Modern Geometry Assignment 2

Robbie Lyman

September 22, 2014

(1) Let Φ ′ = {(Uα,φα) : α ∈ I} be an atlas for M. Then we know that Φ = {(π−1(Uα), ~φα) : α ∈ I} is anatlas for TM, which is smooth because M is. We'd like to show that Φ is an orientation for TM.

So, given α,β ∈ I, we want to show that det(d(~φβ◦ ~φ−1α )x) > 0 for all x ∈ ~φα(π

−1(Uα)∩π−1(Uβ)).Also, note that π−1(Uα)∩π−1(Uβ) =

⋃p∈U∩V TpM so that ~φα(π

−1(Uα)∩π−1(Uβ)) = φα(U∩V)×Rn,and we have ~φβ ◦ ~φ−1

α : φα(U ∩ V)× Rn → φβ(U ∩ V)× Rn.

By de�nition,

~φβ ◦ ~φ−1α (x,u) = (φβ ◦ φ−1

α (x), θ−1Uβ,φβ,φ

−1α (x)

◦ θUα,φα,φ−1α (x)(u))

= (φβ ◦ φ−1α (x), θ−1

Uβ,φβ,φ−1α (x)

([Uα,φα,u])) = (φβ ◦ φ−1α (x), d(φβ ◦ φ−1

α )x(u)),

since (Uα,φα,u) ∼ (Uβ,φβ,u) if d(φβ ◦ φ−1α )x(u) = u. So let x1, . . . , xn be the �rst n coordinates

of R2n, u1, . . . ,un be the last n, φβ ◦ φα = (f1, . . . , fn), and d(φβ ◦ φ−1α ){({) = (g1, . . . ,gn). Then

we have

∂fi

∂xj(x,u) =

∂(φβ ◦ φ−1α )i

∂xj(x),

∂fi

∂uj(x,u) =

∂(φβ ◦ φ−1α )i

∂uj(x) = 0

∂gi

∂xj(x,u) =

∂(d(φβ ◦ φ−1α )x)i

∂xj(u) =

∂

∂xj

[n∑k=1

∂(φβ ◦ φ−1α )i

∂xk(x)uk

]=

n∑k=1

∂2(φβ ◦ φ−1α )i

∂xj∂xk(x)uk (∗)

∂gi

∂uj(x,u) =

∂

∂uj

[n∑k=1

∂(φβ ◦ φ−1α )i

∂xk(x)uk

]=∂(φβ ◦ φ−1

α )i∂xj

(x).

So, in block form, if A = aij is the matrix with entries aij given by (∗), we have

d(~φβ ◦ ~φ−1α )(x,u) =

(d(φβ ◦ φ−1

α )x 0

A d(φβ ◦ φ−1α )x

).

Now, the determinant of this matrix, which we'll call B = (bij) is∑σ∈S2n sgn(σ)

∏2ni=1 biσ(i). Note

that if any of these products contains an element in the upper-right-hand corner of the matrix, the

product is zero. This tells us that our nonzero choices are restricted to the upper-left-hand corner for

the �rst n terms, and thus the lower-right-hand corner for the last n|i.e.

det(d(~φβ ◦ ~φ−1α )(x,u)) = det(d(φβ ◦ φ−1

α )x)2 > 0

(Zero is not a possibility, since the composition of ϕβ ◦ ϕ−1α with ϕα ◦ ϕ−1

β is the identity, and the

chain rule tells us their Jacobians are inverses and thus their determinants must be nonzero.) This

shows that Φ is an orientation for TM.

(2) (a) As a polynomial, p is a smooth map between the smooth manifolds Rk and R. If a is a regular

value of p, then p−1(a) = Xa will be a k− 1-dimensional submanifold of Rk.

1

Now,

dpx =(∂p∂x1

. . . ∂p∂xn

)Fix x ∈ Rn, then g(t) = p(tx) − tmp(x) = 0 for all t, so

g ′(t) = 0 =∂p(tx)

∂t−mtm−1p(x) =⇒ mtm−1p(x) =

n∑k=1

∂p(tx)

∂txk· ∂txk∂t

=

n∑k=1

xk∂p(tx)

∂txk

=⇒ mp(x) =

n∑k=1

xk∂p

∂xk(x)

So suppose p(x) 6= 0, then dpx(x) =∑xi∂p∂xi

(x) = m · p(x) 6= 0. Let α ∈ R, and t = α/mp(x).

Then dpx(tx) = t ·mp(x) = α, so that if a 6= 0, a is a regular value of p, as required.

(b) Let a 6= 0 and t = a−1/m. Then the mapping from Rk to itself given by x 7→ tx is a di�eomorphism.

