Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 Euclidean and Algebraic Geometry Unit Spheres
A Combinatorial Moduli Space David A. Cox Counting Balanced Subspaces Department of Mathematics and Statistics
Two Quickies, Amherst College If Time [email protected] Permits Combinatorial Geometry Problems at the Algebraic Interface IPAM, March 24-28, 2014 Overview
Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 Unit Spheres This lecture will give examples of how algebraic geometry A can be applied to problems in Euclidean geometry. Combinatorial Moduli Space
Counting Balanced Key points: Subspaces
Two Quickies, Algebraic geometry can be a useful tool, but ... If Time Permits There are limitations. Overview
Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 Unit Spheres This lecture will give examples of how algebraic geometry A can be applied to problems in Euclidean geometry. Combinatorial Moduli Space
Counting Balanced Key points: Subspaces
Two Quickies, Algebraic geometry can be a useful tool, but ... If Time Permits There are limitations. Overview
Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 Unit Spheres This lecture will give examples of how algebraic geometry A can be applied to problems in Euclidean geometry. Combinatorial Moduli Space
Counting Balanced Key points: Subspaces
Two Quickies, Algebraic geometry can be a useful tool, but ... If Time Permits There are limitations. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Algebraic Geometry
Euclidean and Algebraic Geometry In its simplest form, algebraic geometry is the study of David A. Cox geometric objects defined by algebraic equations, i.e., Tangents to 4 Unit Spheres polynomials.
A Combinatorial Algebraic geometry is good at: Moduli Space Counting (enumerative algebraic geometry) Counting Balanced Subspaces Giving Structure to Interesting Sets (moduli spaces) Two Quickies, Understanding Structure (minimal model program) If Time Permits The third bullet is beyond the scope of this talk. Later I will give several examples of the second bullet. Let me begin with a classic example of the first bullet. Bezout’s Theorem
Euclidean and Algebraic Let C and D be irreducible plane curves defined by Geometry equations of degrees m and n. Then David A. Cox C ∩ D Tangents to 4 Unit Spheres consists of mn points. A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits Bezout’s Theorem
Euclidean and Algebraic Let C and D be irreducible plane curves defined by Geometry equations of degrees m and n. Then David A. Cox C ∩ D Tangents to 4 Unit Spheres consists of mn points. A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits Fine Print, Part 1
Euclidean and Algebraic Geometry
David A. Cox Count with multiplicity: Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits Work over C: (This will be very important.) Fine Print, Part 2
Euclidean and Algebraic 2 Geometry The curves y = x and x = 1 should meet in 2 1 = 2 points. David A. Cox
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
We need points at ∞! This is why projective space is so important in algebraic geometry. Examples From Euclidean Geometry
Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 1 Tangents to 4 Unit Spheres Unit Spheres
A Combinatorial Moduli Space 2 A Combinatorial Moduli Space Counting Balanced Subspaces
Two Quickies, 3 Counting Balanced Subspaces If Time Permits
4 Two Quickies, If Time Permits Tangents to 4 Unit Spheres
Euclidean and Algebraic Geometry Given four unit spheres (= spheres of radius 1) in R3, how David A. Cox many lines can be simultaneously tangent to all three?
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
How many other tangent lines could there be? Setup
Euclidean and Algebraic Spheres: Assume the spheres have centers 0,c1,c2,c3 Geometry that are not coplanar. David A. Cox Lines: Write a line ℓ in R3 as: Tangents to 4 Unit Spheres ℓ = p + λs | λ ∈ R , where A Combinatorial Moduli Space p = offset vector (unique) Counting s = direction vector (unique up to constant multiple) Balanced Subspaces p s = 0. l Two Quickies, If Time Permits s p
0
ℓ is tangent to the 4 unit spheres ⇐⇒ ℓ has distance 1 to 0,c1,c2,c3. A New Problem
Euclidean and Algebraic When is a line ℓ equidistant from 0,c1,c2,c3? Geometry c2 David A. Cox l c1
Tangents to 4 Unit Spheres s c3 p A Combinatorial Moduli Space 0 Counting Balanced Subspaces
Two Quickies, If Time λ Permits Distance from 0 to ℓ = {p + s} is ||p|| since p s = 0.
Distance from ci to ℓ is ||p|| ⇐⇒ 1 c p = ||c × s||2. i 2||s||2 i Equidistant lines satisfy three linear equations! A New Problem
Euclidean and Algebraic When is a line ℓ equidistant from 0,c1,c2,c3? Geometry c2 David A. Cox l c1
Tangents to 4 Unit Spheres s c3 p A Combinatorial Moduli Space 0 Counting Balanced Subspaces
Two Quickies, If Time λ Permits Distance from 0 to ℓ = {p + s} is ||p|| since p s = 0.
