Topology QR Solutions, January 2015Daniel Irvine

Morning Session

(1)

(2) Let Mn(R) ∼= Rn2

be the space of n× n matrices with real coefficients.(a) Show that

SLn(R) := {A ∈Mn(R) : det(A) = 1}is a smooth submanifold of Mn(R).Proof. This is a special case of the regular value theorem, when applied to the smooth Liegroup homomorphism det : GLn(R)→ R∗. SLn(R) is the kernel of det, and so it is a smoothlyembedded submanifold of GLn(R) and hence of Mn(R). �

(b) Identify the tangent space to SLn(R) at the identity matrix In.

Solution. By definition, this is the Lie algebra to the Lie group SLn(R). This Lie algebra issln(R) = {A ∈Mn(R) : tr(A) = 0}.

Alternative solution: Consider the smooth map f : Rn2

→ R, A 7→ det(A). We shall prove thatSLn(R), namely the preimage f−1(1), is a smooth manifold by using the pre-image theorem. Wewould like to show that the 1×n2 matrix of dfA represents an onto linear map, i.e. not a row of zeros.

Denote dfA = (m1,1,m2,1, ...,mn,1,m1,2,m2,2...). By definition, mi,j is limt→0det(A+tEi,j)−det(A)

twhere Ei,j contians 1 in the i, j spot and zero elsewhere. By direct calculation, we see that mi,j

is (−1)i+j times the minor of the matrix A at the i, j (the determinant of the matrix B obtainedby removing the i’th row and the j’th column from the matrix A). Since not all the minors of aninvertible matrix are zero, 1 is a regular value of f .

The tangent space at I is the kernel of dfI . Notice that all the minors of I are 1 at the i, i spots, andzero elsewhere. So if B is in the tangent space so it must satisfy dfI ·B = 0⇒ b1,1 +b2,2 + ...+bn,n =0⇒ trace(B) = 0

(3)(4) Let X1, X2, ... be a sequence of closed, connected subsets of a compact Hausdorff space Y such that

Xn+1 ⊆ Xn for all n ≥ 1. Show that X = ∩Xi is connected.

Note. I have two proofs of this, but I think the first proof is wrong, because I can’t see where theHausdorff assumption is used. The second proof uses the Hausdorff assumption, but also uses a lotof big theorems.

Proof. Suppose, towards a contradiction, that X = ∩Xi were not connected, i.e. suppose we havea separation X = C ∪ D where C and D are disjoint open sets of X. Then, by definition of thesubspace topology, we can choose disjoint sets U and V which are open in Y , such that C ⊆ U andD ⊆ V . (in order to see why those can be chosen to be disjoint, see the solution for May 2014,Morning, Q3).

Each set Xi \ (U ∪ V ) must be nonempty, because otherwise U and V would form a separation ofXi, contrary to the assumption that Xi is connected. Furthermore, each finite sequence

(X1 \ (U ∪ V )), ..., (Xn \ (U ∪ V ))

must also have nonempty intersection, because

n⋂i=1

(Xi \ (U ∪ V )) = Xn \ (U ∪ V ),

and we just proved that Xn \ (U ∪ V ) is nonempty.The finite intersection theorem applied the compact set Y then implies that

W =⋂i

(Xi \ (U ∪ V ))

is nonempty.This shows that there is a point in each Xi (and hence a point in X) which is not covered by

U ∪ V , contrary to the assumption that C and D form a separation of X. �

Proof. (A more high-powered proof) Every compact Hausdorff space is T4 (c.f. Munkres Thm. 32.3).So Y is T4. Since each Xi is closed, the intersection X = ∩Xi will be closed and hence T4 in Y (c.fMunkres exercise 32.1). Assume there exists a separation X = A ∪ B. By Urysohn’s Lemma (c.f.Munkres §33), A and B can be separated by a continuous function, i.e. there exists a continuousf : X → [0, 1] such that f(a) = 0 ∀a ∈ A and f(b) = 1 ∀b ∈ B. Next, apply the Tietze extensiontheorem (c.f. Munkres §35) to extend the continuous map f : X → [0, 1] to a continuous map

f̃ : Y → [0, 1]. Now, each set Xi is a superset of X, and hence each set Xi contains some point ai of

A and some point bi of B (since A∪B = X). The map f̃ is an extension of f , and so f̃ must satisfy

f̃(ai) = 0 and f̃(bi) = 1 for each i. Since each subspace Xi is connected, and since the continuous

image of a connected set is connected, this calculation proves that for each i, f̃(Xi) is all of [0, 1].Hence, we can apply the intermediate value theorem (c.f. Munkres Thm. 24.3) to each subspace

Xi to find a sequence of points qi ∈ Xi satisfying f̃(qi) = 0.5 for each i. But then X must have an

element q satisfying f̃(q) = 0.5, contrary to our original construction of f . �

Afternoon Session

(1)(2)(3) Let X be a compact metric space and suppose that f : X → X is an isometry. Prove that f : X → X

is a homeomorphism.

Proof. First note that every isometry is a continuous map X → X, because every isometry isdistance-decreasing (alternatively, take δ = ε). The same argument applied to images instead ofpre-images of f shows that f takes open δ-balls to open δ-balls. Hence f is an open mapping. Inorder to show that f is a homeomorphism, all that remains is to show that f is bijective.

