ORIENTATIONS. Contents


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1 ORIENTATIONS Contents 1. Generators for H n R n, R n p 1 1. Generators for H n R n, R n p We ended last time by constructing explicit generators for H n D n, S n 1 by using an explicit nsimplex which depended on a choice of points in S n 1 in general position. One can still wonder about what happens with other choices of points in general position. To discuss this it helps to generalize the question. Consider the inclusion of pairs D n, S n 1 R n, R n 0 where 0 is the origin. Look at the commutative ladder H S n 1 H D n H D n, S n 1 H 1 S n 1 H 1 D n H R n 0 H R n H R n, R n 0 H 1 R n 0 H 1 R n Since D n and R n are both contractible, the maps labeled 2 are isomorphisms. Since S n 1 R n 0 is a homotopy equivalence, the maps labeled 1 are isomorphisms. It follows from the Five Lemma that the maps labeled 3 are isomorphisms. This proves 4 5 Proposition 1.1. The map induced by the inclusion H D n, S n 1 H R n, R n 0 is an isomorphism. Both groups are 0 unless = n in which case they are Z. Corollary 1.2. If h: R n R m is a homeomorphism, n = m. This is a weak version of Invariance of Domain. Next we would like to identify an explicit generator for H n R n, R n 0. Define an n simplex σ : n R n by σt 0,, t n = t i 1 e i Note σ is the composition of the following maps. First map n to { } X n = x 0,, x n R n each x i 0 and x i 1 by t i x i, i 1. This map is a homeomorphism. Note X n R n is the closure of an open, bounded convex set. Then map X n to R n by the translation T 1 n+1 1,1,,1 where T vx = x + v. Since T v is an isometry, σ n is the closure of an open, bounded convex set. As we saw earlier, this means that i=1 Date: Spring
2 2 ORIENTATIONS the compositon n R n 0 F S n 1 is a homeomorphism where F x = x. Just as above, we x see σ : H n n, n H n R n, R n 0 is an isomorphism and so σ = σ 1 n is a generator. There is no particular reason to focus on 0. Given any point p R n, there is a map T p : R n R n defined by T p v = v +p. It is a homeomorphism since its inverse is T p. Since T p : R n, R n 0 R n, R n p is a homeomorphism of pairs, { H R n, R n Z = n p = 0 otherwise S. ince we have a preferred generator at 0 we define the preferred generator at p to be T p of the preferred generator at 0. It can also be defined as the homology class represented by the simplex T p σ : n R n. Given a generator h p for H n R n, R n p and a generator h q for H n R n, R n q we say that they represent the same orientation provided T q p h p = h q. Otherwise we say they represent the opposite orientation. There is also no need for only one set of points. Given points, p 0,..., p n R n, we say that the ordered collection is in general position provided the n vectors p 1 p 0,..., p n p 0 are linearly independent. Given any points in general position, define a simplex γ p0,,p n : n R n by so Check γ p0,,p n t 0,, t n = p 0 + p k = p k p 0 + γ p0,,p n t 0,, t n = p k p k + t 0 1 p 0 = p k p 0 + p 0 p 0 = p k p 0 Since the p k are in general position, the ordered set of p k p 0 is a basis and so there is a unique linear isomorphism α so that αe k = p k p 0. If A is the affine transformation Ax = αx + p 0 then γ p0,,p n = A σ. Proposition 1.3. If the collection p 0,..., p n R n is in general position, then γ p0,,p n is a generator for H n R n, R n p 0. Proof. Since α is invertible, so is A so A : H n R n, R n 0 H n R n, R n p 0 is an isomorphism and under this isomorphism the generator σ goes to γ p0,,p n. We would like to compare the generator in Proposition 1.3 with the standard generator. By applying T p0 we see that α σ = ɛσ, ɛ = ±1 and γ p0,,p n = ɛt p0 σ. Hence it suffices to study α σ. The general linear group over R has two path components given by the sign of the determinant. Proofs may be found in many books on Lie groups. Brian Hall has some arxiv notes which prove this as an exercise.
3 ORIENTATIONS 3 Hence α is homotopic through elements in GL n R to the identity or to the diagonal matrix M n with all 1 s on the diagonal except for the last where the entry if 1. Hence detm n = 1. Since these maps all leave 0 fixed, a homotopy of α induces a homotopy of pairs R n, R n 0. Since homotopic maps induce the same map in homology, ɛ is constant under such a homotopy. Hence if detα > 0, ɛ = 1. It remains to investigate the case α = M n. For several reasons going forward, we need a second method to acquire isomorphisms from H n R n, R n p to H n R n, R n q. Let B be a closed ndisk containing p. Then R n B R n p is a homotopy equivalence so a Five Lemma argument shows ι B,p : H R n, R n B H R n, R n p is an isomorphism. Given p and q find a disk B containing them both. Then H R n, R n q ι B,q H R n, R n B ι B,p = H R n, R n p = gives an isomorphism between the two homology groups. Check that any two disks which contain the two points gives the same isomorphism. The key point is that if W is a disk containing B, then R n W R n B and ι W,p : R n, R n ι W,B W R n, R n ι B,p B R n, R n p so W and B give the same isomorphism. The oneparameter family T tq p starts at the identity and finishes at T q p. Points move a maximum distance of q p during the homotopy so given B, there exists a bigger disk W with B W and T tq p R n W R n B. Hence we have a homotopy of pairs T tq p : R n, R n W [0, 1] R n, R n B between the inclusion ι W,B : R n, R n W R n, R n B and R n, R n W Consider the diagram H n R n, R n W ι W,p H n R n, R n p ι W,B =T q p ι B,p T q p H n R n, R n B ι B,q H n R n, R n q T q p R n, R n B. Using ι W,B shows that the top triangle commutes. Using T q p shows the outer square commutes and hence H n R n, R n p ι B,p = H n R n, R n B = T q p ι B,q = H n R n, R n q commutes. I have added the subscript s to the maps to emphasize that while the diagram commutes it does not commute because there is an underlying commutative diagram of pairs. Proposition 1.4. Let ψ : n R n map the interior of n to an open subset U of R n. Let p and q U. Then ψ is a relative cycle in R n, R n p and in R n, R n q. Let [ψ] p H n R n, R n p and [ψ] q H n R n, R n q be the corresponding homology classes. Then T q p [ψ]p = [ψ]q.
