QUARTIC SPECTRAHEDRA. Bernd Sturmfels UC Berkeley and MPI Bonn. Joint work with John Christian Ottem, Kristian Ranestad and Cynthia Vinzant

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1 QUARTIC SPECTRAHEDRA Bernd Sturmfels UC Berkeley and MPI Bonn Joint work with John Christian Ottem, Kristian Ranestad and Cynthia Vinzant 1 / 20

2 Definition A spectrahedron of degree n in R 3 is a convex body of the form S = { (x, y, z) R 3 : xa + yb + zc + D is positive semidefinite } where A, B, C, D are real symmetric matrices of format n n. n = 1: S is a closed half space. n = 2: S is a quadric cone. Finite intersections of spectrahedra are spectrahedra. 2 / 20

3 Toeplitz Spectrahedron S = { (x, y, z) R 3 : 1 x y z x 1 x y y x 1 x }. 0 z y x 1 is the convex hull of the trigonometric curve {( cos(θ), cos(2θ), cos (3θ) : θ [0, π] }. is the intersection of two quadratic cones. 3 / 20

4 The Pillow... is a quartic spectrahedron: Q: Is it reducible? 1 x 0 x x 1 y 0 0 y 1 z x 0 z 1 Q: Is the convex dual of a spectrahedron is a spectrahedron? 4 / 20

5 A familiar picture The set of 3 3-correlation matrices is the elliptope 1 x y x 1 z y z 1 Q: Does every cubic spectrahedron have four nodes? 5 / 20

6 Symmetroids Fix projective space P 3 with coordinates (x : y : z : w). A symmetroid S of degree n is a surface with equation det(xa + yb + zc + wd) = 0 where A, B, C, D are (complex) symmetric n n matrices. Proposition If A, B, C, D are generic then the singular locus of the symmetroid S consists of ( ) n+1 3 nodes. The web xa+yb+zc+wd contains no matrix of rank n 3. Those of rank n 2 are precisely the nodes. S is transversal symmetroid if this holds. A 3-dimensional spectrahedron is transversal if its algebraic boundary is a transversal symmetroid. Problem Study the geometry and combinatorics of transversal spectrahedra. For instance, how many of the ( ) n+1 3 nodes can lie on its boundary? 6 / 20

7 Cubic Spectrahedra Proposition Every non-empty irreducible 3-dimensional cubic spectrahedron S has either 4 or 2 nodes, and these are connected by edges on S. The cubic surface S in P 3 is a Cayley symmetroid. For an algebraic geometer, this is obtained by blowing up P 2 at six points that are the intersections of four lines defined over R. The two cases correspond to either four real lines, or two real lines and a pair of complex conjugate lines. protrusion_kl.jpg pixels 9/25/13 5:48 AM 7 / 20

8 Cubic Spectrahedra Two combinatorial types for n = 3. How about n = 4? 8 / 20

9 Degtyarev-Itenberg Theorem For quartic spectrahedra, this question was answered in a 2010 paper by Alex Degtyarev and Ilia Itenberg: Theorem There exists a transversal quartic spectrahedron with β nodes in its boundary and σ real nodes in its symmetroid if and only if 0 β σ, 2 σ 10, and both β and σ are even. Their proof is extremely indirect: it rests on the Global Torelli Theorem for K3 surfaces, and on deep topological results of Kharlamov and Nikhulin for moduli spaces of real K3 surfaces. It is impossible to use this to construct matrices A, B, C, D. We give a new proof that is direct, computational and geometric, via projections onto P 2 from rank 2 nodes on S. 9 / 20

10 Twenty Types A B C D (2, 2): (4, 4): (6, 6): (8, 8): (10, 10): (2, 0): (4, 2): etc... etc / 20

11 The View from a Node Figure: A quartic spectrahedron and its projection from an outside node. The ramification curve consists of two cubics totally tangent to a conic. 11 / 20

