Unoriented and oriented conformal completion of 1+1 dimensional spacetime

What we are doing here is playing in the sandbox of a conformally completed 1+1-dimensional spacetime. 

Duality and triality in action

The term “conformal compactification” is often used, but it is somewhat misleading. More accurately, what we perform is a completion: we extend spacetime by admitting infinity itself. In the flat ${}^{}$R1,1 version of spacetime, infinity lingers only as a possibility—an unrealized horizon. But once we complete it into its $S^1\times S^1$ form, infinity ceases to be merely potential: it becomes actual, tangible, part of the fabric.

Now, there are two distinct ways this completion may unfold: oriented and unoriented. In the unoriented version, ${\mathbb{R}}^{1,1}$ is completed by sewing on a single circle at infinity, upon which endless worldlines of the flat spacetime find their closure—though each closes, as a rule,  at a different point of the same circle. In the oriented version, however, we grant a deeper polarity: two circles are added, dividing the worldlines of the future from those of the past. Each kind closes upon its own eternal loop. Yet, such generosity comes with a price—an entire second copy of flat spacetime appears in the completion.

Metaphysically, this picture gains resonance. It reflects duality, echoing matter and antimatter. And beyond duality, it hints at triality: space, its mirror antispace, and the twin circles at infinity—mediators, thresholds, silent connectors between the two realms.

This essay continues the thread begun in Sunday special – projective cone. We pick up precisely where we left off there. 

Identifying points x and λx, with λ>0, on the null cone of R^1,1 we obtain the oriented Grassmannian. To obtain the unoriented one, we additionally identify x with -x. The oriented Grassmannian is  nothing else but the torus S¹⨉S¹. What is then the unoriented one. What becomes of the torus when we identify its opposite points? Different things can happen, depending on what we call "opposite". One way of doing it is by taking the quotient of S¹⨉S¹ by the action of Z₂ defined by

(z,w)⟼(1/z,-w),     |z| = |w| = 1.

This is described in details, for instance, in Conrad's online lecture note "Quotients by group actions", p.4, and following. We obtain the famous unorientable Klein bottle. But we have a different action of Z2 on the torus - our action is

(z,w)⟼(-z,-w).                (1)

Note. Well, we do not have (z,w), we have

(x¹)² + (x²)² = (x³)² + (x⁴)² = 1.                   (2)     

But we can always set z = x¹ + i x², w = x³ + i x⁴. Then |z|= |w| =1, and changing the sign of x = (x¹,x²,x³,x⁴) is the same as changing the signs of z and w.

Proposition 1. The quotient of the torus S¹⨉S¹ by the action of Z₂ defined by (z,w)⟼(-z,-w) is again a torus. The diffeomorphism of the quotient space onto the torus induced by the map

φ(z,w) = (zw,w/z).                (3)

Proof. The map φ is smooth on S¹⨉S¹, and φ(z,w) = φ(-z,-w). Moreover, for any u,v in S¹⨉S¹, the equation φ(z,w)= (u,v) has exactly two solutions for (z,w):

1) z = u^½v-½, w = u^½v^½,
2) z = -u^½v-½, w = -u^½v^½.                (4)

Therefore φ maps the torus onto itself with the required kernel. QED.

Exercise 1. Make sure that you completely understand the proof above.

So torus has the very special property that it can double cover (and even n-cover) itself.

The above special feature holds for the null cone in R^2,2, but does not hold in R^4,2, where we have S³⨉S¹ instead of S¹⨉S¹. For R^4,2 taking the quotient by an analogous Z₂ produces a topologically different manifold.

Parametrization of the unoriented Grassmannian

For the oriented Grassmannian we used Eq. (2) to set

x¹ = cos(α), x² = sin(α),
x³ = cos(β), x⁴ = sin(β) ,               (5)

to define uniquely the angles α and β of a point on the torus. However this parametrization distinguishes the points x and -x, which become identical under the equivalence relation defined by the action (1) of Z₂. Can we find a parametrization of the new torus, obtained by this identification? To find this new parametrization we will use Eq. (3) and Proposition 1.

If we set z = e^iα, w = e^iβ, then u = zw = e^i(α+β), v = w/z =e^i(β-α).  If we define then

φ = (β + α) mod 2π,
ψ = (β - α) mod 2π,                (6),

then (φ,ψ) give us the required parametrization. What we need for defining a point on the torus are cos and sin of φ and ψ. Using (5) and trigonometric formulas for cos and sin of a sum and of difference, we find:

cos(φ) = x¹x³ - x²x⁴,
sin(φ) = x²x³ + x¹x⁴,
cos(ψ)  = x¹x³ + x²x⁴,
sin(ψ) = x²x³ - x¹x⁴.

The required condition sin²+cos² = 1 follows then automatically from (2).

Exercise 2. Verify the last statement.

Sunday special - projective cone

"there are more ways 
of drawing a cat out of a well 
than by the bucket!"

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. R^2,2 is 4-dimensional, and we can't really picture four dimensions. So let us take one dimension out and descent to R^2,1. In R^2,1 we can draw pictures. Of course some things will be different in R^2,1 than in R^2,2, 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 R^2,1.

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 R^2,1 and R^2,2: In R^2,2 the null cone, after removing the origin, becomes consisting of two disconnected parts - on the picture red and blue. In R^2,2 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 Gr_k(Rⁿ) - the Grassmannian of k-planes in Rⁿ (there are also complex Grassmann manifolds). Here we are dealing with Gr₁(R^2,1). 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(R^2,1). 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 R^2,1, 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 R^2,1 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 R^2,1 and graphics After that we will comment on how to deal with the case of interest, namely with R^2,2, 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 R^2,1 is a disjoint union of two circles.   

 

Let us first consider the 3D case, with R^2,1 and graphics After that we will comment on how to deal with the case of interest, namely with R^2,2, 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 R^2,1 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 R^2,1.

In R^2,2, 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!"

So here are these other ways, more cat-friendly:

In the next part we will move to to the twisted torus - the Klein bottle.