MAT 367S – Assignment #3
Solutions
Problem #1:
The Jacobian is given by
2x
1
0
0
2x 2y + 1 2z 2w
This has rank 2 unless z = 0 and w = 0 and xy = 0. (The last equality
is the condition for the first two columns to be linearly dependent.) But if
z = w = x = 0, the defining equations would simplify to y = 0, y 2 + y = 1,
which has no solution. If z = w = y = 0, the defining equations would
simplify to x2 = 0, x2 = 1 which again has no solution. We conclude that
the Jacobian has maximal rank 2 on the set of solutions. The regular value
theorem shows that the solution set is a submanifold of dimension 2.
Problem #2:
a) In terms of the matrix entries, the set S is the solution set of
(1)
a2 + b2 + c2 + d2 = 1, ad − bc = 0.
The Jacobian matrix is
(2)
2a 2b 2c 2d
d −c −b a
.
Let us determine when this has rank less then 2, for elements (a, b, c, d) in
the solution set of (1). Due to a2 + b2 + c2 + d2 = 1, both rows of this
matrix are non-zero. Suppose the rank of (2) is less than 2. Then one row
is a non-zero multiple of the other. In fact, all entries of this matrix are
non-zero: Indeed, if one of a, b, c, d is zero, then by using ad − bc = 0 and
linear dependence of the rows we see that all others would have to be zero
has well. If the two rows are positive multiples of each other, then looking
at the first and second column we must have ad > 0, bc < 0. But this is
inconsistent with ad − bc = 0. If the two rows are negative multiples of each
other, we would similarly get ad < 0, bc > 0, which again is inconsistent
with ad − bc = 0. Having ruled out all cases, we conclude that for (a, b, c, d)
in the solution set of (1), the matrix has rank 2.
b) Let U ⊂ MatR (2) be the open subset where the first row is non-zero. The
map U → RP (1), taking A → (b : −a) is smooth (since it is a composition
of smooth maps: projection to the first row, followed by (a, b) 7→ (−b, a),
followed by the quotient map to RP (1). Hence its restriction to S ∩ U
is smooth. Likewise the restriction to S ∩ V is smooth, where V denotes
matrices whose second row is non-zero.
For a matrix A in S ∩ U , the kernel is spanned by (−b, a). For a matrix
in S ∩ V , the kernel is spanned by (−d, c). (The two are proportional since
ad − bc = 0.) This shows that the fibers π −1 (u : v) (where we may assume
u2 + v 2 = 1) consists of matrices where both rows are multiples of (−v, u):
(a, b) = (−λ v, λ u),
(c, d) = (−µ v, µu).
for some λ, µ. The condition ||A||2 = 1 is equivalent to λ2 + µ2 = 1, thus
(λ, µ) represents a point on the circle S 1 . We hence see that π is surjective,
and the fibers of π are circles.
c) The fact that S maps to RP (1) ∼
= S 1 , with fibers S 1 , implies that it
must be a 2-torus – but this is not entirely obvious. Let us therefore give a
more concrete argument. As shown above, S is parametrized by (u, v) ∈ S 1
and (λ, µ) ∈ S 1 . Let us regard these as complex numbers z = u + iv and
w = λ + iµ of absolute value one. The pairs (z, w) represent an element
of S, but (−z, −w) represents the same element. It is therefore good to
switch from (z, w) to (z, w0 ) with w0 = zw. Then (z, w0 ) ∈ S 1 × S 1 uniquely
describes an element of S, and we conclude that S ∼
= S1 × S1.
Problem #3:
a) We calculate
d
d
|t=0 F (A + tX) = |t=0 (A + tX)J(A> + tX > ) = XJA> + AJX > .
dt
dt
b) Let Y ∈ SkewR (2n) be a given skew matrix. We want to find X such
that XJA> + AJX > = Y . A solution is
1
X = − Y JA,
2
>
2
as one verifies using JAJA = J = −I. Since the differential is surjective,
the regular value theorem show that Sp(2n) has dimension dim MatR (2n) −
dim SkewR (2n), which is the dimension of the space of symmetric 2n × 2n
matrices. The latter is
1
((2n)2 + 2n) = n(2n + 1).
2
Problem #4:
a) We can take
1
F0 (u, eiθ ) = p
(eiθ , ueiθ ).
1 + |u|2
1
F1 (v, eiθ ) = p
(veiθ , eiθ ).
2
1 + |v|
b) On the overlap, we have v = u−1 , by the transition map for the charts of
CP (1). Hence the second map becomes
|u|
1
|u|
|u|
F1 (u−1 , eiθ ) = p
(u−1 eiθ , eiθ ) = p
(eiθ , ueiθ )
2
2
u
u
1 + |u|
1 + |u|
Hence, the desired map K changes the phase factor by
K : (u, eiθ ) 7→ (u, eiθ
|u|
).
u
|u|
u :