12 The View from a Node Our main main technical tool is a result that goes back to Cayley. Theorem Let p be a node on an irreducible quartic surface S P 3. The following are equivalent: S is a symmetroid and p corresponds to a rank 2 matrix. The projection of S from p to P 2 is branched along two cubics C 1 and C 2 that are totally tangent to a common conic Q. The pair (p, S) is real if and only if the pair (C 1 C 2, Q) is real. If this holds, then C 1 and C 2 are both real if and only if p is not on the spectrahedron. Equivalently, p is a node on the spectrahedron if and only if the cubic curves C 1 and C 2 are complex conjugates. Proposition A symmetroid with nonempty spectahedron has no rank 3 nodes. 12 / 20

13 The Kummer Surface... is an irreducible quartic with 16 nodes: Q: Are the convex bodies seen here spectrahedra? 13 / 20

14 Kummer Spectrahedron The following symmetric matrix is found on page 143 of Arthur Coble s 1929 book Algebraic Geometry and Theta Functions. a 6 w 3a 5 w 3a 4 w + z a 3 w y 3wa 5 9a 4 w 2z 9a 3 w + y 3a 2 w + x 3a 4 w + z 9a 3 w + y 9a 2 w 2x 3a 1 w a 3 w y 3a 2 w + x 3a 1 w a 0 w It is associated to polynomials of degree 6 in one variable t: p(t) = a 0 6a 1 t + 15a 2 t 2 20a 3 t a 4 t 4 6a 5 t 5 + a 6 t 6 These symmetroids are the Kummer surfaces. They have 16 nodes, ten of rank 2 and six of rank 3. If a 0,..., a 6 Q then all 16 nodes have rational coordinates. The spectrahedron is non-empty if and only if p(t) 0 on R. Its points are the sum of squares representations of p(t). 14 / 20

15 Transversal Spectrahedra This quartic spectrahedron has β = σ = 10 nodes: 18x+4y +z+9w 6x 6y 3w 6x+6y 3z 6x+4y 6x 6y 3w 2x+13y +4z+10w 2x 9y +9w 2x 8y +6z 6w 6x+6y 3z 2x 9y +9w 2x+9y +9z+9w 2x+6y 6w 6x+4y 2x 8y +6z 6w 2x+6y 6w 4x+5y +9z+4w This quartic spectrahedron has β = 0 and σ = 10: 263x 160y 20z 187w 3x 132y+28z+78w 114x 30y+4z 76w 103x+244y+32z 192w 3x 132y+28z+78w 45x+28y 32z 32w 35x+40y 32z+24w 48x+20y 4z+88w 114x 30y+4z 76w 35x+40y 32z+24w 275x+25y+96z+80w 55x 40y 156z 192w 103x+244y+32z 192w 48x+20y 4z+88w 55x 40y 156z 192w 278x 132y+180z 80w To get bounded convex bodies, set x+y+z+w = 1. Q: What will semidefinite programming do on these matrices? What is the rank of the optimal matrix for random cost functions? 15 / 20

16 Transversal Spectrahedra Out[124]= (β, σ) = (2, 2), (8, 10), (0, 10) 16 / 20

17 Spectrahedral Shadows (β, σ) = (10, 10) Curve of degree 12 with 46 nodes. Problem: Which compact convex semialgebraic subsets of the plane R 2 are projections of 3-dimensional spectrahedra? 17 / 20

18 From Polytope to Spectrahedron Toblerone-of-Switzerland.jpg 6,142 2,457 pixels 9/26/13 6:35 AM 18 / 20

19 Sylvester Spectrahedron Given general linear forms l 0, l 1, l 2, l 3, l 4 in R[x, y, z, w], consider l 1 + l 0 l 0 l 0 l 0 l 0 l 2 + l 0 l 0 l 0 l 0 l 0 l 3 + l 0 l 0 l 0 l 0 l 0 l 4 + l 0 The inequalities l i 0 define the cone over a triangular prism. Sanyal [2013]: the spectrahedron is the Renegar derivative of. The header refers to Sylvester s Pentahedral Theorem: every cubic surface in P 3 admits a unique representation f = l l l l l 3 5 as a sum of five powers of linear forms. Our matrix is the Hessian [ 2 f / x i x j ] in suitable coordinates. The Sylvester symmetroid is transversal. Q: What are the 10 nodes? What are β and σ? 19 / 20

20 Conclusion Happy Birthday, Jean Bernard. Here is your picture postcard. Out[124]= 20 / 20