Assume f were not surjective. Then there is some point y ∈ X such that y /∈ f(X). Choose ε sothat the ε-neighborhood of a is disjoint from f(X). Set x1 = y and then inductively define xn+1 tobe f(xn), assuming that xn is already defined. Then by the choice of ε,

d(x1, xm) ≥ ε ∀m > 1.

Moreover, using the isometry property of f , whenever m > 2, we get

d(x2, xm) = d(f(x1), f(xm−1)) = d(x1, xm−1) ≥ ε,

because this case reduces to the case above. Proceed inductively in this manner. Assuming that forfixed n > 2 we know that d(xn, xm) ≥ ε for all m 6= n, then in the (n+ 1)st case we shall have

d(xn+1, xm) = d(f(xn), f(xm−1)) = d(xn, xm−1) ≥ ε

whenever m 6= n+ 1. All other cases where n 6= m reduce to computations above.In the previous paragraph we showed that d(xn, xm) ≥ ε for all n 6= m. Hence the sequence

{xj}j∈N has no convergent subsequence (since a ball of radius ε/4 around any point can contain atmost one element of the sequence). This shows that X is not sequentially compact, which contradictsthe assumption that X is compact (c.f. Munkres Thm. 28.2). We must have that f is surjective.

Now all that remains to show is that f is injective. This is an immediate consequence of theisometry assumption. For if x, y ∈ X satisfy x 6= y, then we must have d(x, y) =: a > 0. Consequentlyd(f(x), f(y)) = a > 0 so that f(x) 6= f(y).

�

(4) Let A ⊆ Rω be defined by

A := {(xi) ∈ Rω |xi = 0 for all but finitely many i}

(a) Prove that A is dense in Rω with the product topology.Proof. Let ~x = (x1, x2, ...) be a point of Rω. Let U =

∏Ui be a basis element for the product

topology that contains ~x. We show that U intersects A, which will show that ~x ∈ A andhence A = Rω, since ~x was arbitrary. Now, by definition of a basis element in the producttopology, we must have that Ui = R for all i larger than some index N . Then the point~x′ = (x1, x2, ..., xN , 0, 0, ...) is a member of U since xi ∈ Ui and 0 ∈ Ui for i > N , and ~x′ is alsoa member of A.

�(b) Prove that A is not dense in Rω with the box topology.

Proof. By inspecting the above proof, we can see where it will fail if we switch to the boxtopology. Recall that a point ~x ∈ Rω is in A if and only if every basis element B (for thebox topology) containing ~x meets A. That is not the case in this example. For if we let~x = (1, 2, 3, ..., n, ...) then a basis element for the box topology which contains ~x is

B = (1

2,

3

2)× (

3

2,

5

2)× ...× (n− 1

2, n+

1

2)× ...

and this basis element clearly contains no points of A, because no coordinates of any elementof this basic set will ever be zero. Hence this particular ~x is not in A. �

(5) Let S1 = {z ∈ C : |z| = 1}, and let

X = (S1 × S1 t S1 × [−1/2, 1/2])/ ∼where ∼ is the smallest equivalence relation with (z, 1) ∼ (z,−1/2) and (1, z) ∼ (z, 1/2) for allz ∈ S1. Compute the homology of X.

Proof. By thinking about the definition of X, we see that X can be realized as the ∆-complex drawnin the figure below.

v v

v v

a

a

b b

c

c

B

A

Now we can compute the simplicial homology of this ∆-complex. There is one 0-cell, three 1-cells,and two 2-cells. We can choose clever bases for the chain groups C∆

0 , C∆1 , and C∆

2 as follows

C∆0 = Z · v, C∆

1 = Z · (a+ b)⊕ Z · b⊕ Z · c, C∆2 = Z ·A⊕ Z ·B.

Using the figure above we can compute the chain maps

0 −→ C∆2

d2−→ C∆1

d1−→ C00−→ 0,

as follows:d1(a+ b) = d1(b) = d1(c) = v − v = 0 (so d1 ≡ 0),

d2(A) = a+ c+ b− c = a+ b,

d2(B) = a− b− a+ b = 0.

From these computations we see that

image(d1) = 0, ker(d1) = C∆1 , image(d2) = Z · (a+ b), ker(d2) = Z ·B.

Finally, we can compute the simplicial homology of X.

H∆0 (X) = ker(d1)/im(d2) = C∆

0 /0∼= Z.

H∆1 = ker(d1)/im(d2) = (Z · (a+ b)⊕ Z · b⊕ Z · c)/(Z · (a+ b)) ∼= Z2.

H∆2 (X) = ker(d2)/0 = Z ·B/0 ∼= Z.

All other homology groups are trivial. This space has the homology of a torus.�

Lemma 1. An arbitrary product of path connected spaces {Xα}α∈J will be path connected in the producttopology.

Proof. We use the universal property of the product topology (applied to X =∏Xα), which says that if Y

is another topological space with a continuous maps fα : Y → Xα for each α ∈ J , then there exists a uniquecontinuous map f : Y → X such that πα ◦ f = fα for all α ∈ J , where πα is the αth coordinate projectionfunction (which is continuous).

Let Y = [0, 1] and suppose all the spaces Xα are path connected. Then given any two distinct pointsx, y ∈ X there is a path γα : [0, 1]→ Xα connecting their αth coordinates. More precisely, we have γ(0) = xα

and γ(1) = yα, and γα is continuous by path connectedness of Xα. Now invoke the universal property toget a continuous map γ : [0, 1] → X such that πα ◦ γ = γα, which means (γ(0))α = xα and (γ(1))α = yα.Therefore γ is the desired path from x to y in X. �