orthogonal groups

[PWrong]

I read something interesting about orthogonal matrices in my algebra textbook the other day.
Apparently the group of 2x2 orthogonal matrices O(2) is isomorphic to a pair of circles, because there are two of them for each angle theta. According to my textbook, "O(3) is a three dimensional space which is a bit harder to visualise". Does anyone know what it is? A number of spheres maybe?

[pat]

Yes, I believe, O(3) is a double-covering of the sphere just as O(2) is a double-covering of a circle.

In O(2), one of the circles represents the orientation-preserving rotations of R². The other circle represents the orientation-reversing rotations (flip + rotate). In O(3), one of the spheres represents the orientation-preserving rotations of R³. The other sphere represents the orientation-reversing rotations (flip + rotate).

[thigle]

SO(3):http://gregegan.customer.netspace.net.au/APPLETS/01/01.html
SO(4):http://gregegan.customer.netspace.net.au/APPLETS/22/22.html

[pat]

Those rule....

[wendy]

The full space of orthogonal matrices in E3, where each matrix represents a point, can be constructed from the set of angles and great circles.

Consider for example, that any rotation in 3d must produce a north-pole. This means that the set of great arrows corresponds to the set of points on a sphere.

Each of these points then represents any angle from 0 deg to 360 deg, with a convergence at each end. So what you get, is in 4d, the kind of thing like a circle-tegum.

The final stage is a reflection. So you get more-or-less a double-cover of the sphere, with the end points representing half-spheres (because one must accomidate reflection without rotation). So it's not that hard to visualise.

W

[PWrong]

In O(3), one of the spheres represents the orientation-preserving rotations of R3. The other sphere represents the orientation-reversing rotations (flip + rotate).

How many angles are in the general form of the matrix then? You only need two to specify a point on a sphere, but you need three to do a rotation, don't you?

[pat]

Yes, actually, you do need three Euler angles to specify a 3D rotation. I misspoke. Let me see if I can work this out with SO(3).

First, let's start with the difference between O(3) and SO(3). O(3) is the set of real-valued 3x3 orthogonal matrixes. A real-valued matrix M is orthogonal if it's transpose is its inverse.... M^T * M = M * M^T = I. Such matrixes must have determinant +/- 1 because

• det( A * B ) = det( A ) * det( B )
• det( A^T ) = det( A )
• det( I ) = 1

SO(3) is those matrixes in O(3) that have determinant 1. SO(3) is just the rotations. O(3) is the flips+rotations.

Every 3D rotation can be specified by an axis and an angle to rotate around that axis. However, rotating by angle aa around axis xx is equivalent to rotating by angle ( 360 - aa ) around axis -xx. So, there are two axis/angle pairs for every 3D rotation. Further, if the angle is 0, it doesn't matter what the axis is.

One way to visualize this would be to take a solid unit ball without boundary: x² + y² + z² < 1. Now, every point here corresponds to a 3D rotation. Let the angle aa be 360 times the radius to the point. Let the direction to the point from the center be the axis. Now, the problem with this is that are two points in this ball for every rotation (except the identity rotation). For example, a point near the north pole specifies the same rotation as a point just south of the center of the ball.

What can we do about it? We can try limiting the angle. Let's say the angle aa is just 180 times the radius to the point. Now, the only trouble is that we cannot do a 180-degree rotation. So, if we put the boundary back on the ball, what problem do we have? We have the problem that a point on the boundary specifies the same rotation as the point directly opposite it. We could lop off half of those boundary points... say all those for which z < 0, then lop off y < 0 when z = 0. Then lop off x <0>= 0. And, we still have a problem with the equatorial disc. We want to further restrict y >= 0 when z = 0. And, now we still have a problem with one diameter on our boundary, so we also need to restrict x >= 0 when y = 0 and z = 0. So, now we've started with a solid ball without boundary. We've lopped off an open hemisphere. Then, we've lopped off an open half-disc of that hemisphere's boundary. Then, we've lopped off an open radius of that shape's boundary.

Now, it turns out, that both of the above attempts are equivalent to the surface of a unit hypersphere S³ with antipodal points identified (paired up, assumed to be the same point, one-and-the-same). In our above attempts, the center point x = 0, y = 0, z = 0 doesn't have a twin point. But, in S³, it does.... the twins are x = 0, y = 0, z = 0, w = +/- 1.

So, that's SO(3). To get O(3), you need two copies... one copy to represent those without flips, and one copy to represent those with flips. Of course, each copy has its own antipodal points identified. Or, if you want to get really fancy, you can try slicing up hemispheres on S³ and make half of S³ represent rotations and the other half represent flip+rotations.

For example, you might say that the rotations are x,y,z,w where w >= 0, z >= 0 when w = 0, y >= 0 when z = 0 and w = 0, x >= 0 when y = 0 and z = 0 and w = 0. Then, you can say that if [ x, y, z, w ] is a rotation, then [ -x, -y, -z, -w ] is a flip+that-same-rotation.

