ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES. Ren Guo & Feng Luo. Abstract

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1 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES Ren Guo & Feng Luo Dedcated to the memory of Xao-Song Ln. Abstract We establsh the second part of Mlnor s conjecture on the volume of smplexes n hyperbolc and sphercal spaces. A characterzaton of the closure of the space of the angle Gram matrces of smplexes s also obtaned. 1. Introducton Mlnor s conjecture. In [5], John Mlnor conjectured that the volume of a hyperbolc or sphercal n smplex, consdered as a functon of the dhedral angles, can be extended contnuously to the degenerated smplexes. Furthermore, he conjectured that the extended volume functon s non-zero except n the closure of the space of Eucldean smplexes. The frst part of the conjecture on the contnuous extenson was establshed n [4] ([7] has a new proof of t whch generalzes to many polytopes). The purpose of the paper s to establsh the second part of Mlnor s conjecture. To state the result, let us begn wth some notatons and defntons. Gven an n smplex n a sphercal, hyperbolc or Eucldean space wth vertces u 1,..., u n+1, the -th codmenson-1 face s defned to be the (n 1)-smplex wth vertces u 1,..., u 1, u +1,..., u n+1. The dhedral angle between the -th and j-th codmenson-1 faces s denoted by θ j. As a conventon, we defne θ = π and call the symmetrc matrx A = [ cos(θ j )] (n+1) (n+1) the angle Gram matrx of the smplex. It s well known that the angle Gram matrx determnes a hyperbolc or sphercal n-smplex up to sometry and Eucldean n-smplex up to smlarty. Let X n+1, Y n+1, Z n+1 n R (n+1) (n+1) be the subsets of (n + 1) (n + 1) symmetrc matrces correspondng to the angle Gram matrces of sphercal, hyperbolc, or Eucldean n smplexes respectvely. The volume of an n-smplex can be expressed n terms of the angle Gram matrx by the work of Aomoto [1], Kneser [2] and Vnberg [8]. Namely, for a sphercal or hyperbolc n-smplex σ n wth angle Gram matrx A, the volume V s 1

2 2 REN GUO & FENG LUO (1.1) V (A) = µ 1 n det(ad(a)) e xtad(a)x dx. R n+1 0 where R n+1 0 = {(x 1,..., x n+1 ) x 0}, the constant µ k = 0 x k e x2 dx and ad(a) s the adjont matrx of A. In [4], t s proved that the volume functon V : X n+1 Y n+1 R can be extended contnuously to the closure X n+1 Y n+1 n R (n+1) (n+1). The man result of ths paper, whch verfes the second part of Mlnor s conjecture, s the followng theorem. Theorem 1. The extended volume functon V on the closure X n+1 Y n+1 n R (n+1) (n+1) vanshes at a pont A f and only f A s n the closure Z n+1. A characterzaton of angle Gram matrces. We wll use the followng conventons. Gven a real matrx A = [a j ], we use A 0 to denote a j 0 for all, j and A > 0 to denote a j > 0 for all, j. A t s the transpose of A. We use ad(a) to denote the adjont matrx of A. The dagonal k k matrx wth dagonal entres (x 1,...x k ) s denoted by dag(x 1,...x k ). A characterzaton of the angle Gram matrces n X n+1, Y n+1 or Z n+1 s known by the work of [3] and [5]. Proposton 2 ([3],[5]). Gven an (n+1) (n+1) symmetrc matrx A = [a j ] wth a = 1 for all, then (a) A Z n+1 f and only f det(a) = 0, ad(a) > 0 and all prncpal n n submatrces of A are postve defnte, (b) A X n+1 f and only f A s postve defnte, (c) A Y n+1 f and only f det(a) < 0, ad(a) > 0 and all prncpal n n submatrces of A are postve defnte. In partcular, all off-dagonal entres a j have absolute values less than 1,.e., a j < 1 for j. The followng gves a characterzaton of matrces n X n+1, Y n+1 and Z n+1 n R (n+1) (n+1). Theorem 3. Gven an (n + 1) (n + 1) symmetrc matrx A = [a j ] wth a = 1 for all, then (a) A Z n+1 f and only f det(a) = 0, A s postve sem-defnte, and there exsts a prncpal (k + 1) (k + 1) submatrx B of A so that det(b) = 0, ad(b) 0 and ad(b) 0, (b) A X n+1 f and only f ether A s n X n+1 or there exsts a dagonal matrx D = dag(ε 1,..., ε n+1 ) where ε = 1 or 1 for each = 1,..., n + 1, such that DAD Z n+1, (c) A Y n+1 f and only f ether A Z n+1 or det(a) < 0, ad(a) 0 and all prncpal n n submatrces of A are postve sem-defnte.

