplicty of f, with equality iff the tangent cones of f and g have no common lines. Proof. The proof proceeds in two major steps: 1. Assume that f t is a holomorphic family of functions, where t (C, 0). Then for small t, i 0 (f t, g) O i 0 (f 0, g) by continuity: I(f 0,g) is finite-dimensional, and we can find a basis for this vector space; let b 1,..., b k be lifts of these basis vectors in O. Then O = C b 1,..., b k I(f 0, g), and perturbing f 0 preserves some properties by continuity: in particular, we still have C b 1,..., b k I(f t, g) = O, so that k i 0 (f t, g). (Note that this inequality may be strict in general e.g. given f t (z 1, z 2 ) = z z tz 2 2, i 0 (f 0, z 1 ) = 3 but i 0 (f t, z 1 ) = 2 for small t 0.) 2. Write f = f m + f m and g = g n + g n , where now, as before, f i denotes the homogeneous part of f of degree i (rather than a perturbation of f.) Then we have, from the previous, m n i 0 (f, g) i 0 l i +..., l j +... and by applying the implicit function theorem we may turn this into since ( ) is invertible in C{x, y}. i=1 j=1 i 0 (x, y ( )) = i 0 (x, y ) = 1 12

13 { } Recall that (a plane curve) f has an isolated singularity at 0 if Sing V (f) = f z 1 some small neighborhood of zero. If f is given, then we can compute its partial derivatives f z 1 O = dim C ( ) i f 0 z 1, f z 2 I ( f z 1, f z 2 ). Definition The Milnor algebra of f is the quotient ( O I f z, f 1 O ( ) I f z 1, f z 2 { f z 2 } = {0} in and f z 2, and and dim C ) is the Milnor number of f, also deoted µ. z 2 ( ) Example (a) Given f(z 1, z 2 ) = z 1 z 2, I f z 1, f z 2 = I(z 1, z 2 ), and so µ = 1. ( ) (b) Given f(z 1, z 2 ) = z1 n +z2 m, I f z 1, f z 2 = I(nz1 n 1, mz2 m 1 ), and so µ = i 0 (z1 n 1, z2 m 1 ) = (n 1)(m 1). Definition Let f be an irreducible plane curve (or, in general, smooth hypersurface with isolated singularity.) We define a set (actually the semigroup of f) S f := {i 0 (f, g) : g O} N. Example If f(z 1, z 2 ) = z 1, then S f = N. Example If f(z 1, z 2 ) = z z 3 2, then i 0 (f, 1) = 0, i 0 (f, z 2 ) = 2, i 0 (f, z 1 ) = 3, and we may obtain S = N {1}. In general, S f = N if f is smooth at the origin; otherwise S f will have ome gaps. Example The δ-invariant or Serre invariant of f is δ f := #(N S f ). Theorem Let f be an irreducible smooth hypersurface with isolated singularity. Then (a) 2δ = µ 2 ; (b) {µ, µ + 1,... } S f, and µ 1 / S f ; (c) for 0 k µ 1, k S f µ 1 k / S f. (d) s S f t s = (t) 1 t. Note that (b) and (c) together imply (a). Example If f(z 1, z 2 ) = z1 n + z2 m where (n, m) = 1 (so that f is irreducible), then δ f = n, m N is the semigroup generated by n and m. e.g. 3, 4 = {0, 3, 4, 6, 7, 8, 9,... } (and note that µ = (3 1)(4 1) = 6 in this case.) e.g. for (n, m) = (2, 3), S f = {0, 2, 3,..., }, and t s = 1 + t 2 + t 3 + = 1 + t2 1 t = 1 t + t2 = (t) 1 t 1 t s S where the last equality follows from our computation of (t) earlier. 2 In other words: µ is the conductor of the local algebra. 13

