Lecture Notes for 120 - UCLA Department of Mathematics

7.6. THE UPPER HALF PLANE 165
7.6.1. The isometries of H. Our first observation is that the first fundamental
form doesn’t depend on u, so the transformations
F : H ! H
F (u, v) = (u + u 0 ,v)
must be isometries. Let us check what it means for a general transformation
F (u, v) =(F u (u, v) ,F v (u, v)) to be a isometry. Let p =(u, v) and q = F (u, v)
apple 1
apple "
#
v
0
@F
u
@F v
1
apple
0
@F
u
@F u
2
1 = @u @u (F v (u,v)) 2 @u @v
1
0
v
0 2 (F v (u,v)) 2
=
=
@F u
@v
@F v
@v
apple
1
@F
u
(F v (u, v)) 2 @u
"
1
(F v (u, v)) 2
@F v
@u
@F v
@v
@F v
@u
@F v
apple @F
u
@u
@F v
@F u
@v
@F v
@v @u @v
@F u 2
@u +
@F v 2 @F u @F u
@u @v @u
+ @F v @F v
@u @v
@F u
@u
+ @F v @F v @F u 2
@u @v @v +
@F v 2
@v
@F u
@v
@F u
@v
This tells us
✓ ◆ @F
u 2 ✓ ◆ @F
v 2
+ = (F v (u, v)) 2 ✓ ◆ @F
u 2 ✓ ◆ @F
v 2
@v @v
v 2 = + ,
@u @u
@F u @F u
@v @u + @Fv @F v
= 0
@u @v
In particular, we see that the translations F (u, v) =(u + u 0 ,v) really are isometries.
Could there be isometries where F only depends on what happens to v?
We can check an even more general situation: F (u, v) =(u, f (u, v)) where the
equations reduce to
✓ ◆ 2 ✓ ◆ 2 ✓ ◆ 2 @f
f (u, v)
@f
=
=1+ ,
@v
v
@u
@f @f
= 0
@u @v
So first we note that @f
@v
6=0from the first equation, the second then implies that
@f
@u
=0, which then from the first equation gives us that f (u, v) =v is the only
possibility. Next let’s try F (u, v) =(g (u) ,f(v)) . This reduces to
✓ ◆ 2 df
=
dv
✓ ◆ 2 f (v)
=
v
✓ ◆ 2 dg
du
So all transformations of the form F (u, v) =c (u, v) where c>0 is constant are
also isometries. Note that these maps are only similarities in the Euclidean metric,
but have now become genuine isometries. This might give us the idea to check
maps of the form F (u, v) =h (u, v)(u, v) . This is still a bit general so we make
the reasonable assumption that h doesn’t depend on the direction of (u, v) , i.e.,
F (u, v) =h u 2 + v 2 (u, v) . Then
#
apple @F
u
@u
@F v
@u
@F u
@v
@F v
@v
apple h +2u
=
2 h 0 2uvh 0
2uvh 0 h +2v 2 h 0