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The Grassmann Manifold

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The Grassmann Manifold The Grassmann Manifold 1. For vector spaces V and W denote by L(V; W ) the vector space of linear maps from V to W . Thus L(Rk; Rn) may be identified with the space Rk£n of k £ n matrices. An injective linear map u : Rk ! V is called a k-frame in V . The set k n GFk;n = fu 2 L(R ; R ): rank(u) = kg of k-frames in Rn is called the Stiefel manifold. Note that the special case k = n is the general linear group: k k GLk = fa 2 L(R ; R ) : det(a) 6= 0g: The set of all k-dimensional (vector) subspaces ¸ ½ Rn is called the Grassmann n manifold of k-planes in R and denoted by GRk;n or sometimes GRk;n(R) or n GRk(R ). Let k ¼ : GFk;n ! GRk;n; ¼(u) = u(R ) denote the map which assigns to each k-frame u the subspace u(Rk) it spans. ¡1 For ¸ 2 GRk;n the fiber (preimage) ¼ (¸) consists of those k-frames which form a basis for the subspace ¸, i.e. for any u 2 ¼¡1(¸) we have ¡1 ¼ (¸) = fu ± a : a 2 GLkg: Hence we can (and will) view GRk;n as the orbit space of the group action GFk;n £ GLk ! GFk;n :(u; a) 7! u ± a: The exercises below will prove the following n£k Theorem 2. The Stiefel manifold GFk;n is an open subset of the set R of all n £ k matrices. There is a unique differentiable structure on the Grassmann manifold GRk;n such that the map ¼ is a submersion. n£k 3. Show that GFk;n is an open subset of R . Hint: For any k-element subset I = fi1 < i2 < ¢ ¢ ¢ < ikg ½ f1; 2; : : : ; ng; n k define a linear projection pI : R ! R by pI (x1; x2; : : : ; xn) = (xi1 ; xi2 ; : : : ; xik ): For u 2 L(Rm; Rn) the matrix m k uI = pI ± u 2 L(R ; R ) is obtained from u be deleting the columns not corresponding to elements of I. Let n£k VI = fu 2 R : det(uI ) 6= 0g: ¡n¢ The k sets VI cover GFk;n. 1 4. For any subset I ½ f1; 2; : : : ; ng of indices let I0 = f1; 2; : : : ; ng n I denote its complement. Let UI = ¼(VI ). Show that ¡1 VI = ¼ (UI ); that the matrix ¡1 k£(n¡k) ÁI (¸) = uI0 uI 2 R is independent of the choice of u 2 ¼¡1(¸) used to define it, and that the map k£(n¡k) ÁI : UI ! R is a bijection. The inverse is defined by ¡1 n ÁI (A) = fx 2 R : xI0 = AxI g where we have written xI = ¼I (x) and xI0 = ¼I0 (x). ¡ ¢ n 1 5. Show that the k charts (ÁI ;UI ) form a C atlas. Hint: The condition A 2 ÁI (UI \ UJ ) holds if and only if a certain submatrix of A is invertible. 6. Show that the map ¼ : GFk;n ! GRk;n is locally trivial. More precisely, show that for each k-element subset I there is a unique smooth map σI : UI ! VI such that ¼ ± σI is the identity map of Ui and σI (UI ) = fu 2 GFk;n : uI = 1kg: Here 1k denotes the k £ k identity matrix.) Then show that the map ΦI : UI £ GLk ! VI defined by ΦI (¸; a) = σI (¸) ± a is a diffeomorphism satisfying ¼ ± ΦI (¸; a) = ¸. 7. Suppose E, M1, M2 are smooth manifolds and that pi : E ! Mi are surjec- tive submersions. Suppose that f : M1 ! M2 is a bijection such that f ±p1 = p2. Show that f is a diffeomorphism. n 8. Suppose that ¸0 is a k dimensional subspace of R and let ¹0 be any (vector n space) complement to ¸0 in R i.e., n R = ¸0 © ¹0: For A 2 L(¸0; ¹0) let Graph(A) = fx + Ax : x 2 ¸0g: n and let ¼0 : R ! ¸0 be the projection along ¹0. Show that the following conditions on k dimensional subspace ¸ of Rn are equivalent: (a) ¸ = Graph(A) for some A 2 L(¸0; ¹0); (b) ¸ is transverse to ¹0 (meaning that ¸ \ ¹0 = f0g); (c) ¼0j¸ is an isomorphism. 2 n 9. Let R = ¸0 © ¹0 be a splitting as in paragraph 8 and let U be the set of all ¸ 2 GRk;n satisfying the equivalent conditions (a), (b), and (c). Define Á : U ! L(¸0; ¹0) by Á¡1(A) = Graph(A): Prove that U is an open dense subset of GRk;n and that Á is a diffeomorphism. 10. The manifold GRk;n is an example of a homogeneous space. Prove this by showing: (1) For each a 2 GLn the induced map GRk;n ! GRk;n : ¸ 7! a(¸) is a diffeomorphism. (2) The map GLn £ GRk;n ! GRk;n :(a; ¸) 7! a(¸) is smooth. 1 (3) For each ¸0 2 GRk;n the map GLn ! GRk;n : a ! a(¸0) is a C surjective map. k Remark 11. If we take ¸0 = R £ 0 in the map a ! a(¸0) the fibers of this map (i.e., inverse images of points) are left cosets a ¢ GLn;k where GLn;k is the k k subgroup of GLn consisting of all a 2 GLn such that a(R £ 0) = R £ 0. Thus we have a bijective correspondence between GRk;n and the space of left cosets GLn=GLn;k: GRk;n ' GLn=GLn;k: (Warning: GLn;k is not a normal subgroup of GLn. Thus GRk;n is not a group in any natural fashion.) ? ? 12. Prove that the map GRk;n ! GRn¡k;n : ¸ ! ¸ (where ¸ denotes the orthogonal complement to ¸ with respect to the usual inner product on Rn) is a C1 diffeomorphism. 3.
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