14 2.3 February 18: Locally-trivial fibrations Ehresmann s theorem Definition f : (R n, 0) (R m, 0) is a submersion at 0 if df 0 : T 0 R n T 0 R m is onto, i.e. (assuming ( n m) the Jacobian matrix df 0 = fi x j x=0 has maximal rank m. )ij Note we can (and will) replace R with C throughout this definition. Theorem 2.34 (Local version). If df 0 is onto, then we have f (R n, 0) (R m, 0) φ p ψ (R n, 0) (R m, 0) with p(x 1,..., x n ) = (x 1,..., x m ) and φ and ψ isomorphisms, i.e. R n = R n fiber fibration f is locally (at the origin) a trivial R m The global version of this is Ehresmann s theorem: Definition A surjective C map f : M N between C manifolds is a locally-trivial fibration if for all n N there exists a neighborhood U n n and a diffeomorphism ψ n such that we have f 1 (U n ) ψ n U m F f p 1 U n Example A trivial fibration M ψ N F f p 1 N is a locally-trivial fibration. Example A Möbius band is a locally-trivial fibration which is not (globally) trivial. Theorem 2.38 (Ehresmann s Theorem). If f : M N (a C map between C manifolds) is a submersion (i.e. m M : T m f : T m M T f(n) N is onto) and is proper (i.e. the fibers of f are compact), then f is a locally-trivial fibration. Sketch of proof. Use the local theorem, and compactness to pass to a union of finitely-many local neighborhoods. Example ( ) f : C 2 C given by f(z 1, z 2 ) = z 1 (z 1 z 2 1) = z1z 2 2 z 1 has no critical points, since f z 1, f z 2 = (2z 1 z 2 1, z1), 2 and so rank df z = 1 z. However note that f is not proper; we will presently show that f is not a locally-trivial fibration. Note that (it is an exercise to show that) if f : M N is a locally-trivial fibration and N is connected, then all the fibers are diffeomorphic. Now f 1 (0) = {z(z 1 z 2 1) = 0} = {z 1 = 0} {z 1 z 2 = 1} = C C C 2. However (check!) for λ 0, f 1 (λ) = {z1z 2 2 z 1 = λ} is connected (in fact, it is homeomorphic to a twice-punctured CP 1.) Or another way to look at it: consider the projectified fiber F λ = {z1z 2 2 z 1 z3 2 = λz3} 2 CP 2 C 2 ; for (z 3 = 0) [0 : 1 : 0], [1 : 0 : 0], i.e. we have 2 points at infinity for the fiber F (and so f is not locally trivial at λ = 0.) Theorem 2.40 (Relative version). Given f(m, M ) N where M M is a submanifold (e.g. M = M), if f M : M N is a submersion and f is proper then f is a locally-trivial fibration of the pair of spaces (M, M ), i.e. for every n N there exists a neighborhood U n n and F F such that 14

15 Figure 19: Schematic diagram of the exceptional fiber f 1 (0) (right) and of the [topological] transition from the regular fiber f 1 (λ) as λ 0 (left). ( f 1 (U n ), f 1 (U n ) M ) ψ (U n F, U n F ) f (p 1,p 1) (U n, U n ) where ψ is a diffeomorphism. M, M =: M {p, q} I f I F N Figure 20: One archetypal example of a locally-trivial fibration is the Möbius band; here we illustrate how the relative version of Ehresmann s Theorem works on this example with M := M. Note that f a submersion does not necessarily imply f M a submersion: the fibers may not be transversal to M. M Figure 21: The fibers may not be transversal to M. Exercise. Given M M and f : M N a submersion, show F M : M N is a submersion iff f 1 (λ) M λ N, i.e. m M f 1 (λ) : T m M = T m f 1 (λ) T m M. Proposition 2.41 (Properties of locally-trivial fibrations). (f 1 (λ) = F λ.) 1. N connected = the fiber is well-defined 15

16 2. N contractible = a (globally) trivial fibration, i.e. M f N F p 1 N Monodromy If N = S 1, write S 1 = [0, 1]/(0 1). We can cut the locally-trivial fibration at one fiber to get a fibration over [0, 1]; then, by property (2) above, we get the fibration as a product. If we now glue F 0 to F 1 via a diffeomorphism m (the geometric monodromy), we obtain our fibration as F [0, 1]/ ((x, 0) (m(x), 1)). F 0 F f Figure 22 Indeed, this is precisely the mapping torus of m of F. Example With F = [0, 1] and m(x) = 1 x, we get the Möbius band. Example( Consider ) the plane curve f : C 2 C given by f(z 1, z 2 ) = z1 2 + z2. 3 The critical points are given by f z 1, f z 2 = (2z 1, 3z2) 2 = (0, 0), and here the critical value is 0. Thus we have a special fiber at λ = 0; otherwise f 1 : C C is a submersion. It is an exercise to show that f 1 (0) = {z1 2 + z3 2 = 0} is contractible. (Hint: use the weighted homogeneous action for t [0, 1], define t (z 1, z 2 ) = (t 3 z 1, t 2 z 1 ); the exponents of t are chosen to preserve the shape of the level sets.) In fact this exceptional fiber is a cone. We now want to show that f 1 (C ) C is indeed a locally-trivial fibration. Note that we cannot quite use Ehresmann s theorem, because our fibers now are not compact. Proof. Fix λ C. We wish to find U λ λ such that f 1 (U λ ) F λ U λ f p 2 U λ Note that f(z 1, z 2 ) has a weighted homogeneous C -action (as described in the Hint to the Exercise above) given by ξ (z 1, z 2 ) = (ξ 3 z 1, ξ 2 z 2 ), so that f (ξ (z 1, z 2 )) = ξ 6 f(z 1, z 2 ). This induces an action φ on the fibers ξ F λ = F ξ 6 λ; indeed this action is a diffeomorphism, and so the fibers can be identified. Now take a small disc D ɛ := D(1, ɛ) B(λ, ɛ 6 ) =: U λ (see Figure?? above); φ acts between these two sets as ξ ξ 6 λ. Then we have φ D ɛ F λ f 1 (U λ ) D ɛ p 1 φ f U λ 16

