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Assignment # 6
In this assignment you will learn about various constructions of classical examples of manifolds. You may feel that some details are missing, and your task is to fill those gaps. Pay special attention to the Remarks and Examples. You do not need to hand anything, but you should discuss the examples as much as possible with your classmates and your instructor.
- Review of linear algebra
Let V be an n-dimensional vector space over R.^1 The choice of a basis B = {v 1 ,... , vn} for V determines an isomorphism
φB : Rn^
∼= −→ V
by sending x =
x 1 .. . xn
to^ φB (x) =^ x 1 v 1 +^ · · ·^ +^ xnvn.^ Its inverse^ φ− B^1 (v) =
x 1 .. . xn
is denoted by [v]B = (x 1 ,... , xn) and the elements of this^ n-tuple are
called the coordinates of v in the basis B.
Remark 1. Observe that the φ− B^1 can be seen as a chart on V , which makes it into a smooth manifold diffeomorphic to Rn.
Consider two vector spaces V and W of dimensions n and m, respectively. Denote by Hom(V, W ) the set of linear maps from V and W , and recall that this is a vector space of dimension nm. Let B = {v 1 ,... , vn} and B′^ = {w 1 ,... , wm}, be bases for V and W , respectively. To a linear map T : V → W , one associates an m × n matrix [T ]BB′ = (ai,j ) defined by
T (vj ) =
∑^ n
i=
ai,j wi.
This establishes an isomorphism ΦBB′ : Hom(V, W ) → Mm,n(R) between Hom(V, W ) and the vector space Mm,n(R) of m × n matrices with real entries. Given vector spaces L 0 , Λ 1 and Λ 2 of dimensions r, m and n, respectively, the operation of composition gives a bilinear map
(1) Hom(L, Λ 1 ) × Hom(Λ 1 , Λ 2 ) → Hom(L, Λ 2 )
(φ, ψ) 7 → ψ ◦ φ.
(^1) Essentially, all arguments used in this discussion apply to vector spaces over C. You should read the discussion a second time assuming that you are dealing with vector spaces over C , while making the appropriate changes.
If one chooses bases B 0 , B 1 and B 2 for L 0 , Λ 1 and Λ 2 , respectively, then this operation becomes matrix multiplication [ψ ◦ φ]B B^02 = [ψ]B B^12 · [φ]B B^01. In other words, the operation above is equivalent to the smooth map:
(2) Mm,r(R) × Mn,m(R) → Mn,r(R)
(A, B) 7 → BA.
In particular, given fixed φ 0 ∈ Hom(L 0 , Λ 1 ) and ψ 0 ∈ Hom(Λ 1 , Λ 2 ) one defines linear maps
φ 0 ∗ : Hom(Λ 1 , Λ 2 ) → Hom(L 0 , Λ 2 ) ψ 7 → ψ ◦ φ 0 ,
and
ψ∗ 0 : Hom(L 0 , Λ 1 ) → Hom(L 0 , Λ 2 ) φ 7 → ψ 0 ◦ φ.
Given A = (ai,j ) ∈ Mr,n(R), its transpose At^ = (ati,j ) ∈ Mn,r(R) is the matrix satisfying ati,j = aj,i. Recall that A ∈ Mn,n(R) is said to be symmetric if At^ = A. The set M (^) nsym (R) of symmetric n × n matrices with real entries forms a vector subspace of Mn,n(R).
Remark 2. Check that dim M (^) nsym (R) = n(n 2 +1).
Given M = (mi,j ) ∈ Mr,r(R), define its trace to be trace M =
∑r i=1 mi,i^ ∈ R. Now consider the following bilinear map
(3) Mr,n(R) × Mr,n(R) → Mr,r(R)
A, B 7 → ABt.
The combination of the previous two operations defines an inner product on Mr,n(R) by sending A, B to
(4) 〈A, B〉 = trace (ABt).
Remark 3. Verify that this is indeed an inner product on the vector space Mr,n(R).
Recall that the rank of a matrix A ∈ Mr,n(R) is the dimension of the subspace of Rn^ spanned by the rows of A.
Facts 1. 1. The rank of A is also the dimension of the subspace of Rn spanned by the columns of A.
- If one considers A as a linear map from Rn^ to Rr^ then the rank-nullity theorem states that
dim ker A + rank A = n.
at a point P ∈ Mr,n. Remember that the manifolds in question are iso-
morphic (as vector spaces) to Rrn^ and Rr
2 , hence one can interpret tangent spaces in the ”old-fashioned” way. Pick v ∈ TP Mr,n, and note that the previous observation allows one to consider v as an r × n matix. Define γ(t) = A + tv and note that γ(0) = A and Dγ = v = γ′(0). By defini- tion, Φ∗(v) = Φ∗(Dγ ) = DΦ◦γ = (Φ ◦ γ)′(0). However, one can verify that Φ ◦ γ(t) = AAt^ + t(vAt^ + Avt) + t^2 vvt, hence
Φ∗(v) = (Φ ◦ γ)′(0) = vAt^ + Avt.
Example 3. Define Pr,n(R) = {A ∈ Mr,n(R) | AAt^ = I} = Φ−^1 (I). Our goal is to show that this is an embedded compact submanifold of M (^) r,nmax.rk.. We will do this in a few steps.
- Show that Pr,n(R) ⊂ M (^) r,nmax.rk..
Observe that if A ∈ Pr,n(R) then one sees that the composition
Rr^ A
t −→ Rn^ −→A Rr
is precisely the identity. This shows that as a linear map A is surjective. Apply the rank-nullity theorem to conclude that rank A = r.
