[continued from previous message]   
      
   to make the noncommutative torus into a strong r-commutative algebra   
   such that   
      
   R(u x u) = u x u   
   R(v x v) = v x v   
      
   and   
      
   R(u x v) = q v x u.   
      
   (Recall that "strong" simply means that R^2 is the identity, so   
   R(v x u) = q^{-1}u x v.)  One may thus go ahead and define   
   "r-commutative differential forms" for the noncommutative torus, which   
   satisfy   
      
   u(du) = (du)u,        v(dv) = (dv)v,   
   u(dv) = q(dv)u,       (du)v = qv(du)   
      
   and more relations obtained by differentiating these.   One can then   
   calculate the (r-commutative) de Rham cohomology of the noncommutative   
   torus, and, lo and behold, it's isomorphic to that of the usual torus.   
   This fits into the philosophy that the noncommutative torus is obtained   
   from the usual torus by a "continuous deformation" -- no holes have been   
   formed or gotten rid of.   
      
   If you're interested in learning more about noncommutative tori, a good   
   review article is "Noncommutative tori: a case study of noncommutative   
   differential manifolds," Contemp. Math. 105, p. 191, by Rieffels.   
   If you're interested in the r-commutative geometry of noncommutative   
   tori, try my "R-commutative geometry and quantization of Poisson   
   algebras," which will appear in Adv. Math..   
      
   Let me just conclude by saying that noncommutative tori have   
   applications in physics.  This shouldn't really be surprising since   
   they're such simple things.  Let's say you have a charged particle   
   trapped on the xy plane, and there's a magnetic field of constant   
   intensity B perpendicular to the plane.  Then the momentum operators in the x   
   direction and the y direction no longer commute.  Exponentials of these   
   (i.e., translations) generate a noncommutative torus, and this fact has   
   been used by Belissard to do certain calculations of the quantum Hall   
   effect!  See J. Bellisard, "K-theory of C*-algebras in solid state   
   physics," Springer Lecture Notes in Physics 257.   
      
      
   Charged Particles in a Magnetic Field   
      
   I've decided that this is just a bit too brief and enigmatic for those   
   not up on their quantum mechanics.  So I'll explain what happens to a   
   charged particle in the plane when its in a magnetic field, and how one   
   gets a noncommutative torus out of this situation.  Then, since it's the   
   holiday season, I'll let myself digress to discuss a particle on a   
   *sphere* in a magnetic field.   
      
   First, let's set hbar = 1.  CLICK!   
      
   Suppose we have a particle on the plane - with no magnetic field.   
   Then in quantum mechanics the momenta in the x and y directions are   
   given by the operators,   
      
   p_x = -i d/dx   and  p_y =  -i d/dy,   
      
   respectively.  These commute, becuase mixed partials commute.   
      
   Now let's turn on the magnetic field pointing perpendicular to the   
   plane.  Let's say our particle has charge   
   = 1, and the field strength is B.  The curious fact about quantum theory   
   is that (if we neglect the *spin* of the particle) the only effect of   
   the magnetic field is to make us redefine the momentum operators to be   
      
   p_x = -i d/dx + A_x   and p_y = -id/dy + A_y   
      
   where A, the vector potential, has curlA = B.  Now p_x and p_y don't   
   commute, and in fact the commutator   
      
   p_x p_y - p_y p_x   
      
   is just -iB ... as everyone should check for whom it isn't instantly,   
   blindingly obvious.  It's a pity that when I was first learning quantum   
   mechanics the teacher didn't remark on how curious and charming this is:   
   when there's a magnetic field around, the different components of the   
   momentum no longer commute, and the amount by which they fail to   
   commute is precisely the magnetic field!  Recall that the meaning of   
   momentum in quantum theory is that it's the generator of spatial   
   translations.  This means that if you grab your charged particle and   
   move it first along the x direction and then the y direction:   
      
      
   	^   
   	|   
   	|   
    ------->   
      
   the particle winds up in a different state than if you first go in the y   
   direction and then the x direction:   
      
    ^------>   
    |   
    |   
    |   
      
      
   Another way of thinking of it is as follows: take your particle, move it   
   counterclockwise (say) around a rectangle:   
      
    v------<   
    |	|   
    |	|   
    >------^   
      
   and it'll be back in the same place, but its wavefunction will not be   
   the same as it was: it'll differ by a phase (multiplication by a complex   
   number of unit magnitude).  It's easy to calculate that if we call the   
   particles wavefunction to start out with "psi," and when we're done "phi,"   
      
   phi = exp(is) psi   
      
   where  s  is just the line integral of A around the rectangle.   
   routes.   By Stoke's theorem, s is just the integral of B over   
   the rectangle!   
      
   If I had been told this, I might have absorbed it a bit more quickly   
   when I was told in fancier language later on that the momentum   
   operators p_x and p_y were components of a "connection" on a "complex   
   line bundle," and that their commutator was the "curvature" of this   
   connection, so that the magnetic field is really a curvature, and that   
   the difference in phase obtained by taking two routes is called the   
   "holonomy" of the connection.  For that's the modern way of discussing   
   this sort of thing.   
      
   Anyway, now suppose that the magnetic field strength is a constant B.   
   Let  U  denote the unitary operator corresponding to translation by a   
   unit distance in the x direction, and let V be the unitary operator   
   corresponding to a unit translation in the y direction.  Then by what   
   I've said, we have   
      
   UV = qVU   
      
   where  q = exp(iB).  Thus U and V satisfy the relations of a   
   noncommutative torus.   
      
   More generally, if you move a charged particle counterclockwise around   
   any loop, its phase changes by exp(is), where s is the integral of B   
   over the region enclosed by the loop.  Note that the "counterclockwise"   
   bit comes from using Stokes' theorem.  (Also, an oppressive majority of   
   right-handed people have enforced foolish "right-hand rules"  ...   
   don't you think it'd make more sense to call CLOCKWISE the positive   
   orientation?  Oh well.)  If we go clockwise, the phase change   
   is exp(-is).   
      
   This still holds when our particle is not on   
   a plane but on some other sort of two-dimensional surface, but there are   
   some curious consequences.   
      
   Consider, for example, a particle on a sphere (for mathematicians, S^2),   
   with a magnetic field applied that's normal to the sphere at each point.   
   If we move the particle around a loop the phase change will equal the   
   integral of the magnetic field over the region enclosed by the loop.   
   But wait a minute!  There are two different regions, the "inside" and   
   the "outside" of the loop, which count as regions enclosed by the loop!   
   (Just draw a circle on a sphere and look!)  Who is to say   
   which one we should use to calculate the phase change?  There's only one   
   way out: we had better get the same answer each way!   That is, if we   
   call the integral of the magnetic field over the region "inside" the   
   loop B_1, and over the region "outside" the loop B_2, we must have   
      
   exp(iB_1) = exp(-iB_2).   
      
   Where'd that minus sign come from?  Well, if the loop goes   
   counterclockwise around one region, it goes clockwise around the   
   other region, so one of them gets a minus sign. (Draw a circle on a   
   sphere and look!)  In any event, if we write B = B_1 + B_2 for the   
   integral of the magnetic field over the whole sphere, the above equation   
   gives   
      
   exp(iB) = 1   
      
   so B must be an integer multiple of 2 pi!   
      
   This is an odd but true result, and it applies not only to the sphere   
   but to any (compact, oriented) surface: we can only make sense of a   
   charged quantum mechanical particle on such a surface in a magnetic field   
      
   [continued in next message]   
      
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