David Eklund. May 12, PDF Free Download

KTH Stockholm May 12, 2017 1 / 44

Isolated roots of polynomial systems Let f 1,..., f n C[x 0,..., x n ] be homogeneous and consider the subscheme X P n defined by the ideal (f 1,..., f n ). 2 / 44

Isolated roots of polynomial systems Let f 1,..., f n C[x 0,..., x n ] be homogeneous and consider the subscheme X P n defined by the ideal (f 1,..., f n ). We will be concerned with the isolated points of X and how to compute or approximate them. Suppose that we have a closed subscheme X P n C such that π 1 (0) = X where π : X C is the projection. For example, X = {(x, t) : (1 t)f i (x) + tg i (x) = 0, i = 1,..., n} P n C, where g 1,..., g n are homogeneous and deg(g i ) = deg(f i ). 2 / 44

Isolated roots of polynomial systems For a smooth injective curve α : [0, 1] C, a smooth map l : [0, 1] P n with (l(t), α(t)) X for all t [0, 1] is called a homotopy path of (X, α). The points l(0) and l(1) are called the end point and the start point. For t C let {π 1 (t)} 0 denote the number of isolated points of π 1 (t). Theorem For generic t C, {π 1 (t)} 0 is constant. Call this number m. Then {π 1 (t)} 0 m for all t C and there is a dense open subset U C such that if α : (0, 1] U and α(0) = 0, there are m homotopy paths whose endpoints include all isolated points of X. 3 / 44

Examples Consider the minors q 1 = xz y 2, q 2 = xw yz, and q 3 = yw z 2 of the matrix ( ) x y z. y z w 4 / 44

Examples Consider the minors q 1 = xz y 2, q 2 = xw yz, and q 3 = yw z 2 of the matrix ( ) x y z. y z w The corresponding variety C P 3 is the twisted cubic curve. 4 / 44

Examples Consider the minors q 1 = xz y 2, q 2 = xw yz, and q 3 = yw z 2 of the matrix ( ) x y z. y z w The corresponding variety C P 3 is the twisted cubic curve. If c = i l iq i is a general cubic from the ideal, that is l i are general linear forms, then the subvariety X = V (q 1, q 2, c) P 3 consists of C and one isolated point. 4 / 44

Examples One possible start system that we can easily solve is g 1 = x 2 1, g 2 = y 2 1 and g 3 = z 3 1, which has 2 2 3 = 12 isolated roots and no others. 5 / 44

Examples One possible start system that we can easily solve is g 1 = x 2 1, g 2 = y 2 1 and g 3 = z 3 1, which has 2 2 3 = 12 isolated roots and no others. Let F, G : C 3 C 3, F (p) = (f 1 (p), f 2 (p), f 3 (p)) and G(p) = (g 1 (p), g 2 (p), g 3 (p)). A possible homotopy X C 3 [0, 1] is defined by (1 t)f + tγg = 0, where γ C is a random constant (we will return to this). 5 / 44

Examples Show examples with Bertini (www.math.kth.se/~daek/class/cubic/input). 6 / 44

Examples Show examples with Bertini (www.math.kth.se/~daek/class/cubic/input). Show positive dimensional example (www.math.kth.se/ ~daek/class/cubic_posdim/input). 6 / 44

Examples Show examples with Bertini (www.math.kth.se/~daek/class/cubic/input). Show positive dimensional example (www.math.kth.se/ ~daek/class/cubic_posdim/input). Show homogeneous example (www.math.kth.se/~daek/ class/hom_example/input). 6 / 44

Pathtracking of smooth paths How do we track the paths? Two numerical difficulties: 7 / 44

Pathtracking of smooth paths How do we track the paths? Two numerical difficulties: Singular path tracking, tracking paths along which X is of higher multiplicity. 7 / 44

Pathtracking of smooth paths How do we track the paths? Two numerical difficulties: Singular path tracking, tracking paths along which X is of higher multiplicity. The end point might be singular (of higher multiplicity in X) even though the path is otherwise nonsingular. 7 / 44

Pathtracking of smooth paths The method often used for path tracking is a predictor/corrector method. 8 / 44

Pathtracking of smooth paths The method often used for path tracking is a predictor/corrector method. Eulers method (or similar) for differential equations is used for prediction, for example taking a step in the tangent direction of the path. Newtons method is used for correction to bring the predicted point back to the path. Consider H : C n [0, 1] C n given by polynomials and a smooth path l : [0, 1] C n with H(l(t), t) = 0 for all t. 8 / 44