3 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 3 The paper s organzed as follows. In secton 2, we characterze normal vectors of degenerated Eucldean smplexes. In secton 3, we characterze angle Gram matrces of degenerated hyperbolc smplexes. Theorem 1 s proved n secton 4 and Theorem 3 s proved n secton 5. Acknowledgment. We would lke to thank the referee for a very careful readng of the manuscrpt and for hs/her nce suggestons on mprovng the exposton. 2. Normal vectors of Eucldean smplexes As a conventon, all vectors n R m are column vectors and the standard nner product n R m s denoted by u v. In the sequel, for a non-zero vector w R n, we call the set {x R n w x 0} a closed half space, and the set {x R n w x > 0} an open half space. Defne E n+1 = {(v 1,..., v n+1 ) (R n ) n+1 v 1,..., v n+1 form unt outward normal vectors to the codmenson-1 faces of a Eucldean n-smplex }. Followng Mlnor [5], a matrx s called undagonal f ts dagonal entres are 1. An (n + 1) (n + 1) symmetrc undagonal matrx A s n Z n+1 f and only f A = [v v j ] for some pont (v 1,..., v n+1 ) E n+1 (ths s proved n [3],[5]). We clam that an (n + 1) (n + 1) symmetrc undagonal matrx A s n Z n+1 f and only f A = [v v j ] for some pont (v 1,..., v n+1 ) n the closure E n+1 n (R n ) n+1. Indeed, f A = [v v j ] for some pont (v 1,..., v n+1 ) E n+1, then there s a sequence (v (m) 1,..., v (m) n+1 ) E n+1 convergng to (v 1,..., v n+1 ). We have a sequence of matrces A (m) = [v (m) v (m) j ] Z n+1 convergng to A. Conversely f A Z n+1, then there s a sequence of matrces A (m) Z n+1 convergng to A. Wrte A (m) = [v (m) v (m) j ], where (v (m) 1,..., v (m) n+1 ) E n+1. Snce has norm 1 for all, m, by takng subsequence, we may assume lm m (v (m) 1,..., v (m) n+1 ) = (v 1,..., v n+1 ) E n+1 so that A = [v v j ]. A geometrc characterzaton of elements n E n+1 was obtaned n [3]. For completeness, we nclude a proof here. v (m) Lemma 4. A collecton of unt vectors (v 1,..., v n+1 ) (R n ) n+1 s n E n+1 f and only f one of the followng condtons s satsfed. (4.1) The vectors v 1,..., v n+1 are not n any closed half-space. (4.2) Any n vectors of v 1,..., v n+1 are lnear ndependent and the lnear system n+1 =1 a v = 0 has a soluton (a 1,..., a n ) so that a > 0 for all = 1,..., n + 1. Proof. (4.2) (4.1). Suppose otherwse, v 1,..., v n+1 are n a closed half-space,.e., there s a non-zero vector w R n so that w v 0, = 1,..., n + 1. Let a 1,..., a n+1 be the postve numbers gven by (4.2) so

4 4 REN GUO & FENG LUO that n+1 =1 a v = 0. Then n+1 0 = w ( a v ) = =1 n+1 a (w v ). But by the assumpton a > 0, w v 0 for all. Thus w v = 0 for all. Ths means that v 1,..., v n+1 le n the (n 1)-dmensonal subspace perpendcular to w. It contradcts the assumpton n (4.2) that any n vectors of v 1,..., v n+1 are lnear ndependent. (4.1) (4.2). To see that any n vectors of v 1,..., v n+1 are lnear ndependent, suppose otherwse, some n vectors of v 1,..., v n+1 are lnear dependent. Therefore there s an (n 1)-dmensonal hyperplane contanng these n vectors. Then v 1,..., v n+1 are contaned n one of the two closed half spaces bounded by the hyperplane. It contradcts to the assumpton of (4.1). Snce v 1,..., v n+1 are lnear dependent, and any n of them are lnear ndependent, we can fnd real numbers a 0 for all such that n+1 =1 a v = 0. For any j, let H j be the (n 1)-dmensonal hyperplane spanned by the n 1 vectors {v 1,..., v n+1 } \ {v, v j } and u R n {0} be a vector perpendcular to H j. We have =1 n+1 0 = u ( a v ) = a (u v ) + a j (u v j ). =1 By the assumpton of (4.1), v and v j must le n the dfferent sdes