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.

Afternotes

19-09-25 16:50 

From Friday Funnies of Robert W. Malone