Max-Planck-Institut für Mathematik in den Naturwissenschaften Leipzig

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1 Max-Planck-Institut für Mathematik in den Naturwissenschaften Leipzig Random Spectrahedra by Paul Breiding, Khazhgali Kozhasov, and Antonio Lerario Preprint no:

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3 Random spectrahedra PAUL BREIDING, KHAZHGALI KOZHASOV AND ANTONIO LERARIO Abstract Spectrahedral cones are linear sections of the cone of positive semidefinite symmetric matrices We study statistical properties of random spectrahedral cones (intersected with the sphere) { } S l,n = (x 0,, x l ) S l x x 1 R x l R l 0 where R 1,, R l are independent GOE(n)-distributed matrices rescaled by (nl) 1/ We relate the expectation of the volume of S l,n with some statistics of the smallest eigenvalue of a GOE(n) matrix, by providing explicit formulas for this quantity These formulas imply that as l, n on average S l,n keeps a positive fraction of the volume of the sphere S l (the exact constant is Φ( 1) 01587, where Φ is the cumulative distribution function of a standard gaussian variable) For l = spectrahedra are generically smooth, but already when l = 3 singular points on their boundaries appear with positive probability We relate the average number E σ n of singular points on the boundary of a three-dimensional spectrahedron S 3,n to the volume of the set of symmetric matrices whose two smallest eigenvalues coincide In the case of quartic spectrahedra (n = 4) we show that E σ 4 = 6 4 Moreover, we prove that the average 3 number E ρ n of singular points on the random symmetroid surface Σ 3,n = { (x 0, x 1, x, x 3 ) S 3 det(x x 1 R 1 + x R + x 3 R 3 ) = 0 }, equals n(n 1) This quantity is related to the volume of the set of symmetric matrices with repeated eigenvalues 1 Introduction A spectrahedron is an affine-linear section of the cone P n Sym(n, R) of positive semidefinite symmetric matrices On the space Sym(n, R) of n n real symmetric matrices there is a partial order defined by A B, if and only if A B P n Every spectrahedron can then be parametrized as the set of solutions of a linear matrix inequality: M 0 + x 1 M x l M l 0, x = (x 1,, x l ) R l, for some symmetric matrices M 0,, M l Sym(n, R) Optimization of a linear function over a spectrahedron is called semidefinite programming [MR95, Ali95] This is a useful generalization of linear programming, ie optimization of a linear function over a polyhedron Such problems as finding the smallest eigenvalue of a symmetric matrix or optimizing a polynomial function on the sphere can be approached using semidefinite programming The presence of singularities on the boundary of a three-dimensional spectrahedron is relevant for optimization: with a positive probability a linear function constrained on a polyhedron attains its maximum in a vertex of the polyhedron, and, similarly, with a positive probability a linear function constrained on a spectrahedron attains its maximum in a singular point of the boundary of the spectrahedron For example, consider the cubic spectrahedron shown on Figure 11: S = { (x, z, y) R 3 ( 1 x y ) } x 1 z P n y z 1 PB: Max-Planck-Institute for Mathematics in the Sciences (Leipzig), breiding@mismpgde, KK: SISSA (Trieste), kkozhasov@sissait, AL: SISSA (Trieste), lerario@sissait 1

4 Figure 11 On the left is the cubic spectrahedron from the introduction On the right is a quartic spectrahedron (in [GB1] it is called the pillow ) The singularities on the boundaries of both spectrahedra are visible The linear function ψ w (x, y, z) = w, (x, y, z), w S 3, constrained on S attains its maximum at a point (x, y, z) S on the boundary of S at which the normal cone to S contains w At a singular point of the boundary the normal cone has positive dimension and hence the set of w S 3 for which the maximum of ψ w is attained at a singular point of the boundary of the spectrahedron has positive volume in S 3 Besides mentioned applications spectrahedra also appear in modern real algebraic geometry: in [JWH06] Helton and Vinnikov gave a beautiful characterization of two-dimensional spectrahedra and in [AD11] Degtyarev and Itenberg described all generic possibilities for the number of singular points on the boundary of a quartic three-dimensional spectrahedron (see Section 14 for more details) The reader can also look at [Vin14] for a survey article 11 Random spectrahedra In order to perform a probabilisitc study, it is more convenient to work instead with spherical spectrahedra, defined by: x 0 M 0 + x 1 M x l M l 0, x = (x 0, x 1,, x l ) S l A generic spherical spectrahedron has nonempty interior and, after a change of coordinates in the space of symmetric matrices, it can be presented as: (11) S l,n = { x = (x 0,, x l ) S l x x 1 R x l R l 0 } From now on, for simplicity, we abuse the terminology and use the term spectrahedron to refer to a spherical spectrahedron Our choice of random model for spectrahedra is as follows In the representation (11) we sample the matrices R 1,, R l independently and identically distributed from the Gaussian Orthogonal Ensemble GOE(n) [Meh91, Tao1], rescaled by (nl) 1/ : R i = 1 nl Q i, where Q i GOE(n), i = 1,, l By Q GOE(n) we mean that the joint probability density of the entries of the symmetric matrix Q Sym(n, R) is ϕ(q) = 1 C n exp( 1 tr(q )), where C n is the normalization constant with ϕ(q) dq = 1 In other words, the entries of Q are centered gaussian random Sym(n,R) variables, the diagonal entries having variance 1 and the off-diagonal entries having variance 1 The scaling factor (nl) 1/ serves to balance the order of magnitudes of eigenvalues of the two summands x 0 1 and x 1 R x l R l

5 1 Statistical properties of random spectrahedra In the following for a semialgebraic subset X S l of dimension d by X we denote the d-dimensional volume of the set of smooth points of X and by X rel := X we denote the relative volume of X The first statistic we will be S d interested in is the expected value of S l,n rel Furthermore, we will be interested in the expected number of singular points on S 3,n More specifically, the boundary S 3,n of a 3-dimensional spectrahedron S 3,n is in general singular and, generically, has finitely many singular points that are all nodes We denote their number by σ n Note that S 3,n is a semialgebraic subset of (1) Σ 3,n = {x S 3 det(x x 1 R 1 + x R + x 3 R 3 ) = 0}, called the symmetroid surface of S l,n Hence σ n is smaller than the number ρ n of singular points on Σ 3,n Summarizing, we will be interested in: E S l,n rel, E σ n and E ρ n Our main results on those three quantities follow next 13 Main results To state our first main result, let λ min (Q) denote the smallest eigenvalue of the matrix Q For the scaled smallest eigenvalue we write (13) λmin (Q) := λ min(q) n The following is proved in Section 3 below Theorem 11 (Expected volume of the spectrahedron) Let F l denote the cumulative distribution function of the student s t-distribution with l degrees of freedom [NLJ95, Chapter 8] and Φ(x) denote the cumulative distribution function of the normal distribution [JS00, 40:14:] Then: (1) E S l,n rel = E F l ( λmin (Q) ) Q GOE(n) Moreover, we have () E S l,n rel = F l ( 1) + O(n /3 ) uniformly in l, (3) E S l,n rel = Φ( 1) + O(l 1 ) + O(n /3 ) Note that Φ( 1) This means that asymptotically (in both n and l) the average volume of a spectrahedron is at least 15% of the volume of the sphere For the average number of singular points σ n on S 3,n and number of singular points ρ n on Σ 3,n the result is more delicate to state We denote the dimension of Sym(n, R) by N := n(n+1) and the unit sphere there by S N 1 := {Q Sym(n, R) Q = tr(q ) = 1} Let S N 1 be the set of symmetric matrices of unit norm and with repeated eigenvalues and let 1 be its subset consisting of symmetric matrices whose two smallest eigenvalues coincide: := {Q Sym(n, R) S N 1 λ i (Q) = λ j (Q) for some i j}, 1 := {Q Sym(n, R) S N 1 λ 1 (Q) = λ (Q)} Note that and 1 are both semialgebraic subsets of S N 1 of codimension two; is actually algebraic The following theorem relates E σ n and E ρ n to the volumes of 1 and, respectively Theorem 1 (The average number of singular points) The average number of singular points on the boundary of a random 3-dimensional spectrahedron S 3,n S 3 equals (1) E σ n = 1 S N 3 The average number of singular points on the symmetroid Σ 3,n S 3 equals () E ρ n = S N 3 3

