November 2002 - mathfun@mailman.xmission.com

Re: [math-fun] Prodigals (of matrices)
by asimovd＠aol.com 08 Nov '02

08 Nov '02

Shel Kaphan (sjk(a)kaphan.org) asks: << [Extending product integrals to matrices] seems to be related to a way to handle continuous composition of rotations, for example. If you have two rotation matrices A and B, they don't in general commute, but lim n->oo ((A^1/n)(B^1/n))^n = lim n->oo ((B^1/n)(A^1/n))^n. [ One way to see this is that lim n->oo A^1/n = lim n->oo B^1/n = Identity ] How is this related to the product integral? >> It is related to something even simpler than the product integral. If Q is a matrix in SO(n,R), we call it "eligible" if Q does not carry any vector into its negative. Then there is a unique "smallest" skew-symmetric matrix Log(Q) such that Q = exp(Log(Q)) where exp is given by the usual series formula. Then for eligible A and B, we have (*) lim n->oo ((A^1/n)(B^1/n))^n = exp( (Log(A) + Log(B)) / 2 ). (With the understanding that for Q in SO(n,R), for rotation on any invariant plane P of Q, we always take the rotation angle theta to lie in [0,pi), and Q^(1/n) restricted to P is always taken to be rotation by theta / n.) The equation (*) works for other compact matrix groups G but with an appropriate definition of "eligible" that, as with G = SO(n,R), includes all but measure 0 of of G. --Dan

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[math-fun] Uniqueness of projective plane of order 2
by asimovd＠aol.com 08 Nov '02

08 Nov '02

Jim Propp asks about the uniqueness of the projective plane of order 2. LEMMA: Up to isomorphism, there is just one projective plane of order 2. (Def. A proj. plane of order n is a set S of n^2+n+1 points having a collection (denoted Lns) of n^2+n+1 specified subsets of size n+1 called "lines", such that a) any two points lie in a unique line, and b) any two lines intersect in a unique point.) Given a proj. plane (S;Lns) of order 2, then #(S) = 7. Pick any point p, and note that there must be exactly 3 lines containing p. WLOG we denote these lines as {p,a2,a3}, {p,b2,b3}, {p,c2,c3}, where these 6 new elements must all be distinct. WLOG we may assume the line determined by {a2,b2} is {a2,b2,c2}. The remaining 3 lines are now completely determined as {a2,b3,c3},{a3,b2,c3} and {a3,b3,c2}. --Dan

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[math-fun] A request
by asimovd＠aol.com 08 Nov '02

08 Nov '02

I have a favor to ask those posting to math-fun. Could you include your name at the bottom? At least your first name? That would help to orient me to who is posting. Thanks. --Dan

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[math-fun] Re: PSL(2,F_7), SL(3,F_2), and the Fano plane
by James Propp 08 Nov '02

08 Nov '02

I asked: >What's a good way to see that PSL(2,F_7) is isomorphic to SL(3,F_2)? "Good" is of course a subjective term; I was trying to come up with someething suitable for a class that I'm teaching. One slick way might be to 1) give two models of the projective plane over F_2 (one with the usual D_3 symmetry, the other given by translates of {1,2,4} in F_7, with 7-fold rotational symmetry), 2) show (non-constructively) that the 7-point finite projective plane is unique (so that the two models HAVE to be the same), 3) show that the symmetries of the first are given by SL(3,F_2), 4) show that the symmetries of the second are given by PSL(2,F_7), 5) ask them to explicitly construct an element of SL(3,F_2) of order 7. Of course, this approach leaves one with a lingering sense of mystery. Is there a good way to see that the two models of the 7-point finite projective plane are the same? I'd like to wrap things up with a proof of simplicity (especially if there's a sweet proof that takes advantage of the fact that the group has these two different representations). Can anyone think of one? (Is John Conway still reading math-fun? I'd've expected him to know a two or three answers to each of my questions off the top of his head. It seems he hasn't posted since late September.) Jim Propp

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Re: [math-fun] Prodigals
by Mike Stay 07 Nov '02

07 Nov '02

Googling for "product derivatives" (plus appropriate filters for "milk product derivatives" and the like) doesn't turn up anything. Has any work been done there? Or on q-deformed product integrals? -- Mike Stay staym(a)datawest.net

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Re: [math-fun] Prodigals (of Lie groups)
by asimovd＠aol.com 07 Nov '02

07 Nov '02

One can even define a product integral for reasonable subsets of an arbitrary Lie group G. Log is interpreted (almost everywhere) as pulling a group element back to the (linear) tangent space T_e at the identity e. An ordinary integral over a portion of R^n = T_e is calculated. Finally the standard Lie group map exp: T_e -> G is applied to the result. Since all compact Lie groups G are isomorphic to some subgroup of GL(n,R) for some n, (I'm not sure about noncompact G) maybe this turns out to be equivalent to the matrix product integral mentioned by mlb. --Dan

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Re: [math-fun] Prodigals
by James Propp 07 Nov '02

07 Nov '02

Apropos of "prodigals", I once saw a copy of a curious little book called "Non-Newtonian Calculi". Its author (whose name I forget) had the notion that it would be mathematically interesting to study a whole lot of functionals of the form F(D(F^{-1}(f))), where F is some functional (like exponentiation), F^{-1} is its inverse, and D is differentiation. Unfortunately, the most interesting thing that the author could think of to do with this idea was to derive various identities (chain rules, etc.) for these various unmotivated definitions. Jim Propp

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Re: [math-fun] Prodigals
by asimovd＠aol.com 07 Nov '02

07 Nov '02

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[math-fun] Product integration
by Eugene Salamin 07 Nov '02

07 Nov '02

John D. Dollard and Charles N. Friedman, "Product integration with applications to differential equations", Cambridge University Press, 1984. ISBN: 0521302307 __________________________________________________ Do you Yahoo!? U2 on LAUNCH - Exclusive greatest hits videos http://launch.yahoo.com/u2

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Correction: Re: [math-fun] Product integrals
by asimovd＠aol.com 07 Nov '02

07 Nov '02

Correction: I wrote (regarding choosing a domain D in C on which log is defined and continuous): << More generally, D can be chosen to be any simply-connected region of the Riemann surface for log. >> This should read instead: More generally, D can be chosen to be any bounded open subset of the Riemann surface for log. P.S. The 1-dimensional product integral can be adapted to work over curves in C as well. --Dan

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