[PWrong]

First, let's start with the difference between O(3) and SO(3). O(3) is the set of real-valued 3x3 orthogonal matrixes. A real-valued matrix M is orthogonal if it's transpose is its inverse.... MT * M = M * MT = I. Such matrixes must have determinant +/- 1 because
det( A * B ) = det( A ) * det( B )
det( AT ) = det( A )
det( I ) = 1

You just unintentionally answered one of my assignment questions . Too bad I already knew the answer.

So I take it SO(3) is isomorphic to a hemiglome, and O(3) is isomorphic to S³.

I wonder what O(4) would look like then. You need a plane to rotate around (or in), and an angle. It takes three angles to specify a vector with unit length, and we need two vectors to define the plane. So we need 2*3+1 = 7 angles for a rotation. If thats right, O(4) must be a 7D manifold.

[pat]

I wouldn't really say O(3) is isomorphic to S³. The word "isomorphic" implies to me that where things are smooth in S³, they are smooth in O(3). This isn't really the case near the hyperplane w = 0 (in my above formulation) where things go from rotations to flip+rotations.

[PWrong]

I've only recently learned about isomorphisms, but according to my textbook, isomorphic means "the same thing, but with the names changed". Although in vector spaces an isomorphism is just a linear map .

We can think of S¹ as a space of points in the plane, or a collection of angles which we can add. Or as complex numbers of modulus 1 which we can multiply. Or as the set of special orthogonal matrices ... these groups are isomorphic. Which is to say, they are basically the same but we may find it convenient to change the names.

Another example in the book is that the set of all shift maps is isomorphic to R².

So I suppose O(3) is isomorphic to some manifold made up of bits and pieces of S³.

[pat]

SO(3) is isomorphic to S³ with antipodal points identified.

O(3) is isomorphic to two copies of S³ with antipodal points identified within each copy.

[houserichichi]

Stronger still, if I'm not mistaken SO(3) is homeomorphic to S³ with antipodal points identified.

[bo198214]

SO(3) is isomorphic to S³ with antipodal points identified.

which is then our beloved projective plane?

[thigle]

the second one. (is a model of the beloved)

[thigle]

and the SO(3) is a specific case when one of the parameters of the projective plane is 1.

[PWrong]

I wonder what O(4) would look like then. You need a plane to rotate around (or in), and an angle. It takes three angles to specify a vector with unit length, and we need two vectors to define the plane. So we need 2*3+1 = 7 angles for a rotation. If thats right, O(4) must be a 7D manifold.

Now I'm not sure this is right. Alkaline's formulas page has six rotation matrices for 4D. And I think any rotation can be written as a product of these. That would make SO(4) a 6D manifold.

[pat]

Two things... first a plane and an angle is not sufficient to specify every possible 4D rotation. Clifford rotations rotate in two perpendicular planes. For that, you'd need to specify one plane, one angle, and then another angle for the perpendicular plane.

Second, the counting that got to 7D overlooks the fact that when you're specifying the plane, you really need two independent vectors. You can't chose the same unit vector twice. So, you've got an extra degree of freedom. Actually, you've got another extra degree of freedom from the fact that for this exercise, you can call any rotation in that plane the starting rotation.

Hmmm.... that last sentence wasn't very clear. But, consider if you were just doing rotations in the XY plane. By your 7D count, you would need one angle to specify the X axis and one angle to specify how much to rotate. But, really, you get the same answer regardless of which vector you pick to be the X axis. So, that's a bogus degree.

You are correct, any 4d rotation can be written as a product of those six rotation matrices. Alternatively, any rotation can be written as a * x * b where x is the variable for the initial point to be rotated expressed as a quaternion and a and b are unit quaternions. There are three degrees of freedom in specifying a unit quaternion.

Let's count a different way. We want to know how many degrees of freedom are needed to specify any 2-D plane through the origin in 4-D. Let's step back and do the same in 3-D. In 3-D, there is a single normal to the plane. We need to be able to have that normal point in any direction. So, we need two angles to specify the direction of the normal. In 4-D, we can think of a plane as having two normals that are perpendicular to each other.

Start with the XY-plane. It is perpendicular to both the Z and the W axis. What do we need to do to specify how we rotated the XY-plane into a new position? You need three angles to say how the first normal was rotated. You need two angles to say where the second normal ends up in the 3-space perpendicular to the first normal (because it still has to end up being perpendicular to the first normal). But, then you've overspecified by one degree since you can rotate the plane anywhere within the plane and be talking about the same plane. Then, you need one angle for in that plane and one angle for perpendicular to that plane. So, by my count, it's 3 + 2 - 1 + 1 + 1 = 6.

Wadsworth: The game's up, Scarlet. There are no more bullets left in that gun.
Miss Scarlet: Oh, come on, you don't think I'm gonna fall for that old trick?
Wadsworth: It's not a trick. There was one shot at Mr. Boddy in the Study; two for the chandelier; two at the Lounge door and one for the singing telegram.
Miss Scarlet: That's not six.
Wadsworth: One plus two plus two plus one.
Miss Scarlet: Uh-uh, there was only one shot that got the chandelier. That's one plus two plus *one* plus one.
Wadsworth: Even if you were right, that would be one plus one plus two plus one, not one plus *two* plus one plus one.
Miss Scarlet: Okay, fine. One plus two plus one... Shut up! The point is, there is one bullet left in this gun and guess who's gonna get it!