of H j. Thus u v and u v j have dfferent sgn. Ths mples that a and a j have the same sgn. Hence we can make a > 0 for all. E n+1 (4.1). We wll show (v 1,..., v n+1 ) E n+1 f and only f the condton (4.1) holds. In fact, gven an n dmensonal Eucldean smplex σ, let S n 1 be the sphere nscrbed to σ. We may assume after a translaton and a scalng that S n 1 s the unt sphere centered at the orgn. Then the unt vectors v 1,..., v n+1 are the tangent ponts of S n 1 to the codmenson-1 faces of σ. The tangent planes to S n 1 at v s bound a compact regon (the Eucldean smplex σ) contanng the orgn f and only f the tangent ponts v 1,..., v n+1 are not n any closed hemsphere of S n 1. q.e.d. Lemma 5. A collecton of unt vectors (v 1,..., v n+1 ) (R n ) n+1 s n E n+1 f and only f one of the followng condtons s satsfed: (5.1) The vectors v 1,..., v n+1 are not n any open half-space. (5.2) The lnear system n+1 =1 a v = 0 has a nonzero soluton (a 1,..., a n+1 ) so that a 0 for all = 1,..., n + 1. Proof. E n+1 (5.1). To see that elements n E n+1 satsfy (5.1), f (v 1,..., v n+1 ) E n+1, there s a famly of (v (m) 1,..., v (m) n+1 ) E n+1 convergng to (v 1,..., v n+1 ). Snce vectors v (m) 1,..., v (m) n+1 are not n any closed

5 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 5 half-space for any m, by contnuty, vectors v 1,..., v n+1 are not n any open half-space. (5.1) (5.2). Consder the lnear map f : R n R n+1 w f(w) = [v 1, v 2,..., v n+1 ] t w = Statement (5.1) says that v 1 w v 2 w. v n+1 w = {w R n v w > 0, = 1,..., n + 1} = {w R n f(w) > 0} = f(r n ) R n+1 >0.. Snce f(r n ) and R n+1 >0 are convex and dsjont, by the separaton theorem for convex sets, there s a vector a = (a 1,..., a n+1 ) t satsfyng the condtons () and () below. () For all u R n+1 >0, a u > 0. and () For all w R n, 0 a f(w) = a 1 a 2. a n+1 v 1 w v 2 w. v n+1 w n+1 = ( a v ) w. Condton () mples that a 0, for = 1,..., n + 1 and a 0. Condton () mples n+1 =1 a v = 0. Thus (5.2) holds. (5.2) E n+1. To see that a pont (v 1,..., v n+1 ) satsfyng (5.2) s n E n+1, we show that n any ε-neghborhood of (v 1,..., v n+1 ) n (R n ) n+1, there s a pont (v1 ε,..., vε n+1 ) E n+1. Let N k be the set of (v 1,..., v k ) such that v R k 1, v = 1 for all and k =1 a v = 0 has a nonzero soluton (a 1,..., a k ) wth a 0 for all. The goal s to prove N n+1 E n+1. We acheve ths by nducton on n. It s obvous that N 2 E 2. Assume that N n E n holds. For a pont (v 1,..., v n+1 ) N n+1, f any n vectors of v 1,..., v n+1 are lnear ndependent, then each entry a of the non-zero soluton of the lnear system n+1 =1 a v = 0, a 0, = 1,..., n + 1 must be nonzero. Thus a > 0 for all and (v 1,..., v n+1 ) satsfes (4.2), therefore t s n E n+1. In the reman case, wthout loss of generalty, we assume that v 1,..., v n are lnear dependent. We may assume after a change of coordnates =1

6 6 REN GUO & FENG LUO that v R n 1 = R n 1 {0} R n, for = 1,..., n, and v n+1 = (u n+1 cos(θ), sn(θ)) t, where 0 θ π 2 and u n+1 = 1. We clam that there exsts some 1 n + 1 such that (v 1,..., v,..., v n+1 ) N n, where x means deletng the element x. Case 1. If θ > 0.e., v n+1 s not n R n 1, consder the nonzero soluton of the lnear system n+1 =1 a v = 0, a 0, = 1,..., n + 1. The last coordnate gves a a n 0 + a n+1 sn(θ) = 0, whch mples a n+1 = 0. Ths means (a 1,..., a n ) 0,.e., (v 1,..., v n ) N n. Case 2. If θ = 0.e., v n+1 R n 1, then the dmenson of the soluton space W = {(a 1,..., a n+1 ) t R n+1 n+1 =1 a v = 0} s at least 2. Snce (v 1,..., v n+1 ) N n+1, the ntersecton W R n+1 0 {(0,..., 0)} s nonempty. The vector space W must ntersect