# Math Help - Don't know what the group O^+ (1,2) is!

1. ## 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

2. 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.
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

3. Originally Posted by Sooz
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).