6 We prove Theorem 1 in Section 4 As a consequence of the proof of this theorem we will also derive the following interesting proposition Proposition 13 For the generic choice of the matrices R 1, R, R 3 Sym(n, R) there is a one-to-one correspondence between the number of singular points of the projective symmetroid PΣ 3,n = {[x 0 : x 1 : x : x 3 ] RP 3 det(x x 1 R 1 + x R + x 3 R 3 ) = 0} and the number of matrices with multiple eigenvalues in P(span{R 1, R, R 3 }) The following theorem whose proof is given in Section 5 implies that the order of magnitude of E ρ n is O(n ) In particular, this gives an upper bound on E σ n Theorem 14 (The volume of the set of symmetric matrices with repeated eigenvalues) Let 1 S N 1 be as above Then ( ) 1 (1) S N 3 = n 1 n [ E det(q u1) ] 1 {Q u1 0} e u du π n! Q GOE(n ) ( ) () S N 3 = n 1 n [ E det(q u1) ] e u du π n! Q GOE(n ) The last quantity we compute explicitly: ( ) n (3) S N 3 = The theorem is of independent interest: as we explain at the beginning of Section 5 it gives some information on the geometry of the set of matrices with repeated eigenvalues In particular, from Theorem 14 we can extract the following interesting corollary Corollary 15 Let Q S N 1 be a random symmetric matrix uniformly distributed in the sphere S N 1 Let dist S N 1(Q, ) denote the distance in S N 1 between Q and, then Prob {dist S N 1(Q, ) < ɛ} π(1 cos ɛ) ( n ) S N 3 The proof of Corollary 15 is by integrating the volume of a spherical disc of radius ɛ over 14 Quartic spectrahedra Quartic spectrahedra are a special case of our study, corresponding to n = 4 In this case the symmetroid surface Σ 3,4 = {x S 3 det(x x 1 R 1 + x R + x 3 R 3 ) = 0} has degree four, since 1, R 1, R, R 3 Sym(4, R) In [AD11] Degtyarev and Itenberg proved that all possibilities for σ 4 and ρ 4 are realized by some generic spectrahedra S 3,4 and their symmetroids Σ 3,4 under the following constraints: (14) σ 4 is even and σ 4 10; ρ 4 is a multiple of 4 and 4 ρ 4 0 (Degtyarev and Itenberg proved this for the spectrahedron and its symmetroid in projective space, that is why in our condition (14) above we have to double their estimates) An average picture of this result is given in the following proposition Proposition 16 (The average number of nodes on the boundary of a quartic spectrahedron) We have E σ 4 = and E ρ 4 = 1 3 It would be interesting to understand the distribution of the random variables σ 4, ρ 4 and compare it with the deterministic picture (14) We believe that the study of metric properties of spectrahedra such as volume is very promising for future research This lines up with our discussion presented at the beginning of Section 5 E σ In the general case we conjecture that lim n n E ρ n = 0, but it is difficult to predict how small is E σ n compared to E ρ n The main challenge is handling the characteristic function in the integral from Theorem 14 (1) 4

7 15 Another possible random model Another natural model of random spectrahedra is by defining them as linear sections of P n S N 1 by a uniformly distributed (l + 1)-dimensional plane V in Sym(n, R): (15) S l,n (V ) := P n S N 1 V, Before proceeding we argue in favor of the model (11) over the random linear section model (15) The main reason for this is that the expected volume of S l,n (V ) decays to zero for fixed l and n, which we prove in Proposition 17 below In fact, typically a spherical spectrahedron of the form S l,n (V ) is empty (this is essentially due to the fact that the volume of the positive semidefinite cone P n decays exponentially fast as n ), and this model is inaccessible for probabilistic studies For the model introduced in (11) this appears differently: for large n and l the spectrahedron S l,n keeps a fraction of about 15% of the volume of the sphere S l ; cf Theorem 11 In fact, the spectrahedron S l,n is never empty, as it contains an open neighborhood of (1, 0,, 0) S l Proposition 17 (Decay of the random linear section model) Let V be uniformly distributed in the Grassmannian of (l + 1)-dimensional subspaces of Sym(n, R) Then for every c > 0 we have P{S l,n (V ) } O(n c ) Proof Let us denote by µ(v ) the maximum number of positive eigenvalues that a matrix in V has For n 4, we have the simple bound P{S l,n (V ) } = P{µ(V ) = n} P{µ(V ) n + n} By [AL16, Lemma 4] the last quantity is smaller than O(n c ) for every c > 0 16 Notation Throughout the article some symbols are repeatedly used for the same purposes: Sym(n, R) stands for the space of n n real symmetric matrices By the symbols Q = (Q 1,, Q l ) Sym(n, R) l and R = (R 1,, R l ) Sym(n, R) l we denote a collection of l symmetric matrices and its rescaled version respectively, ie R i = 1 nl Q i The k-dimensional sphere endowed with the standard metric is denoted S k The symbol 1 stands for the unit matrix (of any dimension) For x = (x 0, x 1,, x l ) S l we denote the matrices Q(x) = x 1 Q x l Q l and A(x) = x Q(x) By S l,n, S l,n and Σ l,n we denote a (random) spectrahedron, its boundary and a symmetroid hypersurface respectively Letters α, λ and µ are used to denote eigenvalues and λ = 1 n λ stands for the rescaled eigenvalue λ 17 Organization of the article The organization of the paper is as follows In the next section we recall some known deviation inequalities for the smallest eigenvalue of a GOE(n)- matrix In Sections 3 5 we prove our main theorems, Section 6 deals with the case of quartic spectrahedra In the Appendix we prove a technical result, based on the computation of the expectation of the square of the characteristic polynomial of a GOE matrix Acknowledgements The authors wish to thank P Bürgisser, S Naldi and B Sturmfels for helpful suggestions and remarks on the paper Deviation inequalities for the smallest eigenvalue In this section we want to summarize known inequalities for the deviation of λ min (Q) from its expected value in the GOE(n) random matrix model The results that we present are due to [ML10] Note that in that reference, however, the inequalities are given for the largest eigenvalue λ max (Q) Since the GOE(n)-distribution is symmetric around 0, we have λ max (Q) λ min (Q) Using this we translate the deviation inequalities for λ max (Q) from [ML10] into deviation inequalities for λ min (Q) Furthermore, note that in [ML10, (1)] the variance for the GOE(n)-ensemble is defined differently than it is here: eigenvalues of a random matrix in [ML10] are times eigenvalues in our definition 5