- Show that I is a regular value of Φ.
This means that if A ∈ Φ−^1 (I) = Pr,n(R) then Φ∗ : TAMr,n → TI M (^) rsym (R) is surjective. This ammounts to show that given w ∈ TI M (^) rsym (R) ≡ M (^) rsym (R) (recall this identification) one can find v ∈ Mr,n such that Φ∗(v) = w. How- ever, we have seen that Φ∗(v) = vAt^ + Avt. Hence if one defines v := 12 wA, then
Φ∗(v) = (
wA)At^ + A(
wA)t
=
wAAt^ +
AAtwt^ =
w +
wt = w,
where the second to last inequality follows from the fact that A satisfies AAt^ = I and the last one follows from the fact that w is a symmetric matrix. We will prove the following result in class:
Theorem 4. Let Φ : M → N denote a smooth map between smooth mani- folds of dimensions m and n respectively, and let y ∈ N be a regular value such that Φ−^1 (y) 6 = ∅. Then Φ−^1 (y) is an embedded (m − n)-dimensional submanifold of M.
It follows from this theorem and the above observations that Pr,n(R) is an embedded submanifold of M (^) r,nmax.rk. and that its dimension is
dim Pr,n(R) = rn −
r(r + 1) 2
r 2
(2n − r − 1).
- Pr,n(R) is compact.
Just observe that since Pr,n(R) = Φ−^1 (I) ⊂ Mr,n, then Pr,n(R) is a closed subspace of Mr,n. On the other hand, if A ∈ Pr,n(R), then |A|^2 = 〈A, A〉 = trace (AAt) = trace I = r. Hence, Pr,n(R) is a bounded subspace of Mr,n. This shows that Pr,n(R) is compact.
A particular case of the above example is when r = n.
Example 4. Note that when r = n then every matrix in Pn,n(R) is invert- ible, and that is is closed under matrix multiplication. We denote Pn,n(R) by O(n) and call it the orthogonal group. This is a compact subgroup of the general linear group GLn(R), and it is another example of a Lie group, since it is a manifold and a group with smooth operations. It follows from
the formula above that dim O(n) = n(n 2 − 1).
2.2. Grassmann manifolds. Let Gr,n(R) denote the set of r-dimensional linear subspaces of Rn. The techniques used to show the following result form a beautiful application of linear algebra combined with the construc- tions made above. Pay lots of attention to the details.
Proposition 5. The set Gr,n(R) has a natural topology which makes it into a compact smooth manifold of dimension r(n − r).
Remark 6. Note that G 1 ,n+1(R) is the set of lines through the origin in Rn+1. There is a surjective map Sn^ → G 1 ,n+1(R) which sends x ∈ Sn^ to the line containing x and the origin. You have seen this before. In fact, G 1 ,n+1(R) is nothing else than the projective space RPn.
- Topologizing Gr,n(R).
This follows the idea in the previous remark. Consider the manifold Pr,n(R) studied above. Given A ∈ Pr,n(R) we know that the subspace span(A) of Rn, generated by the rows of A, has dimension r. Hence this defines a surjective map
π : Pr,n(R) → Gr,n(R).
Give Gr,n(R) the quotient topology and observe that Gr,n(R) becomes a compact space.
- Understanding the map π.
Given a linear subspace L ⊂ Rn^ of dimension r, the fiber π−^1 (L) of the map π consist of the r × n matrices A ∈ Pr,n(R) whose rows span L. Prove the following.
if and only if
x = (π 1 ◦ i 0 )|L (u) and y = π 2 ◦ i 0 (u) for a unique u ∈ L,
if and only if (x, y) = j(u)
for a unique u ∈ L. It is easy to see now that the map L ∈ UΛ 7 → φL defines a homeomorphism between UΛ and Hom(L 0 , Λ).
Remark 9. At this point, it is not very difficult for you to prove that Gr,n(R) is 2nd countable.
Given L 0 ∈ UΛ 1 ∩ UΛ 2 , consider the isomorphisms i 1 : Rn^ → L 0 ⊕ Λ 1 and i 2 : Rn^ → L 0 ⊕ Λ 2 as in the proof of the lemma. Verify that the composition
L 0 ⊕ Λ 1
i− 11 −−→ Rn^ −i→^2 L 0 ⊕ Λ 2
has the form i 2 ◦ i− 1 1 (v, w) = (v, Ψ 12 (w)), where Ψ 12 : Λ 1 → Λ 2 is an isomorphism. Now, let us denote by φi : UΛi → Hom(L 0 , Λi), for i = 1, 2, the isomor- phisms described in the Lemma above. It follows from the above consider- ations that the composition
φ 2 ◦ φ− 1 1 : φ 1 (UΛ 1 ∩ UΛ 2 ) → φ 2 (UΛ 1 ∩ UΛ 2 )
is given by taking ϕ ∈ φ 1 (UΛ 1 ∩ UΛ 2 ) ⊂ Hom(L 0 , Λ 1 ) to Ψ 12 ◦ ϕ ∈ Hom(L 0 , Λ 2 ). (Verify this assertion.) In other words, φ 2 ◦ φ− 1 1 is the restriction of Ψ 12 ∗ : Hom(L 0 , Λ 1 ) → Hom(L 0 , Λ 2 ) to the appropriate subsets. Since this is a linear map, it is smooth.
In conclusion, the collection {(UΛ, φΛ)} forms an atlas for Gr,n(R). There- fore, Gr,n(R) is a compact, smooth manifold of dimension r(n − r).