[wendy]

Consider first, 3D.

A great arrow is a great circle + direction.

For every great arrow, you can define a "north pole", and since any point represents a "north pole" for some great arrow, the plot of great arrows is the same as the points on the sphere.

Now, this is a dynamic thing: a sphere rotating. A given matrix represents a rotation through an angle, ie 38° on a great arrow, or 360-38° on its opposite. So a pair of antipodal points, representing the great circle, is to be replaced by a circle. So the set of points represented by, say SO3 (the matrices of even sign), represents the product of points, being a circle * sphere / antipodes.

For SO4, the situation is rather more complex, but we start with the clifford parallels, for example.

The first space we derive is that of clifford parallels. This corresponds to a 3-sphere. That is, a set of clifford parallels forms a set of points that correspond to the points of a 3d sphere. We suppose that these form in space, the surface formed by X1, X2, X3, such that rss(X1,X2,X3) = 1.

There is a matching set of right-parallels, and every great arrow is right-parallel to a single great arrow on the left-arrow. So what happens, is that the set of possible great arrows in 4d, corresponds to a point in six dimensions, (X1, X2, X3, Z1, Z2, Z3).

We now look at the point say (X,Z). The points X,-Z, and -X,Z represent the orthogonal rotation, which is going either one way or the other. The point -X,-Z is the same great circle, going the opposite direction.

We then have the four points, XZ, X-Z, -Z,X and -Z,-X. Between each of these, we might construct a zigzag X,Z to X,-Z to -X,-Z to -X,Z and back. Since each pair of line goes from speed ratio 1:1, to 1:-1, to -1:-1 to -1:1, we can plot the relative speed of any rotation phase on a single point of this zigzag. For example, we have the point X,Z + X,-Z. The first represents a great arrow in left-turning, X,*, and right-turning *,Z. The second represents a great-arrow that is left-turning X,* and right-turning *,-Z. When we add these together 50%X + 50%Y, the right turning cancells out, and we are left with a single rotation: a great arrow.

But we can go from 0 to 100%, and thence the shape is not just four points, but the zigzag formed by four sides of a tetrahedron. The remaining opposite sides form the diameters of the spheres X1,X2,X3 and Y1,Y2,Y3.

The shape so far formed is then a bi-glomochoric prism: a petix in 7D.

But this is the dymanic rotations in 4D, not just a single rotation. So opposite pairs of points are now represented by a phase angle: that is, dynamic --> and <--- become a full circle of rotations.

So you now have 8D.

[PWrong]

Ok, I can see know why it must be 6D. In defining the plane, we can assume the vectors are orthonormal. You need 3 angles for the first vector, two for the next vector, and one for the amount of rotation.

By the same logic, you need 4+3+1 = 8 for SO(5). For SO(n), n>3, you need (n-1) + (n-2) + 1 = 2n - 2

But following the pattern on Alkaline's formulas page, it should be (n-1)C2 = (n-1)n/2. So which is it ?

If we can answer this, I'd also like to prove that every element of SO(4) has an invariant plane. One of my assignment questions in algebra was to prove that every element of SO(3) has an invariant axis. Here's the proof in my textbook:

The characteristic equation of the matrix is a cubic polynomial. Every cubic has one real root, since the limit of a0 + a1x + a2^2 + a3^3 as x -> infinity is ±infinity depending on the sign of a3, and the limit as x -> -infinity is also infinity multiplied by the opposite sign. There is therefore some value of x for which the function takes the value zero by the intermediate value theorem. So there is at least one real eigenvalue and a corresponding eigenvector. The eigenvalue gives the stretching value along the eigenvector and since a special orthogonal matrix leaves distances invariant, the eigenvalue must be ±1. If all the eigenvalues are real, then they must all be ±1 and since the determinant is their product and is +1, at least one of the eigenvalues is +1. If two are complex conjugates, their product is positive so the real eigenvalue is also positive and must be +1, So there has to be an eigenvalue of +1.

So there are three cases: one real eigenvalue of 1 and a complex conjugate pair, in which case there is an eigenvector with eigenvalue 1, or three real eigenvalues one being 1 and the other two being −1, and again there is an eigenvector for the eigenvalue 1. The last case is when every eigenvalue is 1, and again at least one eigenvector with eigenvalue 1. (In fact the only possibility here is that A is the identity matrix, but we don’t need to prove this). The eigenspace for this eigenvalue 1 is therefore a line through the origin which is left fixed by the map, that is, it is the axis of a rotation. It follows that every special orthogonal map on R3 is a rotation.

I'd like to do the same for SO(4), but I can't get past the first bit, because a quartic polynomial isn't symmetric. I think I have to prove that there is an eigenvalue of 1, and its eigenspace is a plane.

EDIT: Just realised that the clifford rotations are a counterexample. I suppose two pairs of complex conjugate eigenvalues correspond to a clifford rotation.