[math-fun] Symplectic apoplexy
Thought my embryonic GA glossary could do with mentioning the symplectic group. Bad decision. Turns out there are several ... but for now, I'm just interested in the real one: Sp(2n) sometimes, or less ambiguously, Sp(2n, R). Trying to get a handle on this beast, I scratched around for a matrix representation, or better still an algebra along the lines of the Clifford algebras for the Spin groups PSO(p,q) etc. Kick off with the symplectic plane SP(2), defined as 2x2 matrices M such that M' K M = K, where M' denotes transpose, and K = [[1,0],[0,-1]] is the canonical skew-symmetric matrix. [Why transpose rather than inverse? And shouldn't there be a quotient by scalar matrices in there somewhere? Never mind, press on ...] In Wikipedia there lurks following linear basis: I = [[1,0],[0,1]], X = [[1,1],[0,1]], Y = [[1,0],[1,1]]. OK, it is a basis --- every matrix M = a I + b X + c Y preserves K, and it has the correct dimension (2n+1)n = 3. [Nice --- tho' it's not obvious to me why a basis should exist at all ...] But the corresponding algebra is not very enticing: the relations are x x = 2 x - 1, y y = 2 y - 1, x y + y x = 2 x + 2 y - 1 and I can't see how to reduce x y at all. Instead, here's a much nicer presentation: take X = [[1,-1],[1,0]], Y = [[0,1],[-1,1]] = -X^2, I = [[1,0],[0,1]] = X^6; now the algebra becomes x x = -y, y y = -x, x y = 1. Fired up by this success, I proceed to n = 2 --- and hit the buffers. Is there an analogous basis of 5 matrices for Sp(4)? I can't see how to construct one at all, let alone one with nice symmetry. So before I waste any more time re-inventing the wheel, would somebody out there like to shoot my fox for me (or maybe just run over it)? Fred Lunnon
Rats --- tho' it's early for All Fools' Day, I've been royally conned! "Sp(2)" turns out to be just twice SL(2) --- it's 2x2 real matrices with unit determinant (duh!). And if I want to make an algebra out of it, I have to deal instead with GSp(2) --- which is just GL(2) --- then consider the "spinor" subgroup with unit norm (determinant). So my little algebra was a mirage --- which comes as a relief! Questions remain. I now have no idea what the Wikipedia page was trying to say about this --- tho' it did come with a health warning. I'm in the dark about what's so special about the "symplectic plane" that it gets given this name in Porteous' "Classical Groups" --- not helped by what I take to be the egregious typo reducing his definition of it to (literally) a cipher --- see props 6.5, 7.3. And I'm intrigued by the use of transpose instead of inverse --- of course, there's an suggestion of the orthogonal group involved in this. Finally, is there anything in the idea of constructing Clifford-style algebras for symplectic groups --- test case seems to be GSp(4, |R), and central problem is whether these matrices comprise a vector space ... anybody know the answer? If so, (algebraic) dimension is at least 10 + 1 = 11 (not 5 as earlier wittered); grade (number of generators) must be 4; and scalar part (constant polynomial coefficient) of product of two vectors (linear polynomials) A = a x + b y + c u + d v, A`, should equal the (quasi-pseudo-metric) bilinear form A' K A` = a d` + b c` - c b` - d a` . Fred Lunnon [27/02/10] On 2/27/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
Thought my embryonic GA glossary could do with mentioning the symplectic group.
Bad decision. Turns out there are several ... but for now, I'm just interested in the real one: Sp(2n) sometimes, or less ambiguously, Sp(2n, R).
Trying to get a handle on this beast, I scratched around for a matrix representation, or better still an algebra along the lines of the Clifford algebras for the Spin groups PSO(p,q) etc.
Kick off with the symplectic plane SP(2), defined as 2x2 matrices M such that M' K M = K, where M' denotes transpose, and K = [[1,0],[0,-1]] is the canonical skew-symmetric matrix.
[Why transpose rather than inverse? And shouldn't there be a quotient by scalar matrices in there somewhere? Never mind, press on ...]
