[math-fun] A cautionary tale of 6 polynomials in 10 variables
A cautionary tale of 6 polynomials in 10 variables _______________________________________ Some relevant, useful, or just pretty facts concerning Sp(2n) --- (0) Given n, all nonsingular real skew-symmetric 2n x 2n matrices are equivalent; so fix on the convenient canonical choice K == [[O_n, -I_n], [I_n, O_n]], where O_n, I_n denote n x n zero, unit matrices resp. (1) The "symplectic group" Sp(2n) comprises real 2n x 2n matrices M such that M' K M = K, where M' denotes transpose of M. (2) For M in Sp(2n), the determinant |M| = +1 always, -1 never; (3) Sp(2n) is generated by products of at most 4n "transvections" I + t U' U K, where U ranges over real nonzero 2n-vectors |R^(2n), t over reals |R. (4) The dimension of Sp(2n) equals (2n+1)_C_2 = (2n+1)n; (5) If M = Q S denote the polar decomposition of symplectic M into a unique product of orthogonal Q matrix and symmetric matrix S, then Q,S must also be symplectic; (6) Sp(2n) is a connected Lie group, though not compact; (7) PSp(2n), its "projective symplectic" quotient with scalars factored out, is simple; (8) Mp(2n), its "metaplectic" double cover, is simply connected. (9) Mp(2n) has no (faithful, finite order) matrix representation. I became diverted from an earlier investigation by the unexpected and rather pretty result (6). This familiar situation was doubtless exacerbated by (6) being just about the only item in section 1 of Eckhard Meinrenken's "Symplectic Geometry" (free download online) that I could grasp at sight, and by its proof appearing simple and elegant but quite incomprehensible --- what do diffeomorphisms have to do with a theorem in elementary matrix algebra? So I sat down to investigate (6) by myself. And the first question that occurred to me was: what are the dimensions of the orthogonal and symmetric factors in the polar decomposition of a symplectic matrix? Now physicists by and large don't seem to be terribly interested in Lie groups as such: they are often perfectly content with the Lie algebra --- the vector-space tangent to the identity transformation --- the dimension of which must equal that of the group manifold. Standard results in the literature assert that (10) Lie algebra of Sp(2n) comprises T with T' K + K T = 0; so T = [[A, B], [C, -A']] with B,C symmetric, dimension = 2 (n+1)_C_2 + n^2 = (2n+1)_C_2 = 2 n^2 + n; and (11) Lie algebra of O(2n) comprises T with T' I + I T = 0; so T skew-symmetric; dimension = (2n)_C_2 = 2 n^2 - n; hence (12) Lie algebra of orthogonal symplectic SpO(2n) comprises T = [[A, -B][B, A]], B symm, A skew-symm; dimension = (n+1)_C_2 + n_C_2 = n^2; and (13) By taking the complement of vector space (12) in (11), symmetric symplectic matrices have dimension = 2 n^2 + n - n^2 = n^2 + n = (n+1)_C_2. Much as I mistrust all this high-falutin' abstraction, I'm beginning to wonder if they have a point --- though it should be said in my defence that symmetric matrices, not being a group, are less than amenable to the method. So, just as as a check, I pressed on to try to verify (13) directly. Studiously avoiding an earlier misfortune [involving much earnest investigation of the case n = 1, prior to eventual realisation that actually Sp(2) = SL(2)], I started off with n = 2, for which (13) predicts dimension 6. Setting M = [[z,a,b,d], [a,y,c,e], [b,c,x,f], [d,e,f,w]] and evaluating M' K M = K yields 6 constraints, all of which must vanish: P_1 = b^2 + c*d - f*a - x*z + 1, P_2 = e^2 + c*d - f*a - w*y + 1, P_3 = b*c + e*c - a*x - f*y, P_4 = b*d + e*d - a*w - f*z, P_5 = b*f - e*f + c*w - d*x, P_6 = b*a - e*a - c*z + d*y. It's easy to see informally that these are functionally independent: for each P1,...P6 in turn, set to zero all but b; e; f,y; f,z; d,x; d,y respectively, to yield an expression varying while the other 5 vanish. Or more formally, verify that the 6 x 10 differential matrix [d P_j / d V_i] of linear polynomials in the variables V = [a,b,c,d,e,f,z,y,x,w] has full rank 6. Hence the manifold should have dimension 10-6 = 4; and more generally, dimension = (2n+1)_C_2 - 4 n_C_2 = 2n (?) Comparison with (13) now reveals to the astute observer a trifling discrepancy --- a shortfall of 6-4 = 2 in my bull-at-the-gate estimate for the symmetric symplectic dimension when n = 2; and more generally, of (n+1)n - 2n = (n-1)n = n_C_2. No prizes for guessing which argument is shambolically addled. But just exactly at what point did its misguided author's powers of ratiocination desert him this time round? Fred Lunnon