the boundary of the cone R n+1 0 {(0,..., 0)}. Let (a 1,..., a n+1 ) be a pont n both W and the boundary of the cone R n+1 0 {(0,..., 0)}. Then there s some a = 0. Then {v 1 (t),..., v (t),..., v n+1 (t)} N n. By the above dscusson, wthout lose of generalty, we may assume that (v 1,..., v n ) N n. By the nducton assumpton N n E n,.e., n any ɛ 2 -neghborhood of (v 1,..., v n ), we can fnd a pont (u 1,..., u n ) E n, where u R n 1 for all. Recall we wrte v n+1 = (u n+1 cos(θ), sn(θ)) t. Let us defne a contnuous famly of n + 1 unt vectors v 1 (t),..., v n+1 (t) by settng v (t) = (u cos(t 2 ), sn(t 2 )) t, 1 n, v n+1 (t) = (u n+1 cos(θ + t), sn(θ + t)) t. We clam that there s a pont (v 1 (t),..., v n+1 (t)) E n+1 for small t > 0 wthn ɛ 2 -neghborhood of ((u 1, 0) t,..., (u n, 0) t, v n+1 ). By trangle nequalty, ths pont s wthn ε-neghborhood of (v 1,..., v n+1 ). We only need to check that (v 1 (t),..., v n+1 (t)) E n+1 for suffcently small t > 0 by verfyng the condton (4.2). To show any n vectors of v 1 (t),..., v n+1 (t) are lnear ndependent, t s equvalent to show that det[v 1 (t),..., v (t),..., v n+1 (t)] 0 for each = 1,..., n + 1. Frst, [ u1 cos(t det[v 1 (t),..., v n (t)] = det 2 ) u 2 cos(t 2 )... u n cos(t 2 ] ) sn(t 2 ) sn(t 2 )... sn(t 2 ) [ ] = sn(t 2 ) cos(t 2 ) n 1 u1 u det 2... u n To see the determnant s nonzero, suppose there are real numbers a 1,..., a n such that n =1 a (u 1, 1) t = 0. Then we have n =1 a u = 0 and n =1 a = 0. By assumpton (u 1,..., u n ) E n, we know ether a = 0 for all or a 0 and have the same sgn for all. Hence n =1 a = 0 n n

7 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 7 mples a = 0 for all. Thus the vectors (u 1, 1) t,..., (u n, 1) t are lnear ndependent. Hence det[v 1 (t),..., v n (t)] 0 for t (0, π 2 ]. Second, we calculate the determnant of the matrx whose columns are v n+1 (t) and some n 1 vectors of v 1 (t),..., v n (t). Wthout loss of generalty, consder f(t) = det[v 2 (t),..., v n (t), v n+1 (t)] [ u2 cos(t = det 2 )... u n cos(t 2 ] ) u n+1 cos(θ + t) sn(t 2 )... sn(t 2 ) sn(θ + t) If θ > 0, by the assumpton (u 1,..., u n ) E n, then [ ] u2... u f(0) = det n u n+1 cos(θ) = sn(θ) det[u sn(θ) 2,..., u n ] 0. It mples f(t) 0 holds for suffcently small t > 0. If θ = 0, then f(0) = 0. By expandng the determnant, f(t) = sn(t 2 )g(t) + sn(t) det[u 2 cos(t 2 ),..., u n cos(t 2 )], for some functon g(t), therefore f (0) = det(u 2,..., u n ) 0. Hence f(t) 0 holds for suffcently small t > 0. Next, let a (t) = ( 1) 1 det[v 1 (t),..., v (t),..., v n+1 (t)], 1 n + 1, then n+1 =1 a (t)v (t) = 0. Snce det[v 1 (0),..., v n (0)] = 0, we have a n+1 (0) = 0. Ths shows that n =1 a (0)v (0) = 0, therefore n =1 a (0)u = 0. By the assumpton (u 1,..., u n ) E n, we obtan a (0) a j (0) > 0 for 0, j n. By the contnuty we obtan a (t) a j (t) > 0 for 0, j n, for suffcent small t > 0. Consder the last coordnate of n+1 =1 a (t)v (t) = 0 we obtan n sn(t 2 ) a (t) + sn(θ + t)a n+1 (t) = 0. =1 Thus a n+1 (t) has the same sgn as that of a (t). Thus (a 1 (t),..., a n+1 (t)) or ( a 1 (t),..., a n+1 (t)) s a soluton requred n condton (4.2). q.e.d. 3. Degenerate hyperbolc smplexes Let R n,1 be the Mnkowsk space whch s R n+1 wth an nner product, where (x 1,..., x n, x n+1 ) t, (y 1,..., y n, y n+1 ) t = x 1 y x n y n x n+1 y n+1. Let H n = {x = (x 1,..., x n+1 ) t R n,1 x, x = 1, x n+1 > 0} be the hyperbolod model of the hyperbolc space. The de Stter space s {x R n,1 x, x = 1}. For a hyperbolc smplex σ n H n, the center and the radus of the smplex σ are defned to be the center and radus of ts nscrbed ball.