Distance from ci to ℓ is ||p|| ⇐⇒ 1 c p = ||c × s||2. i 2||s||2 i Equidistant lines satisfy three linear equations! A New Problem
Euclidean and Algebraic When is a line ℓ equidistant from 0,c1,c2,c3? Geometry c2 David A. Cox l c1
Tangents to 4 Unit Spheres s c3 p A Combinatorial Moduli Space 0 Counting Balanced Subspaces
Two Quickies, If Time λ Permits Distance from 0 to ℓ = {p + s} is ||p|| since p s = 0.
Distance from ci to ℓ is ||p|| ⇐⇒ 1 c p = ||c × s||2. i 2||s||2 i Equidistant lines satisfy three linear equations! A Theorem
Euclidean and Algebraic Solving these equations by Cramer’s rule, we obtain: Geometry David A. Cox Theorem Tangents to 4 Suppose that the line ℓ = {p + λs} is equidistant from the Unit Spheres
A centers 0,c1,c2,c3. Define the vectors Combinatorial Moduli Space n1 = c2 × c3 Counting Balanced Subspaces n2 = c3 × c1 Two Quickies, n3 = c1 × c2. If Time Permits Then 3 1 2 p = 2 ∑ ||ci × s|| ni , 2||s|| D i=1
where D = det(c1,c2,c3) = 0 (non-coplanar). Key Point #1
Euclidean and Algebraic If ℓ is equidistant from the centers, the offset vector is Geometry 3 David A. Cox 1 2 p = 2 ∑ ||ci × s|| ni . Tangents to 4 2||s|| D i=1 Unit Spheres
A Combinatorial Then p s = 0 gives the equation in s: Moduli Space
Counting 3 Balanced 2 Subspaces ∑ ||ci × s|| s ni = 0. i=1 Two Quickies, If Time Permits The direction vector s is unique up to multiplication by a nonzero scalar and hence gives a point in the projective plane. Thus we have proved:
equidistant lines ←→ points on a cubic plane curve. Key Point #1
Euclidean and Algebraic If ℓ is equidistant from the centers, the offset vector is Geometry 3 David A. Cox 1 2 p = 2 ∑ ||ci × s|| ni . Tangents to 4 2||s|| D i=1 Unit Spheres
A Combinatorial Then p s = 0 gives the equation in s: Moduli Space
Counting 3 Balanced 2 Subspaces ∑ ||ci × s|| s ni = 0. i=1 Two Quickies, If Time Permits The direction vector s is unique up to multiplication by a nonzero scalar and hence gives a point in the projective plane. Thus we have proved:
equidistant lines ←→ points on a cubic plane curve. Key Point #2
Euclidean and Algebraic Geometry λ David A. Cox When ℓ = {p + s} is equidistant, the distance is
Tangents to 4 3 Unit Spheres 1 2 ||p|| = 2 ∑ ||ci × s|| ni A 2||s|| D i 1 Combinatorial = Moduli Space
Counting where D = det(c1,c2,c3). Balanced Subspaces
Two Quickies, Hence the distance is 1 ⇐⇒ If Time Permits 3 2 4 2 2 4||s|| D = ∑ ||ci × s|| ni
i=1
This defines a plane curve of degree 4. Key Point #2
Euclidean and Algebraic Geometry λ David A. Cox When ℓ = {p + s} is equidistant, the distance is
Tangents to 4 3 Unit Spheres 1 2 ||p|| = 2 ∑ ||ci × s|| ni A 2||s|| D i 1 Combinatorial = Moduli Space
Counting where D = det(c1,c2,c3). Balanced Subspaces
Two Quickies, Hence the distance is 1 ⇐⇒ If Time Permits 3 2 4 2 2 4||s|| D = ∑ ||ci × s|| ni
i=1
This defines a plane curve of degree 4. At Most 12 Tangents!
Euclidean and Algebraic Geometry Theorem (MacDonald-Pach-Theobald, 2001) David A. Cox There are at most 12 lines tangent to 4 unit spheres when Tangents to 4 the centers are not collinear. Unit Spheres
A Combinatorial Proof for non-coplanar centers. If ℓ = {p + λs} is tangent Moduli Space to unit spheres with centers 0 c c c , then in the Counting , 1, 2, 3 Balanced projective plane, the direction vector s lies in Subspaces
Two Quickies, If Time degree 3 curve degree 4 curve. Permits \ equidistant distance 1 | {z } | {z } By Bezout, this intersection has 12 points over C. QED Note that {direction vectors}/rescaling is more than a just a set: it is the projective plane, which is an algebraic variety. At Most 12 Tangents!