There are quite a few places in the saga where a neophyte might have fallen into error through carelessness or misunderstanding, but after double-checking and numerical verification all stand up; except at the final point where it was assumed that the rank of the differential matrix (number of independent constraints), plus the dimension of the manifold (of their zero set), necessarily equals the dimension of the embedding space (number of variables). This assertion conceals a trap, in that it is only almost always true: but it might fail if all maximal (polynomial) cofactors of the differential matrix --- 10_C_6 = 210 determinants above --- happen to vanish for every point of the manifold. For some, vanishing is easy to establish: selecting columns corresponding to a,b,c,d,e,f, the 6 x 6 determinant factorises as (e^2-b^2+x*z-w*y)*(quartic) = (P_2 - P_1)*(quartic) = 0 on the manifold; on the other hand, b,c,d,e,f,x yields an irreducible sextic. In algebraic terms, a determinant vanishes on the manifold just when it is a member of the root ideal generated by the constraints [Hilbert Nullstellensatz]: in order to decide this question, it is necessary [though not in general always sufficient] to compute a Groebner basis for the constraints, then reduce each determinant to normal form (zero) modulo this basis: 16 degree-5 polynomials when n = 2 above, with total degree ordering. Born of desperation, an alternative strategy is simply to feed constraints P_1,...,P_6 to Maple's equation solver --- and pray.
From a fog of special solutions involving radicals, for n = 2 there emerges unexpectedly this rational septic(!) general solution, with c,d,e,f,x,y free --- establishing incontrovertibly that the dimension really is 6 after all, rather than 4 --- a = -(e^2*c^2+c^3*d-2*y*e*f*c+f^2*y^2-y*d*x*c+c^2) / (f*(-c^2+y*x)), b = -(-e*f*c^2-e*f*y*x+f^2*y*c+x*e^2*c+x*c^2*d-y*d*x^2+x*c) / (f*(-c^2+y*x)), z = (-2*e*f*c^4*d+5*e^2*f^2*c^2*y-4*e*f^3*c*y^2+2*f^2*y*c^3*d +2*x*e^2*c^3*d-2*x^2*c^2*d^2*y+y^2*d^2*x^3-2*y*d*x^2*c +f^2*y^2*x-2*e^3*f*c^3-2*e*f*c^3+f^4*y^3+f^2*y*c^2+x*e^4*c^2 +2*x*e^2*c^2+x*c^4*d^2+2*x*c^3*d+x*c^2-2*e^3*f*y*x*c+e^2*f^2*y^2*x +2*e*f*y^2*x^2*d-2*e*f*c*y*x-2*f^2*y^2*c*d*x-2*x^2*e^2*c*y*d) / (f^2*(-c^2+y*x)^2), w = (-2*e*f*c+f^2*y+x*e^2+x) / (-c^2+y*x).
[More tedious detail: the 4 permutations (), (b,e)(a,f)(x,y)(z,w), (b,e)(c,d)(x,w)(y,z), (x,z)(y,w)(a,f)(c,d), are symmetrical on the P_i, and yield further isomorphic solutions; which presumably provide charts overlapping on the same manifold? Then remaining singularities may be removed, yielding these special solutions with c,d,x,y free a = f = 0, b = -e, x*z = y*w = (e^2+c*d+1), c*z = d*y; with c,d free a = f = w = x = y = z = 0, b = -e, e^2 = -c*d-1; with a,f free c = d = w = x = y = z = 0, b = e, e^2 = f*a-1.] Geometrically, what happens here is that pairwise intersections of primal manifolds defined by individual equations are each reducible, with some component being common to them all. What about a simpler example of this phenomenon? I don't think any is nontrivially possible in 2-space. On a general 3-space ellipsoid lie two parallel families of circles [see David Hilbert, S. Cohn-Vossen "Geometry and the Imagination", para 3]; containing a given circle imposes 6-1 = 5 constraints, leaving freedom 9-5 = 4. At most 3 quadric polynomials defining such ellipsoids can be functionally independent --- otherwise some 3 equations may be solved, fixing values of the variables for the remainder. So the construction yields examples with 3 equations in 3 variables, but a 1-dimensional solution. Old hat to an ideal theory expert, no doubt; but a salutory experience to encounter unexpectedly in the setting of an apparently innocuous application! Fred Lunnon On 3/26/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
A cautionary tale of 6 polynomials in 10 variables _______________________________________
Some relevant, useful, or just pretty facts concerning Sp(2n) ---
(0) Given n, all nonsingular real skew-symmetric 2n x 2n matrices are equivalent; so fix on the convenient canonical choice K == [[O_n, -I_n], [I_n, O_n]], where O_n, I_n denote n x n zero, unit matrices resp.