8 We express the deviation inequalities in terms of the scaled eigenvalue λ min (Q), cf (13) The following Proposition is [ML10, Theorem 1] We will not need this result in the rest of the paper directly, but we decided to recall it here because it gives an idea of the behavior of the smallest eigenvalue of a random GOE(n) matrix, in terms of which our theorem on the volume of random spectrahedra is stated Proposition 1 For some constant C > 0, all n 1 and 0 < ɛ < 1, we have { λmin (Q) (1 + ɛ)} Ce C 1 nɛ 3 and Prob Q GOE(n) Prob { λmin (Q) (1 ɛ)} Ce C 1 n ɛ 3 Q GOE(n) Proposition 1 shows that for large n the mass of λ min (Q) concentrates exponentially around 1 Thus E λ min (Q) converges to 1 as the following proposition shows Proposition For some constant C > 0 and all n 1 we have E λ min (Q) + 1 E λ min (Q) + 1 Cn 3 Proof By [ML10, Equation after Corollary 3] we have (1) lim sup p n p 3 E λmin (Q) + 1 p < n The assertion follows from monotonicity of the integral: E λ min (Q) + 1 E λ min (Q) + 1 and (1) with p = 1 Remark The distribution of the scaled largest eigenvalue of a GOE(n) matrix for n is known as the Tracy-Widom distribution [CT96] Suprisingly, this distribution appears in branches of probability that at first sight seem unrelated For instance, the length of the longest increasing subsequence in a permutation that is chosen uniformly at random in the limit follows the Tracy-Widom distribution [JB99] In the survey article [CAT0] Tracy and Widom give an overview of appearances of the distribution in growth processes, random tilings, statistics, queuing theory and superconductors The present article adds spectrehedra to that list 3 Expected volume of the spectrahedron 31 Proof of Theorem 11 (1) Note that due to the rotational invariance of the standard Gaussian distribution N(0, 1) the volume of a spectrahedron S l,n can be computed as follows: { } (31) S l,n rel = S l,n S l = Prob ξ l ξ i Q i 0 iid ξ 0,,ξ l N(0,1) nl Using this and the following shorthand notation l (3) Q(x) = x i Q i, A(x) = x nl Q(x) i=1 we now write the expectation E S l,n rel of the relative volume of the random spectrahedron as: { } E S l,n rel = E Prob ξ l ξ i Q i 0 = E E 1 {A(ξ) 0} =: ( ) Q GOE(n) l nl Q ξ iid ξ 0,,ξ l N(0,1) where 1 Y denotes the characteristic function of the set Y Using Tonelli s theorem the two integrations can be exchanged: { (33) ( ) = E E 1 {A(ξ) 0} = E Prob ξ } Q(ξ) 0 ξ Q ξ Q GOE(n) l nl 6 i=1 i=1

9 For a unit vector x = (x 1,, x l ) S l 1 by the orthogonal invariance of the GOE-ensemble we have Q(x) GOE(n) which leads to { } ξ Q 0 nl ( ) = E ξ Prob Q GOE(n) = E Prob Q GOE(n) ξ (ξ1 + + ξ l ) 1 { ξ 0 l (ξ ξ l ) 1 } Q 0, n where in the second equality we again used Tonelli s theorem Let us put t l := ξ 0 l (ξ 1 + +ξ l ) 1 Note that by [NLJ95, (81)] the random variable t l follows the Student s t-distribution with l degrees of freedom Since this distribution is symmetric around the origin and t l n Q 0 is equivalent to t l 1 n λ min (Q), we have ( ) = E Prob Q GOE(n) t l { t l 1 n λ min (Q) } = E F l ( λmin (Q) ), Q GOE(n) where F l is the cumulative distribution function of the random variable t l This proves Theorem 11 (1) since ( ) = E S l,n rel 3 Proof of Theorem 11 () The random variable t l is absolutely continuous Moreover, its density F l is continuous and bounded uniformly in l [NLJ95, (8)] This combined with the following lemma proves Theorem 11 (): Lemma 31 Let f l : R R be a sequence of smooth functions such that there exists a constant c > 0 with f l c, l 1 Then E Q GOE(n) f l ( λmin (Q)) = f l ( 1)+O(n /3) uniformly in l Proof Write f l ( λ min ) as f l ( λ min ) = f l ( 1) + f l (x)( λ min + 1) for some x = x( λ min ) R Taking expectation we obtain E f l ( λ min ) = f l ( 1) + E(f l(x)( λ min + 1)) f l ( 1) + c E λ min + 1 = f l ( 1) + O(n 3/ ) where the last inequality follows from Proposition 33 Proof of Theorem 11 (3) In the preceding subsection we have shown the (uniform in l) asymptotic E Σ l,n rel = F l ( 1) + O(n 3 ), where F l (x) = Prob {t l x} and the random variable t l follows the student s t-distribution with l degrees of freedom By [NLJ95, (815)] for fixed x we have F l (x) = Φ(x)(1 + O(l 1 )) where Φ is the cumulative distribution function of the standard normal distribution Plugging in x = 1 settles Theorem 11 (3) 4 The average number of singular points For the study of the average number of singular points on the boundary of a random spectrahedron and on a symmetroid surface we will rely on the following proposition, which implies that this number is finite Proposition 41 Let S (k) l,n be the set of matrices of corank k in the spectrahedron S l,n and Σ (k) l,n the set of matrices of corank k in the symmetroid hypersurface Σ l,n For a generic choice of R = (R 1,, R l ) Sym(n, R) l the sets S (k) l,n, Σ(k) l,n Sl are semialgebraic of codimension ( ) k+1 Proof In the space Sym(n, R) consider the semialgebraic stratification given by the corank: Sym(n, R) = n k=0 Z(k), where Z (k) denotes the set matrices of corank k, and the induced stratification on the cone P n of positive semidefinite matrices P n = n k=0 (Z(k) P n ) These are Nash stratifications [AA1, Proposition 9] and the codimensions of both Z (k) and Z (k) P n are equal to ( ) k+1 7