8 8 REN GUO & FENG LUO Lemma 6. For an n-dmensonal hyperbolc smplex σ H n wth center e n+1 = (0,..., 0, 1) t, ts unt outward normal vectors n the de Stter space are n a compact set ndependent of σ. Proof. Let v 1,..., v n+1 be the unt outward normal vectors of σ,.e., σ = {x H n x, v 0 and v, v = 1 for all }. Let v be the totally geodesc hyperplane n H n contanng the (n 1) dmensonal face of σ perpendcular to v for each = 1,..., n + 1. The radus of σ s the dstance from the center e n+1 to v for any = 1,..., n + 1 whch s equal to snh 1 ( e n+1, v ) (see for nstance [9] p26). It s well known that the volume of an n dmensonal hyperbolc smplex s bounded by the volume of the n dmensonal regular deal hyperbolc smplex whch s fnte (see for nstance [6] p539). It mples that the radus of a hyperbolc smplex σ s bounded from above by a constant ndependent of σ. Hence e n+1, v 2 s bounded from above by a constant c n ndependent of σ for any = 1,..., n + 1. It follows that v 1,..., v n+1 are n the compact set X n = {x = (x 1,..., x n+1 ) t x, x = 1, e n+1, x 2 c n } = {x = (x 1,..., x n+1 ) t x x 2 n = x 2 n+1 + 1, x 2 n+1 c n } ndependent of σ. q.e.d. Lemma 7. If A Y n+1 and det(a) = 0, then A Z n+1. Proof. Let A (m) be a sequence of angle Gram matrxes n Y n+1 convergng to A. By Proposton 2 (c), for any m, all prncpal n n submatrces of A (m) are postve defnte. Thus all prncpal n n submatrces of A are postve sem-defnte. Snce det(a) = 0, we see that A s postve sem-defnte. Let σ (m) be the n dmensonal hyperbolc smplex n the hyperbolod model H n whose angle Gram matrx s A (m) and whose center s e n+1 = (0,..., 0, 1) t. By Lemma 6, ts unt outward normal vector v (m) s n a compact set. Thus by takng a subsequence, we may assume (v (m) 1,..., v (m) A (m) = [ v (m) n+1 ) converges to (v 1,..., v n+1 ) wth v, v = 1. Snce, v (m) ] and A (m) converges to A, we obtan j A = [ v, v j ] = [v 1,..., v n+1 ] t S[v 1,..., v n+1 ], where S s the dagonal matrx dag(1,..., 1, 1). Snce det(a) = 0, the vectors v 1,..., v n+1 are lnear dependent. Assume that the vectors v 1,..., v n+1 span a k dmensonal subspace W of R n,1, where k n.

9 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 9 For any vector x W, wrte x = n+1 =1 x v. Then x, x = (x 1,..., x n+1 )[v 1,..., v n+1 ] t S[v 1,..., v n+1 ](x 1,..., x n+1 ) t = (x 1,..., x n+1 )A(x 1,..., x n+1 ) t 0 due to the fact that A s postve sem-defnte. Now for any x, y W, the nequalty x+ty, x+ty 0 for any t R mples the Schwartz nequalty x, y 2 x, x y, y. To prove that A Z n+1, we consder the followng two possbltes. Case 1. If x, x > 0 holds for any non-zero x W, then the Mnkowsk nner product restrcted on W s postve defnte. Snce the Mnkowsk nner product restrcted on R k = R k 0 R n,1 s postve defnte, by Wtt s theorem, there s an sometry γ of R n,1 sendng W to R k (see [9] p14-p15). By replacng v (m) by γ(v (m) ) for each and m, we may assume that v 1,..., v n+1 are contaned n R k. Thus v, v = v v j for all, j. Therefore A = [v 1,..., v n+1 ] t S[v 1,..., v n+1 ] = [v 1,..., v n+1 ] t [v 1,..., v n+1 ]. To show A Z n+1, by Lemma 5, we only need to show that v 1,..., v n+1 are not contaned n any open half space of R k. Ths s the same as that v 1,..., v n+1 are not contaned n any open half space of R n. Suppose otherwse, there exsts a vector w R k such that v w > 0 for all. Thus v, w = v w > 0. By takng m large enough, we obtan v (m), w > 0 for all. It s well known that for the unt normal vectors v (m) of a compact hyperbolc smplex n H n, the condtons v (m), w > 0 for all mples w, w < 0. But ths contradcts the assumpton that w R k whch mples w, w 0. Case 2. If there