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"there are more ways |
This note is special. It is in reply to Anna's comment under the previous post. Let me quote it here:
" I do not understand about embedding...
In particular, about the "more geometric picture of it". What is the
relation between the null cone of R2,2 and the Grassmann manifold ~
S1xS1 = torus?
Probably, the torus is a 2d section of the 3d cone?
And why the image of embedding flat space R1,1 into a curved manifold
S1xS1 covers only a part of it? On the contrary, if we try to embed R1
into a circle, it needs to be wrapped around the circle many times
making a multiple cover ..."
I see from this question that I did not explain the "order of things" with sufficient clarity - so far. So, let us start from scratch.
It is always good to have in mind some picture. Pictures can be misleading, but they can be extremely useful. To have a picture in mind we need to take out one dimension. R2,2 is 4-dimensional, and we can't really picture four dimensions. So let us take one dimension out and descent to R2,1. In R2,1 we can draw pictures. Of course some things will be different in R2,1 than in R2,2, but I will stress similarities and we will also note the differences. In fact we will be using simply our 3D space R3, but we will call the coordinates x,y,t, and we will be particularly interested in the cone x2 + y2 = t2, the natural "null cone" of R2,1.
In fact, our object of interest is not the whole null cone - we remove from the null cone the "origin", the point with coordinates x = y = t = 0. Here is a symbolic image:
And here I want to note the first difference between R2,1 and R2,2: In R2,2 the null cone, after removing the origin, becomes consisting of two disconnected parts - on the picture red and blue. In R2,2 the null cone has the equation (x1)2 + (x2)2 = (x3)2 + (x4)2, and even after removing the origin it is still a connected surface - all one color. We can't imagine it, but we can prove it algebraically: we can construct a continues path joining a point (x1,x2,x3,x4) on the cone, with (-x1,-x2,-x3,-x4), a path always staying on the cone and never passing through the origin! This is important.
But we are not so much interested in the points of the null cone, we are interested in straight lines through the origin that span the cone - generating lines. In 2+1 Minkowski space they would be light rays through the origin. Each such line represent a point in the manifold of lines. This is a particular case of what is being called a Grassmanian. In general we have Grassmann manifolds of lines, planes, and more-generally the Grassmannian Grk(Rn) - the Grassmannian of k-planes in Rn (there are also complex Grassmann manifolds). Here we are dealing with Gr1(R2,1). The points on one line, for one reason or another, are indistinguishable for the observer, and the whole family of such points is being interpreted as just one "event" - metaphorically speaking.
Two kinds of Grassmannians
There are two kinds of Grassmannian: unoriented and oriented ones. Usually only unoriented are being discussed. For instance Wikipedia article on Grassmannians has only four lines, very incomplete, about the oriented case. I opt for the oriented one. It i richer in structure, which we will see later on. Let us discuss the case of Gr1(R2,1). We are interested in null lines through the origin. We first notice that all these lines meet at the origin, and yet they are being considered as different - disjoint points in the manifold of lines. There is another view of Gr1 that avoids this potential confusion. Namely we remove the origin from R2,1, and introduce equivalence relation between remaining points, the two points x,x' are considered equivalent if x' is proportional to x with a non-zero proportionality constant. This is equivalent to x and x' being on the same line through the origin. This is another, equivalent, definition of the unoriented Grassmann manifold of lines. Its points are the equivalence classes of so defined equivalence relation. But we can as well demand the proportionality constant to be positive. Doing so we obtain the oriented manifold of lines. Now one equivalence class of the first relation contains two equivalence classes of the second relation. On our picture in R2,1 it means that, roughly speaking, we distinguish between the future and the past points on the light cone, even if the are on the same light ray.
Topology of unoriented and oriented Grassmannians
Let us first consider the 3D case, with R2,1 and graphics After that we will comment on how to deal with the case of interest, namely with R2,2,
when we have to do it only analytically. Let us take the case of
oriented null lines. We have lines through the origin extending into the
future, t>0, and null lines extending into the the past. If we take
the plane t = 1, it will cut each null line extending the future at
exactly one point, and never cut the line extending into the past. The
common points of the plane and the cone form a circle. So we can
uniquely parametrize the future null lines by points of this circle. But
the lines on the past light cone are still missing. For them we need
another circle, at t = -1. Thus, topologically, the oriented Grassmann
manifold of null lines in R2,1 is a disjoint union of two circles.
For unoriented null lines the two circles merge into just one, since every point x on one of the circles is identified with -x on the second circle We are getting then only one circle. That is the case with the projective null cone in R2,1.
In R2,2, the topologies of oriented and
unoriented null Grassmannians become more interesting. Consider the
oriented case first case. Points on th null cone with origin removed
satisfy equation:
(x1)2 + (x2)2 = (x3)2 + (x4)2 = r2, r>0.
Since r>0, for each such point there is a unique representative of
the equivalence class of the oriented Grassmannian, for which r=1.
Thus we obtain unique representation by a point satisfying:
(x1)2 + (x2)2 = (x3)2 + (x4)2 = 1.
This is an equation of a torus. We can set:
x1 = cos(α), x2 = sin(α),
x3 = cos(β), x4 = sin(β).
with α,β in [0,2π). We can represent the torus as a square, with opposite edges identified.
To turn a square with edges labeled α,β into a torus in three dimensions, first imagine bending the square so that one pair of opposite edges (say, where alpha is 0 an2π) are joined seamlessly, forming a cylinder with the beta edges as open ends(see the generated image above). Next, curve this cylinder around in space so the two open ends (where beta is 0 and 2π) meet and merge, creating a smooth doughnut shape where both the alpha and beta coordinates wrap around without any boundary(see the animation below).
Of course "there are more ways of drawing a cat out of a well than by the bucket!"




This post is especially for those who is fond of cones like me. I should mention once again the famous and mysterious Koide formula that has recently captivated me. It describes the masses of the three families of leptons: the electron, the muon, and the tau meson. It can also be applied to other triads, such as quarks and neutrinos. The point is that its geometric image is exactly the null cone in R(2,1). And nobody knows why it is so. I can't show the picture here, but I'll email it to Ark.
ReplyDeleteSo, looking ahead, it seems that my guess was not far from the truth: the 1Sx1S torus is Gr_2, i.e., a 2d section of the 3d null cone in 4d R(2,2), isn't it right?
ReplyDeletePoor cat! Luckily, it is only virtual and can survive in all the dangerous transformations that Ark put it through.
ReplyDeleteIt is now quite clear - the torus is indeed the section of a null-surface in R(2,2), just like the circles in the case of R(2,1), with 1 extra dimension. Equating (x1)^2 + (x2)^2 = (x3)^2 + (x4)^2 we impose one condition, which eats one dimension, to obtain 3d null-surface; and putting (x1)^2 + (x2)^2 = (x3)^2 + (x4)^2 = 1 we impose the second condition to obtain the toroidal cross section of it.