10 Consider now the semialgebraic map (41) F : S l (Sym(n, R)) l Sym(n, R), (x, R) x x 1 R x l R l Then Σ (k) l,n = {x Sl F (R, x) Z (k) } and S (k) l,n consequently they are semialgebraic = {x Sl F (R, x) Z (k) P n } and We now prove that F is transversal to all the strata of these stratifications Then the parametric transversality theorem [Hir94, Chapter 3, Theorem 7] will imply that for a generic choice of R the set S l,n is stratified by the S (k) l,n and the same for the set Σ l,n To see that F is transversal to all the strata of the stratifications we compute its differential At points (x, R) with x e 0 = (1, 0,, 0) we have D (x,r) F (0, R) = x 1 R x l R l and the equation D (x,r) F ( x, Q) = P can be solved by taking x = 0 and R = (0,, 0, x 1 i P, 0,, 0) where x 1 i P is in the i-th entry and i is such that x i 0 (in other words, already variations in R ensure surjectivity of D (x,r) F ) All points of the form (e 0, R) are mapped by F to the identity matrix 1 which belongs to the open stratum Z (0), on which transversality is automatic (because this stratum has full dimension) This concludes the proof Proposition 4 For generic R Sym(n, R) 3 the number of singular points ρ n on the symmetroid Σ 3,n and hence the number of singular points σ n on S 3,n is finite and satisfies n(n + 1)(n 1) σ n ρ n 3 Moreover, for any n 1 there exists a generic symmetroid Σ 3,n with ρ n = n(n+1)(n 1) 3 singular points on it Proof The fact that σ n ρ n are generically finite follows from Proposition 41 with k =, as remarked before Observe that ρ n is bounded by twice (since Σ 3,n is a subset of S 3 ) the number #Sing(Σ C 3,n) of singular points on the complex symmetroid projective surface Σ C 3,n = {x CP 3 det(x x 1 R 1 + x R + x 3 R 3 )) = 0} Since Sing(Σ C 3,n) is obtained as a linear section of the set Z () C of n n complex symmetric matrices of corank two (using similar transversality arguments as in Proposition 41) we have that generically #Sing(Σ C 3,n) = deg(z () n(n+1)(n 1) C ) The latter is equal to 6 ; see [JH84] Now comes the proof of the second claim, we are thankful to Bernd Sturmfels and Simone Naldi for helping us with this For a generic collection of n + 1 linear forms L 1,, L n+1 in l + 1 variables we denote by p(x) := L 1 (x) L n+1 (x) their product and by P = {x R l+1 L i (x) > 0, i = 1,, n + 1} the polyhedral cone Let e int(p ) be any interior point of P Then [San13, Thm 11] implies that the derivative p, e of p along the constant vector field e R l+1 is a hyperbolic polynomial in direction e and that the closure of the connected component of R l+1 \ { p, e = 0} containing e is a spectrahedral cone Let s consider the intersection of this spectrahedral cone with the generic linear 4-space V R l+1 and denote by S 3,n, Σ 3,n the corresponding spectrahedron and its symmetroid surface respectively It is straightforward to check that the triple intersections of the hyperplanes L 1,, L n+1 when intersected with V produce ( ) n+1 3 = (n+1)n(n 1) 3 singular points on Σ 3,n This completes the proof since the above number coincides with the complex bound Now we prove Theorem 1 (1) 41 Proof of Theorem 1 (1) Recall that 1 S N 1 Sym(n, R) denotes the set of unit symmetric matrices such that λ 1 = λ The ordered eigenvalues α 1 (x) α n (x) of A(x) = x nl i=1 x iq i satisfy α i (x) = x n µ i (x), 8

11 e 0 = (1, 0) (x 0, 1 x 0 ) W e 0 = ( 1, 0) Figure 41 Two-dimensional depiction of R 4 : The north pole e 0 is identified with the unit matrix 1, while the south pole e 0 is identified with 1 By continuity there must be a point (x 0, 1 x 0 ) on the arc joining e0 and e0 that has 0 as eigenvalue where µ 1 (x) µ n (x) are the ordered eigenvalues of Q(x) = 3 i=1 x iq i Then Sing( S 3,n ) = {x S 3 α 1 (x) = α (x) = 0} Consider now the vector space: = {x = (x 0, x 1, x, x 3 ) S 3 µ 1 (x) = µ (x) and µ 1 (x) = 6nx 0 } W = span{q 1, Q, Q 3 } Sym(n, R) Because Q = (Q 1, Q, Q 3 ) GOE(n) 3, with probability one W is three-dimensional In the sequel, for fixed Q we will naturally identify R 4 with V := span{1, Q 1, Q, Q 3 } by using the isomorphism A : R 4 V, x A(x) By Proposition 41 (for k = ) the set of symmetric matrices with repeated eigenvalues has codimension two in Sym(n, R) Hence, with probability one there are finitely many half lines L 1,, L s where µ 1 (x) = µ (x) Because on span{1, L i } we have α 1 = α, every singular point in S 3,n maps into span{1, L i } for some i {1,, s} We examine now the condition for one of the half lines L 1,, L s actually contributing to a singular point The construction we make for this is depicted in Figure 41 Let V 0 := A(S 3 ) (an ellipsoid), and let A(y) be the point of intersection of some fixed L {L 1,, L s } with V 0 Note that, because L W, we have A(y) W, ie, the first coordinate of y is zero Moreover, by orthogonal invariance of the GOE(n) distribution we can assume that y = (0, 1, 0, 0) The arc on the sphere S 3 through y joining the north pole e 0 = (1, 0, 0, 0) S 3 and the south pole e 0 = ( 1, 0, 0, 0) S 3 is then parametrized by [ 1, 1] S 3, x 0 (x 0, 1 x 0, 0, 0) For the smallest eigenvalue we get x 0 α 1 (x 0 ) := x x 0 6n λ 1, where λ 1 is the smallest eigenvalue of the first matrix Q 1 This function is strictly monotone with α 1 ( 1) = 1 and α 1 (1) = 1 Hence it has exactly one zero, which correspond to a point on this arc where α 1 = α = 0, ie a singular points The above considerations show that the number of singular points on S 3,n = S 3,n (Q) equals the number s of half-lines on W = span{q 1, Q, Q 3 } that intersect 1 Let us write s(q) := s to put emphasis on its dependence on Q Observe that s(q) is equal to the number of lines (not half-lines) that intersect 1, unless there exists a matrix in W, where both the two smallest and the two largest eigenvalues coincide Since the latter is a non-generic property, s(q) = #(W 1 ) with probability one By the kinematic formula from [How93] we have: 1 E σ n = s(q) dq = E #(W 1) = GOE(n) 3 W G(3,Sym(n,R)) S n(n+1) 3 since the GOE(n) distribution of the Q i induces the uniform distribution on W This shows the assertion 9

12 4 Proof of Theorem 1 () and Proposition 13 Recall that S N 1 Sym(n, R) denotes the set of unit symmetric matrices with repeated eigenvalues By the same argument as in the proof of Theorem 1 (1) singular points of the symmetroid Σ 3,n = Σ 3,n (Q) are in one-to-one correspondence with matrices in span{q 1, Q, Q 3 } Invoking again the kinematic formula from [How93] we obtain: (4) E ρ n = E W G(3,Sym(n,R)) #(W ) = S n(n+1) 3 This finishes the proof of Theorem 1 and at the same time proves Proposition 13 5 The volume of the set of symmetric matrices with repeated eigenvalues In this section we give a proof of Theorem 14 That is, we compute the volume of the set of symmetric matrices with repeated eigenvalues and of its subset consisting of matrices whose two smallest eigenvalues coincide Before, however, we want to discuss some of it s consequences 51 The degree of the set of complex symmetric matrices with repeated eigenvalues Let us consider the complex hypersurface C P Sym(n, C) consisting of complex symmetric matrices with repeated eigenvalues Its ideal is generated by the discriminant D of the characteristic polynomial of a matrix: D(Q) = i<j (λ i(q) λ j (Q)) The discriminant is a homogeneous polynomial of degree n(n 1) in the entries of Q and it s a sum of squares of real polynomials [Par0, Chapter ] By definition of the degree for a generic line L P Sym(n, C) we have # ( L P C) = deg( C ) = n(n 1) On the other hand, from the integral geometry formula [How93] and Theorem 14 (3) it follows that E # (W P ) = n(n 1) = 1 deg( C ) where W is a uniformly distributed projective -plane in P Sym(n, R) Note, however, that since the real zero locus P P Sym(n, R) of the discriminant hypersurface C P Sym(n, C) is of codimension two the degree deg( C ) does not give an upper bound on the number #(W P ) of real symmetric matrices with repeated eigenvalues in a generic projective -plane W Indeed, by Proposition 13 for generic R 1, R, R 3 Sym(n, R) matrices in P span(r 1, R, R 3 )) P are in one-to-one correspondence with singular points on the projective symmetroid PΣ 3,n = {x RP 3 det(x x 1 R 1 + x R + x 3 R 3 ) = 0} This together with Proposition 4 implies that the number #(W P ) of matrices with repeated eigenvalues in the generic projective -plane W is bounded by ( ) n+1 3 Moreover, this bound is sharp: for some generic W we have #(W P ) = ( ) n+1 3 which is bigger than deg( C ) = n(n 1) for n > 5 The number of symmetric matrices with repeated eigenvalues in a generic projective -plane can be estimated using general Milnor-type bounds, ie estimates for the sum of Betti numbers of a real algebraic variety in terms of the degree of defining it polynomial Indeed, for any 3- dimensional space W Sym(n, R) the set W S 3 = W S N 1 is given by a homogeneous equation of degree n(n 1) Therefore by [Ler16, Proposition 14] we have b( W ) 4n 4 +O(n 3 ), where b( ) denotes the sum of Betti numbers with Z coefficients In particular, when W is finite we have #( W ) = b( W ) 4n 4 + O(n 3 ) Hence, for generic projective -plane W general techniques give the fourth order bound #(W P ) n 4 + O(n 3 ) whereas the above discussion shows that ( ) n+1 3 is the sharp cubic bound for #(W P ) 10

