# Is poincare grouo simply connected?

1. Sep 15, 2009

### andrea.dapor

We call a group G "simply connected" if every curve C(t) in G which is closed (that is, C(0) = C(1) = I) can be continuously deformed into the trivial curve C'(t) = I (where I is the unit element in G). This is formalised saying that, for each closed C(t), there exists a continuous function F: [0, 1]x[0, 1] -> G such that

1) F(0, t) = C(t), for all t
2) F(1, t) = I, for all t
3) F(s, 0) = F(s, 1) = I, for all s

Now, Wald (General Relativity, 1984) says that the Poincare group is not simply connected, beacuse in particular for a rotation of $$2\pi$$ about an axis - say z - such a function F does not exist.

My question follows.
Consider the function

F(s, t) := sI + (1 - s)C(t),

where C(t) is the closed curve in Poincare group G associated to a rotation of $$2\pi$$ about z, that is,

C(t) =
(1 0 0 0)
(0 cos2\pi t -sin2\pi t 0)
(0 sin2\pi t cos2\pi t 0)
(0 0 0 1)

with t in [0, 1].
This F seems to verify (1)-(3)... where is my mistake?

I thank you for your help, and apologize for the "matrix" above...

2. Sep 15, 2009

### hamster143

Your F is not a member of Poincare group ... I and C are, but neither the addition nor multiplication by scalar are valid operations in SO(3,1). You can easily do the math and see that the determinant of your "rotation" det F is, generally speaking, not equal to 1.

Last edited: Sep 15, 2009
3. Sep 15, 2009

### andrea.dapor

Oh, yes! Even sI is not a poincare transformation (if s is not 1, then its determinant is different from 1, that is, it is not a Lorentz transformation). We must build an F just by composing (matrix product) elements of poincare group. Thank you.

-split the $$2\pi$$ rotation C(t) = R_z($$2\pi$$t) into two $$\pi$$ rotations,

R_z($$2\pi$$t) = R_z($$\pi$$t) R_z($$\pi$$t)

-then rotate the axis of the first in such a way that it becomes a rotation of -$$\pi$$t, that is, a rotation about -z; this can be done rotating the first by $$\pi$$ about x:

[R_x($$\pi$$) R_z($$\pi$$t) R_x($$\pi$$)^T] R_z($$\pi$$t) = R_z(-$$\pi$$t) R_z($$\pi$$t) = I.

-So we see that apparently the function

F(s, t) := R_x($$\pi$$s)R_z($$\pi$$t)R_x($$\pi$$s)^T R_z($$\pi$$t)

satisfies (1)-(3).
However, (3) is not verified, since when t = 1 the rotation R_z($$\pi$$t) is not the identity, and so we cannot write

F(s, 1) = R_x($$\pi$$s)R_z($$\pi$$)R_x($$\pi$$s)^T R_z($$\pi$$) = R_x($$\pi$$s)R_x($$\pi$$s)^T = I.

This, instead, is true for a rotation of $$4 \pi$$ about z, since R_z($$2\pi$$t) is the identity when t = 1. Indeed, Wald specifies that a $$4\pi$$ rotation can be continuously deformed to I: the function F is then

F(s, t) := R_x($$\pi$$s)R_z($$2\pi$$t)R_x($$\pi$$s)^T R_z($$2\pi$$t).

What do you think? Am I correct?

4. Sep 15, 2009

### genneth

Yes --- 2 pi is not deformable to the identity, but 4 pi is.

5. Sep 15, 2009

### javierR

Re: Is poincare group simply connected?

I don’t know if I see the essence of the non-simply-connectedness in your posts. I'll give a long response, hoping to help out somewhere.

First, a technical point: The Lorentz group (Poincare group minus the translations) itself is not connected, but can be broken into 4 connected components, one of which contains the identity. If $$\Lambda$$ is a general Lorentz group element, then the component with $$det\Lambda=+1\;\;\;\;sign\Lambda^{0}_{0}=+1$$ forms a subgroup called the “restricted Lorentz group” (if we tag on spatial and temporal inversions, looking at this subgroup is sufficient for understanding the full Lorentz group).

While the restricted Lorentz group is connected, it’s not *simply* connected, which is the statement you are examining. It happens that, to have spacetime spinors, you need to have a simply connected spacetime group that contains the restricted Lorentz group. This group is, therefore, called Spin(3,1), and happens to be isomorphic to SL(2,C) the group of 2x2 complex matrices of determinant 1. How is this related to the restricted Lorentz group? If U is an element of SL(2,C), it describes a Lorentz transformation. The element –U is a distinct element of the group, but it ends up describing the same Lorentz transformation when acting on spacetime vectors (generally, on tensors). It’s not surprising that SL(2,C) is called a “double cover” of the restricted Lorentz group. The restricted Lorentz group is then obtained by identifying U with –U: U== -U. A way of writing this is $$SL(2,C)/Z_{2}$$, where $$Z_{2}$$ has two elements that act as +1 or -1.

