Unless otherwise specified, V denotes an arbitrary finite-dimensional vector space.

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1 MAT 90 // 0 points Exam Solutions Unless otherwise specified, V denotes an arbitrary finite-dimensional vector space..(0) Prove: a central arrangement A in V is essential if and only if the dual projective point configuration A in P(V ) is not contained in any projective hyperplane. Solution: Let A be a central arrangement in V. Let T = A. Since the assignment W ann(w ) is an order-reversing bijection from the subspace lattice of V to that of V, and T is the meet of the set of hyperplanes A, the subspace ann(t ) is the join of the set of lines {ann(h) H A} in V, that is, ann(t ) = H A ann(h). Then T = 0 iff H A ann(h) = V. Then T = 0 iff the smallest projective subspace of P(V ) containing A is P(V ), iff A is not contained in a a projective hyperplane. SInce T = 0 iff A is essential, this proves the result..() Let V = R. Let A be the arrangement in V defined by the polynomial Q = xyz(x + y)(x y)(y + z)(y z). Write A = {H,..., H } with the labelling determined by the left-to-right order of factors in Q. (a) Sketch the projective arrangement A = {H,..., H } in RP. Here H i denotes the projective hyperplane P(H i ) in P(V ) = RP. Sketch pictures in each of the three coordinate charts, indicating the line at infinity with a large circle. Solution: x=0 y=0 z=0 (b) Sketch a Hasse diagram of the lattice of flats L(A), and for each flat in L(A), determine the associated element of the intersection lattice. Label the (projectivizations of) elements of the intersection lattice in the illustration from part (a) with their associated flats. Solution: For convenience we drop set braces and commas from the notation, writing, e.g., instead of {a, b, c}.

2 { } The flat {} corresponds to the subspace R. The flat corresponds to the subspace {0}. The correspondences between rank-one and rank-two flats and subspaces are listed below.: x = 0 y = 0 z = 0 x = y x = y y = z y = z x = 0 x = 0 x = 0 x = y x = y = 0 x = y y = z = 0 x = y x = y x = y x = y z = 0 y = z y = z z = 0 z = 0 x = z x = z x = z x = z Here are the decorated pictures from part (a). x=0 y=0 z=0

3 (c) Sketch the dual point configuration A in V = RP. Label the points according to the labelling of the corresponding hyperplanes in A. Indicate any nontrivial (projective) collinearities with line segments or curves in your illustration. Solution: Here are depictions in each of the three coordinate charts. z z x=0 y=0 y x y z=0 x (d) Find an affine chart containing all the points of A, and sketch A in this affine chart (by changing coordinates), again labelling points and indicating any nontrivial collinearities with line segments. Solution: Experimenting with the Mathematica notebook from the web page, we arrived at the matrices 0 P = and D = Diag(,,,,,, ), we get 0 A = DAP =, 0 0 where A = 0 0 0

4 is the defining matrix of A. Since the third entry in each row of A is one, we drop it and get the resulting affine point configuration representing A : which is graphed below. {(0, ), (, ), (, ), (, ), (, ), (0, ), (, 0)}, (e) Find an affine chart which intersects every hyperplane of A, and sketch the resulting affine arrangement. Solution: Since the matrix A above, projectively equivalent to A, has no row equal to (0,, 0), each hyperplane in this realization meets the affine chart y 0. The intersections of the planes of A with this chart are obtained by setting y = in the defining equations, so we plot the function z = ax b for each row (a, b, c) of A. (See the course web page for the Mathematica notebook

5 .(0) Let B be the arrangement in V = R defined by the polynomial P = z(x + y)(x y)(x + z)(x z)(y + z)(y z). (a) Repeat parts (a)-(e) of problem for the arrangement B. Solution: x=0 y=0 z=0 z z x=0 y=0 y x y z=0 x { }