13 5 Proof of Theorem 14 As before let us denote by λ 1 (Q) λ n (Q) the ordered eigenvalues of a symmetric matrix Q Sym(n, R), by N := dim(sym(n, R)) = n(n+1) the dimension of the set of n n real symmetric matrices and by S N 1 the unit sphere in the space Sym(n, R) Let j, j = 1,, n 1 denote the set of n n real symmetric matrices of unit norm, whose j-th and (j + 1)-th eigenvalues are equal: (51) j := {Q S N 1 λ j (Q) = λ j+1 (Q)}, j = 1,, n 1 It is easily seen that j is a semialgebraic subset of S N, and from Section 4 one can deduce that it is of codimension two The smooth locus ( j ) sm of j consists of matrices of unit norm whose j-th and (j + 1)-th eigenvalues are equal and all other eigenvalues are of multiplicity one: (5) ( j ) sm = {Q S N 1 λ 1 (Q) < < λ j (Q) = λ j+1 (Q) < < λ n (Q)} Recall that denotes the algebraic set of n n real symmetric matrices of unit norm that have at least one repeated eigenvalue Observe that is a union of the sets j, j = 1,, n 1 and its smooth locus sm is a disjoint union of ( j ) sm, j = 1,, n 1 Furthermore, denote by Z n the normalization constant for the density of eigenvalues of the GOE(n)-ensemble: Z n := R n e λ (λ) dλ where (λ) = 1 i<j n (λ j λ i ) is the Vandermonde determinant It is equal to (53) Z n = π n n i=1 Γ(1 + i ) Γ( 3 ), see, eg, [Meh91, (1759) for γ = 1 ] The proof of Theorem 14 is based on the following proposition, which we prove in the subsequent subsection Proposition 51 Let 1 j < n Then j S N 3 = 4 ( ) ( ) n n Z n j 1 µ 1,,µ j 1<u u<µ j,,µ n n (µ i u) e µ u (µ) d(µ, u), Before proving the proposition we first use it to finish the proof of Theorem 14: Proof of Theorem 14 According to Proposition 51 we have 1 S N 3 = 4 ( ) n n (µ i u) e µ u (µ) d(µ, u) Z n u<µ 1,,µ n i=1 Interpreting µ 1,, µ n as the eigenvalues of a GOE(n ) matrix we can rewrite this as follows: 1 S N 3 = 4Z ( ) n n [ E 1{Q u1 0} det(q u1) ] e u du Z n Q GOE(n ) From (53) it s easy to see that Z n = 8 Γ( n+1 n+ )Γ( ) Z n or, using the duplication formula for Gamma function, Z n = n+3 π n! Z n From this we get ( ) 1 (54) S N 3 = n 1 n [ E 1{Q u1 0} det(q u1) ] e u du π n! Q GOE(n ) which proves Theorem 14 (1) For Theorem 14 () note that since sm = n 1 j=1 ( j) sm is a disjoint union we have that = n 1 j=1 j and hence, by Proposition 51, ) S N 3 = 4 Z n ( n ) n 1 j=1 ( n j 1 i=1 µ 1,,µ j 1<u u<µ j,,µ n 11 n (µ i u) e µ u (µ) d(µ, u), i=1

14 This, together with the summation lemma [Bre17, Lemma E35], gives S N 3 = 4 ( ) n n (µ i u) e µ u (µ) dµdu, Z n µ R n i=1 Again, treating µ 1,, µ n as the eigenvalues of a GOE(n ) matrix and then proceeding as we did to get (54) we obtain ( ) (55) S N 3 = n 1 n [ E det(q u1) ] e u du π n! Q GOE(n ) This proves Theorem 14 () Finally, Theorem A1 entails that the integral on the right-hand-side of (55) is equal to π n!, n 1 which implies that S N 3 = ( n ) This finishes the proof 53 Proof of Proposition 51 Recall that ( j ) sm = {Q S N 1 λ 1 (Q) < < λ j (Q) = λ j+1 (Q) < < λ n (Q)} In the following, we denote for brevity λ i := λ i (Q) In order to compute j = ( j ) sm define for δ > 0 (56) X j (δ) := {Q ( j ) sm λ j λ j 1 > δ, λ j+ λ j+1 > δ} Then ( j ) sm = δ>0 X j(δ) and by continuity of the Lebesgue measure (57) j = lim δ 0 X j (δ) For a fixed δ > 0 and for any ε > 0 let T (X j (δ), ε) S N 1 denote the ε-tube around X j (δ) S N 1 Weyl s formula [Wey39] gives the expansion of the volume of the ε-tube around a submanifold of the sphere Here it is enough to have it in the following simplified form Theorem 5 (Weyl s tube formula for X j (δ)) For any ε > 0, such that the fibres of T (X j (δ), ε) do not intersect, the volume of the ε-tube around X j (δ) is T (X j (δ), ε) = πε X j (δ) + O(ε 3 ) In the lemma below we describe the ε-tube around X j (δ) and show that for a sufficiently small ε > 0 its fibers do not intersect We postpone its proof to the next subsection Lemma 53 For 0 < ε < arctan( δ) we have λ 1 < < λ j λ j+1 < < λ n, T (X j (δ), ε) = Q S N 1 λ j+1 λ j < sin ε, λ j+ λ j + λ j+1 > δ cos ε, λ j + λ j+1 and the fibers of T (X j (δ), ε) do not intersect λ j 1 > δ cos ε Combining this lemma with Weyl s formula we are allowed to compute the volume of X j (δ) as X j (δ) = lim ε 0 1 πε T (X j (δ), ε) and, consequently, by (57): 1 (58) j = lim lim δ 0 ε 0 πε T (X j (δ), ε) To actually compute this limit, we will rewrite the volume of T (X j (δ), ε) in terms of a GOE(n) random variable The following formula gives the relation between the volume of a measurable set E S N 1 and the GOE(n)-measure of the homogeneous cone C(E) = {A Sym(n, R) A/ A E} over E E S N 1 = Prob Q GOE(n) C(E) 1