17 f 1 (0) f U λ Figure 23 and so f 1 (C ) C f 1 (S 1 f ) S 1 We may verify that the fiber F 1 is given by a punctured torus S 1 S 1 {pt}: write F 1 := {z 2 1+z 3 2 = 1} C 2 and consider the second projection p = p 2 : F 1 C. p 1 (z 2 ) consists of 2 points if z 3 2 1, and 1 point if z 3 2 = 1. Thus F 1 is a double cover of the complex plane C branched, with ramification (at the branch points) given by z z 2, and χ(f 1 ) = χ(c {p, q, r}) 2 + χ({p, q, r}) = = 1. Moreover, F = S 1 since we have a non-trivial double cover of S 1. E F B A F E A B C D D C Figure 24: The [Milnor] fiber F 1 as a branched double cover of the complex plane. The ramification around each of the branch points (circled on left) is given by z z 2. It may be observed, by following the boundary as it moves between the two sheets (see right), that F 1 is a trefoil. We may further find that the geometric monodromy in this case is given by multiplication by e 2πi/3 in the first coordinate and by e πi in the second. In this case H 1 (F 1 ; Z) = Z 2, and H 1 (m) : H 1 (F 1 ) H 1 (F 1 ) is the algebraic monodromy. Note that the characteristic polynomial of the algebraic monodromy det(1 th 1 (m)) = t 2 t + 1 is the (normalized) Alexander polynomial (and this is true in general!) To compute the monodromy we can e.g. use a change of coordinates: let C : C 2 C 2 be given by C(z 1, z 2 ) = (c = z1, 2 d = z2). 3 If we let L be the line {c + d = 1}, then C 1 (L) = F, and C F : F L = C. Thus F and L are homotopy equivalent, and we may simplify L to a complete bipartite graph K 2,3 (see diagrams above); we see that the homotopy type of F is given by S 1 S 1, and in particular H 1 (F ) = Z 2. We see directly what the geometric monodromy should be (it acts cyclically on the vertices in each partite set); to compute the algebraic monodromy, note that H 1 (F ) has a basis e 1, e 2, and note the monodromy H 1 (m) = m alg characteristic polynomial is as desired. sends e 1 e 2, e 2 e 1 + e 2, and so has matrix ( det(tm 1) = 1 t t t 1 = t2 t + 1 = φ 6 ). It then follows that the 17

18 F C lift Figure 25: One way to compute the monodromy on the [Milnor] fiber F February 20: Algebraic knots / links a) Recall the distance function r : C n [0, ] given by r(z) = z 2. If we separate the real and complex parts of z as z j = x j + iy j, then r(z) = j x2 j + j y2 j, and dr dz = 0 = z = 0, that is the only critical point is the origin. Hence r : C n {0} (0, ) is regular and proper (and its fibers are (2n 1)-spheres); by Ehresmann s theorem, r is a (locally-)trivial fibration; indeed, C n {0} = (0, ) S 2n 1. b) Fix f C{z 1,..., z n } with an isolated singularity at the origin (i.e. { f(z) = 0} U = {0} for some small neighborhood U 0. By the Inverse Function Theorem, f 1 (0) {0} is smooth, i.e. it is a smooth submanifold of C n {0}. Consider r : (C n 0, V 0) (0, ) and define N(z) = f(z). The fibers or r S 2n 1 V T z0 V T z0 S 2n 1 λ C : λz = f(z). The middle condition can be interpreted in a purely algebraic way: it states simply that some ranks drop. The critical set in C n is given by the equations f(z) = 0 f(z) = λz λ C and so the critical values (the image of the critical set) are the zero set of a real analytic equation. We see a certain rigidity of algebraic links here. Now since the singularity at the origin is isolated, there exists c 0 > 0 such that (0, c 0 ) contains only regular values, and from the above we see that we must have r 1 )ɛ) V ɛ (0, c 0 ). By Ehresmann s theorem, we have a trivial fibration r 1 (0, c 0 ) V = (r 1 (ɛ) V ) (0, c 0 ). Thus: given V = {f = 0} where f : (C n, 0) (C, 0) has only isolated singularities, there exists ɛ 0 > 0 such that for all ɛ (0, ɛ 0 ], S 2n 1 ɛ V = K ɛ is a well-defined smooth oriented manifold K ɛ independent of ɛ. Definition K ɛ S 2n 1 ɛ (which is independent of ɛ) is the algebraic knot associated with f. Note dim R K ɛ = 2n 3. Remark. V B ɛ0 B ɛ0 The cone of the sphere is the ball. = = cone(k) cone ( S 2n 1 ɛ 0 ) Example If n = 3, f : (C 3, 0) (C, 0) and the algebraic link is a K 3 S 5, i.e. it is a 3-manifold. (i) Let f(z 1, z 2, z 3 ) = z 1. Then V = {z 1 = 0} = C 2 and V S 5 = S 3. 18