Euclidean and Algebraic Geometry Theorem (MacDonald-Pach-Theobald, 2001) David A. Cox There are at most 12 lines tangent to 4 unit spheres when Tangents to 4 the centers are not collinear. Unit Spheres
A Combinatorial Proof for non-coplanar centers. If ℓ = {p + λs} is tangent Moduli Space to unit spheres with centers 0 c c c , then in the Counting , 1, 2, 3 Balanced projective plane, the direction vector s lies in Subspaces
Two Quickies, If Time degree 3 curve degree 4 curve. Permits \ equidistant distance 1 | {z } | {z } By Bezout, this intersection has 12 points over C. QED Note that {direction vectors}/rescaling is more than a just a set: it is the projective plane, which is an algebraic variety. At Most 12 Tangents!
Euclidean and Algebraic Geometry Theorem (MacDonald-Pach-Theobald, 2001) David A. Cox There are at most 12 lines tangent to 4 unit spheres when Tangents to 4 the centers are not collinear. Unit Spheres
A Combinatorial Proof for non-coplanar centers. If ℓ = {p + λs} is tangent Moduli Space to unit spheres with centers 0 c c c , then in the Counting , 1, 2, 3 Balanced projective plane, the direction vector s lies in Subspaces
Two Quickies, If Time degree 3 curve degree 4 curve. Permits \ equidistant distance 1 | {z } | {z } By Bezout, this intersection has 12 points over C. QED Note that {direction vectors}/rescaling is more than a just a set: it is the projective plane, which is an algebraic variety. 12 Real Tangents
Euclidean and Algebraic For 12 lines tangent to four unit spheres, all 12 can be real: Geometry
David A. Cox
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
Picture by F. Sottile and T. Theobald In this picture, the spheres are not disjoint. An Open Question
Euclidean and Algebraic If the spheres are allowed to have unequal radii, then we Geometry can have 12 real lines with disjoint spheres: David A. Cox
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
Picture by F. Sottile and T. Theobald Open Question Can four disjoint unit spheres have 12 real tangent lines? An Open Question
Euclidean and Algebraic If the spheres are allowed to have unequal radii, then we Geometry can have 12 real lines with disjoint spheres: David A. Cox
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
Picture by F. Sottile and T. Theobald Open Question Can four disjoint unit spheres have 12 real tangent lines? A Combinatorial Moduli Space
Euclidean and Algebraic Geometry 2 A line arrangement An = {L ,...,Ln} in P is combinatorially David A. Cox 1 equivalent to Bn, written An ∼ Bn, if there is an inclusion- Tangents to 4 Unit Spheres preserving bijection between the their intersection lattices
A formed by all possible intersections of the lines. Combinatorial Moduli Space B B A Counting The moduli space MAn = { n | n ∼ n}/PGL(3) Balanced Subspaces measures the difference between combinatorial and
Two Quickies, projective equivalence. If Time Permits These moduli spaces are well understood for n ≤ 10.
We will assume that An has only double and triple points and that every line of An contains at least three triple points. A Combinatorial Moduli Space
Euclidean and Algebraic Geometry 2 A line arrangement An = {L ,...,Ln} in P is combinatorially David A. Cox 1 equivalent to Bn, written An ∼ Bn, if there is an inclusion- Tangents to 4 Unit Spheres preserving bijection between the their intersection lattices
A formed by all possible intersections of the lines. Combinatorial Moduli Space B B A Counting The moduli space MAn = { n | n ∼ n}/PGL(3) Balanced Subspaces measures the difference between combinatorial and
Two Quickies, projective equivalence. If Time Permits These moduli spaces are well understood for n ≤ 10.
We will assume that An has only double and triple points and that every line of An contains at least three triple points. A Combinatorial Moduli Space
Euclidean and Algebraic Geometry 2 A line arrangement An = {L ,...,Ln} in P is combinatorially David A. Cox 1 equivalent to Bn, written An ∼ Bn, if there is an inclusion- Tangents to 4 Unit Spheres preserving bijection between the their intersection lattices
A formed by all possible intersections of the lines. Combinatorial Moduli Space B B A Counting The moduli space MAn = { n | n ∼ n}/PGL(3) Balanced Subspaces measures the difference between combinatorial and
Two Quickies, projective equivalence. If Time Permits These moduli spaces are well understood for n ≤ 10.
We will assume that An has only double and triple points and that every line of An contains at least three triple points. Moduli for A10
Euclidean and Algebraic For n = 10, there are 18 cases where the moduli space is Geometry disconnected: David A. Cox 1 moduli space of dimension one. Tangents to 4 Unit Spheres 10 moduli spaces of dimension zero that cannot be
A realized over R. Combinatorial Moduli Space 7 moduli spaces of dimension zero realizable over R, Counting three of which have exactly two points. Balanced Subspaces A B Two Quickies, For a moduli spaces of the form { 10, 10}, a natural If Time Permits question to study is whether the pairs
2 2 (P ,A10) and (P ,B10)
are homeomorphic. (They are.)