15 Applying this to the measurable set T (X j (δ), ε) we get λ 1 < < λ j λ j+1 < < λ n, T S N 1 (X j (δ), ε) S N 1 λ j+1 λ j < Q sin ε, = Prob Q GOE(n) λ j+ λ j + λ j+1 > δ Q cos ε (59) λ j + λ j+1 λ j 1 > δ Q cos ε In the following we denote the event λ 1 < < λ j λ j+1 < < λ n, = ( ) λ j+1 λ j < Q sin ε, E(λ) := λ j+ λ j + λ j+1 > δ Q cos ε λ j + λ j+1 λ j 1 > δ Q cos ε Writing (59) in terms of the density of eigenvalues of the GOE(n) ensemble it becomes ( ) = n! 1 E(λ) e λ (λ 1,, λ n ) dλ, Z n R n see [Mui8, Theorem 317] Here, (λ 1,, λ n ) = 1 i<j n (λ j λ i ) is the Vandermonde determinant, 1 E(λ) is the characteristic function of the set E(λ), and Z n is the normalization constant from (53) We express now the integral in terms of the following event: λ 1,, λ j 1 < λ j, λ j+1 < λ j+1,, λ n, λ j+1 λ j < λ sin ε, Ẽ(λ) := λ i λ j + λ j+1 > δ λ cos ε for i j +, λ j + λ j+1 λ i > δ λ cos ε for i j 1 There are (j 1)! possibilities to arrange the first j 1 eigenvalues, possibilities to arrange λ j and λ j+1 and (n (j + 1))! possibilities to arrange the last n (j + 1) eigenvalues Hence, ( ) = n! 1 1Ẽ(λ) e λ (λ 1,, λ n ) dλ Z n (j 1)!(n (j + 1))! R n = 1 ( ) ( ) n n λ (λ 1,, λ n ) dλ Z n j 1 R n 1Ẽ(λ) e Next, we perform the following orthogonal change of variables µ 1 := λ 1,, µ j 1 := λ j 1, µ j := λ j+,, µ n := λ n and x = λ j + λ j+1, y = λ j+1 λ j (µ 1,, µ n now become the eigenvalues of a new GOE(n ) matrix and we treat the variables x, y separately) We get ( ) = 1 ( )( ) n n 1Ê(µ,x,y) g(µ, x, y)e µ +y +x (510) (µ) d(µ, x, y) Z n j 1 (µ,x,y) R n R R where (511) g(µ, x, y) = y n i=1 13 ((µ i x ) y )

16 and µ 1,, µ j 1 < 1 (x y), 1 (x + y) < µ j,, µ n, y < (µ, x, y) sin ε, Ê(µ, x, y) := µ i x > δ (µ, x, y) cos ε for i j, x µ i > δ (µ, x, y) cos ε for i j 1 We perform another change of varables: (51) y sin ε (µ, x) t = sin ε (µ,x,y) dy = (1 (sin ε) t ) 3/ x, µ 1,, µ n are as before Note that after this change a factor of (sin ε) appears and the function y(t, x, µ, ε) 0 in the limits ε 0 We multiply the integral in (510) by 1 πε and, thereafter, invoke the dominated convergence theorem that allows us to pass to the limit ε 0 under the integral: 1 lim ε 0 πε 1Ê(µ,x,y) g(µ, x, y)e µ +y +x (µ) d(µ, x, y) (µ,x,y) R n R R 1 n ( = 1 π Ē(µ,x) t µ, x µ i x ) e µ +x (µ) dt d(µ, x), where (µ,x) R n R t= 1 Using that 1 t dt = 1 we have t= 1 1 lim ε 0 πε T (X j (δ), ε) S N 1 ( ) ( ) n n = πz n j 1 i=1 µ 1,, µ j 1 < x < µ j,, µ n, µ Ē(µ, x) := i x > δ (µ, x) for i j, x µ i > δ (µ, x) for i j 1 R n R n 1 Ē(µ,x) µ, x i=1 ( µ i x ) e µ +x (µ) d(µ, x), Plugging this into (58) and again using the dominated convergence theorem we get j = lim πε T (X j (δ), ε) S N 1 ( ) ( ) n n = πz n j 1 1 lim δ 0 ε 0 D n µ, x i=1 ( µ i x ) e µ +x (µ) d(µ, x), where the region of integration is D = {µ 1,, µ j 1 < x < µ j,, µ n } Next, we use the fact that for a homogeneous function f of degree d in m variables one has x f(x)e x dx = (d + m) f(x)e x dx x R m x R m (the proof is straightforward using polar coordinates) In our case, we have m = n 1 and d = (n ) + (n )(n 3) = (n )(n+1) Thus d + m = n +n 4 and j = (n + n 4) S N 1 ( ) ( ) n n n ( µ i x ) e µ +x (µ) d(µ, x) πzn j 1 D i=1 14

17 Finally, we make a change of variables u := x and use (n + n 4) S N 1 = 4π S N 3 to conclude that j S N 3 = 4 Z n ( n ) ( ) n j 1 µ 1,,µ j 1<u u<µ j,,µ n n (µ i u) e µ u (µ) d(µ, u) i=1 This completes the proof of Proposition Proof of Lemma 53 We can assume without loss of generality that Q X j (δ) is diagonal: Q = diag(λ 1, λ,, λ n ) and λ 1 < < λ j = λ j+1 < < λ n Then, the fiber N Q T Q S N 1 of the normal bundle to X j (δ) S N 1 at Q is described as follows For a, b R let V a,b = (v i,j ) Sym(n, ( R) be the matrix that has zeros everywhere except for the following vj,j v block on the diagonal: j,j+1 ) ( v j+1,j v j+1,j+1 = 1 a b ) b a Note that Va,b S N 1 if and only if a + b = 1 We claim that (513) N Q = {V a,b a, b R} It is easy to see that V a,b is orthogonal to Q, ie, V a,b T Q S N 1, and that the tangent space T Q X j (δ) T Q S N 1 to X j (δ) at Q is spanned by the following ( ) n+1 3 vectors: diag(λ e 1 λ 1 e ), diag(λ j 1 e j λ j e j 1 ) diag(λ j+ e j 1 λ j 1 e j+ ) diag(λ j+3 e j+ λ j+ e j+3 ) diag(λ n e n 1 λ n 1 e n ), and diag ( λ j λ i e i + λ i (e j + e j+1 ) ), i j,j+1 i j,j+1 and r s r 0 λ s λ r, r, s = 1, n, r s, {r, s} {j, j + 1} s λ s λ r 0 It is immediate to see that these vectors are all orthogonal to V a,b Thus, N Q = {V a,b a, b R} Now we prove that T (X j (δ), ε) has the asserted form and that the fibers of the normal ε-tube T (X j (δ), ε) do not intersect provided that ε < arctan( δ) The fibers are swept out by geodesics of length less than ε starting at Q in the direction of some V a,b S N 1, in formulas: 15