exsts some non-zero vector x 0 W such that x 0, x 0 = 0, then by the Schwartz nequalty we have x 0, y 2 x 0, x 0 y, y = 0 for any y W. Thus x 0, y = 0 for any y W. Ths mples that the subspace W s contaned n x 0, the orthogonal complement of x 0. Snce the vector u = (0,..., 0, 1, 1) t R n,1 satsfes u, u = 0, there s an sometry γ of R n,1 sendng x 0 to u. Thus γ sends x 0 to u. By replacng v (m) by γ(v (m) ) for each and m, we may assume that v 1,..., v n+1 are contaned n u. For any, snce v, u = v, (0,..., 0, 1, 1) t = 0, we can wrte v as v = w + a u

10 10 REN GUO & FENG LUO for some w R n 1 and a R. Snce w, u = 0, thus v, v j = w w j for all, j. Therefore A = [v 1,..., v n+1 ] t S[v 1,..., v n+1 ] = [w 1,..., w n+1 ] t [w 1,..., w n+1 ]. To show A Z n+1, by Lemma 5, we only need to show that w 1,..., w n+1 are not contaned n any open half space of R n 1 whch s equvalent to that w 1,..., w n+1 are not contaned n any open half space of R n. Suppose otherwse, there exsts a vector w R n 1 such that w w > 0 for all. Then v, w = w, w + (0,..., 0, a, a ) t, w = w w + 0 > 0 for all. By takng m large enough, we obtan v (m), w > 0 for all. By the same argument above, t s a contradcton. q.e.d. 4. Proof of Theorem 1 Sphercal case. We begn wth a bref revew of the relevant result n [4]. For any postve sem-defnte symmetrc matrx A, there exsts a unque postve sem-defnte symmetrc matrx A so that ( A) 2 = A. It s well know that the map A A s contnuous on the space of all postve sem-defnte symmetrc matrces. Suppose A X n+1 = {A = [a j ] R (n+1) (n+1) A t = A, all a = 1, A s postve defnte}, the space of the angle Gram matrces of sphercal smplexes (by the Proposton 2). By makng a change of varables, the Aomoto-Kneser-Vnberg formula (1.1) s equvalent to (4.1) V (A) = µ 1 n χ( Ax)e xtx dx, R n+1 where χ s the characterstc functon of the set R n+1 0 n R n+1. It s proved n [4] that volume formula (4.1) stll holds for any matrx n X n+1 ={A = [a j ] R (n+1) (n+1) A t = A, all a = 1, A s postve sem-defnte}. Suppose V (A) = 0, by formula (4.1), we see the functon χ h : R n+1 R s zero almost everywhere, where h : R n+1 R n+1 s the lnear map sendng x to Ax. Equvalently, the (n + 1)-dmensonal Lebesque measure of h 1 (R n+1 0 ) s zero. We clam h(rn+1 ) R n+1 >0 =. For otherwse, h 1 (R n+1 >0 ) s a nonempty open subset n Rn+1 wth postve (n + 1)-dmensonal Lebesque measure. Ths s a contradcton. Now let A = [v 1,..., v n+1 ] t (n+1) (n+1), where v R n+1 s a column vector for each. Frst h(r n+1 ) R n+1 >0 = mples that det A = 0. Therefore {v 1,..., v n+1 } are lnear dependent. We may assume, after a

11 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 11 rotaton r O(n + 1), the vectors v 1,..., v n+1 le n R n {0}. Now = h(r n+1 ) R n+1 >0 = { Aw w R n+1 } R n+1 >0 = {(v 1 w,..., v n+1 w) t w R n+1 } R n+1 >0 Ths shows that there s no w R n+1 such that v w > 0 for = 1,..., n + 1,.e., the vectors v 1,..., v n+1 are not n any open half space. By lemma 5, we have (v 1,..., v n+1 ) E n+1, therefore A = [v v j ] Z n+1. Hyperbolc case. Let A Y n+1. If det(a) 0, t s proved n [4] that the volume formula (1.1) stll holds for A. In formula (1.1), snce x t ad(a)x s fnte, the ntegrant e xtad(a)x > 0. Therefore the ntegral R n+1 0 e xt ad(a)x dx > 0. Hence V (A) > 0. It follows that f the extended volume functon vanshes at A, then det A = 0. By Lemma 7, we have A Z n Proof of Theorem 3 Proof of (a). If A Z n+1, then A = [v v j ] for some pont (v 1,..., v n+1 ) E n+1. By Lemma 5, the lnear system n+1 =1 a v = 0, a 0, = 1,..., n + 1 has a nonzero soluton. Let (a 1,..., a n+1 ) be a soluton wth the least number of nonzero