19 (ii) Let f(z 1, z 2, z 3 ) = z 1 z 2 z 2 3. Then V = {z 1 z 2 z 2 3 = 0} C 3. We claim that this is in fact a RP 3. To see this, define an involution ι : C[u, v] C[u, v] by (u, v) ( u, v) (this is a Galois action of Z 2 on C 2 (0, 0).) The invariants of C[u, v] under ι are generated by u 2, v 2, uv; label these as z 1, z 2, z 3 resp. (and note that z 1 z 2 = z 2 3.) Now define a map ϕ : C 2 V C 3 by (u, v) (u 2, v 2, uv). We note that the restriction of ϕ to the unit sphere in C 2 = R 4 is a non-trivial double cover S 3 V S 5, and so its image is RP 3, as desired. 2.5 February 25: Milnor fibers Let f : (C n, 0) (C, 0) be a holomorphic germ of function with isolated singularity, and let (V, 0) = ({f = 0}, 0). Recall Theorem There exists ɛ 0 > 0 such that for any ɛ [0, ɛ 0 ], S 2n 1 ɛ (a) K ɛ := S 2n 1 ɛ (b) the embedding K 2n 3 ɛ V (f), so is an oriented (2n 3)-manifold independent of the choice of ɛ (by Ehresmann s theorem.) S 2n 1 ɛ is also independent of ɛ. (c) (B ɛ, B ɛ V (f)) cone(s ɛ, K ɛ ) ( the analogue of Seifert surfaces in the algebraic context ) Definition B ɛ0 is called the Milnor ball of f. Theorem 2.48 (Milnor fiber). Let f : (C n, 0) (C, 0) be a holomorphic germ of function with isolated singularities, and let B ɛ0 be a Milnor ball for f. Then 0 < δ ɛ 0 such that (a) f 1 (λ) S 2n 1 ɛ 0 λ δ. (b) the map ψ : ( f 1 (D δ ) B ɛ 0, f 1 (D δ ) S ɛ 0 ) (D δ, S δ ) given by ψ ( f 1 (λ) B ɛ0, f 1 (λ) S ɛ0 ) = λ is a locally-trivial fibration of a pair of spaces. (c) we can extend this to a locally trivial fibration over the whole of D δ using f 1 (D δ ) S2n 1 D δ f 1 (D δ ) S 2n 1 D δ B 1 Bɛ0 F f D δ B ɛ0 C V (f) f 1 (λ 0 ) f 1 (λ 1 ) F λ F 0 Figure 26: The Milnor fibers f 1 (λ) are the regular fibers obtained when λ is taken from a small punctured disc Dδ around the origin (left); in this construction the variety V (f) = F (0) is an exceptional fiber (right). 19

