The Real Grassmannian Gr(2, 4)
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1 The Real Grassmannian Gr(2, 4) We discuss the topology of the real Grassmannian Gr(2, 4) of 2-planes in R 4 and its double cover Gr + (2, 4) by the Grassmannian of oriented 2-planes They are compact four-manifolds 0 A Remark on Four-Manifolds By applying the universal coefficients theorem and Poincaré duality to a general closed orientable four-manifold M, one finds that its homology and cohomology is restricted to take the following form: H p (M, Z) Z Z a T Z b T Z a Z 0 H p (M, Z) Z Z a Z b T Z a T Z 0 Here a and b are unknown integers and T is a finite abelian group determined by the fundamental group and the Euler characteristic They are H 1 = π ab 1 χ = 2 2a + b We will compute the homology and cohomology of Gr(2, 4) by finding these two invariants 1 A Cell Decomosition Assume for now that the Grassmannian Gr(2, 4) is orientable Any 2-plane can be represented as the row space of a 2 4 matrix, and there is always a unique row-reduced representative This decomposes Gr(2, 4) into cells according to the shape of the row-reduced matrix 0-cells 1-cells 2-cells 3-cells 4-cells ( ( ( ( ( ( So the Euler characteristic is χ = = 2
2 Since π 1 is always generated by the 1-cells, we see that π 1 = H 1 is a cyclic group generated by the loop γ corresponding to the 1-cell in the decomposition above We will be finished if we can establish the following claim: Proposition The element γ has order two Proof 1 The Grassmannian admits a connected double cover Gr + (2, 4) Gr(2, 4) by the Grassmannian of oriented 2-planes The existence of such a covering implies that π 1, and hence, γ is nontrivial To see that γ has order two, observe that it lies in the subspace Gr(2, 3) = {2-planes contained in the hyperplane (0,,, )} Gr(2, 4) The fundamental group of Gr(2, 3) = RP 2 is Z/2Z, so γ 2 is nullhomotopic Proof 2 It is possible to circumvent the use of Gr + (2, 4) by identifying the attaching maps of the 2-cells directly For example, the Gr(2, 3) we just considered is a union of cells ( ) ( ) ( ) Therefore, the attaching map of the 2-cell in this decomposition has degree 2 For the other 2-cell, observe that ( ) ( ) ( ) is the subspace of 2-planes containing L This is the same as the space of lines in R 4 /L, which forms another RP 2 = Gr(1, 3) So the attaching map of this 2-cell has degree two as well In any case, we have determined the homology and cohomology of Gr(2, 4) 2 A Splitting H p (M, Z) Z Z/2Z Z/2Z 0 Z 0 H p (M, Z) Z 0 Z/2Z Z/2Z Z 0 In this section, we will obtain identifications Gr + (2, 4) = S 2 S 2 Gr(2, 4) = S 2 S 2 / (τ, τ) where τ : S 2 S 2 is the antipodal map This justifies our previous assumption that Gr(2, 4) is orientable It is also gives another way to compute the homology and cohomology of Gr(2, 4), since we can read off the invariants π 1 (Gr(2, 4)) = Z/2Z χ(gr(2, 4)) = 1 2 χ(s2 ) 2 = 2
3 Recall that SU(2) can be identified with the group of unit norm quaternions In what follows, we will not need to view elements of SU(2) as unitary matrices, so we may as well take this to be the definition This allows us to define a map f : SU(2) SU(2) SO(4) (p, q) p( )q 1 To be precise, we use the orthonormal basis 1, i, j, k for H to identify H with R 4 and hence the group of orthogonal transformations of H with SO(4) Lemma 1 The map f induces an isomorphism f : SU(2) SU(2)/{±(I, I)} SO(4) Proof If (p, q) belongs to the kernel of f, then x = f(p, q) x = pxq 1 for all x H Taking x = 1 shows that p = q Then this equation says that p is in the center Z(H) = R So the kernel of f is ±(I, I) But now f is an injective map between Lie groups of the same dimension, so it is open Its image is closed because SU(2) SU(2) is compact Therefore, f is surjective because SO(4) is connected Lemma 2 There is a homeomorphism SO(4)/SO(2) SO(2) Gr + (2, 4) A Ae 1, Ae 2, where SO(2) SO(2) SO(4) (A, B) ( ) A 0 0 B Now we identify the preimage of SO(2) SO(2) in SU(2) SU(2) We will make use of the identification ( ) U(1) cos θ sin θ = SO(2) cos θ + i sin θ sin θ cos θ Lemma 3 We have a cartesian square U(1) U(1) SU(2) SU(2) SO(2) SO(2) SO(4) The upper horizontal map is the composition U(1) = SO(2) SU(2) on each factor, and the vertical map on the left is (z, w) (zw 1, zw)
4 Proof The commutativity of the diagram amounts to the identity ( ) zw 0 x = zxw 0 zw In fact, this equality holds true for any pair of complex numbers z and w To verify this, note that it suffices to consider pairs of the form (z, 1) and (1, w) Furthermore, the equation is bilinear over R, so we only need to check (i, 1) and (1, i) f(i, 1) : 1 i i 1 j k k j f(1, i) : 1 i i 1 j k k j The square is cartesian because the arrow on the left is surjective and U(1) U(1) contains the kernel of f Combining these lemmata gives a chain of homeomorphisms Gr + (2, 4) = SO(4)/SO(2) SO(2) = SU(2) SU(2)/U(1) U(1) Of course is the Hopf fibration, so SU(2) SU(2)/U(1) = P 1 Gr + (2, 4) = S 2 S 2 We can go further and identify explicitly the involution giving the double cover Proposition We have where τ is the antipodal map on S 2 Proof Observe that Gr + (2, 4) Gr(2, 4) Gr(2, 4) = S 2 S 2 / (τ, τ) Gr(2, 4) = J \SO(4)/SO(2) SO(2) J = diag(1, 1, 1, 1) The pre-image of J in SU(2) SU(2) is the subgroup generated by (j, j) f(j, j) : 1 1 i i j j k k So we need to identify the action of j on SU(2)/U(1) as the antipodal action Viewing H = C Cj as a vector space over C, we have an explicit formula for the Hopf fibration SU(2) P 1 x 0 + x 1 j [x 0 : x 1 ] We compute that j(x 0 + x 1 j) = x 1 + x 0 j, so multiplication by j descends to the antipodal map [x 0 : x 1 ] [ x 1 : x 0 ]
5 3 Lie Theory The result SO(4)/SO(2) SO(2) = P 1 P 1 of the previous section can be obtained without calculation if we are willing to use the theory of compact Lie groups The subgroup T = SO(2) SO(2) is a maximal torus of G = SO(4), and we need to identify the space G/T Since SU(2) SU(2) is simply connected and has the same complexified Lie algebra as G, it must be the universal cover G of G Since T = U(1) U(1) is a maximal torus of G, G/T = G/ T = SU(2)/U(1) SU(2)/U(1) = P 1 P 1 We could also use the fact that G/T = G C /B is the flag variety of the complexified group In our case, we the Borel subgroups of SO(4, C) are the stabilizers of flags taking the form 0 L W = W L C 4 Every line with L L determines two such flags Furthermore, two flags have the same stabilizer precisely when they start with the same line L Therefore, the flag variety SO(4, C)/B is the space of lines in C 4 such that L L This is a smooth quadric surface in P 3
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