Rotations and reflections

Bobhawke

In R^2, a reflection can be achieved equally well by a rotation, because the group space of U(1) is connected. Visually I think of this as being able to peform a rotation that moves the positive y axis into the negative y axis and vice versa and without changing the x axis by imagining R^2 as embedded in R^3 and rotating about the x axis, so it is unchanged but the y axis sweeps through the extra dimension and eventually gets turned upside down.

In R^3, rotations and reflections are not connected. My question is, at what point does the analogue of the above reasoning fail for R^3, and also does the dis-connectedness of the group space have anything to do with the fact that the group is now non-abelian?

Thanks

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jdougherty

I'm not entirely sure what the analogue of the above reasoning would be, but it seems to me that it would involve rotation through a 4th spatial dimension, which I have a difficult time imagining. Also, I'm not entirely sure I understand your visualization of the R^2 situation anyway. This might be faulty reasoning on my part, but I always imagine the 2-D rotation group as actions on a point on the unit circle. Rotating this point through 180 degrees is indistinguishable from reflecting it to the antipodal point on the circle, so you can't tell the difference between rotations and reflections.

As for the difference between reflections and rotations in R^3, the non-abelian-ness is what is meant, mathematically, by the physical fact rotations are different from reflections. This, in turn, is related to the fact that as a topological space, SO(3) - the rotation group in three dimensions - is connected, but not simply connected.

cf. http://en.wikipedia.org/wiki/Rotation_group#Topology

Bobhawke

I understand the reasoning you gave with reflections and rotations being equivalent on the group space of U(1)~SO(2), but I was trying to get a more "physical" picture, in real space as it were.

To explain my reasoning before a bit better, I was just imagining rotating the y axis about the x axis by 180 degrees. This would mean it goes through a 3rd dimension before coming into coincidence with itself but with the opposite orientation.

I was imagining this because it seems that you cannot rotate the axes staying in R^2 in such a way to flip one axis without flipping the other. Consider trying to perform such a rotation. If you rotated the both axes by 180 degrees (and this would be a rotation about the z axis if we were in R^3), you would certinaly flip the y axis, but you would also flip the x axis. Thus you wouldnt have achieved a reflection.

Also, I just read that article you linked but then I found something that said that the cirlce is connected but not simply connected. If that is the source of differences in rotations and reflections, then why is there not such difference for U(1), whose group space is a circle?

yyat

In R^2, a reflection can be achieved equally well by a rotation, because the group space of U(1) is connected.
U(1) consists only of rotations, not reflections. The group of isometries of R^2 is O(2), the orthogonal group. It has two connected components, in particular a reflection can not be achieved by a rotation (reflections reverse orientation, rotations preserve orientation).

It is true however, that any rotation can be achieved as a composition of reflections (in arbitrary dimension), in other words the reflections generate O(n).

dslowik

When you imagine 'rotating' the positive y-axis to the negative in 3-D, that's a 3-D rotation, not 2-D.
The parity of an elementary reflection is -1, rotations are +1. even number of reflections give a rotation. But rotations dont' generally compose to give a reflection. It works in 2-d for the case of reflecting both x and y to the negatives(reflection through origin) since the space is even dimensional. In 3-d that would be 3 reflections which has wrong parity to be a rotation.

I think: O(2) is the group of all isometries of R^2. It is disconnected into 2 pieces having det +1 (rotations) and -1(reflections). SO(2) is the connected sub-group having det = +1 (rotations); it's isomorphic with U(1). It's connected but not 'simply connected' ; a straight line segment is simply connected; has to do with shrinking it to a point..

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wofsy

In R^2, a reflection can be achieved equally well by a rotation, because the group space of U(1) is connected. Visually I think of this as being able to peform a rotation that moves the positive y axis into the negative y axis and vice versa and without changing the x axis by imagining R^2 as embedded in R^3 and rotating about the x axis, so it is unchanged but the y axis sweeps through the extra dimension and eventually gets turned upside down.

In R^3, rotations and reflections are not connected. My question is, at what point does the analogue of the above reasoning fail for R^3, and also does the dis-connectedness of the group space have anything to do with the fact that the group is now non-abelian?

Thanks
Rotations and reflections are not connected in R^2. O(2) is not connected. U(1) is the group of rotations of R^2. It does not contain any reflections. O(2) is the group of all orthogonal linear transformation of R^2.

If you go to R^4 you can just as well turn a reflection in R^3 into a rotation in R^4.

sundried

Maybe a bit of confusion about reflections in points versus reflections in lines. In R^2 the former are also rotations, as they preserve orientation. In R^3 they do not preserve orientation, so can't be rotations.

Cantab Morgan

In R^2, a reflection can be achieved equally well by a rotation, because the group space of U(1) is connected. Visually I think of this as being able to peform a rotation that moves the positive y axis into the negative y axis and vice versa and without changing the x axis by imagining R^2 as embedded in R^3 and rotating about the x axis, so it is unchanged but the y axis sweeps through the extra dimension and eventually gets turned upside down.
I wouldn't call that a rotation. I'd call it an orthogonal transformation. It's a linear Euclidean transformation, but it doesn't have a positive determinant. Orientation is not preserved.

Similarly, in $$R^3$$, there is an orthogonal transformation that maps my right hand onto my left hand. It does not preserve orientation and does not have a positive determinant.

I'm not sure, but I think it's generally true that if you embed a shape in $$R^n$$ inside $$R^{n+1}$$, then your orthogonal transformations are isomorphic to rotations (with determinant 1) in the larger space. I agree that it's hard to imagine the physical picture here where $$n \geq 3$$ because we are only three dimensional creatures.

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