20 Example For a weighted homogeneous curve (e.g. f(z 1, z 2 ) = z z 3 2), we can choose ɛ 0 = 1; then {z z 3 2 = λ} B 1 is well-defined for λ D δ with δ 1. Proof. f : B ɛ C n C holomorphic implies that its critical points (values) are discrete; in particular, 0 is not in the set of critical values. Set δ 1 such that 0 λ δ 1 implies that λ is a regular value ( = B ɛ0 f 1 ( D δ 1 ) D δ1 is a locally-trivial fibration. Furthermore, f Bɛ0 : B ɛ0 C has 0 as a regular value (since f 1 (0) B ɛ0 ); so there exists δ 2 > 0 such that λ δ 2 = f 1 (λ) S ɛ. Now take δ = min(δ 1, δ 2 ) (the idea is simply that whereas the exceptional fiber is not regular at the origin, away from the origin it still looks regular.) Definition F λ := f 1 (λ) B ɛ0 (where λ D δ ) is called the Milnor fiber of f. (F λ, F λ ) is a smooth manifold with boundary F λ = Kf. Note dim R F λ = 2 2 and dim R F λ = 2n 3, consistent with what we computed before. Note that (F λ, F λ ) (f 1 ( D δ ) B ɛ0, f 1 ( D δ ) S ɛ0 ) D δ = S 1 (here working in n = 2 complex dimensions) is a locally-trivial fibration. The geometric monodromy is a map m = m g : (F δ ), F δ ) (F δ, F δ ); up to isotopy, WMA m g Fδ = id. The algebraic monodromy m a,q is the induced map on H q (F, F δ ); recall the characteristic polynomial of this gives the Alexander polynomial. Theorem Given a hypersurface f : (C n, 0) (C, 0) with isolated singularities, F has the homotopy type of a bouquet of spheres (more precisely, F µ Sn 1 where µ is the Milnor number. Proof. See Milnor s book (reference here...) Example We consider a few examples of hypersurfaces f : (C 3, 0) (C, 0) with isolated singularities: f (1) If f is regular (i.e. z (0) 0), then f is isotopic to (z 1, z 2, z 3 ) p 1, and K f = S 3 S 5 is the trivial (unknotted) embedding It is a deep theorem of Mumford that the converse holds: Theorem 2.53 (Mumford). If K f is diffeomorphic to S 3 (in fact, as long as π 1 (K f ) is trivial), then f is regular. (2) See Example 6.2(ii) above. (3) We may generalize the above example: consider f(z 1, z 2, z 3 ) = z 1 z 2 z n 3, and the map ϕ : C 2 V = {z 1 z 2 = z n 3 } given by ϕ(u, v) = (u n, v n, uv). Then ϕ is a regular n-fold cover C 2 (0, 0) V (0, 0, 0), and Z n acts on C 2 (0, 0) by ξ (u, v) = (ξu ξv; in particular, the restriction of ϕ to S 3 is a Z n -regular covering of our link K f, so that K f = S 3 /Z n = L(n, n 1) (a lens space.) Remark. The fibration S 1 S 3 which we may also think of as C C 2 0 has Euler number -1. S 2 CP 1 20

21 The fibration S 1 RP 3 has Euler number -2. In general, the fibration S 2 S 1 M 3 is an algebraic link iff it has negative Euler number. link. e.g. the trivial fibration S 2 S 1 S 2 S 1 S 2 has Euler number 0 and hence S 2 S 1 is not a algebraic Example Given f : (C 2, 0) (C, 0), F λ : f 1 (λ) B has complex dimension 1 and real dimension 2. K f = F λ is a 1-dimensional compact manifold; as an abstract link it is equal to the disjoint union of r circles S 1 S 1. We (more usually) would like consider the embedded link K f S 3. We will show (in the next section) that the number of components r of the link K f is equal to the number of irreducible components of f. Example Consider g : (C 3, 0) (C, 0) given by g(z 1, z 2, z 3 ) = f(z 1, z 2 ) z3 n (these are known as suspension singularities. K g is a 3-manifold which we may describe: consider the projection map p : V (g) = {f z3 n = 0} C 2 given by p(z 1, z 2, z 3 ) = (z 1, z 2 ). Then (z 1, z 2 ) V (f) = p 1 (z 1, z 2 ) = 0 (z 1, z 2 ) / V (f) = p 1 (z 1, z 2 ) = Z n and so we have a Z n -regular cover V (g) {z 3 = 0} C 2 {f = 0}. V (g) is a branched cover of C 2 with branch locus V (f); K g is a branched cover of S 3 with branch locus K f S 3. Note that K g {z 3 = 0} S 3 K f is a homeomorphism. We have also the monodromy representation π 1 (S 3 K f ) p Z n H 1 (S 3 K f ) = Z r (n 1,...,n r) n i 2.6 February 27: Normalisation of a domain; Newton diagrams We return for a moment to a more algebraic mode 3. (1) Let C{t} be the local ring of convergent power series in a single variable (or consider C[[t]], the ring of formal power series.) We may write a typical element as ϕ(t) = a 0 + a 1 t + a 2 t Note that ϕ(t) is invertible iff a 0 0, and we may write the unique maximal ideal of C{t} as m = {a 1 t + a 2 t }, so that C{t}/m = C. Given ϕ(t) = a m t m + a m+1 t m with a m 0, we write ϕ(t) = t m (a m + a m+1 t +... ) and let φ(t) = a m + a m+1 t From the above we have that φ is invertible, and there exists ψ : C C such that ψ m = φ. Thus we write ϕ(t) = t m φ(t) = (tψ(t)) m and thus we have 3 Because it s good to hear the melodies of the language [of commutative algebra]. 21