Combinatorial Symmetry of Line Arrangements and Applications, by Amram, Cohen, Sun, Teicher, Ye & Zarkh, arXiv:1310.0700 Moduli for A10
Euclidean and Algebraic For n = 10, there are 18 cases where the moduli space is Geometry disconnected: David A. Cox 1 moduli space of dimension one. Tangents to 4 Unit Spheres 10 moduli spaces of dimension zero that cannot be
A realized over R. Combinatorial Moduli Space 7 moduli spaces of dimension zero realizable over R, Counting three of which have exactly two points. Balanced Subspaces A B Two Quickies, For a moduli spaces of the form { 10, 10}, a natural If Time Permits question to study is whether the pairs
2 2 (P ,A10) and (P ,B10)
are homeomorphic. (They are.)
Combinatorial Symmetry of Line Arrangements and Applications, by Amram, Cohen, Sun, Teicher, Ye & Zarkh, arXiv:1310.0700 Moduli for A10
Euclidean and Algebraic For n = 10, there are 18 cases where the moduli space is Geometry disconnected: David A. Cox 1 moduli space of dimension one. Tangents to 4 Unit Spheres 10 moduli spaces of dimension zero that cannot be
A realized over R. Combinatorial Moduli Space 7 moduli spaces of dimension zero realizable over R, Counting three of which have exactly two points. Balanced Subspaces A B Two Quickies, For a moduli spaces of the form { 10, 10}, a natural If Time Permits question to study is whether the pairs
2 2 (P ,A10) and (P ,B10)
are homeomorphic. (They are.)
Combinatorial Symmetry of Line Arrangements and Applications, by Amram, Cohen, Sun, Teicher, Ye & Zarkh, arXiv:1310.0700 Schubert Varieties
Euclidean and n Algebraic The set of d-dimensional subspaces of C has a natural Geometry structure as an algebraic variety, the Grassmannian G(d,n). David A. Cox Inside of G(d,n) are Schubert varieties, defined as follows.
Tangents to 4 Given a flag of subspaces Unit Spheres A W1 W2 Wk , Combinatorial Moduli Space
Counting the corresponding Schubert variety is Balanced Subspaces {V ∈ G(d,n) | dim(V ∩ Wi ) ≥ i, 1 ≤ i ≤ k}. Two Quickies, If Time Permits Many problems in enumerative algebraic geometry can be formulated in terms of counting points in intersections of Schubert varieties. This led to the Schubert calculus. I will discuss an example from Real Solutions of a Problem in Enumerative Geometry by Fehér and Matszangosz (arXiv:1401.4638). They use work of Vakil from 2006. Schubert Varieties
Euclidean and n Algebraic The set of d-dimensional subspaces of C has a natural Geometry structure as an algebraic variety, the Grassmannian G(d,n). David A. Cox Inside of G(d,n) are Schubert varieties, defined as follows.
Tangents to 4 Given a flag of subspaces Unit Spheres A W1 W2 Wk , Combinatorial Moduli Space
Counting the corresponding Schubert variety is Balanced Subspaces {V ∈ G(d,n) | dim(V ∩ Wi ) ≥ i, 1 ≤ i ≤ k}. Two Quickies, If Time Permits Many problems in enumerative algebraic geometry can be formulated in terms of counting points in intersections of Schubert varieties. This led to the Schubert calculus. I will discuss an example from Real Solutions of a Problem in Enumerative Geometry by Fehér and Matszangosz (arXiv:1401.4638). They use work of Vakil from 2006. The Result over C
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is C in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ C2n satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space We say that W is balanced with respect to V1,...,V4. Counting Balanced Subspaces This problem reduces to the intersection of four Schubert
Two Quickies, varieties. By the Schubert calculus, one gets the answer If Time Permits n . m To do: Explain the combinatorics behind this answer. Ask the same question over R. The Result over C
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is C in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ C2n satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space We say that W is balanced with respect to V1,...,V4. Counting Balanced Subspaces This problem reduces to the intersection of four Schubert
Two Quickies, varieties. By the Schubert calculus, one gets the answer If Time Permits n . m To do: Explain the combinatorics behind this answer. Ask the same question over R. The Result over C
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is C in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ C2n satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space We say that W is balanced with respect to V1,...,V4. Counting Balanced Subspaces This problem reduces to the intersection of four Schubert