Let f be the restriction of this mapping to Xa. f will be an embedding as soon as we can show

that it is an immersion. But dfx = t · Ik, which is injective, so f(Xa) is a submanifold of Rk. Forx ∈ Xa, p(x) = a =⇒ p ◦ f(x) = p(tx) = tmp(x) = tm · a = a/|a| = ±1. This tells us that Xa is

di�eomorphic to X1 if a > 0, and to X−1 if a < 0.

(3) (a) det : Mn(R) ∼= Rn2 → R is eminently a smooth function, given that it's de�ned as a sum of

products of the variables in its domain. So once we can show that 1 is a regular value of det, we'll

have det−1(1) = SL(n,R) as an (n2 − 1)-dimensional submanifold of Rn2 . So suppose A = (aij)

is a matrix such that det(A) 6= 0. Then

d(det)A =(∂det∂x11

(A) . . . ∂det∂xnn

(A)),

where ∂det∂xij

(A) = ∂∂xij

∑σ∈Sn sgn(σ)

∏nk=1 akσ(k) =

∑σ∈Sn,σ(i)=j sgn(σ)

∏nk=1,k6=i akσ(k) = Cij,

where Cij is the i, jth cofactor of A, i.e. the determinant of the i, jth minor of A times (−1)i+j.

Now, since detA =∑ni=1 a1iC1i, we have a vector a in Rn2 , a = (a11, . . . ,a1n, 0, . . . , 0) for

which d(det)A(a) 6= 0, so by linearity, det is a submersion at A if det(A) 6= 0. In particular,

then, 1 is a regular value of det, as required.

(b) We know that, since 1 is a regular value of det, that TInSL(n,R) = Ker(d(det)In). But since

the i, jth minor of In is always just In−1,∂det∂xij

(In) = (−1)i+j, so d(det)In(x) = 0 =⇒∑i,j(−1)

i+jxij = 0. It is clear that this is a linear subspace of Mn(R), since the sum of the

elements of any two matrices whose elements sum to zero (after multiplying by a constant de-

termined by the position of the element) is still zero, as ismultiplying the matrix by a constant.

Further, choosing the �rst n2 − 1 elements, we have one well-de�ned choice for the last one to

make the sum zero|i.e. this linear subspace has dimension n2 − 1, as desired.

∴ TInSL(n,R) = {(aij) ∈Mn(R) :∑i,j(−1)

i+jaij = 0}.

(c) As we saw in part (a), ∂det∂xij

= Cij, the i, jth cofactor of A, and generalising the argument above,

we have x ∈ TASLn(R) if∑i,j Cijxij = 0. Therefore,

TSL(n,R) =

A× B ∈Mn(R)×Mn(R) : detA = 1,∑i,j

Cijbij = 0

,

where, as usual, Cij is the i, jth cofactor of A.

(4) Given {(Ui,φi)}, the atlas for Pn(R) given in class, de�ne ~φi : π−1(Ui) → Rn+1 as ~φi(`, v) = (φi(`), t).

We choose t as follows: since Ui = {[x1, . . . , xn+1] : xi 6= 0}, we can choose a representative x of ` with

norm 1 and positive ith coordinate, and then let t ∈ R such that v = tx.

2

With this de�nition, the transition function ~φj ◦ ~φ−1i is just ~φj ◦ ~φ−1

i (y, t) = ~φj(φ−1i (y), tx) =

(φj ◦φ−1i (y),±t), where x is the representative discussed above. The plus or minus comes as follows:

φ−1i (y) ∈ Ui ∩Uj =⇒ both xi and xj are not zero, but it could be the case that one is positive while

the other is negative. In such a case, then, our choice of t will di�er in sign between the two functions.

Since this sign depends only on i, j and `, and Ui and Uj exclude the crossover region (where xi or

xj could be zero), this last coordinate function is continuous, and therefore smooth, showing that E is

a smooth manifold (of dimension Rn+1). So to show that π : E → M is a vector bundle, we need an

open cover with accompanying local trivialisations that satisfy a transition function condition.

Choose our cover to be {Ui : 1 6 i 6 n + 1}, the domains from the atlas for Pn(R) given in

class, and de�ne hi : π−1(Ui) → Ui × R as hi(`, v) = (`, t), where t is chosen as above. (Given

the chart ~φi above, the �nal coordinate function ends up being just the identity, so hi is certainly

a di�eomorphism.) For the transition function condition, suppose Ui ∩ Uj 6= ∅. Then, similarly as

above, hj ◦ h−1i (`, t) = (`,±t). The choice of positive or negative, since it depends only on i, j and `

can be thought of as a function from Ui ∩Uj to GL1,R(= R∗) is smooth because it is constant on the

connected components of Ui ∩Uj.

3