Now, since you’re concerned with “simple connectedness”, we want to know about the topological properties of the group and curves connecting elements of groups. Focusing on the compact part of SL(2,C) (the non-compact part will be irrelevant), it consists of unitary matrices SU(2), which has the topology of a 3-sphere (one step up from everyday spheres that we’re used to). Any closed curves (loops) on the sphere can be contracted to a point, so SU(2) is simply connected. Let u be the element of SU(2) corresponding to the element U of SL(2,C) we were talking about before. Then identifying u and –u like we did before, the transformations are points of the topological space (3-sphere)/$$Z_{2}$$. That is, identify opposing points of a 3-sphere. You can now imagine paths on the 3-sphere between two group elements (ok, you can try to draw these paths on a 2-sphere): let’s say path 1 goes between the identity element u=1 in the northern hemisphere and an element u’ in the same hemisphere that acts as a rotation about the z-axis by some angle theta (generally, you need more parameters to cover the sphere, but we're just looking at theta). Now, we could instead go on path 2 from u=1 to the direct opposite side of the sphere from u’; that is, going to –u’ on the southern hemisphere. Now, in our space, the points u’ and –u’ of the northern and southern hemispheres are identified, so these two paths start and end at the same point, from u=1 to u’ == –u’, so it is technically a loop. So can we contract this loop? I’ve tried here to diagrammatically show why you cannot:

*---<----u====>=====u’----<-----*

Here there’s the path 1 from u to u’ denoted by ===>===. Then there’s path 2 denoted by --->--- lines, and which is broken up by * symbols because at that point the path *jumps* across to the opposite side of the sphere and then continues toward u’ (that is, instead of drawing path 2 from u=1 to –u’ in the southern hemisphere, I drew the *reflected path* once you cross into the “–u’s”, so the entire path is drawn on the northern hemisphere). The path is stuck to the *’s, so the loop cannot contract. This is why the (restricted) Lorentz group is not simply connected.

How does this fit in with the 2*pi and 4*pi rotations you were talking about? Like I mentioned before when we want spacetime spinors, we need to consider the full group SL(2,C), in which case u and –u are again distinct elements. Even though they represent the same Lorentz transformation on tensors, they are not the same for spinors. Consider the SL(2,C) element
$$U=\left(\begin{array}{cc}e^{i\theta /2} & 0 \\ 0 & e^{-i\theta /2}\end{array}\right)$$
describing a rotation by angle theta about the z-axis. As you vary theta, you move along a path on the sphere. Putting in theta=2*pi, you get U=-1. Putting in theta=4*pi you get U=+1. This may look familiar because it’s the rotation of a spacetime spin-1/2 object, and spinors do indeed have that strange property that a 2*pi rotation leads to a different state. So the SL(2,C) group handles spinors appropriately, not SO(3,1).
We can see the paths for this special case, with u=1 (theta=0), u’=-1 (theta=2*pi). In the SL(2,C) (or SU(2)) picture this is just moving from one point on the 3-sphere to the point opposite it, and these represent distinct points. But for (3-sphere)/$$Z_{2}$$, the points are identified so that we are forming a closed loop:
*<-----u=1----<------*
But as we’ve seen that loop isn’t contractible; at the *'s the matrices that make up the points of out path (parametrized by theta) show a jump. So by trying to enforce an equivalence btwn “no rotation” and “2*pi rotation”, we end up with the feature that some special closed loops and can’t be contracted…something strange is happening, and that’s because we’re technically not supposed to make these equivalent. Rotating further to 4*pi is the equivalent of overlapping two of these:
*<-----u=1----<------*,
one undoing the other (that’s called “doubly connected”). It’s equivalent to combining one of these paths with its reverse, which is equivalent to a fully contracted loop. Can you see how you would write a double-layered path in terms of the matrix-parametrization above so that the jumps "cancel" each other?

6. Sep 15, 2009

### marcus

javierR, that is beautifully explained.
I post here because I want to be certain to catch your attention. You said something on some thread, I forget where, that made me think you are a European particle physicist. I think you probably know about the MAGIC telescope on the island of Las Palmas in the Atlantic.

I want to know if the MAGIC telescope(s) found anything about particle physics.

This is a different topic, so I will start a new thread here in the same forum (High energy, nuclear, particle physics forum).

I am interested in these telescopes because they are a new type, that sees incoming gamma from their Cherenkov trails. Do you know anything you can tell us about this new telescope technology?