Pathtracking of smooth paths Using d dth(l(t), t) = 0 and the chain rule we get dh(l(t), t)l (t) + H t (l(t), t) = 0, for all t, where H t is the partial derivative wrt t. 9 / 44

Pathtracking of smooth paths Using d dth(l(t), t) = 0 and the chain rule we get dh(l(t), t)l (t) + H t (l(t), t) = 0, for all t, where H t is the partial derivative wrt t. Hence l (t) = dh(l(t), t) 1 H t (l(t), t) and we can use this for prediction. The first order approximation H(x + x, t) = H(x, t) + dh(x, t) x, with H(x + x, t) = 0 given Newtons method: x = dh(x, t) 1 H(x, t). 9 / 44

Complexity basics Gauss Jordan elimination, which can be used for the matrix inversion in path tracking has complexity O(n 3 ) (one can actually do better but we will not go into this). 10 / 44

Complexity basics Gauss Jordan elimination, which can be used for the matrix inversion in path tracking has complexity O(n 3 ) (one can actually do better but we will not go into this). Except for this the important elements of complexity is how many matrices have to be inverted and the number of paths to track. We will focus on the second issue here. Interesting work relating to the first issue and the length of paths may be found in various papers by Shub and Smale on the Complexity of Bézout s theorem. As we shall see for the number of paths we can expect something like O(d n ), where d is the degree of the defining equations. 10 / 44

Software for numerical Bertini: Bates, Hauenstein, Sommese, Wampler (https://bertini.nd.edu). 11 / 44

Software for numerical Bertini: Bates, Hauenstein, Sommese, Wampler (https://bertini.nd.edu). PHCpack: Verschelde (http: //homepages.math.uic.edu/~jan/download.html). 11 / 44

Software for numerical Bertini: Bates, Hauenstein, Sommese, Wampler (https://bertini.nd.edu). PHCpack: Verschelde (http: //homepages.math.uic.edu/~jan/download.html). Hom4PS-3: Chen, Lee, Li, Ovenhouse (http://www. hom4ps3.org/store/c1/featured_products.html) 11 / 44

Bézout s theorem Let f 1,..., f n C[x 0,..., x n ] be homogeneous polynomials and let X P n be the subscheme defined by the ideal (f 1,..., f n ). Put d i = deg(f i ) and let n b = d i, i=1 the Bézout number or total degree. 12 / 44

Bézout s theorem Let f 1,..., f n C[x 0,..., x n ] be homogeneous polynomials and let X P n be the subscheme defined by the ideal (f 1,..., f n ). Put d i = deg(f i ) and let b = n d i, i=1 the Bézout number or total degree. Theorem Then X has at most b isolated points. If X is finite, b is the number of points counted with multiplicity. If f 1,..., f n are generic of the given degrees, X consists of b points of multiplicity 1. 12 / 44

Examples Two generic curves in P 2 of degree d and e, respectively, intersect in de points of multiplicity 1. 13 / 44

Examples Two generic curves in P 2 of degree d and e, respectively, intersect in de points of multiplicity 1. A smooth conic and a line in P 2 generally intersect in two points, but if the line is tangent to the conic the intersection is 1 point of multiplicity 2. A general line through the singular point of the nodal cubic zy 2 = x 2 (x z) intersects in 1 smooth point and a point of multiplicity 2. In example above two quadrics and a cubic surface in P 3 intersect in the twisted cubic curve and one point of multiplicity 1. The Bézout number is 12 and so the curve eats up 11 points. 13 / 44

Total degree homotopy Let X C n be defined by f 1,..., f n C[x 1,..., x n ] with d i = deg(f i ). 14 / 44

Total degree homotopy Let X C n be defined by f 1,..., f n C[x 1,..., x n ] with d i = deg(f i ). One of the simplest is X C n C is defined by (1 tγ 1 )f i (x) + t(x d i i 1) = 0, where γ S 1 C. 14 / 44

Total degree homotopy Let X C n be defined by f 1,..., f n C[x 1,..., x n ] with d i = deg(f i ). One of the simplest is X C n C is defined by (1 tγ 1 )f i (x) + t(x d i i 1) = 0, where γ S 1 C. According to the above theorem (in the affine case), the end points of the Π n i=1 d i solution paths include the isolated points of X if α : (0, 1] C : s sγ avoids a finite number of points in C. This is true for all but finitely many γ and hence if γ S 1 is chosen randomly true with probability 1. 14 / 44