22 t ϕ(t) (C, 0) (C, 0) t t (C, 0) t t m where t =: tψ(t); in other words, any ϕ(t) C{t} looks (locally, in suitable coordinates) like a z m. Let x(t), y(t) C{t} be such that x(0) = y(0) = 0. Define n : (C, 0) (C 2, 0) by n(t) = (x(t), y(t)). We expect that there is a f(w, z) C{w, z} such that In = {f = 0}, and we now wish to find this f. Example If (x, y) = (t 2, t 3 ), then In = {w 3 z 2 = 0}. For a slightly more illustrative example, consider n(t) = (t 2, t 3 + t 4 ), so that x(t) = t 2 and y(t) = t 3 + t 4. One way to find a suitable f (which works in general) is to write t 2 x = 0 t 3 + t 4 y = 0 and then compute the resolvent, which is the matrix formed by the coefficients of the first equation repeated deg y(t) times followed by the coefficients of the second equation repeated deg x(t) times. 4 Example Given x = t 2 and y = t, we write t 2 x = 0 t y = 0 and compute the resolvent to be 1 0 x 1 y y = y2 x. (2) Suppose we have n : (C, 0) (C 2, 0) with n(t) = (x(t), y(t)) as above (which will turn out to be a useful parametrisation for the study of algebraic links.) Write Sɛ 3 := { x 2 + y 2 = ɛ} and N ɛ := {t ( x(t) 2 + y(t) 2) ɛ} C is a topological disc (in suitable local coordinates, it is a disc) ; N ɛ = Sɛ 1, and n S 1 ɛ : Sɛ 1 Sɛ 3 is a knot. For ɛ 1, n S 1 ɛ is independent of ɛ. Question: can we characterize the knot from x(t) and y(t) alone? (3) Given f(x, y) C{x, y} irreducible, can we find a paramterisation t (x(t), y(t)) of {f = 0}? In other words, can we find a homeomorphism between the pairs (C, 0) ({f = 0}, 0)? Theorem Given f irreducible, there exists a bijection (in fact a birational map) n : (C, 0) {f = 0} given by t (x(t), y(t)). Two sketches of proof. Geometrically, we have {f = 0} (C 2, 0) ϕ f=0 ϕ and (C, 0) {f = 0} (C 2, 0) Algebraically, we have 4 The resolvent will appear again near the beginning of Section??. (C, 0) 22

23 A = C{x, y}/(f) C{x, y} and C{x,y} (f) C{x, y}? C{t} To determine the? that appears in the last commutative diagram, we define the notion of the normalisation of a domain. If f is irreducible, then (f) is a prime ideal and A/(f) is a domain (i.e. it does not have any nontrivial zero divisors.) Given a domain A, let Q(A) be its field of fractions; we have the natural inclusion A Q(A) given by a a 1. Consider the integral closure of A, defined as Ā := {r Q(A) : r n + a 1 r n a n = 0, a i A} and note that we have an inclusion A Ā since A = {r : r a = 0, a A}. Thus we have the chain of inclusions A Ā Q(A) (note Ā is not always Q(A); this is the case iff A is a UFD.) Thus our theorem reduces to the claim that, in our case with A = C{x, y}/(f), whenever dim Ā = 1 and Ā is normal [in Q(A)] then Ā = C{t}: for then A C{t} and so we can find x(t), y(t) such that f (x(t), y(t)) = 0. ( ) 2 Example For {x 2 y 3 = 0}, A = C{x,y} (x 2 y 3 ). We can write x y = y; if we define r = x y, then we may write r 2 y = 0, and so r Ā; now y = r2 and x = ry = r 3 further imply that A C{r} as desired. Now note that r = x y : V {0} (C, 0) is continuous, and can be continued over the origin by defining its value there to be zero. Hence we normalize our domain of functions by adding the function ϕ : V C defined by { x y if (x, y) (0, 0) ϕ(x, y) = 0 if (x, y) = (0, 0) In short: these paramterisations are given by the implicit function theorem in the smooth case; without smoothness, we can still get near-paramterisations (which are a little wonky at the origin) in the case of algebraic curves. The method of Newton Given f(x, y) = 0, we wish to find an expression of the form y = ϕ(x). (e.g. for f(x, y) = x 2 y 3, we have y 3 = x 2 and so y = x 2/3 ; in general we might have something like y = x 5/7 + 55x 100/ ) In the single-variable case, we can simply look at the ordering of the exponents: e.g. given f(x) = 3x 2 + 5x 5 x 10, the largest power x 10 is dominant if we consider f C[x] (here we are looking at global behavior), but the smallest power 3x 2 is dominant if we consider f C{x} or f C[[x]] (here we are looking at local behavior near the origin.) When we have more variables, this becomes tricker / more subtle: for instance, how should we order xy 2 and x 2 y relative to each other? Newton diagrams provide a method for doing so. Here we demonstrate this in 2 dimensions: Given f(x, y) = i,j a ijx i y j C{x, y} where (i, j) N 2 and a ij 0, the Newton diagram N(f) is the convex closure of a ij 0 ( (i, j) + R 2 0 ) (the intuition being that x 2 y 3, for instance, dominates x 22 y 23 locally, and that convex combinations also do so.) Example If f(x, y) is homogeneous of degree d, then [the boundary of] N(f) is part of the line i + j = d. e.g., the Newton diagram for f(x, y) = y 6 + x 2 y 4 + x 3 y 3 + x 5 y is as follows: 23