Two Quickies, varieties. By the Schubert calculus, one gets the answer If Time Permits n . m To do: Explain the combinatorics behind this answer. Ask the same question over R. The Result over C
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is C in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ C2n satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space We say that W is balanced with respect to V1,...,V4. Counting Balanced Subspaces This problem reduces to the intersection of four Schubert
Two Quickies, varieties. By the Schubert calculus, one gets the answer If Time Permits n . m To do: Explain the combinatorics behind this answer. Ask the same question over R. The Result over R
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is R in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ R2n are balanced, i.e., satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space For n = 7, m = 5 Counting 7 Balanced The possibilities are = 21, 11, 5, 3. Subspaces 5
Two Quickies, If Time Permits For General n, m c c n − 2c ∑ , i=0 i m − 2i
where c is an integer 0 ≤ c ≤ n/2 depending on V1,...,V4. Furthermore, all possible values of c can occur. The Result over R
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is R in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ R2n are balanced, i.e., satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space For n = 7, m = 5 Counting 7 Balanced The possibilities are = 21, 11, 5, 3. Subspaces 5
Two Quickies, If Time Permits For General n, m c c n − 2c ∑ , i=0 i m − 2i
where c is an integer 0 ≤ c ≤ n/2 depending on V1,...,V4. Furthermore, all possible values of c can occur. The Result over R
Euclidean and 2n Algebraic Consider four n-dimensional subspaces V1,...,V4 is R in Geometry general position. How many 2m-dimensional subspaces David A. Cox W ⊆ R2n are balanced, i.e., satisfy Tangents to 4 Unit Spheres dim(W ∩ Vi )= m, 1 ≤ i ≤ 4? A Combinatorial Moduli Space For n = 7, m = 5 Counting 7 Balanced The possibilities are = 21, 11, 5, 3. Subspaces 5
Two Quickies, If Time Permits For General n, m c c n − 2c ∑ , i=0 i m − 2i
where c is an integer 0 ≤ c ≤ n/2 depending on V1,...,V4. Furthermore, all possible values of c can occur. The Map ϕ : V1 → V1
Euclidean and 2n Algebraic Given n-dimensional subspaces V1,...,V4 ⊆ C , we will Geometry assume Vi ∩ Vj = {0} for all i = j. Then use V3 to define an David A. Cox isomorphism Tangents to 4 ϕ3 : V1 → V2 Unit Spheres A using Combinatorial Moduli Space V3 V2 Counting Balanced → ϕ ( ) 3 v Subspaces ↑ Two Quickies, If Time Permits V1 v ← graph ϕ of 3
Similarly, V4 gives an isomorphism ϕ4 : V1 → V2. Then we ϕ ϕ ϕ−1 ϕ define : V1 → V1 to be = 3 ◦ 4. The Key Theorem
Euclidean and Algebraic Geometry General Position David A. Cox Vi ∩ Vj = {0} for i = j and ϕ : V1 → V1 has distinct eigenvalues. Tangents to 4 Unit Spheres
A Combinatorial Theorem (Fehér & Matszangosz) Moduli Space The map W → W ∩ V induces a bijection Counting 1 Balanced Subspaces 2m-dimensional m-dimensional subspaces ≃ Two Quickies, balanced subspaces of V1 invariant under ϕ If Time n o n o Permits Key Step of Proof.
For W1 ⊆ V1 m-dimensional and ϕ-invariant, show that
W = W1 + ϕ3(W1)= W1 + ϕ3(ϕ(W1)) = W1 + ϕ4(W1) is balanced of dimension 2m. The Key Theorem
Euclidean and Algebraic Geometry General Position David A. Cox Vi ∩ Vj = {0} for i = j and ϕ : V1 → V1 has distinct eigenvalues. Tangents to 4 Unit Spheres
A Combinatorial Theorem (Fehér & Matszangosz) Moduli Space The map W → W ∩ V induces a bijection Counting 1 Balanced Subspaces 2m-dimensional m-dimensional subspaces ≃ Two Quickies, balanced subspaces of V1 invariant under ϕ If Time n o n o Permits Key Step of Proof.
For W1 ⊆ V1 m-dimensional and ϕ-invariant, show that
W = W1 + ϕ3(W1)= W1 + ϕ3(ϕ(W1)) = W1 + ϕ4(W1) is balanced of dimension 2m. The Key Theorem
Euclidean and Algebraic Geometry General Position David A. Cox Vi ∩ Vj = {0} for i = j and ϕ : V1 → V1 has distinct eigenvalues. Tangents to 4 Unit Spheres
A Combinatorial Theorem (Fehér & Matszangosz) Moduli Space The map W → W ∩ V induces a bijection Counting 1 Balanced Subspaces 2m-dimensional m-dimensional subspaces ≃ Two Quickies, balanced subspaces of V1 invariant under ϕ If Time n o n o Permits Key Step of Proof.