Other approaches As we have seen, if X has special structure the number of isolated points might be lower than the Bézout s bound. 15 / 44

Other approaches As we have seen, if X has special structure the number of isolated points might be lower than the Bézout s bound. For example, consider the affine case and a sparse system f 1,..., f n. We then have the BKK-bound given by the mixed volume of the Newton polytopes of f 1,..., f n, which is a bound for the number of isolated roots in (C ) n. Mixed volume is the basis for other with fewer paths to track but with the added expense of computing the mixed volume. 15 / 44

Exercises Compute the intersection of the curves x 4 + y 4 = 1 and x 2 + y 2 + xy = 2 in C 2. Compute the singular locus of the cubic surface xyz + xyw + xzw + yzw = 0 in P 3. What is the degree of the dual of the cubic xyz + xyw + xzw + yzw = 0? Compare to a smooth cubic. Find the closest point on the curve x 4 + x 2 y 2 + y 4 = x(x 2 + y 2 ) in R 2 from the point (2/3, 0). Consider graph below and its embeddings in R 2 (up to rigid motion) with unit edge lengths. How many such embeddings can you find? 16 / 44

Exercises What is the minimum of the function x 3 + y 3 + z 3 y + x on the variety in R 3 defined by the polynomials x 4 + 3x 2 y 2 + y 4 + z 4 1, (9/7)x 2 + 2xy + (9/2)y 2 + (3/2)xz + yz + (4/9)z 2 1? Compute the lines on a smooth cubic surface in P 3. What about the lines on a general quintic hypersurface in P 4? And a heptic in P 5? 17 / 44

P n P m Let us look at the case of X P n P m defined by f 1,..., f n+m C[x 0,..., x n, y 0,..., y m ] with f i of bidegree (d i, e i ). 18 / 44

P n P m Let us look at the case of X P n P m defined by f 1,..., f n+m C[x 0,..., x n, y 0,..., y m ] with f i of bidegree (d i, e i ). The f i could come from n + m polynomials in C[x 1,..., x n, y 1,..., y m ] homogenized a certain way. One approach to counting the number of isolated points of X is intersection theory. For this we need the intersection ring A(Y ) of a smooth projective variety Y. The ring A(Y ) consists of cycles on Y modulo rational equivalence and is equipped with an intersection product. 18 / 44

P n P m Rational equivalence corresponds to deformation of subvarieties of Y in a certain sense. For example, if W Y P 1 is a subvariety that dominates P 1, π 1 W (s) and π 1 W (t) are rationally equivalent for s, t P1. We define the Chow group as A(Y ) = Z(Y )/R(Y ), where R(Y ) is the subgroup of cycles rationally equivalent to 0. 19 / 44

P n P m For a subvariety C Y we will write [C] A(Y ) for the class r i=1 C i where C 1,..., C r are the components of C. 20 / 44

P n P m For a subvariety C Y we will write [C] A(Y ) for the class r i=1 C i where C 1,..., C r are the components of C. To define the intersection product first consider the case where of subvarieties A, B Y which intersect generically transversely. That is for each component C of A B, the projective tangentspaces T p A and T p B meet transversely in T p Y at a generic point p C. 20 / 44

P n P m The classical approach to dealing with non-proper non-transversal intersections is the moving lemma: for cycle classes α, β A(Y ) there are cycles i n ia i and i m ib i representing α and β such that A i and B j meet generically transversely for all i and j. The intersection product then has to be αβ = i,j n i m i [A i B j ]. That it does not depend on A and B is part of the moving lemma. 21 / 44

P n P m In the cases Y = P n and Y = P n P m the moving lemma is easier to prove as we then have many automorphisms acting on Y. 22 / 44

P n P m In the cases Y = P n and Y = P n P m the moving lemma is easier to prove as we then have many automorphisms acting on Y. For example, if Y = P n we have the group PGL(n + 1) which acts transversely on P n and the moving lemma follows from Kleimans theorem. 22 / 44

P n P m We have that A(P n ) = Z[h]/(h n+1 ) where h is the class of a hyperplane. 23 / 44

