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Old June 6th, 2009, 06:36 AM
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Default Don't know what the group O^+ (1,2) is!

Hi,
I'm told that symmetries in hyperbolic space is o^+ (1,2) - I'm not sure what that is!
Thanks,
Sooz
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Old July 27th, 2009, 06:26 AM
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I'm assuming o is a flip and r is a rotation. Meaning:

r( 123 ) → 231
o( 123) → 321

The great thing about it is that this is the set of symmetries of a triangle. r represents a rotation, o represents a flip over an altitude of the triangle.

See: http://en.wikipedia.org/wiki/Dihedral_gr…

o and r are the generators of the group:
http://en.wikipedia.org/wiki/Generating_…

So to find these inverses, you basically just do what you did, backwards, in the opposite order.

I'm using ' as the inverse symbol.
Start with the following:
o² = e
r³ = e
o' = o
r' = r²
r² ' = r
e is the identity

This is because o has order 2 and r has order 3. See:
http://en.wikipedia.org/wiki/Order_%28gr…

Using those rules, we get:

(or)' = r'o' = r'o = r²o

o'r' = or' = or²

r'o' = r²o

What's nice about these is that if you rotate, then flip, you could have flipped, then rotated in the opposite direction. Does that make sense? So:

(or)² = oror = oor'r = o²e = o² = e

You can test that out if you don't believe it:

123 → 321 → 213 → 312 → 123

And finally:

o²r² = er² = r²
herpes testing
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Old July 28th, 2009, 12:11 PM
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Quote:
Originally Posted by Sooz View Post
Hi,
I'm told that symmetries in hyperbolic space is o^+ (1,2) - I'm not sure what that is!
I think it should probably be a capital O (for orthogonal). Then the group of symmetries of hyperbolic space, O^+ (1,2), ought to mean the group of 3×3 matrices with positive determinant (that's what the + is for) that preserve the quadratic form x_1^2-x_2^2-x_3^2 (which has 1 positive and 2 negative terms).
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