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MATH 465/565: Grassmannian Notes MATH 465/565: Grassmannian Notes The Grassmannian G(r; n) is the set of r-dimensional subspaces of the k-vector space kn; it has a natural bijection with the set G(r − 1; n − 1) of (r − 1)-dimensional linear subspaces Pr−1 ⊆ Pn. We write G(k; V ) for the set of k-dimensional subspaces of an n-dimensional k-vector space V . We'd like to be able to think of G(r; V ) as a quasiprojective variety; to do so, we consider the Pl¨uckerembedding: r γ ∶ G(r; V ) → P V Span(v1; : : : ; vr) ↦ [v1 ∧ ⋯ ∧ vr] If w a v is another ordered basis for Λ Span v ; : : : ; v , where A a is an invertible i = ∑j ij j1≤i≤r = ( 1 r) = ( ij) matrix, then w1 ∧⋯∧wr = (det A)(v1 ∧⋯∧vr). Thus the Pl¨ucker embedding is a well-defined function r from G(k; V ) to P ⋀ V . We would like to show, in analogy with what we were able to show for the Segre embedding σ∶ P(V ) × P(W ) → P(V ⊗ W ), that • the Pl¨ucker embedding γ is injective, • the image γ(G(r; V )) is closed, and • the Grassmannian G(r; V ) \locally" can be given a structure as an affine variety, and γ restricts to an isomorphism between these \local" pieces of G(r; V ) and Zariski open subsets of the image. r Given x ∈ ⋀ V , we say that x is totally decomposable if x = v1 ∧ ⋯ ∧ vr for some v1; : : : ; vr ∈ V , or equivalently, if [x] is in the image of the Pl¨ucker embedding. 1 Example. Every non-zero element of ⋀ V is trivially totally decomposable. 2 Example. If dim V = 3, then every non-zero element of ⋀ V is totally decomposable. 2 Proof. Given a sum v1 ∧ v2 + v3 ∧ v4 of two non-zero elements of ⋀ V , the two-dimensional subspaces Span(v1; v2) and Span(v3; v4) must intersect. If w1 is in the intersection, then we can rewrite v1 ∧v2 = w1 ∧ w2 for some w1 ∈ Span(v1; v2). Similarly we can rewrite v3 ∧ v4 = w1 ∧ w3 for some w3. Then v1 ∧ v2 + v3 ∧ v4 = w1 ∧ w2 + w1 ∧ w3 = w1 ∧ (w2 + w3) is totally decomposable. Proceeding in the same way by induction, we can show that any element of 2 ⋀ V is totally decomposable. Example. On the other hand, if v1; v2; v3; v4 ∈ V are linearly independent, then v1 ∧ v2 + v3 ∧ v4 is not totally decomposable. This follows (for chark =~ 0) from the observation that since v ∧ v = 0, if r x ∈ ⋀ V is totally decomposable, then x ∧ x = 0. Since (v1 ∧ v2 + v3 ∧ v4) ∧ (v1 ∧ v2 + v3 ∧ v4) = v1 ∧ v2 ∧ v3 ∧ v4 + v3 ∧ v4 ∧ v1 ∧ v2 = 2v1 ∧ v2 ∧ v3 ∧ v4; which is non-zero since v1; v2; v3; v4 ∈ V are linearly independent (certainly v3 ∧ v4 ∧ v1 ∧ v2 = ±v1 ∧ v2 ∧ v3 ∧ v4, and the sign is positive because (13)(24) is an even permutation). 1 r If e1; : : : ; en is a basis for V , then the set of eI = ei1 ∧ ei2 ∧ ⋯ ∧ eir is a basis for ⋀ V , where r I = {i1; : : : ; ir}, with 1 ≤ i1 < i2 < ⋯ < ir ≤ n. Thus any element of x ∈ ⋀ V has a unique representation in the form x = Q aI eI = Q ai1;i2;:::;ir (ei1 ∧ ⋯ ∧ eir ); I⊆{1;:::;n} 1≤i1<⋯<ir≤n SIS=r r n−1 and we call the homogeneous coordinates aI the Pl¨uckercoordinates on P V ≅ P r associated to the choice of ordered basis (e1; : : : ; en) for V . 2 Example. If V has basis e1; e2; e3; e4, then every element x ∈ ⋀ V can be uniquely written as x = a12(e1 ∧ e2) + a13(e1 ∧ e3) + a14(e1 ∧ e4) + a23(e2 ∧ e3) + a24(e2 ∧ e4) + a34(e3 ∧ e4): If x is totally decomposable, then we know x∧x = 0; we compute in terms of the Pl¨ucker coordinates that x ∧ x = (a12a34 − a13a24 + a14a23)(e1 ∧ e2 ∧ e3 ∧ e4); Hence the image of the Pl¨ucker embedding of G(2; 4) into P5 satisfies the homogeneous quadric equation a12a34 − a13a24 + a14a23 = 0. (In fact, in this particular case the image is precisely the zero locus of this polynomial.) To show that the Pl¨ucker embedding is an injection, we must describe how to recover Span(v1; : : : ; vr) from x = v1 ∧ ⋯ ∧ vr. We observe first that vi ∧ x = 0 for 1 ≤ i ≤ r, and more generally, if v ∈ Span(v1; : : : ; vr), then v ∧ x is zero. In fact, this property determines Span(v1; : : : ; vr): r r+1 Proposition. Given a non-zero x ∈ ⋀ V , let 'x∶ V → ⋀ V be the linear map 'x(v) = v ∧ x: Then dim ker('x) ≤ r, with equality if and only if x is totally decomposable. If x = v1 ∧ ⋯ ∧ vr, then ker('x) = Span(v1; : : : ; vr). Proof. Chose a basis e1; : : : ; en for V so that e1; : : : ; es is a basis for ker('x), where s = dim ker('x). Let x = Q aI eI : Then we have SIS=r 'x(ei) = ei ∧ x = Q aI (ei ∧ eI ) = Q ±aI eI∪{i}; SIS=r SIS=r; I∋~i so if ei ∧ x = 0, then aI = 0 whenever i ∈~ I, or equivalently, every non-zero term of x involves ei. Since this is true for every i with 1 ≤ i ≤ s, we have s ≤ r and we can write x = (e1 ∧ ⋯ ∧ es) ∧ y for some r−s y ∈ ⋀ V . In the case that r = s, we get that x = a1;:::;re1 ∧ ⋯ ∧ er, and x is totally decomposable. As for the second statement, in the case x = v1 ∧ ⋯ ∧ vr, we know that Span(v1; : : : ; vr) ⊆ ker('x), and we have just shown that both spaces have the same dimension. k Corollary. The Pl¨uckerembedding is an injection and its image is closed in P ⋀ V 2 Proof. Given x = v1 ∧ ⋯ ∧ vr, we've shown Span(v1; : : : ; vr) = ker('x), so we can recover the r- dimensional subspace Span(v1; : : : ; vr) from x, and the Pl¨ucker embedding is injective. To show that its image is closed, we note that since dim ker('x) ≤ r, with equality if and only if x is totally decomposable, Rank('x) ≥ n − r, with equality if and only if x is totally decomposable. r+1 If Mx is a matrix for 'x in a given basis e1; : : : en for V and the corresponding basis for ⋀ V , with x = Q aI eI ; then the entries of Mx are all 0 or ±aI . SIS=r The condition that 'x have rank at most n−r is expressed by the vanishing of all the (n−r +1)× (n − r + 1) minors of the matrix Mx. Since the entries of Mx are homogeneous linear in the Pl¨ucker coordinates aI , we find that the image of the Pl¨ucker embedding is the set of common zeros of a collection of homogeneous polynomials of degree n − r + 1 in the aI . Thus the image is closed. Example. We return to the case of G(2; 4), with x = a12(e1 ∧ e2) + a13(e1 ∧ e3) + a14(e1 ∧ e4) + a23(e2 ∧ e3) + a24(e2 ∧ e4) + a34(e3 ∧ e4): 3 In the bases (e1; e2; e3; e4) for V and (e1 ∧ e2 ∧ e3; e1 ∧ e2 ∧ e4; e1 ∧ e3 ∧ e4; e2 ∧ e3 ∧ e4) for ⋀ V , the matrix of 'x is ⎡a a a 0 ⎤ ⎢ 23 − 13 12 ⎥ ⎢ ⎥ ⎢a24 −a14 0 a12⎥ Mx = ⎢ ⎥ : ⎢a34 0 a14 a13⎥ ⎢ − ⎥ ⎢ 0 a a a ⎥ ⎣ 34 − 24 24⎦ Here, it's easy to see directly that if any entry of this matrix is non-zero, the matrix must have rank at least 2. The image of G(2; 4) under the Pl¨ucker embedding is the common zero locus of the sixteen 3 × 3 minors of this matrix, such as Ra a a R R 23 − 13 12 R Ra a 0 R a a a a a a a a a a a a a a a a ; R 24 − 14 R = 23(− 14)(− 14) − 24(− 13)(− 14) − 34(− 14) 12 = 14( 14 23 − 13 24 + 12 34) Ra 0 a R R 34 − 14R Ra a a R R 23 − 13 12 R Ra a 0 R a a a a a a a a a a a a a a a a , and R 24 − 14 R = 23(− 14)(− 24) + 12 24 34 − 24(− 13)(− 24) = 24( 14 23 + 12 34 − 13 24) R 0 a a R R 34 − 24R R a a 0 R R− 13 12 R R a 0 a R a a a a a a 0: R− 14 12R = −(− 13)( 12)(− 14) − 12(− 14) 13 = R 0 a a R R − 14 13R Indeed, four of the 3 × 3 minors are zero, and the rest are all multiples of the degree two polynomial a12a34 − a13a24 + a14a23 that we found earlier by ±aI . In general, while the equations we found for the Pl¨ucker image have degree n − r + 1, in fact in characteristic 0 the homogeneous ideal of the Pl¨ucker image is always generated by degree 2 polynomials. Local coordinates on Grassmannians Given an r × n matrix B = (bij) of rank r, the row space of B maps under the Pl¨ucker embedding to (b11e1 + ⋯ + b1nen) ∧ ⋯ ∧ (br1e1 + ⋯ + brnen) = Q aJ eJ ; SJS=r 3 where the aI are the usual Pl¨ucker coordinates. In this product, only the terms bijej with j ∈ J contribute to the term aJ eJ , and (b1j1 ej1 + ⋯ + b1jr ejr ) ∧ ⋯ ∧ (brj1 ej1 + ⋯ + brjn ejn ) = (det(bijl )1≤i;l≤r)(ej1 ∧ ⋯ ∧ ejr ); i.e. the Pl¨ucker coordinate aJ is the r × r minor of the matrix B obtained by taking all r rows and the r columns with indices in J.
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