P n P m We have that A(P n ) = Z[h]/(h n+1 ) where h is the class of a hyperplane. The class of a hypersurface H P n of degree d is dh. The product of the classes of n hypersurfaces H 1,..., H n with deg(h i ) = d i is n i=1 d ih = h n n i=1 d i, the Bézout number times the class of a point h n. For P n P m we have that A(P n P m ) = Z[h 1, h 2 ]/(h n+1 1, h m+1 2 ), where h 1 and h 2 are the classes of H 1 P m and P n H 2 for hyperplanes H 1 P n and H 2 P m. 23 / 44

P n P m Return to the bihomogeneous polynomials f 1,..., f n+m with f i of bidegree (d i, e i ). The hypersurface in P n P m defined by f i has class d i h 1 + e i h 2. 24 / 44

P n P m Return to the bihomogeneous polynomials f 1,..., f n+m with f i of bidegree (d i, e i ). The hypersurface in P n P m defined by f i has class d i h 1 + e i h 2. Taking the product of these n + m classes we get (d i h 1 + e i h 2 ) = kh n 1 h m 2, n+m i=1 for some integer k depending on (d i, e i ) which is called the multihomogeneous Bézout number. 24 / 44

P n P m If the f i s are generic of the given bidegree, X is finite and the number of points is the multihomogeneous Bézout number. 25 / 44

P n P m If the f i s are generic of the given bidegree, X is finite and the number of points is the multihomogeneous Bézout number. can be based on this multihomogeneous root count, which generally is smaller than the total degree, and therefore there are less paths to follow. For example if d i = d and e i = 1 for all i, the multihomogeneous Bézout number is ( ) n+m n d n whereas the Bézout number is (d + 1) n+m. 25 / 44

Example To illustrate consider P 1 P 1. For context, observe that the product may be embedded as the smooth quadric surface {xw yz = 0} P 3 via the Segre embedding P 1 P 1 P 3 : (a, b; c, d) (ac, ad, bc, bd). The quadric is ruled by lines. In fact it has two rulings given by {p} P 1 and P 1 {q}. Distinct lines from the same ruling are disjoint while lines from different rulings meet at a point (the pair of lines is the intersection of the quadric with the tangent space at that point). 26 / 44

Example Show example with Bertini (www.math.kth.se/~daek/class/quadric/input). 27 / 44

Example Show example with Bertini (www.math.kth.se/~daek/class/quadric/input). Show example with Bézout bound (www.math.kth.se/ ~daek/class/quadric_bezout/input). 27 / 44

Example Let X P 3 be a smooth cubic surface defined by F = 0. We will compute the 27 lines on the cubic in a simple way. 28 / 44

Example Let X P 3 be a smooth cubic surface defined by F = 0. We will compute the 27 lines on the cubic in a simple way. Let H 1, H 2 P 3 be generic planes. Then every line on X meets H 1 and H 2 in one point each, away from the line H 1 H 2. We can set up equations on P 2 P 2 for the lines on X, letting (x, y) P 2 P 2 and require that the line through x and y is contained in X. 28 / 44

Example Take for example the Fermat cubic F = x 3 + y 3 + z 3 + w 3 = 0. Show cubic surface example (www.math.kth.se/~daek/class/cubic_surf). 29 / 44

Example Take for example the Fermat cubic F = x 3 + y 3 + z 3 + w 3 = 0. Show cubic surface example (www.math.kth.se/~daek/class/cubic_surf). The multihomogeneous root count is 45 which is smaller than the the Bézout number 3 4 = 81. But there are only 27 lines on X. There are some fake lines among the solutions, namely points with x = y and x X. The line H 1 H 2 intersects X in 3 points and each has multiplicity 6 in the intersection on P 2 P 2. This gives 3 6 = 18 = 45 27 extra solutions. 29 / 44

Chow group of a quadric Consider the smooth quadric Q P 7. 30 / 44

Chow group of a quadric Consider the smooth quadric Q P 7. If Q has an affine stratification then A(Q) is free and a basis is given by the closures of the strata. 30 / 44

Chow group of a quadric Consider the smooth quadric Q P 7. If Q has an affine stratification then A(Q) is free and a basis is given by the closures of the strata. We want to find a cell-decomposition Q = k i U i of locally closed subsets each isomorphic to an affine space such that Ūi is a union of cells for all i. 30 / 44