For W1 ⊆ V1 m-dimensional and ϕ-invariant, show that
W = W1 + ϕ3(W1)= W1 + ϕ3(ϕ(W1)) = W1 + ϕ4(W1) is balanced of dimension 2m. The Result over C and R
Euclidean and Algebraic We need to count m-dimensional ϕ-invariant subspaces. Geometry Proof over . David A. Cox C V1 is a direct sum of n one-dimensional eigenspaces of ϕ. Tangents to 4 Unit Spheres An m-dimensional ϕ-invariant subspace is a direct sum of m n A of these. Hence there are such subspaces. Combinatorial m Moduli Space Counting Proof over R. Balanced Subspaces Let c =# complex-conjugate pairs of eigenvalues of ϕ. Two Quickies, If Time Then V1 is a direct sum of c two-dimensional invariant Permits subspaces and n − 2c one-dimensional eigenspaces. An m-dimensional ϕ-invariant subspace is built from i of the former and m − 2i of the latter. This gives c c n − 2c ∑ . i=0 i m − 2i The Result over C and R
Euclidean and Algebraic We need to count m-dimensional ϕ-invariant subspaces. Geometry Proof over . David A. Cox C V1 is a direct sum of n one-dimensional eigenspaces of ϕ. Tangents to 4 Unit Spheres An m-dimensional ϕ-invariant subspace is a direct sum of m n A of these. Hence there are such subspaces. Combinatorial m Moduli Space Counting Proof over R. Balanced Subspaces Let c =# complex-conjugate pairs of eigenvalues of ϕ. Two Quickies, If Time Then V1 is a direct sum of c two-dimensional invariant Permits subspaces and n − 2c one-dimensional eigenspaces. An m-dimensional ϕ-invariant subspace is built from i of the former and m − 2i of the latter. This gives c c n − 2c ∑ . i=0 i m − 2i A Moduli Space
Euclidean and 4 Algebraic The generic V1,...,V4 form an open subset U ⊂ G(n,2n) . Geometry Hence the moduli space is U/GL(2n). To understand this David A. Cox n quotient, consider the map U → C sending (V1,...,V4) to Tangents to 4 the coefficients of det(ϕ − xI). Unit Spheres A Lemma (Fehér & Matszangosz) Combinatorial Moduli Space The nonempty fibers of the map U → Cn are precisely the Counting Balanced orbits of GL(2n) acting on U. Subspaces Two Quickies, Note that 0 is not an eigenvalue since ϕ is an isomorphism; If Time Permits the same is true for 1 since V3 ∩ V4 = {0}. If we write n n n−1 det(ϕ − xI) = (−1) x + a1x + + an, n then U/GL(2n) consists of all (a1,...,an) ∈ C satisfying n−1 disc(det(ϕ − xI)) = 0, an = 0, a1 + + an = (−1) . So U/GL(2n) is an algebraic variety. A Moduli Space
Euclidean and 4 Algebraic The generic V1,...,V4 form an open subset U ⊂ G(n,2n) . Geometry Hence the moduli space is U/GL(2n). To understand this David A. Cox n quotient, consider the map U → C sending (V1,...,V4) to Tangents to 4 the coefficients of det(ϕ − xI). Unit Spheres A Lemma (Fehér & Matszangosz) Combinatorial Moduli Space The nonempty fibers of the map U → Cn are precisely the Counting Balanced orbits of GL(2n) acting on U. Subspaces Two Quickies, Note that 0 is not an eigenvalue since ϕ is an isomorphism; If Time Permits the same is true for 1 since V3 ∩ V4 = {0}. If we write n n n−1 det(ϕ − xI) = (−1) x + a1x + + an, n then U/GL(2n) consists of all (a1,...,an) ∈ C satisfying n−1 disc(det(ϕ − xI)) = 0, an = 0, a1 + + an = (−1) . So U/GL(2n) is an algebraic variety. Two Examples
Euclidean and Algebraic Geometry n = 2, m = 1
David A. Cox 4 Here, V1,...,V4 and W are planes through the origin in C , 3 Tangents to 4 which correspond to lines in projective 3-space P . Since Unit Spheres 2 1 = 2, we recover the classical fact that there are two A 3 Combinatorial lines meeting four given lines in P in general position. Moduli Space
Counting Balanced Subspaces n = 1 Two Quickies, 2 If Time Here, V1,...,V4 are lines through the origin in C , which Permits correspond to points in the projective line P1. Since ϕ : V1 → V1 is multiplication by a = 0,1, the map (V1,...,V4) → ϕ is (up to sign) the classical cross-ratio of four distinct points in P2. The moduli space is U/GL(2)= P1 \ {0,1,∞}. Two Examples
Euclidean and Algebraic Geometry n = 2, m = 1
David A. Cox 4 Here, V1,...,V4 and W are planes through the origin in C , 3 Tangents to 4 which correspond to lines in projective 3-space P . Since Unit Spheres 2 1 = 2, we recover the classical fact that there are two A 3 Combinatorial lines meeting four given lines in P in general position. Moduli Space
Counting Balanced Subspaces n = 1 Two Quickies, 2 If Time Here, V1,...,V4 are lines through the origin in C , which Permits correspond to points in the projective line P1. Since ϕ : V1 → V1 is multiplication by a = 0,1, the map (V1,...,V4) → ϕ is (up to sign) the classical cross-ratio of four distinct points in P2. The moduli space is U/GL(2)= P1 \ {0,1,∞}. Quickie #1. The Polynomial Method
Euclidean and n n Algebraic Let L be a collection of N lines in R (or F for any field F). Geometry A joint for L is a point p such that there are n lines in L David A. Cox which contain p and do not line in a hyperplane. Tangents to 4 Unit Spheres Theorem (Joints Conjecture) A Combinatorial n/(n−1) Moduli Space The number of joints is at most nN .