18 {cos t Q + sin t V a,b 0 t < ε} We write explicitly the matrix cos t Q + sin t V a,b : λ 1 cos t λj 1 cos t λ j cos t + a sin t b sin t b sin t λ j+1 cos t a sin t λ j+ cos t λn cos t Provided that ε < arctan( δ) the eigenvalues of this matrix are (514) λ 1 cos t < < λ j 1 cos t < λ j cos t ± sin t < λ j+ cos t < < λ n cos t since Q X j (δ) (see (56)) Moreover, for 0 t < ε these eigenvalues satisfy the inequalities (λ j cos t + sin t ) + (λ j cos t sin t ) λ j 1 cos t = (λ j λ j 1 ) cos t > δ cos ε, (λ j cos t + sin t ) (λ j cos t sin t ) = sin t < sin ε, λ j+ cos t (λ j+1 cos t + sin t ) + (λ j+1 cos t sin t ) = (λ j+ λ j+1 ) cos t > δ cos ε This shows that T (X j (δ), ε) is contained in the set we claim it to be To show the other inclusion let A S N 1 be a matrix whose eigenvalues α 1 < < α j 1 < α j α j+1 < α j+ < < α n satisfy α j + α j+1 α j 1 > δ cos ε, α j+1 α j < sin ε, and α j+ α j + α j+1 > δ cos ε We can assume again that A = diag(α 1, α,, α n ) is diagonal Let 0 t < ε be such that α j+1 α j = sin t One can easily verify that A = cos t Q + sin t V 1,0 for Q = 1 cos t diag(α 1,, α j 1, 1 (α j + α j+1 ), 1 (α j + α j+1 ), α j+,, α n ) X j (δ) This implies that A T (X j (δ), ε) and T (X j (δ), ε) has the claimed form It remains to show that the fibers of the normal ε-tube T (X j (δ), ε) do not intersect when ε < arctan( δ) For this assume there is another representation A = cos t Q 0 +sin t V of the matrix A = diag(α 1,, α n ) T (X j (δ), ε), where Q 0 X j (δ), V N Q0 and 0 t < ε We will prove that actually Q 0 = Q, V = V 1,0 and t = t To show this, we consider the diagonalization of Q 0 ; that is, Q 0 = C1 T Q 1 C 1, where Q 1 = diag(λ 1,, λ n ) is diagonal and C 1 is orthogonal We may assume λ 1 < < λ j 1 < λ j = λ j+1 < λ j+ < < λ n Note that the normal bundle N Q0 to X j (δ) at Q 0 = C1 T Q 1 C 1 is given by N C T 1 Q 1C 1 = C1 T N Q1 C 1 = {C1 T V a,b C 1 a, b R} It follows that V = C1 T V a,b C 1 for some a, b R and we can write A = C1 T (cos t Q 1 + sin t V a,b )C 1 Note that the eigenvalues of the inner matrix are given as in (514) Therefore, we can write 16

19 A = C1 T C T Q C C 1, where the orthogonal matrix C commutes with Q 1 and ( Q = diag λ 1 cos t,, λ j 1 cos t, λ j cos t sin t, λ j cos t + sin ) t, λ j+ cos t,, λ n cos t = cos t Q 1 + sin t V 1,0 The condition ε < arctan( δ) together with Q 1 X j (δ) ensures λ j 1 cos t < λ j cos t sin t and λ j cos t + sin t < λ j+ cos t Now since the diagonal matrices A and Q both have ordered entries it follows that C C 1 can be taken to be the identity matrix Therefore α i = λ i cos t for i = 1,, j 1, j +,, n, and α j = λ j cos t sin t and α j+1 = λ j cos t + sin t It is straightforward now to see that t = t, Q 0 = Q and V = V 1,0 6 Proof of Proposition 16 The identity E ρ 4 = 1 follows immediately from Theorem 1 and Theorem 14 for n = 4 For the other identity we apply again Theorem 1 and Theorem 14 and write: E σ 4 = 1 S N 3 ( ) = π 4! = 4 ( 1 π = ( ) R Z E Q GOE() [ 1{Q u1 0} det(q u1) ] e u du R (λ 1 u) (λ u) λ 1 λ e λ 1 +λ 1 {λ1 u>0,λ u>0}dλ 1 dλ ) e u du We apply now the change of variable α 1 = λ 1 u and α = λ u in the innermost integral, obtaining: ( ) = 4 ( ) 1 (α 1 α ) α 1 α e α 1 +α e u u(α 1+α ) dα 1 dα e u du π R Z = 4 1 π Z R + = 4 Γ ( ) 3 π π = 1 π R + R + (α 1 α ) α 1 α e α 1 +α R + ( ( (α 1 α ) α 1 α e α 1 +α π ) e u u(α 1+α ) du R ) e (α1+α) 8 (α 1 α ) α 1 α e α 1 +α + (α 1 +α ) 8 dα 1 dα dα 1 dα = (α 1 α ) π R α 1 α e α 1 +α + (α 1 +α ) 8 dα 1 dα + {α1<α} dα 1 dα In the last equality we have used the fact that the integrand is invariant under the symmetry (α 1, α ) (α, α 1 ) Consider now the map F : R + {α 1 < α } R[x] R given by F (α 1, α ) = (x α 1 )(x α ) = x (α 1 + α )x + α 1 α Essentially, F maps the pair (α 1, α ) to a monic polynomial of degree two whose ordered roots are (α 1, α ) Observe that F is injective on the region R + {α 1 < α } with never-vanishing Jacobian JF (α 1, α ) = α 1 α What is the image of F in the space of polynomials R[x]? Denoting by a 1, a the coefficients of a monic polynomial p(x) = x a 1 x + a R[x], we see first that the conditions α 1, α > 0 imply a 1, a > 0 Moreover the polynomial p(x) = F (α 1, α ) has by construction real roots, hence its discriminant a 1 4a must be positive Viceversa, given (a 1, a ) such that a 1, a > 0 and a 1 4a > 0, the roots of p(x) = x a 1 x + a are real and 17

20 positive Hence, F (R + {α 1 < α }) = {(a 1, a ) R a 1, a > 0, a 1 4a > 0} Thus we can write the above integral as ( ) = a 1 4 a e a 1 a + a 1 8 da da 1 = e 3a a a e a da da 1 π π and performing elementary integration we obtain ( ) = E σ 4 = Appendix In this section we will give a proof for the following formula, that is needed in the proof of Theorem 14 Theorem A1 For a fixed positive integer k we have E det(q Q GOE(k) u1) e u du = (k + )! π k+1 Before proving Theorem A1 below we give definition of Hermite polynomials, recall their properties that we will need and then prove an auxiliary proposition A1 Hermite polynomials The (physicist s) Hermite polynomials H i (x), i = 0, 1,, form a family of orthogonal polynomials on the real line with respect to the measure e x dx They are defined by (A1) and satisfy (A) H i (x) = ( 1) i e x H i (u)h j (u)e u du = di dx i e x, i 0 { i i! π, if i = j 0, else A Hermite polynomial is either odd (if the degree is odd) or even (if the degree is even) function: (A3) and its derivative satisfies (A4) H i ( x) = ( 1) i H i (x); H i(x) = ih i 1 (x) (see [JS00, (4:5:1)], [IG15, (8951)] for these properties) A The expected value of the square of the characteristic polynomial of a GOEmatrix The following proposition is crucial for the proof of Theorem A1 Proposition A (Expected value of the square of the characteristic polynomial) For a fixed positive integer k and a fixed u R the following holds (1) If k = m is even, then where E det(q Q GOE(k) u1) = (m)! m ( H X j (u) = j (u) H j+1 (u) H j (u) 18 m H j+1 j 1 (j)! det X j (u), H j (u) ) (u) H j (u)