Chow group of a quadric As decomposition we can take: U 0 = {x 0 0} Q (dim 6) 31 / 44

Chow group of a quadric As decomposition we can take: U 0 = {x 0 0} Q (dim 6) U 1 = {x 0 = 0, x 1 0} Q (dim 5) 31 / 44

Chow group of a quadric As decomposition we can take: U 0 = {x 0 0} Q (dim 6) U 1 = {x 0 = 0, x 1 0} Q (dim 5) U 2 = {x 0 = x 1 = 0, x 2 0} Q (dim 4) 31 / 44

Chow group of a quadric As decomposition we can take: U 0 = {x 0 0} Q (dim 6) U 1 = {x 0 = 0, x 1 0} Q (dim 5) U 2 = {x 0 = x 1 = 0, x 2 0} Q (dim 4) U 3 = {x 0 = x 1 = x 2 = x 4 = 0, x 3 0} and U 4 = {x 0 = x 1 = x 2 = x 3 = 0, x 4 0} (dim 3) U 5 = {x 0 =... = x 4 = 0, x 5 0} (dim 2) 31 / 44

Chow group of a quadric Lets look at A 3 (Q) and the basis g 1 = [Ū3] and g 2 = [Ū4]. 32 / 44

Chow group of a quadric Lets look at A 3 (Q) and the basis g 1 = [Ū3] and g 2 = [Ū4]. Let f be the involution of Q given by x i x 7 i. This induces an automorphism of A(Q). 32 / 44

Chow group of a quadric Lets look at A 3 (Q) and the basis g 1 = [Ū3] and g 2 = [Ū4]. Let f be the involution of Q given by x i x 7 i. This induces an automorphism of A(Q). You can check that Ū3 f(ū3) = Ū4 f(ū4) =. Moreover, Ū 3 and f(ū4) intersect transversely in one point. 32 / 44

Chow group of a quadric It is interesting to note that the decomposition above is induced by an action of of the multiplicative group C on Q. 33 / 44

Chow group of a quadric It is interesting to note that the decomposition above is induced by an action of of the multiplicative group C on Q. Define C Q Q by (t, x) tx = (x 0, tx 1, t 2 x 2, t 3 x 3, t 3 x 4, t 4 x 5, t 5 x 6, t 6 x 7 ). The fixed points of this action are the coordinate vertices p 0 = (1, 0,..., 0), p 1 = (0, 1, 0,..., 0) and so on. 33 / 44

Example Consider the Segre embedding s : P 1 P 1 P 1 P 7. We will make sure that the image S is contained in Q by letting s(a 0, a 1, b 0, b 1, c 0, c 1 ) = a 0 b 0 c 0 a 1 b 1 c 0 a 1 b 0 c 1 a 0 b 1 c 1 a 1 b 0 c 0 a 0 b 1 c 0 a 0 b 0 c 1 a 1 b 1 c 1. 34 / 44

Example We want to compute deg([s] 2 ): deg((ag 1 + bg 2 ) 2 ) = deg(a 2 g 2 1 + 2abg 1 g 2 + b 2 g 2 2) = 2ab. 35 / 44

Example We want to compute deg([s] 2 ): deg((ag 1 + bg 2 ) 2 ) = deg(a 2 g 2 1 + 2abg 1 g 2 + b 2 g 2 2) = 2ab. The intersection Ū4 S is transverse and consists of 4 points with following coordinates equal to 0: (a 0, b 0, c 0 ), (a 0, b 1, c 1 ), (a 1, b 0, c 1 ), (a 1, b 1, c 0 ). 35 / 44

Example We want to compute deg([s] 2 ): deg((ag 1 + bg 2 ) 2 ) = deg(a 2 g 2 1 + 2abg 1 g 2 + b 2 g 2 2) = 2ab. The intersection Ū4 S is transverse and consists of 4 points with following coordinates equal to 0: This means that (a 0, b 0, c 0 ), (a 0, b 1, c 1 ), (a 1, b 0, c 1 ), (a 1, b 1, c 0 ). 4 = deg([s] g 2 ) = deg(ag 1 g 2 + bg 2 2) = a. 35 / 44

Example Moreover, S P 7 has degree 3! = 6. 36 / 44

Example Moreover, S P 7 has degree 3! = 6. Let h A 5 (Q) be the hyperplane class. Then 6 = deg(h 3 [S]) = a deg(h 3 g 1 ) + b deg(h 3 g 2 ) = a + b, and hence b = 2. 36 / 44

