I am trying to figure out the fundamental group (actually simply connected or not will suffice) of the following quotient space of $SO(3)$:

Let $X = SO(3)/E$, where $E$ is the equivalence relation defined as follows:

$E \equiv M \sim {S_{A}}^{i} * M * {S_{B}}^{j}$ where ${S_{A}}^{i} \in $ Crystallographic point group $A$ and ${S_{B}}^{j} \in $ Crystallographic point group $B$, $M \in SO(3)$.

$*$ represents the multiplication operation (matrix multiplication if rotations are represented as $3 \times 3$ special orthogonal matrices)

Crystallographic point group is defined here: http://en.wikipedia.org/wiki/Crystallographic_point_group

"In crystallography, a crystallographic point group is a set of symmetry operations, like rotations or reflections, that leave the crystal invariant (hence a symmetry)."

Only consider the point groups with rotational symmetries. There exist 11 crystallographic point groups for three-dimensional crystals. Let us start with just considering cyclic point groups. They are the following:

(a) $C_1 = \{I \}$ where I is the identity rotation.

If $Z_{\omega}$ is a rotation of angle $\omega$ about the $Z-$axis.

(b) $C_2 = \{I, Z_{\pi} \}$

(c) $C_3 = \{I, Z_{\frac{2\pi}{3}}, Z_{\frac{4\pi}{3}} \}$

(d) $C_4 = \{I, Z_{\frac{\pi}{2}}, Z_{\pi}, Z_{\frac{3\pi}{2}} \}$

(e) $C_6 = \{I, Z_{\frac{\pi}{3}}, Z_{\frac{2\pi}{3}}, Z_{\pi}, Z_{\frac{4\pi}{3}}, Z_{\frac{5\pi}{3}} \}$

For example if Crystallographic point group of A is $C_2$ and B is $C_3$, the equivalence relations are:

$ M \sim M * Z_{\frac{2\pi}{3}} \sim M * Z_{\frac{4\pi}{3}} \sim Z_{\pi}* M \sim Z_{\pi} * M * Z_{\frac{2\pi}{3}} \sim Z_{\pi} * M * Z_{\frac{4\pi}{3}} $

There exists literature for cases when one of the point groups is $C_1 = \{ I \}$. In this case it is a group action on $SO(3)$ and the space $X = SO(3) / G$ where $G$ is one of the 11 crystallographic point groups. These spaces fall under the so-called spherical 3-manifolds ( http://en.wikipedia.org/wiki/Spherical_3-manifold ). When $G$ is one of the cyclic groups above, $X$ is a lens space. $L(2n,1) \cong SO(3)/ C_{n}$. I am not able to figure out how to think of these spaces when there are two point groups involved.

Progress so far: Even when there are two point groups acting on $SO(3)$, which if we refer to as $ G_{1} \backslash SO(3)/ G_{2}$ where $G_{1}$ is crystallographic point group of $A$ and $G_2$ refers to system $B$, there exists a finite subgroup of $\Gamma$ of $SO(4)$ such that $ G_{1} \backslash SO(3)/ G_{2} \cong S^{3}/ \Gamma$. In the case $G_{1} = C_{1}$, it turns out that $\Gamma$ acts properly discontinuously and hence the fundamental group of that space is $\Gamma$ itself (from literature). But I am not sure how to check if $\Gamma$ acts "properly discontinuously" or not. And I am not sure how to check if the space is simply connected or not if $\Gamma$ does not act properly discontinuously.

Any help appreciated.

Thank you.

To answer Aaron's question: do you know how to obtain this group $\Gamma$ explicitly?

I use the quaternion representation for rotations. Let $M = (q_0, q_1, q_2, q_3)$. I also use the following fact: Each 4D rotation $R$ is in two ways the product of left- and right-isoclinic rotations $R_L$ and $R_R$. $R_L$ and $R_R$ are together determined up to the central inversion, i.e. when both $R_L$ and $R_R$ are multiplied by the central inversion their product is $R$ again. (from wiki: http://en.wikipedia.org/wiki/SO(4) ). So for any operation ${S_{A}}^{i} * M * {S_{B}}^{j}$, I can find $\Gamma_{ij}$ such that ${S_{A}}^{i} * M * {S_{B}}^{j} = \Gamma_{ij}*[q_0 \ q_1 \ q_2 \ q_3]'$.