Counting Balanced This was proved independently in 2009/10 by Kaplan, Subspaces Sharir, Shustin and also by Quilodrán. Two Quickies, If Time Permits The proof uses polynomials that vanish at the joints and a variant of Bezout’s Theorem. All known proofs involve algebraic geometry in some way. Tao presents the proof in the context of what he calls the polynomial method.
Algebraic Combinatorial Geometry: The Polynomial Method in Arithmetic Combinatorics, Incidence Combinatorics, and Number Theory, by Tao, arXiv:1310.6482 Quickie #1. The Polynomial Method
Euclidean and n n Algebraic Let L be a collection of N lines in R (or F for any field F). Geometry A joint for L is a point p such that there are n lines in L David A. Cox which contain p and do not line in a hyperplane. Tangents to 4 Unit Spheres Theorem (Joints Conjecture) A Combinatorial n/(n−1) Moduli Space The number of joints is at most nN .
Counting Balanced This was proved independently in 2009/10 by Kaplan, Subspaces Sharir, Shustin and also by Quilodrán. Two Quickies, If Time Permits The proof uses polynomials that vanish at the joints and a variant of Bezout’s Theorem. All known proofs involve algebraic geometry in some way. Tao presents the proof in the context of what he calls the polynomial method.
Algebraic Combinatorial Geometry: The Polynomial Method in Arithmetic Combinatorics, Incidence Combinatorics, and Number Theory, by Tao, arXiv:1310.6482 Quickie #2. Ellipses and Elliptic Curves
Euclidean and Algebraic The next example is modest but amusing. Geometry
David A. Cox An ellipse centered at the origin is defined by an equation
Tangents to 4 x 2 y 2 Unit Spheres 2 + 2 = 1 A a b Combinatorial Moduli Space of degree 2, while an elliptic curve in Weierstrass form is Counting defined by an equation Balanced Subspaces 2 3 Two Quickies, y = 4x − g2x − g3 If Time Permits of degree 3. Clearly different – why such similar names?
Brief answer: arc length elliptic elliptic elliptic ellipse ⇒ ⇒ ⇒ ⇒ Greeks of ellipse integral function curve 17th C 18th C 19th C 20th C Ellipses in the Plane and Sphere
Euclidean and Algebraic Here is another way to Geometry think about the ellipse: David A. Cox
Tangents to 4 Unit Spheres
A This construction Combinatorial Moduli Space works nicely in
Counting spherical geometry. Balanced Subspaces
Two Quickies, people.richland.edu/james/lecture/m116/conics/elldef.html If Time Permits Given two points A = B on the unit sphere S2, a spherical ellipse is the locus of points P ∈ S2 such that
dist(P,A)+ dist(P,B)= 2a,
where “dist” is the great circle distance between the points. Pictures of Spherical Ellipses
Euclidean and Algebraic Geometry
David A. Cox
Tangents to 4 Unit Spheres
A Combinatorial Moduli Space
Counting Balanced Subspaces
Two Quickies, If Time Permits
Images from: virtualmathmuseum.org//SpaceCurves/spherical_ellipse/spherical_ellipse.html etc.usf.edu/clipart/galleries/275-spheres . Equation of a Spherical Ellipse
Euclidean and 2 2 Algebraic Fix A = B in S . Then P ∈ S satisfies Geometry David A. Cox (1) dist(P,A)+ dist(P,B)= 2a.
Tangents to 4 Unit Spheres Recall that A Combinatorial 1 1 Moduli Space dist(P,A)= cos− (P A) and sin(cos− (x)) = 1 − x 2. Counting p Balanced Subspaces Applying cos to (1) and doing some algebra gives Two Quickies, If Time 2 2 2 Permits sin (2a) − (P A) − (P B) + 2cos(2a)(P A)(P B)= 0.
Since P P = 1, we obtain the homogeneous equation
sin2(2a)(P P)−(P A)2 −(P B)2 +2cos(2a)(P A)(P B)= 0.
This defines a quadric cone in R3. Equation of a Spherical Ellipse
Euclidean and 2 2 Algebraic Fix A = B in S . Then P ∈ S satisfies Geometry David A. Cox (1) dist(P,A)+ dist(P,B)= 2a.