Example Moreover, S P 7 has degree 3! = 6. Let h A 5 (Q) be the hyperplane class. Then 6 = deg(h 3 [S]) = a deg(h 3 g 1 ) + b deg(h 3 g 2 ) = a + b, and hence b = 2. Since [S] = 4g 1 + 2g 2 we have that [S] 2 = 2 4 2 = 16. We took a special embedding of P 1 P 1 P 1 in this example but there is a linear group SO 8 (C) that acts transitively on Q and [f(s)] = [S] for all f SO 8 (C). 36 / 44

Parameter Note that if we find the isolated points of a generic enough member π 1 (t) of the family X, namely one that has the generic number m of isolated points, we can solve for the isolated points of any other member of the family by tracking m paths. This means that you only have to solve the generic problem once and then track the optimal number of paths to solve any other problem in the family. 38 / 44

Squaring systems What about non-square systems? That is a scheme X P n defined by a homogeneous ideal I = (f 1,..., f k ) C[x 0,..., x n ]. 39 / 44

Squaring systems What about non-square systems? That is a scheme X P n defined by a homogeneous ideal I = (f 1,..., f k ) C[x 0,..., x n ]. If k < n, X has no isolated points. 39 / 44

Squaring systems What about non-square systems? That is a scheme X P n defined by a homogeneous ideal I = (f 1,..., f k ) C[x 0,..., x n ]. If k < n, X has no isolated points. If k > n we can square the system, that is pick n elements from the ideal I. In connection with this we have the following Bertini type theorem. Theorem Let d i = deg(f i ) and put d = max{d 1,..., d k }. For general elements g 1,..., g n I of degree d, the isolated points of X are included in the isolated points of the scheme Y P n defined by the ideal (g 1,..., g n ). In fact, set theoretically, Y is the union of X and finitely many points. 39 / 44

Squaring systems Proof sketch: Let h 0,..., h r be the polynomials x d d i j f i for j {0,..., n} and i {1,..., k}. The scheme X is cut out by the ideal generated by h 0,..., h r as well. Now consider the blow up Bl X P n of P n in X. The morphism P n \ X P r : x (h 0 (x),..., h r (x)) extends to a morphism φ : Bl X P n P r. Since Bl X P n is irreducible of dimension n, φ(bl X P n ) is irreducible of dimension e n. 40 / 44

Squaring systems Thus for the scheme Y defined by n general linear combinations of h 0,..., h r we have that Y \ X = φ 1 (L) where L is a general linear space of codimension n. If e < n, φ 1 (L) = and if e = n, φ is generically finite-to-one and L φ(bl X P n ) is finite. In the latter case, φ 1 (L) is finite. This shows the statement for a general member of a particular subspace < h 0,..., h r > I d of the degree d elements of the ideal I = (f 1,..., f k ). To get the statement in the theorem you can for example do the same thing for the whole of I d. 41 / 44

Squaring systems Observe that in general the multiplicity of a point in X might change as we square the system. 42 / 44

Squaring systems Observe that in general the multiplicity of a point in X might change as we square the system. Consider the set of reducible conics in P 2 consisting of two lines that meet at (0, 0, 1). Their degree 2 equations are spanned {x 2, y 2, xy}. The intersection of two generic such conics is (0, 0, 1), with multiplicity 4 by Bézout s theorem. But the ideal (x 2, y 2, xy) C[x, y, z] defines the point (0, 0, 1) with multiplicity 3 (you can check this with Macaulay2 or compute the Hilbert polynomial by hand). 42 / 44

Squaring systems In the affine case we get the following. Let X C n be defined by an ideal I = (f 1,..., f k ) C[x 1,..., x n ]. 43 / 44

Squaring systems In the affine case we get the following. Let X C n be defined by an ideal I = (f 1,..., f k ) C[x 1,..., x n ]. Theorem For general linear combinations g 1,..., g n of f 1,..., f k, the isolated points of X are included in the isolated points of the scheme Y C n defined by the ideal (g 1,..., g n ). 43 / 44

Making your own The software Bertini allows you to easily formulate your own. 44 / 44

Making your own The software Bertini allows you to easily formulate your own. Show user homotopy (www.math.kth.se/~daek/class/user/input). 44 / 44

David Eklund. May 12, 2017