21 () If k = m + 1 is odd, then where E det(q Q GOE(k) u1) = π(m + 1)! 4m+ Γ(m + 3 ) m j (j)! det Y j (u), (j)! j! H j (u) H j (u) Y j (u) = 0 H j+1 (u) H j (u) H j+1 (u) H j (u) (m+)! (m+1)! H m+ (u) H m+(u) For the proof of Proposition A we need the following lemma Lemma A3 Let P m = 1 m π m m i=0 (i)! and let Z m denote the normalization constant from (53) Then P m = 1 m Z m Proof By (53) we have Z m = (π) m m Γ( i +1) i=0 Γ( 3 Use the duplication formula for Gamma ) functions to transform this into m Z m = (π) m Γ( i 1 + 1)Γ( i + 1) Γ( 3 ) i=0 i=0 Now we use the formula Γ(z)Γ(z + 1 ) = 1 z Γ(z) π [JS00, 43:5:7] with z = i 1 to get m Z m = (π) m 1 (i 1) π Γ(i 1 + ) m Γ( 3 ) = (π) m i π Γ(i + 1) Γ( 3 ) Using furthermore Γ( 3 ) = π we get Z m = m m π m m i=0 (i)! Therefore, 1 m m m π i=0 (i)! as claimed P m Z m = Now we prove Proposition A i=0 m m m m = 1 m π i=0 (i)! Proof of Proposition A In [Meh91, Section ] one finds two different formulas for the cases k = m even and k = m + 1 odd If k = m, we have by [Meh91, (38)] that E det(q u1) = (m)! P m m j 1 Z m (j)! ( Rj (u) R j det (u) ) R j+1 (u) R j+1 (u), where P m = 1 m π m m i=0 (i)! is as in Lemma A3, Z m is the normalization constant (53) and where R j (u) = j H j (u) and R j+1 (u) = (j+1) (H j+1 (u) H j (u)) Using the multilinearity of the determinant we get E det(q u1) = (m)! P m m j det X j (u) Z m (j)! By Lemma A3 we have Pm Z m = 1 m Putting everything together yields the first claim In the case k = m + 1 we get from [Meh91, (38)] that E det(q u1) = (m + 1)! P m Z m+1 m j 1 (j)! 19 g j R j (u) R j (u) det g j+1 R j+1 (u) R j+1 (u), g m+ R m+ (u) R m+(u)

22 where P m, R j (u), R j+1 (u) are as above and g i = R i (u) exp( u ) du Note that by (A3) H j+1 (u) is an odd function Hence, we have g j+1 = 0 For even indices we use [IG15, (7373)] to get g j = j π (j)! j! By the multilinearity of the determinant: π(m + 1)! (A5) E det(q u1) Pm m j = m+ det Y j (u) Z m+1 (j)! By (53) we have Z m+1 = Γ(m + 3 ) Z m Using also Lemma A3 we see that Plugging this into (A5) we get This finishes the proof E det(q u1) = P m m = Z m+1 Γ(m + 3 ) π(m + 1)! 4m+ Γ(m + 3 ) m j (j)! det Y j (u) A3 Proof of Theorem A1 Due to the nature of Proposition A we also have to make a distinction in the proof of Theorem A1 In the case k = m we use the formula from Proposition A (1) to write E det(q u1) e u du = (m)! m j 1 m det X j (u) du (j)! By (A4) we have H i (u) = ih i 1(u) Hence, X j (u) can be written as ( ) H j (u) 4jH j 1 (u) H j+1 (u) 4jH j 1 (u) (j + 1)H j (u) 8j(j 1)H j (u) From (A) we can deduce that det X j (u) du = (j + 1) j (j)! π + 16j j 1 (j 1)! π From this we see that m j 1 (A6) (j)! and hence, u = j+1 (j)! π(4j + 1) det X j (u) du = π m (4j + 1) = π (m + 1)(m + 1) E det(q u1) e u du = (m)! m π (m + 1)(m + 1) = (m + )! m+1 π Plugging back in m = k finishes the proof of the case k = m In the case k = m + 1 we use the formula from Proposition A () to see that π(m + 1)! m E det(q u1) j e u du = 4m+ Γ(m + 3 ) det Y j (u) e u du (j)! Note that the top right -submatrix of Y j (u) is X j (u), so that det Y j (u) is equal to (m + )! (A7) det X j (u) + (j)! ( Hj+1 (u) H j det (u) H j+1 (u) H j (u) ) (m + 1)! j! H m+ (u) H m+(u) 0 u

23 Because taking derivatives of Hermite polynomials decreases the index by one (A4) and because the integral over a product of two Hermite polynomials is only non-vanishing, if their indices agree, the integral of the determinant in (A7) is only non-vanishing for j = m, in which case it is equal to H m+1 (u)h m+(u) e u du = (m + ) m+1 (m + 1)! π, by (A) and (A4) Hence, det Y j (u) e u du (m+)! (m+1)! det X m(u) e u du + (m)! m! m+ (m + )! π, if j = m, = (m+)! (m+1)! det X j(u) e u du, else We find that m j (j)! u det Y j (u) e u du (m + )! (m + )! = π + m! (m + 1)! (m + )! (m + )! = π + m! (m + 1)! π (m + 3)! = ; m! m j (j)! u π (m + 1)(m + 1) det X j (u) e u du the second-to-last line by (A6) It follows that π(m + 1)! π E det(q u1) (m + 3)! e u du = 4m+ Γ(m + 3 ) m! = π(m + 1)!(m + 3)! 4m+3 Γ(m + 3 ) m! Using the duplication formula for Gamma function, it is not difficult to verify that the last term is m π (m + 3)! Substituting m + 1 = k shows the assertion in this case References [AA1] A Agrachev, A Lerario: Systems of quadratic inequalities Proc Lond Math Soc (3), 105(3):6 660, 01 [AD11] A Degtyarev, I Itenberg: On real determinantal quartics Proceedings of the Gökova Geometry Topology Conference 010, 011 [AL16] A Lerario, E Lundberg: Gap probabilities and Betti numbers of a random intersection of quadrics Discrete Comput Geom, 55():46 496, 016 [Ali95] Alizadeh, F: Interior Point Methods in Semidefinite Programming with Applications to Combinatorial Optimization SIAM Journal on Optimization, 5(1):13 51, 1995 [Bre17] Breiding, P: Numerical and statistical aspects of tensor decompositions, PhD Thesis Technische Universität Berlin, [CAT0] C A Tracy, H Widom: Distribution functions for largest eigenvalues and their applications ArXiv Mathematical Physics e-prints, 00 [CT96] CA Tracy, H Widom: On orthogonal and symplectic matrix ensembles Communications in Mathematical Physics, 177(3):77 754, 1996 [GB1] G Blekherman, P A Parrilo, R R Thomas: Semidefinite Optimization and Convex Algebraic Geometry SIAM Series on Optimization 13, 01 [Hir94] Hirsch, M W: Differential topology Graduate Texts in Mathematics Springer Verlag, New York, 1994 Corrected reprint of the 1976 original [How93] Howard, R: The kinematic formula in Riemannian homogeneous spaces Mem Amer Math Soc, 106(509):vi+69,