The collection of all such $\Gamma_{ij}$ (and $\Gamma_{ij} * -I_4$, where $I_4$ is the $4 \times 4$ identity matrix) forms a finite subgroup $\Gamma$ of $SO(4)$. I used the definition provided by Aaron below for free action and found that at least for the cases where crystallographic point groups of $A$ and $B$ are the same (e.g. $C_2 \backslash SO(3) / C_2$), the action of finite subgroup $\Gamma_{C_2,C_2}$ is not properly discontinuous. Even though I tried to generalize the problem I am more interested in the cases when the crystallographic point groups of A and B are the same. So, how can I say whether this space $C_2 \backslash SO(3) / C_2$ is simply connected or not, especially now that i know that the action of $\Gamma$ is not properly discontinuous ??

Any ideas ?


1 Answer 1


This is cool. Are you doing chemistry?

So, a slightly more standard notation might be $G_1\backslash SO(3)/G_2$ -- this indicates that you're left-multiplying by elements of $G_1$ and right-multiplying by elements of $G_2$.

A "properly discontinuous" action of a group $G$ on a space $X$ is one where every point $x\in X$ has a neighborhood $U$ such that $g(U)\cap U=\emptyset$ unless $g=1$; that is, $G$ not only acts freely (no nonidentity element has fixed points), but the nonidentity elements of $G$ take every point "sufficiently far away from itself". This may seem vacuous, but in fact when you're working with Kleinian groups and things this can be an important condition. However, in your case what is immediately true (since your space $SO(3)$ is Hausdorff and your groups are finite) is that any free action is automatically properly discontinuous. So the only thing you need to check is that your actions have no fixed points.

So I guess the question becomes, do you know how to obtain this group $\Gamma$ explicitly? If it's cyclic, it probably will be taken to be a subgroup of the circle, which acts freely on $S^3$. (This is $S^1\subseteq \mathbb{C}$ acting on $S^3\subseteq \mathbb{C}^2$ by multiplication.)

If $\Gamma$ doesn't act properly discontinuously, things are harder.

  • 1
    $\begingroup$ Materials Science to be more precise. And the space $X$ is called the misorientation space. It is the rotational difference between two crystals $A$ and $B$ of different orientations. Since each crystal system has symmetries, there exist symmetrically equivalent descriptions of orientations and hence symmetrically equivalent descriptions of misorientations as given by equivalence relations $E$ $\endgroup$
    – Srikanth
    Dec 10, 2010 at 19:02
  • $\begingroup$ FYI, I edited the question showing how I explicitly find $\Gamma$ $\endgroup$
    – Srikanth
    Dec 10, 2010 at 20:14
  • $\begingroup$ Sorry to take so long. Hopefully you'll get a notification about this. Anyways, this is a silly question, but are your copies of $C_n$ acting in the same way? If so, you may as well consider it as just a quotient by a single action. In any case, if I had to guess I'd say that this is probably not simply-connected, just because it's an (admittedly non-free) finite quotient of a non-simply-connected space. Intuitively, I don't see how quotienting could de-essentialize an essential loop... $\endgroup$ Dec 17, 2010 at 11:36
  • 1
    $\begingroup$ Sorry I'm way late to the game, but quotienting can turn a nonsimply connected space into a simply connected one. Take your space as $S^1$ and your group $\mathbb{Z}/2\mathbb{Z}$ acting by reflection. Then $S^1/(\mathbb{Z}/2\mathbb{Z})$ is homeomorphic to $[0,1]$, so is contractible. I'll try and think about Srikanth's general problem a bit. $\endgroup$ Mar 13, 2012 at 19:00
  • $\begingroup$ You're right, I'm not sure what I was thinking (all those years ago). Obviously, any transitive group action on a space has contractible quotient! Thank goodness people invented stacks... $\endgroup$ Apr 14, 2012 at 13:04

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