Download - q 6= 0. Let R qR - math.ucsd.edubprhoades/190w18/190w18HW4.pdf · 2J be a family of topological spaces and consider the ... as the set of lines through the ... Prove that we have

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Page 1: q 6= 0. Let R qR - math.ucsd.edubprhoades/190w18/190w18HW4.pdf · 2J be a family of topological spaces and consider the ... as the set of lines through the ... Prove that we have

Math 190: Winter 2018Homework 4

Due 5:00pm on Thursday 2/15/2018

Problem 1: Let (X, d) be a metric space. Prove that the function d : X ×X → R≥0is continuous.

Problem 2: Let (X, dX) and (Y, dY ) be metric spaces. Let f : X → Y be a func-tion satisfying dX(x1, x2) = dY (f(x1), f(x2)) for all x1, x2 ∈ X. Prove that f is animbedding.

Problem 3: Let {Xα}α∈J be a family of topological spaces and consider the disjointunion qα∈JXα of the sets Xα. The disjoint union topology on qα∈JXα has open setsqα∈JUα, where Uα ⊂ Xα is open for all α ∈ J .

• For each α0 ∈ J , let ια0 : Xα0 ↪→ qα∈JXα be the inclusion. Prove that ια0 iscontinuous.• Prove the following universal property:

If A is a topological space and fα : Xα → A is a family of con-tinuous maps (for all α ∈ J), there exists a unique continuous mapf : qα∈JXα → A such that fα = f ◦ ια for all α ∈ J .

You have proven that qα∈JXα serves as the coproduct in the category of topo-logical spaces.

Problem 4: Give R the standard topology and let R/Q be the quotient space obtainedby identifying Q to a point. Prove that R/Q is not Hausdorff.

Problem 5: Consider the disjoint union R1 qR2 of two copies of the real line. Let ∼be the equivalence relation on R1 q R2 given by setting x1 ∼ x2 if xi is the copy of xin Ri for i = 1, 2 and x 6= 0. Let X := R1 q R2/ ∼ be the associated quotient space.Prove that X is not Hausdorff.X is called the ‘line with two origins’.

Problem 6: Let X = Rn+1 − {0} be (n + 1)-dimensional Euclidean space with theorigin removed. Define an equivalence relation ∼ on X by the rule v ∼ λv for allvectors v ∈ X and scalars λ 6= 0. The n-dimensional real projective space is thequotient Pn(R) := X/ ∼. Prove that Pn(R) is Hausdorff.

We can think about Pn(R) as the set of lines through the origin in Rn+1. We candefine complex projective space Pn(C) in the same way; it is also Hausdorff.

Problem 7: Let Dn be the closed n-dimensional unit disc

Dn = {(x1, . . . , xn) ∈ Rn : x21 + · · ·+ x2n ≤ 1}

and let Sn−1 be its boundary. Exhibit a continuous bijection f : Dn/Sn−1 → Sn. (Wewill soon have the technology to prove that any such continuous bijection f must be ahomeomorphism – Sn is Hausdorff and Dn/Sn−1 is compact.)

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Page 2: q 6= 0. Let R qR - math.ucsd.edubprhoades/190w18/190w18HW4.pdf · 2J be a family of topological spaces and consider the ... as the set of lines through the ... Prove that we have

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Problem 8: Let M be the Mobius strip and let D = D2 be the closed 2-dimensionaldisc. Prove that the quotient space X = (MqD)/ ∼ obtained by glueing the boundarycircle of M to the boundary circle of D is homeomorphic to 2-dimensional real projec-tive space P2(R). (Hint: Model P2(R) as the sphere S2 with antipodal points identified.Constructing the actual maps here is tedious; intuitive ‘cut and paste arguments’ areOK. Consider the ‘arctic and antarctic circles’ on S2.)

Problem 9: (Optional - not to be handed in.) Let k ≤ n be positive integers. The(real) Grassmannian Gr(k, n) is the family of all k-dimensional subspaces of Rn:

Gr(k, n) = {V a subspace of Rn : dim(V ) = k}.(a) Explain why Gr(1, n) is the same as projective space Pn−1(R).We give Gr(k, n) a topology as follows. Let Uk,n ⊂ Matk×n(R) be the family of full

rank k × n real matrices.(b) Prove that Uk,n is an open subset of Matk×n(R). (Hint: Having full rank means

something about the minors, i.e. sub-determinants, of M .)The group GLk(R) acts on Uk,n by the rule A.M = AM for all A ∈ GLk(R) and

M ∈ Uk,n.(c) Prove that Gr(k, n) may be identified with the set of orbits

GLk(R)\Uk,n := {GLk(R).M : M ∈ Uk,n}.(Hint: Linear algebra.) We may new endow Gr(k, n) with the quotient topology.

A partition is a way to shove boxes into a corner. We will consider shoving 1×1 boxesinto the northeast corner of a k × (n − k) room (this corner choice is called Japanesenotation). If we have a matrix M ∈ Un,k, there is a unique matrix A ∈ GLk(R) suchthat AM is in row reduced echelon form (why?). If n = 9 and k = 4, such a RREFmatrix might look like this:

0 1 ∗ ∗ 0 0 ∗ 0 ∗0 0 0 0 1 0 ∗ 0 ∗0 0 0 0 0 1 ∗ 0 ∗0 0 0 0 0 0 0 1 ∗

As you can see, the stars in the above picture form a (Japanese) partition:

∗ ∗ ∗ ∗∗ ∗∗ ∗∗

We might call this partition (4, 2, 2, 1). The open Schubert cell Cλ ⊂ Gr(k, n) cor-responding to a partition λ ⊂ k × (n − k) is the family of all orbits which can berepresented by matrices with RREF corresponding to λ.

(d) Prove that we have a disjoint union decomposition Gr(k, n) =⊔λ⊂k×(n−k)Cλ.

(e) Prove that Cλ is homeomorphic to R|λ|, where |λ| is the number of boxes in λ.(This requires ingenuity – how can you write down your homeomorphism?)