The Real Grassmannian Gr(2,4)

The Real Grassmannian Gr(2, 4)

We discuss the topology of the real Grassmannian Gr(2, 4) of 2-planes in R4 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:

0 1 2 3 4 ≥ 5 a b a Hp(M, Z) ZZ ⊕ T Z ⊕ T Z Z 0 Hp(M, Z) ZZa Zb ⊕ T Za ⊕ T Z 0

Here a and b are unknown integers and T is a ﬁnite abelian group. They are determined by the fundamental group and the Euler characteristic.

ab H1 = π1 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 0 1 0 0-cells 0 0 0 1

0 1 ∗ 0 1-cells 0 0 0 1

0 1 0 ∗ 1 ∗ ∗ 0 2-cells 0 0 1 ∗ 0 0 0 1

1 ∗ 0 ∗ 3-cells 0 0 1 ∗

1 0 ∗ ∗ 4-cells 0 1 ∗ ∗

So the Euler characteristic is

χ = 1 − 1 + 2 − 1 + 1 = 2. Since π1 is always generated by the 1-cells, we see that π1 = H1 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 π1, 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) =∼ RP2 is Z/2Z, so γ2 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 0 0 1 0 0 1 ∗ 0 0 1 0 ∗ ∪ ∪ . 0 0 0 1 0 0 0 1 0 0 1 ∗ Therefore, the attaching map of the 2-cell in this decomposition has degree 2. For the other 2-cell, observe that 0 0 1 0 0 1 ∗ 0 1 ∗ ∗ 0 ∪ ∪ 0 0 0 1 0 0 0 1 0 0 0 1 is the subspace of 2-planes containing L. This is the same as the space of lines in R4/L, which forms another RP2 = 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).

0 1 2 3 4 ≥ 5 Hp(M, Z) ZZ/2ZZ/2Z 0 Z 0 Hp(M, Z) Z 0 Z/2ZZ/2ZZ 0

2. A Splitting In this section, we will obtain identifications Gr+(2, 4) =∼ S2 × S2 Gr(2, 4) =∼ S2 × S2/h(τ, τ)i where τ : S2 → S2 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 1 π (Gr(2, 4)) = Z/2Z χ(Gr(2, 4)) = χ(S2)2 = 2. 1 2 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) 7−→ p(−)q−1.

To be precise, we use the orthonormal basis 1, i, j, k for H to identify H with R4 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 7−→ hAe1, Ae2i, where A 0 SO(2) × SO(2) ,→ SO(4) (A, B) 7−→ . 0 B

Now we identify the preimage of SO(2) × SO(2) in SU(2) × SU(2). We will make use of the identiﬁcation cos θ − sin θ U(1) =∼ SO(2) cos θ + i sin θ 7−→ . 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) 7−→ (zw−1, zw). 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 7→ i i 7→ −1 j 7→ k k 7→ −j f(1, i) : 1 7→ −i i 7→ 1 j 7→ k k 7→ −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 SU(2) −→ SU(2)/U(1) =∼ P1 is the Hopf ﬁbration, so Gr+(2, 4) =∼ S2 × S2.

We can go further and identify explicitly the involution giving the double cover Gr+(2, 4) −→ Gr(2, 4).

Proposition. We have Gr(2, 4) =∼ S2 × S2/h(τ, τ)i where τ is the antipodal map on S2.

Proof. Observe that Gr(2, 4) = hJi\SO(4)/SO(2) × SO(2) J = diag(1, −1, 1, −1). The pre-image of hJi in SU(2) × SU(2) is the subgroup generated by (j, j). f(j, j) : 1 7→ 1 i 7→ −i j 7→ j k 7→ −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 1 SU(2) −→ P x0 + x1j 7−→ [x0 : x1]. We compute that j(x0 + x1j) = −x1 + x0j, so multiplication by j descends to the antipodal map [x0 : x1] 7→ [−x1 : x0]. 3. Lie Theory The result SO(4)/SO(2) × SO(2) =∼ P1 × P1 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 Ge of G. Since Te = U(1) × U(1) is a maximal torus of Ge,

G/T =∼ G/e Te =∼ SU(2)/U(1) × SU(2)/U(1) =∼ P1 × P1.

∼ We could also use the fact that G/T = GC/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⊥ ⊂ C4. 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 C4 such that L ⊂ L⊥. This is a smooth quadric surface in P3.