6 (b) Prove there is no isomorphism of posets L(A) L(B). Solution: There is not even a bijection of sets from L(A) to L(B) because L(A) has 0 elements and L(B) has 8. Hence there is no isomorphism of posets from L(A) to L(B). (c) Prove there is no linear isomorphism of V to itself carrying the hyperplanes of A to those of B (in any order). Solution: If there were a linear automorphism of V carrying A to B, it would induce an isomorphism of posets from the lattice of intersections of A to that of B. SInce these lattices are isomorphic to the lattice of flats computed above, and those lattices are not isomorphic, there is no linear automorphism of V carrying A to B. (d) Show that R A and R B have the same number of components. Solution: Since x = 0 is a hyperplane of A, and z = 0 is a hyperplane of B, we can count chambers using the pictures above. One sees there are components of the complement of A satisfying x > 0, and, by symmetry, there are the same number satisfying x < 0, so there are components of R A. Similarly, from the picture of B on the right there are components of the complement satisfying z > 0, and the same number with z < 0. Thus both A and B have complements with connected components..(0) Recall the definition and characterization of modular element from Exercise : (i) an element X of a geometric lattice L is modular if r(x Y ) + r(x Y ) = r(x) + r(y ), for all Y L; (ii) if X or Y is an atom, then equality holds; in particular, every atom is modular; (iii) if L = L(A) is the intersection lattice of an arrangement A, then X L is modular if and only if X +Y L for all Y L. (a) Find a modular element of rank in the lattice L(A) from problem. Indicate this element in both illustrations of A and in the illustration of A from problem. Solution: The flats and are each modular and of rank two. We verify this for : The subspace corresponding to is X = H H H H, which is the subspace x = y = 0, spanned by (0, 0, ). If Y = V or 0, then X + Y = V or X, respectively, hence X + Y L. If Y is a hyperplane then X + Y = Y if X Y and X + Y = V if X Y, so X + Y L. The remaining cases can be computed using the pictures from Problem - see the table below. For instance, the flat corresponds to the subspace H H, which has defining equations x = z = 0. This line is spanned by (0,, 0), so X + Y is spanned by {(0, 0, ), (0,, 0)}, which is the plane H, with equation x = 0. Thus X + Y L. The are listed in the table below. Y X + Y H H H H H H H H H (b) Formulate an equivalent statement to statement (iii) for the lattice of flats of a projective point configuration. Solution: The flat associated with X Y is A X+Y, which equals A X A Y. Moreover, (X, Y ) is a modular pair iff codim(x + Y ) = codim(x Y ). Dually, (X, Y ) is a modular pair iff dim(ann(x + Y )) = dim(span(a X A Y )). It follows that R A and R B are homeomorphic. meaning r(x) + r(y ) = r(x Y ) + r(x Y ).

7 (c) Use the illustration of B from Problem to verify that there is no modular element of rank in the lattice L(B). Solution: By statement (i) above, if (X, Y ) is not a modular pair, then both X and Y must have rank two. If X Y then X Y has rank strictly bigger than two, hence has rank. By the submodular law, r(x Y ). Then (X, Y ) is not a modular pair iff r(x Y ) = 0, iff A X A Y =. In the table below we list for each rank-two flat X a rank-two flat Y for which A X A Y =, implying X is not modular. We reproduce the matroid diagram (based on the pictures of B above). X Y.(0) Let A be the arrangement in R defined by the polynomial Q = xyzw(x + y + z)(x + y + w), and let B be the graphic arrangement {H ij ij edge(γ)} in R associated with the graph Γ with vertex set [] and edge set edge(γ) = {,,,,, }. (a) Identify the circuits (= minimal dependent subsets) of A. Solution: Number the hyperplanes of A according to the order of factors in Q. Then the set of circuits of A is C(A) = {,, }. Note that A has rank, so every -element subset is dependent. But every five-element subset contains a four-element circuit, so none of these is minimal. (b) Find a bijection f : A B that induces a bijection from the circuits of A to the circuits of B. (Thus A and B are matroid-equivalent arrangements.) Solution: The set of circuits of B (abbreviating H ij to ij) is Define f : A B by the table below: C(B) = {{,,, }, {,,, }, {,,, }} H H H H H H H f(h) H H H H H H Then f({,,, }) = {,,, }, f({,,, }) = {,,, }, and f({,,, }) = {,,, }.

8 (c) Sketch the projective point configuration A in some affine chart on RP containing all the points. Indicate the dependent flats by decorating your illustration is some way. Solution: Pictured below is the affine diagram of A, along with a Mathematica graph of A using the coordinate change 0 P = 0. It is admittedly difficult to see the planar (i.e., rank-three) dependencies in the actual graph. x x y w x+y+w z x+y+z.() Let L be a geometric lattice, with rank function r, and let A be the set of atoms of L. { r(x) if a x; (a) Prove: for any x L and a A, r(x a) = r(x) + if a x. Solution: If a x then x a = x, so r(x a) = r(x). Suppose a x. Then x a < a, so x a = 0 L since r(a) = and 0 L is the unique element of rank zero. Also, x < x a, so r(x) < r(x a), by the lemma in the footnote. By the submodular law, r(x)+r(a) r(x a)+r(x a), so r(x)+ r(x a), since r(a) = and r(x a) = 0. Since r(x) < r(x a) r(x) +, we conclude r(x a) = r(x) +. For x L define A x by A x = {a A a x}. An atom in a ranked lattice is an element of rank one. Lemma: in a ranked lattice, if x < y then r(x) < r(y). Proof: Choose a maximal chain from 0 L to x and a maximal chain from x to y. The latter has length one or greater. The union of these chains is a maximal chain from 0 L to y, so r(y) > r(x). 8