Tangents to 4 Unit Spheres Recall that A Combinatorial 1 1 Moduli Space dist(P,A)= cos− (P A) and sin(cos− (x)) = 1 − x 2. Counting p Balanced Subspaces Applying cos to (1) and doing some algebra gives Two Quickies, If Time 2 2 2 Permits sin (2a) − (P A) − (P B) + 2cos(2a)(P A)(P B)= 0.
Since P P = 1, we obtain the homogeneous equation
sin2(2a)(P P)−(P A)2 −(P B)2 +2cos(2a)(P A)(P B)= 0.
This defines a quadric cone in R3. Elliptic Ellipses are Elliptic!
Euclidean and 2 Algebraic It follows that our spherical ellipse in S is the intersection Geometry David A. Cox quadric cone sphere. Tangents to 4 \ Unit Spheres degree 2 degree 2
A However: | {z } | {z } Combinatorial Moduli Space Theorem (Algebraic Geometry) Counting Balanced The intersection of two quadric surfaces in 3-dimensional Subspaces space is an elliptic curve. Two Quickies, If Time Permits Spherical ellipses are elliptic curves. Spherical Geometry is also called Elliptic Geometry. Hence the title of the slide! A small lie: a spherical ellipse is half of an elliptic curve, so elliptic ellipses are half elliptic!
Reference: virtualmathmuseum.org/SpaceCurves/spherical_ellipse/Spherical_Ellipse.pdf. Elliptic Ellipses are Elliptic!
Euclidean and 2 Algebraic It follows that our spherical ellipse in S is the intersection Geometry David A. Cox quadric cone sphere. Tangents to 4 \ Unit Spheres degree 2 degree 2
A However: | {z } | {z } Combinatorial Moduli Space Theorem (Algebraic Geometry) Counting Balanced The intersection of two quadric surfaces in 3-dimensional Subspaces space is an elliptic curve. Two Quickies, If Time Permits Spherical ellipses are elliptic curves. Spherical Geometry is also called Elliptic Geometry. Hence the title of the slide! A small lie: a spherical ellipse is half of an elliptic curve, so elliptic ellipses are half elliptic!
Reference: virtualmathmuseum.org/SpaceCurves/spherical_ellipse/Spherical_Ellipse.pdf. Elliptic Ellipses are Elliptic!
Euclidean and 2 Algebraic It follows that our spherical ellipse in S is the intersection Geometry David A. Cox quadric cone sphere. Tangents to 4 \ Unit Spheres degree 2 degree 2
A However: | {z } | {z } Combinatorial Moduli Space Theorem (Algebraic Geometry) Counting Balanced The intersection of two quadric surfaces in 3-dimensional Subspaces space is an elliptic curve. Two Quickies, If Time Permits Spherical ellipses are elliptic curves. Spherical Geometry is also called Elliptic Geometry. Hence the title of the slide! A small lie: a spherical ellipse is half of an elliptic curve, so elliptic ellipses are half elliptic!
Reference: virtualmathmuseum.org/SpaceCurves/spherical_ellipse/Spherical_Ellipse.pdf. Conclusion
Euclidean and Algebraic Geometry These days, it is fashionable to give talks about the David A. Cox unreasonable effectiveness of some type of math Tangents to 4 Unit Spheres (such as number theory) in other areas of math or
A science. Combinatorial Moduli Space
Counting In this talk, we have seen that algebraic geometry Balanced Subspaces cannot claim to be unreasonably effective in Euclidean Two Quickies, geometry – it has definite strengths and weaknesses. If Time Permits Nevertheless, I hope I have convinced you that algebraic geometry is pretty darn good!
Thank you! Conclusion
Euclidean and Algebraic Geometry These days, it is fashionable to give talks about the David A. Cox unreasonable effectiveness of some type of math Tangents to 4 Unit Spheres (such as number theory) in other areas of math or
A science. Combinatorial Moduli Space
Counting In this talk, we have seen that algebraic geometry Balanced Subspaces cannot claim to be unreasonably effective in Euclidean Two Quickies, geometry – it has definite strengths and weaknesses. If Time Permits Nevertheless, I hope I have convinced you that algebraic geometry is pretty darn good!
Thank you! Conclusion
Euclidean and Algebraic Geometry These days, it is fashionable to give talks about the David A. Cox unreasonable effectiveness of some type of math Tangents to 4 Unit Spheres (such as number theory) in other areas of math or
A science. Combinatorial Moduli Space
Counting In this talk, we have seen that algebraic geometry Balanced Subspaces cannot claim to be unreasonably effective in Euclidean Two Quickies, geometry – it has definite strengths and weaknesses. If Time Permits Nevertheless, I hope I have convinced you that algebraic geometry is pretty darn good!
Thank you!