(1) The "symplectic group" Sp(2n) comprises real 2n x 2n matrices M such that M' K M = K, where M' denotes transpose of M.
(2) For M in Sp(2n), the determinant |M| = +1 always, -1 never;
(3) Sp(2n) is generated by products of at most 4n "transvections" I + t U' U K, where U ranges over real nonzero 2n-vectors |R^(2n), t over reals |R.
(4) The dimension of Sp(2n) equals (2n+1)_C_2 = (2n+1)n;
(5) If M = Q S denote the polar decomposition of symplectic M into a unique product of orthogonal Q matrix and symmetric matrix S, then Q,S must also be symplectic;
(6) Sp(2n) is a connected Lie group, though not compact;
(7) PSp(2n), its "projective symplectic" quotient with scalars factored out, is simple;
(8) Mp(2n), its "metaplectic" double cover, is simply connected.
(9) Mp(2n) has no (faithful, finite order) matrix representation.
I became diverted from an earlier investigation by the unexpected and rather pretty result (6). This familiar situation was doubtless exacerbated by (6) being just about the only item in section 1 of Eckhard Meinrenken's "Symplectic Geometry" (free download online) that I could grasp at sight, and by its proof appearing simple and elegant but quite incomprehensible --- what do diffeomorphisms have to do with a theorem in elementary matrix algebra?
So I sat down to investigate (6) by myself. And the first question that occurred to me was: what are the dimensions of the orthogonal and symmetric factors in the polar decomposition of a symplectic matrix?
Now physicists by and large don't seem to be terribly interested in Lie groups as such: they are often perfectly content with the Lie algebra --- the vector-space tangent to the identity transformation --- the dimension of which must equal that of the group manifold. Standard results in the literature assert that
(10) Lie algebra of Sp(2n) comprises T with T' K + K T = 0; so T = [[A, B], [C, -A']] with B,C symmetric, dimension = 2 (n+1)_C_2 + n^2 = (2n+1)_C_2 = 2 n^2 + n; and (11) Lie algebra of O(2n) comprises T with T' I + I T = 0; so T skew-symmetric; dimension = (2n)_C_2 = 2 n^2 - n; hence (12) Lie algebra of orthogonal symplectic SpO(2n) comprises T = [[A, -B][B, A]], B symm, A skew-symm; dimension = (n+1)_C_2 + n_C_2 = n^2; and (13) By taking the complement of vector space (12) in (11), symmetric symplectic matrices have dimension = 2 n^2 + n - n^2 = n^2 + n = (n+1)_C_2.
Much as I mistrust all this high-falutin' abstraction, I'm beginning to wonder if they have a point --- though it should be said in my defence that symmetric matrices, not being a group, are less than amenable to the method. So, just as as a check, I pressed on to try to verify (13) directly.
Studiously avoiding an earlier misfortune [involving much earnest investigation of the case n = 1, prior to eventual realisation that actually Sp(2) = SL(2)], I started off with n = 2, for which (13) predicts dimension 6. Setting M = [[z,a,b,d], [a,y,c,e], [b,c,x,f], [d,e,f,w]] and evaluating M' K M = K yields 6 constraints, all of which must vanish: P_1 = b^2 + c*d - f*a - x*z + 1, P_2 = e^2 + c*d - f*a - w*y + 1, P_3 = b*c + e*c - a*x - f*y, P_4 = b*d + e*d - a*w - f*z, P_5 = b*f - e*f + c*w - d*x, P_6 = b*a - e*a - c*z + d*y.
It's easy to see informally that these are functionally independent: for each P1,...P6 in turn, set to zero all but b; e; f,y; f,z; d,x; d,y respectively, to yield an expression varying while the other 5 vanish. Or more formally, verify that the 6 x 10 differential matrix [d P_j / d V_i] of linear polynomials in the variables V = [a,b,c,d,e,f,z,y,x,w] has full rank 6. Hence the manifold should have dimension 10-6 = 4; and more generally, dimension = (2n+1)_C_2 - 4 n_C_2 = 2n (?)
Comparison with (13) now reveals to the astute observer a trifling discrepancy --- a shortfall of 6-4 = 2 in my bull-at-the-gate estimate for the symmetric symplectic dimension when n = 2; and more generally, of (n+1)n - 2n = (n-1)n = n_C_2.
No prizes for guessing which argument is shambolically addled. But just exactly at what point did its misguided author's powers of ratiocination desert him this time round?
Fred Lunnon
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Fred lunnon