Download We consider the projection : SO(3) ! S 2, where we send a matrix to

We consider the projection ⇡ : SO(3) ! S 2 , where we send a matrix to its first row vector.
Since matrices in SO(3) are such that all the row and column vectors are of unit length, we see
that this map is well-defined.
(a) We will show that ⇡ is a principal S 1 bundle. For this, we first need to define the action of S 1
on SO(3). We define it as follows:
0 1
0
1
a
a
@ v A · (z1 + iz2 ) := @ z1 v + z2 w A ,
w
z2 v + z1 w
where we denote with a, v, w row-vectors and z1 + iz2 2 S 1 ⇢ C an arbitrary element. Note that
using that we can characterize z1 + iz2 = ei✓ , we find that
!
✓ ◆
✓
◆
a
a
cos(✓) sin(✓)
i✓
^
i✓ =
·e = ^
,
with
A
·
e
Ã.
sin(✓) cos(✓)
Ã
A · ei✓
Note that we have the following computations:
00 1
1
0
1
a
a
det @@ v A · (z1 + iz2 )A = det @ z1 v + z2 w A = z12 det(A) + z22 det(A) = det(A) = 1,
w
z 2 v + z1 w
with some standard rule of taking the determinant,
ha, z1 v + z2 wi = z1 ha, vi + z2 ha, wi = 0 + 0 = 0,
ha, z2 v + z1 wi =
z2 ha, vi + z1 ha, wi =
0 + 0 = 0,
hz1 v + z2 w, z1 v + z2 wi = z12 hv, vi + z22 hw, wi + 0 = 1,
h z2 v + z1 w, z2 v + z1 wi = z22 hv, vi + z12 hw, wi + 0 =
1.
These computation show that A · (z1 + iz2 ) 2 SO(3). Let us check that this is indeed an action
on SO(3). Firstly, if ei✓ = 1, then we matrix multiply with the identity matrix so that A · 1 = A
for all A. Secondly, if we have g, h 2 S 1 , represented by ✓g , ✓h , then:
0
1
a
✓
◆
A·h
(A · g) · h = @ cos(✓g ) sin(✓g )
Ã
sin(✓g ) cos(✓g )
0
1
a
✓
◆✓
◆
A
cos(✓g ) sin(✓g )
= @ cos(✓h ) sin(✓h )
Ã
sin(✓h ) cos(✓h )
sin(✓g ) cos(✓g )
0
1
a
✓
◆
A
= @ cos(✓h + ✓g ) sin(✓h + ✓g )
Ã
sin(✓h + ✓g ) cos(✓h + ✓g )
= A · (g · h),
where we use goniemetric identities and that S 1 is additive with respect to the ✓. So indeed, this
is an action.
Now let us check whether this action makes SO(3) into a principal bundle over S 2 . First of all, ⇡
is surjective. Given an element a 2 S 2 ⇢ R3 , we can use Gramm-Schmidt to get an orthonormal
0 1
a
basis {a, v, w} of R3 such that a is the first basis elements. Then, we find that either @ v A or
w
0
1
a
@ v A has determinant 1, so that at least one of these two are elements of SO(3). Note that ⇡
w
1
maps them exactly to a.
Secondly, we find that ⇡ is S 1 invariant by the following computation:
0
1
✓ ◆
a
✓
◆
A = a = ⇡ a = ⇡(A)
⇡(A · g) = ⇡ @ cos(✓) sin(✓)
Ã
Ã
sin(✓) cos(✓)
Finally, we will show that locally we can construct a section of ⇡, which is equivalent with the local
triviality statement. First we let U = S 2 \ {±e3 }. If a 2 U , we know that a3 6= ±1. Hence we can
define ṽa := e3 a3 a and va := |ṽṽaa | using Gramm-Schmidt. Note that since a3 6= ±1, we find that
0 1
a0
va depends smoothly on a. Now, for any a0 2 U , we pick the unique w0 such that @va0 A 2 SO(3).
w0
We do this by construction, pick any element w 2
/ span(a0 , av00
) and1use Gramm-schmidt to make
a0
an orthonormal basis {a0 , va0 , w0 }. Then the determinant of @va0 A is either plus or minus one,
w0
so after switching w0 with w0 when the determinant was negative, gives the wanted w0 .
a
Now, we define w̃a := w0 hw0 , aia hw0 , va iva and wa := |w̃
w̃a | . Let us see whether we can
actually do this. We see that w̃a = 0 if and only if w0 = hw0 , aia + hw0 , va iva . Since this is not
the case for a = a0 and the innerproduct and va all depend smoothly on a, we see that there
exists some small neighborhood V around a0 such that also wa is well-defined and hence smoothly
depending on a. Now note that we can choose V to be connected, so that by continuity of the
determinant and since it has only values ±1 on O(3):
0 1
0 1
a
a0
det @ va A = det @va0 A = 1.
wa
w0
0 1
a
We conclude that on V , we have the section which sends a 7! @ va A. Of course we can do the
wa
2
same for all points inside S \ {±e2 }, so we conclude that around all points there exists a small
neighbourhood and a local section on it. We conclude that SO(r) ! S 2 is a principal S 1 -bundle.