4 4 ORIENTATIONS Proof. Fix p and let V U be the set of all q U such that T q p [ψ]p = [ψ]q. Note p V. Since U is open, there is a small disk B U centered at p. Note ψ is a relative cycle in R n, R n B and since commutes, q V for all q B. Hence V is open. Let W U be the set of all q U such that T q p [ψ]p [ψ]q. If W is nonempty let q 0 W. Then as above, there is a small disk around q 0 in W so W is open. Since the interior of n is path connected, so is U and therefore either V or W is empty so V = U as required. To investigate how M n acts, it helps to return to the model for generators from Handout 10. There we took p 0 = e 1 and p k = e k, k = 1,..., n and G p0,,p n t 0,, t n = t k p k. Let β p0,,p n = 1 p k be the barycenter of these points. Then the simplex G p0,,p n is G p0,,p n t 0,, t n = p k + β p0,,p n = γ p0,,p n + β p0,,p n p 0. The simplex G p0,,p n is a relative cycle in R n, R n β p0,,p n which represents the same generator as γ p0,,p n. Hence we need to compute ɛ = ±1 such that [G e1,e 1,e n 1,e n ] and ɛ[g e1,e 1,e n 1, e n ] represent the same orientation. Let the coordinates of R n be x 1,, x n. To save on notation let ψ + = G e1,e 1,e n 1,e n and ψ = G e1,e 1,e n 1, e n. The relative cycle ψ + represents a generator of H n R n, R n β + and ψ represents a generator of H n R n, R n β where β ± is the image of the barycenter of n under π ±. In the hyperplane Z = {x n = 0} the images of ψ ± are both the face n, n the face opposite the last vertex. Call this simplex φ 0 : n 1 Z R n. The remaining faces of ψ + all have interiors with x n > 0 and the remaining faces of ψ s all have interiors with x n < 0. Let K + be the faces in ψ + with x n > 0 and let K be the faces in ψ with x n < 0. Then Rn n ψ + = 1 k k n ψ n φ 0 where the sum is over the simplices in K +. Similarly n Rn ψ = 1 k k n ψ + 1 n φ 0 where the sum is over the simplices in K. This picture for n = 2 may help. The green triangle is ψ + and the red one is ψ. Z is the xaxis, y = x 2 = 0 and both p 0 and p 1 lie in Z.
5 ORIENTATIONS 5 p 2 p 0 p 1 p 2 We next replace ψ ± by simplices which have an interior point in common. Let L = Imψ + Imψ R n. Prove L is convex and then Imψ + L is a convex subset. Radial projection from the barycenter of ψ + gives a new simplex ˆψ + whose image is L. As subsets of R n, K = ˆψ + Imφ0. Then 1 + n Rn ˆψ + = 1 k k n ψ n ˆψ+ φ 0 There is also an nsimplex, c defined as follows. The image of ψ is the cone on Imφ 0 coned from 0,, 0, 1. There is a map which just pushes up the cone lines from n to the union of the faces not the last. This is c. Then 2 n Rn c = 1 k k n ψ + 1 n ˆψ+ φ 0 It is important to note that the simplex c is in R n 0. Reversing roles defines ψ ˆ and c + with formulas 1 and 2 + similar to 1 + and 2. Then the classes represented by ψ + and by ˆψ + in H n R n, R n β + are the same. The classes represented by ˆψ + in H n R n, R n β + and H n R n, R n 0 are the same. Hence it suffices to compare the class represented by ˆψ + in H n R n, R n 0 with that represented by ˆψ in H n R n, R n 0. To do this we will calculate the images of these classes in H n 1 R n 0. More precisely, we will compute n ˆψ+ + ˆψ. From equations 1 ± it follows that n Rn ˆψ + + ˆψ = 1 k k n ψ n ˆψ+ φ 0 + From equations 2 it follows that Rn 0 n c + + c = 1 k k n ψ n ˆψ φ k k n ψ + 1 n ˆψ φ 0 1 k k n ψ + 1 n ˆψ+ φ 0
6 6 ORIENTATIONS Since these two sums are the same n Rn ˆψ + + ˆψ is a boundary in R n 0 and so the sum of our two classes is 0 H n 1 R n 0. Since the map H n R n, R n 0 H n 1 R n 0 is an isomorphism, the two generators have opposite sign in H n R n, R n 0. We have proved the following Theorem 1.5. If the collection p 0,..., p n R n is in general position, then γ p0,,p n is the preferred generator for H n R n, R n p 0 if and only if the basis p 1 p 0,... p n p 0 is positive.
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