24 Example More generally, if f(x, y) is weighted-homogeneous, there exist w x, w y with (w x, w y ) = 1 such that a ij = 0 only if iw x + jw y = d, so that [the boundary of] N(f) lies on the line iw x + jw y = d. e.g., the Newton diagram for f(x, y) = y 6 + x 2 y 3 + x 4 is as follows: Indeed (it is an exercise to show that) if f is weighted-homogeneous, then we may write f in the form f(x, y) = x a y b j (x wy + α j y wx ). Example Consider f(x, y) = y 4 2x 3 y 2 4x 5 y + x 6 x 7. The Newton diagram N(f) has boundary consisting of a single line segment and we may define f (x, y) = (i,j) a ijx i y j where denotes the boundary of the Newton diagram; then f (x, y) = y 4 2x 3 y 2 + x 6 = (y 2 x 3 ) 2, i.e. (y, x) = (x 3 1, x 2 1). Since we do not yet have y = x n with n Z, we introduce a perturbation (y, x) = ( x 3 1(1 + y 1 ), x 2 1) and substitute these into f (x, y) to get for which we get the Newton diagram f 1 (x 1, y 1 ) = x 12 1 (4y y y x 1 4x 1 y 1 x

25 from which f 1 (x 1, y 1 ) = 4y 2 1 4x 1, and so y 2 1 = x 1, and (y 1, x 1 ) = (x 2, x 2 2). Hence we have y = x 3 1(1 + y 1 ) = x 6 2(1 + x 2 ) x = x 2 1 = x 4 2 and so the parametrisation is (x(t), y(t)) = (t 4, t 6 + t 7 ). We can compute invariants such as the Milnor number: given a Newton diagram with a the y-intercept, b the x-intercept, and A the area (in the first quadrant) under the diagram, we have µ = 2A (a + b) 1. Example Given f(x, y) = x 2 y 2 + x 5 + y 5, we have the Newton diagram and so µ = 20 (5 + 5) + 1 = March 4: Puiseux pairs If (C, 0) (C 2, 0) is [the variety of] an irreducible germ [of a holomorphic function f], then there exists η : (C, 0) (C, 0) a (holomorphic) bijection, i.e. n sends t (C, 0) to (x(t), y(t)) such that {f(x(t), y(t)) = 0} = (C, 0), and x(t) and y(t) are holomorphic. Remark. η is invertible as a continous function but not neccesarily as a holomorphic / algebraic function. e.g. C = {x 2 y 3 = 0}; η is given by t x(t) = t 3, y(t) = t 2. Then η 1 : (C, 0) (C, 0), given by { x y if (x, y) (0, 0) (x.y) 0 if x = y = 0 is not holomorphic (algebraic) but meromorphic (rational), or indeed weakly holomorphic. As x(t) and y(t) are holomorphic we may write We may simplify this further by x(t) = a m t m + a m+1 t m y(t) = b n t n + b n+1 t n (1) rewriting x(t) and y(t) using the m th root of the power series of x(t) (or, equivalently, using the implicit function theorem); (2) recalling that the multiplicity µ of C = {f = 0} is the degree of the lowest-degree monomial in f, and µ = # ({l = ɛ} C) for l = αx+βy a generic linear form 5, and that on the other hand # ({l = ɛ} C) = #{t : αx(t) + βy(t) = ɛ}, which is equal to the order of t in αx(t) + βy(t), which is equal to min(m, n); and (3) subtracting any nonzero multiple of t m from the resulting power series of y(t) 5 Since { fm(x, y) + f m+1 (x, y) + = 0 αx + βy = ɛ where the f k represent the homogenous parts of degree k has, locally, n solutions. 25