(b) Here we will prove that ✓
E := E(SO(3), R◆2 ) is isomorphic to T S 2 . We use the reprecos(✓)
sin(✓)
sentation ⇢ : S 1 ! GL2 ; g 7!
. Note that ⇢(g) 1 = ⇢(g)T , what we will use
sin(✓) cos(✓)
later on. Consider the following isomorphism:
: E ! T S 2 ; [A, ⇠] 7! ⇠ T Ã.
Here we view ⇠ T as a row vector and we use normal matrix multiplication. Let us first check
whether everything is well-defined. So, we have to check whether [A · g, ⇢(g) 1 ⇠] gives the same
answer. Here we find that
([A · g, ⇢(g)
1
⇠]) = (⇢(g)
T
1
⇠)T · Ag
· g) = ⇠ T · (⇢(g)
= ⇠ ⇢(g)⇢(g)
1
T
à = ⇠ à =
1 T
) ⇢(g)
1
Ã
([A, ⇠]).
2
So indeed, it is well-defined on the quotient. Now let us check whether
the image
✓ ◆
✓ ◆is in T S . For
a
v
this to hold, we need that ([A, ⇠]) is orthogonal to a, with A =
and à =
. We compute
w
Ã
that:
ha, ([A, ⇠])i = ha, ⇠ T Ãi = ha, ⇠1 v + ⇠2 wi = ⇠1 ha, vi + ⇠2 ha, wi = ⇠1 0 + ⇠2 0 = 0.
2
Here we use that A 2 SO(3), so that the row columns form an orthonormal basis. We need to
check a couple more things to conclude that is an isomorphism. Note that it is clear that is
smooth, since it is just polynomial. Secondly, it is linear, since:
([A, ⇠] + [A, ⌘]) =
([A, ⇠ + ⌘]) = (⇠ + ⌘)T Ã = ⇠ T Ã + ⌘ T Ã =
([A, ⇠]) +
([A, ⌘]).
And by the construction of E and the independence of choice of representative of the quotient
class, this is all we need to check. We also find that this is indeed a vectorbundle morphism, that
is, ⇡
= ⇡ by construction.
So we are left to check whether all a are isomorphisms as linear maps. We start with injectivity.
Suppose that ([A, ⇠]) = 0, this means that ⇠ T Ã = 0 2 R3 . Now suppose without loss of generality
that ⇠1 6= 0. Then we find that since 0 = ⇠1 v + ⇠2 w, that v = ⇠⇠21 w. Note that immediately
implies that ⇠2 6= 0, or alse v = 0 which is not possible since it has to have length 1. Now we find
that:
0 = hv, wi =
⇠2
hw, wi =
⇠1
⇠2
6= 0.
⇠1
Hence we find a contradiction, so that ⇠ = 0 and hence [A, ⇠] = 0.
x
Now let us prove surjectivity. Let x = (x1 , x2 , x3 ) 2 Ta S 2 . That is, ha, xi = 0. Now let v = |x|
and use Gramm-Schmidt once more to find an w such that {a,
w} is an orthonormal basis.
0v, 1
a
Once more, we might have to change w into w, so that A = @ v A 2 SO(3). Now finally, let
w
⇠ = (|x|, 0). Then:
([A, ⇠]) = ⇠ T Ã = |x| ·
We conclude that
x
+ 0 = x.
|x|
is an isomorphism.
(c) Proposition 4.43 tells us that SO(3) is isomorphic to a SO(2) structure on S 2 . Now
let us consider what this exactly means. We know that a O(n) structure is the same as a metric,
and that a GL+ (n) structure is the same as an orientation. Since O(n) \ GL+ (n) = SO(n), we
find that finding a SO(n)-structure is the same as picking a metric and an orientation at the
same time. Here use that in the linear case, a SO(n) structure is a choice of basises of V which
are orthogonal and change of basis between two of them has positive determinant, so that the
orthogonal basises induce an orientation. We conclude that S 2 is a oriented riemannian manifold.
3