In Wikipedia there lurks following linear basis: I = [[1,0],[0,1]], X = [[1,1],[0,1]], Y = [[1,0],[1,1]]. OK, it is a basis --- every matrix M = a I + b X + c Y preserves K, and it has the correct dimension (2n+1)n = 3.
[Nice --- tho' it's not obvious to me why a basis should exist at all ...]
But the corresponding algebra is not very enticing: the relations are x x = 2 x - 1, y y = 2 y - 1, x y + y x = 2 x + 2 y - 1 and I can't see how to reduce x y at all.
Instead, here's a much nicer presentation: take X = [[1,-1],[1,0]], Y = [[0,1],[-1,1]] = -X^2, I = [[1,0],[0,1]] = X^6; now the algebra becomes x x = -y, y y = -x, x y = 1.
Fired up by this success, I proceed to n = 2 --- and hit the buffers. Is there an analogous basis of 5 matrices for Sp(4)? I can't see how to construct one at all, let alone one with nice symmetry.
So before I waste any more time re-inventing the wheel, would somebody out there like to shoot my fox for me (or maybe just run over it)?
Fred Lunnon
A typographical eagle points out that the earlier
where K = [[1,0],[0,-1]] is the canonical skew-symmetric matrix
should obviously have read
where K = [[0,-1],[1,0]] is the canonical skew-symmetric matrix
Sp(4) of course needs instead K = [[0,0,0,-1],[0,0,-1,0],[0,1,0,0],[1,0,0,0]]; or perhaps K = [[0,-1,0,0],[1,0,0,0],[0,0,0,-1],[0,0,1,0]]. And while I'm at it, I have just realised that
[the] central problem is whether these matrices comprise a vector space
is a gross over-simplification. Rather, it's whether, somewhere within the dimension-11 GSp(4), there exists a dimension-4 vector space, which then generates the whole group via multiplication. (Sigh ...) WFL
Update on Sp(4, |R) algebra (like, so much more than you needed to know) --- I did find a 4-subspace: M = [[a,-c,-d,-b],[c,-a,b,d],[d,-a,-b,-c],[b,-d,c,a]] satisfies M' K M = (a^2 + b^2 - c^2 - d^2) K for all real a,b,c,d. Algebra generators x,y,u,v correspond to setting a,b,c,d = 1 individually with other three zero; algebra relations are x x = y y = 1, u u = v v = -1, x y = u v (= K), v y = y v = u x = - x u, u y = y u = x v = -v x. Alas, simple computation now shows that the spinor dimension (of products, reduced modulo scalars) is only 3, against target 10. By analogy with Clifford algebra, the best to be hoped for anyway would be 4_C_2 = 6; achieving 10 almost certainly requires 5 generators. The extra generator cannot contribute to the inner product, for which the skew-symmetric bilinear form is given: so the larger algebra is degenerate. Geometrically, this would correspond to compactification of the space, via attachment of some subspace "at infinity"; the corresponding extra matrix attached to the basis is presumably singular, and annihilates K. [By the way, contrary to a hint dropped earlier, the "spinor norm" (for a vector, the inner product with itself) is distinct from the matrix determinant; though either may be used for normalisation. At least, the norm might be used, were it not for the inconvenient circumstance that in this algebra, the norm of every vector is zero!] I reckon there's a good chance that something useful and interesting is waiting to be unearthed here; and given the notable reluctance of most respectable mathematicians to engage with degenerate algebras, it's quite likely that nobody has yet got around do doing so! Fred Lunnon On 2/27/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
A typographical eagle points out that the earlier
where K = [[1,0],[0,-1]] is the canonical skew-symmetric matrix
should obviously have read
where K = [[0,-1],[1,0]] is the canonical skew-symmetric matrix
Sp(4) of course needs instead K = [[0,0,0,-1],[0,0,-1,0],[0,1,0,0],[1,0,0,0]]; or perhaps K = [[0,-1,0,0],[1,0,0,0],[0,0,0,-1],[0,0,1,0]].
And while I'm at it, I have just realised that
[the] central problem is whether these matrices comprise a vector space
is a gross over-simplification. Rather, it's whether, somewhere within the dimension-11 GSp(4), there exists a dimension-4 vector space, which then generates the whole group via multiplication.
(Sigh ...) WFL
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Fred lunnon