Homework 6 (Solutions)
Problem 1.
Let M = P2 be the projective plane (thought of as the set of all lines in
3
R passing through the origin).
Let U = {[x : y : z] ∈ P2 |z 6= 0} and V = {[x : y : z] ∈ P2 |x 6= 0}.
Consider the charts φ = (x1 , x2 ) : U → R2 given by φ([x : y : z]) =
(x/z, y/z) for [x : y : z] ∈ U , and ψ = (y 1 , y 2 ) : U → R2 given by φ([x : y :
z]) = (y/x, z/x) for [x : y : z] ∈ V .
Let α be a 1-form on P2 such that in the chart φ = (x1 , x2 ) we have
α = ex
1 x2
dx1 + (3(x1 )2 − 2x2 )dx2 .
(1) Express dx1 and dx2 in terms of dy 1 and dy 2 .
(2) Express α in the (y 1 , y 2 ) coordinates.
(3) For a point p = [3 : 1 : 5] ∈ U and ~v = 2 ∂x∂ 1 |p − 10 ∂x∂ 2 |p ∈ Mp compute
the value α|p (~v ).
(4) Consider the function f : P2 → R given by f ([x : y : z]) =
Express df in the chart φ = (x1 , x2 ).
x2 +y 2 +z 2
.
x2 +2y 2 +3z 2
(5) For p and ~v as in part (3), consider the function r~v : Mp∗ → R defined
as r~v (ω) := ω(~v ) for every ω ∈ Mp∗ .
Compute r~v (a1 dx1 +a2 dx2 ) where a1 , a2 ∈ R are arbitrary. Is the function
r~v : Mp∗ → R linear?
Solutions.
(1) We have (ψ ◦ φ−1 )(x1 , x2 ) = ψ([x1 : x2 : 1]) = (x2 /x1 , 1/x1 ) so that
1
y (x1 , x2 ) = x2 /x1 and y 2 (x1 , x2 ) = 1/x1 .
∂xi
We’ll need to compute ∂y
j . We can do that by either first computing
1
2
∂(y ,y )
the matrix ∂(x
1 ,x2 ) and then using the Inverse Function Theorem, or, in
this case, by explicitly expressing (x1 , x2 ) in terms of (y 1 , y 2 ). Thus from
y 1 = x2 /x1 , y 2 = 1/x1 we get x1 = 1/y 2 , x2 = y 1 /y 2 . Hence
∂(x1 , x2 )
0
−1/(y 2 )2
=
.
1/y 2 −y 1 /(y 2 )2
∂(y 1 , y 2 )
We have
dx1 =
∂x1 ∂
∂x1 ∂
∂
+
= −1/(y 2 )2 2
∂y 1 ∂y 1
∂y 2 ∂y 2
∂y
and
∂x2 ∂
∂x2 ∂
∂
∂
+
= (1/y 2 ) 1 − y 1 /(y 2 )2 2
1
1
2
2
∂y ∂y
∂y ∂y
∂y
∂y
1
2
Thus for p = [x : y : z] ∈ U ∩ V we have y (p) = y/x, y (p) = z/x and so
∂
∂
dx1 |p = −1/(y 2 )2 2 |p = −(x/z)2 2 |p
∂y
∂y
dx2 =
1
2
and
dx2 |p = (x/z)
∂
∂
|p − (yx/z 2 ) 2 |p .
∂y 1
∂y
1 2
(2) We have α = ex x dx1 + (3(x1 )2 − 2x2 )dx2 = α1 dx1 + α2 dx2 with
1 2
α1 = ex x , α2 = (3(x1 )2 − 2x2 ).
When expressing α in the chart ψ as α = α10 dy 1 + α20 dy 2 we have
α10 =
X ∂xi
∂x1
∂x2
α
=
α
+
α = 0 + (1/y 2 )(3(x1 )2 − 2x2 ) =
i
1
1
1
1 2
∂y
∂y
∂y
2
i=1
(1/y 2 )((1/y 2 )2 − 2y 1 /y 2 ) = (1 − 2y 1 y 2 )/(y 2 )3
Similarly,
α20 =
∂x1
∂x2
1 2
α
+
α2 = (−1/(y 2 )2 )ex x + (−y 1 /(y 2 )2 )(3(x1 )2 − 2x2 ) =
1
2
2
∂y
∂y
= −1/(y 2 )2 )ey
1 /(y 2 )2
+ (2(y 1 )2 y 2 − 3y 1 )/(4y 2 ).
Hence with respect to the chart ψ we have
α = (1 − 2y 1 y 2 )/(y 2 )3 dy 1 + [−1/(y 2 )2 )ey
1 /(y 2 )2
+ (2(y 1 )2 y 2 − 3y 1 )/(4y 2 )]dy 2 .
(3) For the point p = [3 : 1 : 5] ∈ U we have x1 (p) = 3/5, x2 (p) = 1/5.
1 2
Since α = ex x dx1 + (3(x1 )2 − 2x2 )dx2 , we have
α|p = e3/25 dx1 |p + (17/25)dx2 |p
Therefore, by multi-linearity, for ~v = 2 ∂x∂ 1 |p − 10 ∂x∂ 2 |p ∈ Mp we have
α|p (~v ) = 2e3/25 − 10(17/25) = 2e3/25 − 34/5.
(4) In the chart φ we have
(f ◦ φ−1 )(x1 , x2 ) = f ([x1 : x2 : 1]) =
(x1 )2 + (x2 )2 + 1
(x1 )2 + 2(x2 )2 + 3
Therefore
∂f
2x1 ((x1 )2 + 2(x2 )2 + 3) − ((x1 )2 + (x2 )2 + 1)2x1
2x1 ((x2 )2 − 1)
=
=
∂x1
[(x1 )2 + 2(x2 )2 + 3]2
[(x1 )2 + 2(x2 )2 + 3]2
and
∂f
2x2 ((x1 )2 + 2(x2 )2 + 3) − ((x1 )2 + (x2 )2 + 1)4x2
2x2 (1 − (x1 )2 )
=
=
.
∂x2
[(x1 )2 + 2(x2 )2 + 3]2
[(x1 )2 + 2(x2 )2 + 3]2
Thus
df =
2x1 ((x2 )2 − 1)
1
[(x )2 + 2(x2 )2 + 3]2
dx1 +
2x2 (1 − (x1 )2 )
1
[(x )2 + 2(x2 )2 + 3]2
dx2 .
3
(5) We have
r~v (a1 dx1 + a2 dx2 ) = (a1 dx1 |p + a2 dx2 |p )(~v ) =
∂
∂
(a1 dx1 + a2 dx2 )(2 1 |p − 10 2 |p ) = 2a1 − 10a2 .
∂x
∂x