entres among all solutons. By rearrange the ndex, we may assume a 1 > 0,..., a k+1 > 0, a k+2 =... = a n+1 = 0. We clam rank[v 1,..., v k+1 ] = k. Otherwse rank[v 1,..., v k+1 ] k 1, then the dmenson of the soluton space W = {(x 1,..., x k+1 ) t R k+1 k+1 =1 x v = 0} s at least 2. Thus Ω = W R k+1 >0 s a nonempty open convex set n W whose dmenson s at least 2. Hence Ω contans a boundary pont (b 1,..., b k+1 ) Ω {(0,..., 0)} wth some b j = 0, due to dmw 2. Now we obtan a soluton (b 1,...b j 1, 0, b j+1,..., b k+1, 0,..., 0) whch has lesser number of nonzero entres than (a 1,..., a n+1 ). Ths s a contradcton. Let B = [v v j ] (k+1) (k+1). Snce rank[v 1,..., v k+1 ] = k, we have det(b) = 0. We clam that ad(b) 0 and ad(b) 0. Ths wll verfy the condton (a) n Theorem 3 for A. Let ad(b) = [b j ] (k+1) (k+1). Evdently, due to rank(b) = k, ad(b) 0. It remans to prove that ad(b) 0. By the constructon of B, we see b jj 0, for all j. Snce rank[v 1,..., v k+1 ] = k, t follows the dmenson of the soluton space of k+1 =1 a v = 0 s 1. Snce k+1 =1 b jv j = 0, (b 1,..., b k+1 ) s proportonal to (a 1,..., a k+1 ), where a > 0 for 1 k + 1. Ths shows that f b jj > 0, then b j > 0 for all, f b jj = 0, then b j = 0 for all. Ths shows ad(b) 0. Conversely, f A s postve sem-defnte so that det(a) = 0 and there exsts a prncpal (k + 1) (k + 1) submatrx B so that det(b) = 0, ad(b) 0 and ad(b) 0, we wll show that A Z n+1. Snce

12 12 REN GUO & FENG LUO A s postve sem-defnte and undagonal, there exst unt vectors v 1,..., v n+1 n R n such that A = [v v j ]. We may assume B = [v v j ] (k+1) (k+1), 1, j k + 1 and ad(b) = [b j ]. Due to det(b) = 0, ad(b) 0, we have rank(v 1,..., v k+1 ) = k. We may assume v 2,..., v k+1 are ndependent. Thus the cofactor b 11 > 0. By the assumpton ad(b) 0, we have b 1s 0 for s = 1,...k + 1. Snce k+1 s=1 b 1s(v s v j ) = 0 for all j = 2,..., k + 1 and v 2,..., v k+1 are ndependent, we get k+1 s=1 b 1sv s = 0. Thus we get a nonzero soluton for the lnear system n+1 =1 a v = 0, a 0, = 1,..., n + 1. Proof of (b). If A X n+1 X n+1, then A = [v v j ] where v 1,..., v n+1 are lnear dependent. We can assume v 1,..., v n+1 le n R n {0}. By change subndex, we may assume n+1 =1 a v = 0 has a non-zero soluton wth a 0 f = 1,..., k whle a < 0 f = k + 1,..., n + 1. Thus vectors v 1,..., v k, v k+1,..., v n+1 satsfy the condton (5.2) n Lemma 5. Let D be the dagonal matrx dag(1,..., 1, 1,..., 1) wth k dagonal entres beng 1 and n k + 1 dagonal entres beng 1. Thus by Lemma 5, DAD Z n+1. On the other hand, f for some dagonal matrx D n Theorem 3 (b), we have DAD Z n+1, then by Theorem 3 (a), DAD s postve sem-defnte. Therefore A s postve sem-defnte. Take B X n+1 and consder the famly A(t) = (1 t)a+tb for t [0, 1]. Then lm t 0 A(t) = A and A(t) X n+1 for t > 0. Thus A X n+1. Proof of (c). Frst we show that the condtons are suffcent. Suppose A = [a j ] (n+1) (n+1) s a symmetrc undagonal matrx wth all prncpal n n submatrces postve sem-defnte so that ether A Z n+1 or det(a) < 0 and ad(a) 0. We wll show A Y n+1. If A Z n+1, t s suffcent to show that Z n+1 Y n+1,.e., we may assume A Z n+1. In ths case, let J = [c j ] (n+1) (n+1) so that c = 1 and c j = 1 for j. Consder the famly A(t) = (1 t)a + tj, for 0 t 1. Evdently lm t 0 A(t) = A. We clam that A(t) Y n+1 for small t > 0. Snce all prncpal n n submatrces of A are postve defnte, by contnuty, all prncpal n n submatrces of A(t) are postve defnte for small t > 0. It remans to check det(a(t)) < 0 for small t > 0. To ths end, let us consder d dt t=0 det(a(t)). We have d dt t=0 det(a(t)) = ( a j 1)cof(A) j < 0, j due to ad(a) = [cof(a) j ] > 0 and a j 1 < 0 for all j. Snce det(a) = 0 t follows that det(a(t)) < 0 for small t > 0. In the second case that det(a) < 0 and ad(a) 0 and all prncpal n n submatrces of A are postve sem-defnte. Then A has a unque negatve egenvalue λ, where λ > 0. Consder the famly A(t) = A +