26 and so we may write x(t) = t m y(t) = c n t n + c n+1 t n (1) where m is the multiplicity of (C, 0) and n > m. Question can we eliminate more terms from the series for y(t) (in particular, can we eliminate all but finitely many of them) if we are only interested in the topology of C and K f? Now let B ɛ be a Milnor ball, and suppose x 2 + y 2 < ɛ. Then Dɛ t := {t C : η(t) B ɛ C} is topologically a disc (i.e. homeomorphic to a disc.) One way to think of this is that we have x 2 + y 2 = t 2m +..., and the first term dominates the local behaviour, so that locally Dɛ t looks (topologically) like t 2m = ɛ. So C B ɛ is homeomorphic to a disc D ɛ, and then the link C S ɛ is homeomorphic to D ɛ = S 1. In particular, given a general holomorphic function f, the number of irreducible components of f is equal to the number of components of the algebraic link K f. Summarising the discussion above, we obtain the following Fact. If f = f 1 f ralg is the irreducible decomposition of f, and (f = 0) Sɛ 3 =: K f = S 1 S 1 is the disjoint union of r top circles, then r alg = r top. In particular, if f is irreducible, L f = S 1 S 3 is an algebraic knot. Thus we see that our parametrisation η, and in particular the power series in (??), are sufficient to characterise L f = K f in this case. Our goal now is to obtain a complete topological classification of algebraic knots / links. This is different from a purely analytic classification: in an analytic classification, we would identify two algebraic knots K f and K g, associated to maps f : C 2 C and g : C 2 C resp., if there exists a locally analytic isomorphism φ : (C 2, 0) (C 2, 0) such that we have (C 2, 0) (C, 0) φ (C 2, 0) g f Note in this case that φ is holomorphic and det dφ 0 0, i.e. a local change-of-variables. On the other hand, in a topological classification we only require φ to be a homeomorphism (and not a rigid diffeomorphism.) In the topological classification we expect something discrete rather than continuious... Example Consider an embedding of 4 lines through the origin L 1 L 2 L 3 L 4 C 2,say f λ (x, y) = xy(x + y)(x + λy). This represents 4 points in CP 1. Now f λ f µ topologically for any pair of λ, µ, since K f = S 1 S 1 is a set of 4 fibers of the Hopf fibration S 1 S 3 S 2 = CP 1 However, we may not have f λ f µ analytically, since the locally analytic isomorphisms are precisely the linear transformations, and these must preserve the cross-ratio. 26

27 So: given a parametrisation of {f = 0} as in (??), we want to ask: which of the (c k ) k n are important for the topological classification? To answer this question we first write, formally, t = x 1/m and so y(x) = a n x n m + an+1 x n+1 m +... (2) and now we want a finite list of important {k/m}: these will be given by the Puiseux pairs (which will be defined presently.) To motivate their definition consider the following examples: Example Suppose we have { x = t 2 y = t 3 + t 4 Consider first the approximation { x = t 2 y = t 3 If we write x = ɛe 2πis where s [0, 1], then we have t = ± ɛe πis and y = ±ɛ 3/2 e 3πis. This gives us our now-familiar (2,3)-torus knot (i.e. the trefoil), as the closure of a braid of two strands moving at speed 3/2. Now let us go back to the original parametrisation. Again with x = ɛe 2πis and t = ± ɛe πis, we now obtain y = ±ɛ 3/2 e 3πis + ɛ 2 e 4πis. We note that ɛ 2 ɛ 3/2 since we have 0 < ɛ 1 by assumption; and we see that we do not get any new strands, and that this can be isotoped to the knot obtained from our previous approximation (intuitively speaking, the additional term corresponds to small perturbations of the strands in the approximation), so that this is topologically the same link as before. Example Consider now { x = t 4 y = t 6 + t 7 The approximation { x = t 4 y = t 6 yields the same knot as before (indeed, it is a double cover of the knot obtained from x = t 2, y = t 3.) When we now go back to the original parametrisation, however, we obtain x = ɛe 2πis, t = 4ɛζe πi/2 where ζ {±1, ±i} is a 4 th root of unity, and y = ±ɛ 3/2 e 3πis + ɛ 7/4 ζ 7 e 4πis. Now the additional term splits each of the two strands in the approximation into two new strands (since there are four choices for ζ); our new knot is a trefoil, but an iterated torus knot (in particular, this one is a (2,7)-torus knot satellite of the trefoil.) Thus we see that each additional term which introduces a new denominator is important. We quantify this precisely thus: given y(x) as in (??), going from left to right, and considering only those pairs (k, m) for which the coefficient of y k/m is nonzero, integer powers are not important,since they represent covers / perturbations of the trivial braid; (n 1, m 1 ) is the first Puiseux pair if (n 1, m 1 ) = 1 and m 1 > 1; The next Puisuex pair is decided by the next n 2 /m 1 m 2 with (n 2, m 2 ) = 1 and m 2 a new denominator. Repeat until the algorithm terminates. The algorithm must eventually terminate since m has finitely many divisors. (Otherwise, the power of every monomial in y(t) is divisible by some factor m 1 of m. Then our link is a m 1 -fold cover; divide through all exponents by m 1, and start again.) 27