13 ON A CONJECTURE OF MILNOR ABOUT VOLUME OF SIMPLEXES 13 tλi, for 0 t 1, where I s the dentty matrx, so that 1 lm A(t) = A. t λt We clam there s a dagonal matrx D whose dagonal entres are ±1 so that (1) DAD = A, 1 (2) 1+λt DA(t)D Y n+1 for 0 < t < 1. As a consequence, t follows A = DAD 1 = lm t λt DA(t)D Y n+1. To fnd ths dagonal matrx D, by the contnuty, det(a(t)) < 0 for 0 < t < 1 and det(a(1)) = 0. Furthermore, all prncpal n n submatrces of A are postve defnte for t > 0 due to postve defnteness of tλi. Let us recall the Lemma 3.4 n [4] whch says that f B s a symmetrc (n + 1) (n + 1) matrx so that all n n prncpal submatrces n B are postve defnte and det(b) 0, then no entry n the adjacent matrx ad(b) s zero. It follows that every entry of ad(a(t)) s nonzero for 0 < t 1. Let ad(a(1)) = [b j ] (n+1) (n+1) and D to be the dagonal matrx wth dagonal entres beng b 1 b 1 = ±1 for = 1,..., n + 1. Then the entres of the frst row and the frst column of Dad(A(1))D are postve. Snce det(a(1)) = 0 and ad(a(1)) 0, we see the rank of ad(a(1)) s 1. Thus any other column s propostonal to the frst column. But b > 0 for all, hence ad(a(1)) > 0. Now snce every entry of Dad(A(t))D s nonzero for t > 0, by contnuty Dad(A(t))D > 0 for t > 0 and Dad(A)D = Dad(A(0))D 0. By the assumpton ad(a) 0, t follows Dad(A)D = ad(a). On the other hand Dad(A)D = ad(d 1 AD 1 ), and det(a) 0. Thus D 1 AD 1 = A or the same A = DAD. Ths shows A = DAD = lm t 0 DA(t)D 1 = lm t λt DA(t)D. By the constructon above 1 1+λt DA(t)D Y n+1 for 0 < t < 1, ths shows A Y n+1. Fnally, we show the condton n (c) s necessary. Suppose A = lm m A (m) where A (m) Y n+1. By Proposton 2, det(a (m) ) < 0, ad(a (m) ) > 0 and all prncpal n n submatrces of A (m) are postve

14 14 REN GUO & FENG LUO defnte. We want to show that A satsfes the condtons stated n (c). Evdently, all prncpal n n submatrces of A are postve semdefnte, ad(a) 0 and det(a) 0. If det(a) < 0, then we are done. If det(a) = 0, by Lemma 7, we see that A Z n+1. References 1. K. Aomoto, Analytc structure of Schläfl functon. Nagoya Math J. 68 (1977), 1 16, MR , Zbl H. Kneser, Der Smplexnhalt n der nchteukldschen Geometre. Deutsche Math. 1(1936), , Zbl F. Luo, On a problem of Fenchel. Geom. Dedcata 64 (1997), no. 3, , MR , Zbl F. Luo, Contnuty of the Volume of Smplces n Classcal Geometry. Commun. Contemp. Math. 8 (2006), no. 3, , MR , Zbl J. Mlnor, The Schläfl dfferental equalty. Collected papers, vol. 1. Publsh or Persh, Inc., Houston, TX, 1994, MR , Zbl J. G. Ratclffe, Foundatons of Hyperbolc Manfolds. Second Edton. GTM 149, Sprnger, 2006, MR , Zbl I. Rvn, Contnuty of volumes on a generalzaton of a conjecture of J. W. Mlnor. Geom. Dedcata 131 (2008), 73 85, MR , Zbl E. B. Vnberg, Volumes of non-eucldean polyhedra. (Russan) Uspekh Mat. Nauk 48 (1993), no. 2(290), 17 46; translaton n Russan Math. Surveys 48 (1993), no. 2, 15 45, MR , Zbl E. B. Vnberg (Ed.), Geometry II, Spaces of Constant Curvature. Encyclopaeda of Mathematcal Scences, Vol. 29, Sprnger-Verlag Berln Hedelberg 1993, MR , Zbl Department of Mathematcs, Rutgers Unversty, Pscataway, NJ, Current address: School of Mathematcs, Unversty of Mnnesota, Mnneapols, MN, E-mal address: guoxx170@math.umn.edu The Center of Mathematcal Scence, Zhejang Unversty, Hangzhou, Chna, Department of Mathematcs, Rutgers Unversty, Pscataway, NJ, E-mal address: fluo@math.rutgers.edu