9 A subset A of A is closed if A = A x for some x L. The closure cl(s) of an arbitrary subset S of A is defined to equal A x, where x = S. (b) Prove the following: (i) S cl(s), with equality if and only if S is closed. Solution: Let a S. Then a S = x, hence a A x = cl(s). If S is closed, then S = A y for some x L. Since a y for all a S, S y. Suppose the inequality is strict. Since L is geometric, y is a join of atoms, y = T for some T A. Since S < y, T S, hence there exists a T \ S = T \ A y, which implies a y, a contradiction. Thus S = A y with y = S, so S = cl(s). (ii) If A is a closed set and S A, then cl(s) A. Solution: By hypothesis, A = A y for some y L. Since S A, a y for all a S. Then x = S y, which implies cl(s) = A x A y = A. (iii) If A and A are closed subsets of A, then A A is closed. Solution: Write A = A x and A = A y, for x, y L. We claim A A = A x y. To see this, observe a A x A y iff a x and a y iff a x y, by definition of meet (= greatest lower bound), iff a A x y. (c) Prove: if b cl(a {a}) cl(a), then a cl(a {b}). Hint: Consider cl(a {a, b}) and use (a). Solution: Suppose b cl(a {a}) \ cl(a). Let x = A, y = x a and z = x b. Then cl(a) = A x, cl(a {a}) = A y, and cl(a {b}) = A z, by definition of closure. By hypothesis, b A y \A x, so b y and b x. Also a x since A y \ A x. Then r(z) = r(x b) = r(x) + = r(x a) = r(y), by (a)(i), and y b = y. Then r(z a) = r(y b) = r(y) = r(z). Then z a = z, so a z. Then a A z = cl(a {b})..() Let A be a central arrangement in V, and let W be a linear subspace of V. The restriction of A to W is the arrangement A W = {H W H A, H W } in the vector space W. Assume H W for all H A. (a) Show that the surjective map f : A A W defined by f(h) = H W induces an orderpreserving map ˆf : L(A) L(A W ). Give an example to show that f need not be injective. Solution: Let X Y in L(A). Then X Y. Then X W Y W, so X W Y W in L(A W ). Thus ˆf is order-preserving. Let A be the arrangement with defining equation Q = xyz with hyperplanes labelled according to the order of factors in Q, and let W be the hyperplane defined by x = y. Then each of H W and H W is equal to the line x = y = 0 in W, so ˆf is not injective in this example. (b) Prove: if A is a closed subset of A W, then f (A) is a closed subset of A. Solution: By hypothesis, A = (A W ) X, with X = A L(A W ). Then f (A) = {H A H W X}. Since X W, H W X iff H X. Thus f (A) = A X. By HW., A X = A Y where Y = A X. Thus f (A) is closed in A. This establishes the exchange axiom for matroid closure, with the consequence that every geometric lattice is the lattice of closed sets (called flats) of a matroid. This exercise shows that f is a strong map of the underlying matroid of A to that of A W. The statement holds with some modification in case H W for some H A. 9

10 (c) Describe how the dual point configurations A and (A W ) are related. Hint: Show that W can be identified with a quotient of V. Solution: Let ϕ H V with ker(ϕ H ) = H. Let ψ H = (ϕ H ) W. Then ψ H W and ker(ψ H ) = H W. Thus the dual point configuration (A W ) is equal to {ψ H H A}. (Note that ψ H and ψ H maybe equal or proportional for H H.) The dual space W is isomorphic to V / ann(w ), by the first isomorphism theorem applied to the restriction map V W. Then the assignment ϕ H ψ H can be identified with the mapping P(V ) P(V / ann(w )) induced by quotient mapping V V / ann(w ), and (A W ) is the image of A under this mapping. 8.() Let G l (k n ) be the Grassmannian of l-dimensional linear subspaces in k n. Then each P G l (k n ) is equal to the row space row(a) of an l n matrix A = [a ij ] of rank l. Let ( ) l denote the collection of subsets of of cardinality l. If J ( ) l, denote by AJ the l l submatrix [a ij ] j J of A. (a) Let N = ( ) n l, and label the coordinates of k N by the set ( ) l, ordered lexicographically. Show that the mapping P = row(a) (det(a J ) J ( ) l ) determines a well-defined function ρ: G l (k n ) P N. Solution: Firstly, if since row(a) = P has dimension l by hypthesis, some l l minor det(a J ) is nonzero, hence (det(a J ) J ( ) l ) determines a point of P n. By linear algebra, if A and B are l n matrices with row(a) = row(b), then there is an invertible l l matrix P such that B = P A. Then, for each J ( ) l, BJ = P A J, and hence det(b J ) = det(p ) det(a J ). Since det(p ) is a nonzero constant, the vectors (det(a J ) J ( ) l ) and (det(bj ) J ( ) l ) in k N represent the same point of P N. (b) By considering reduced row-echelon forms, prove that the function ρ is injective. Solution: For every P G l (k n ) there is a unique reduced row-echelon form matrix A of rank l such that row(a) = P ; that is, P is uniquely determined by A. Suppose A is such a matrix, and let J 0 be the set of pivot columns of A. Fix i and j and let i 0 J 0 index the pivot column with nonzero entry in row i. Consider the submatrix A K indexed by K = J 0 {i 0 } {j}. The determinant of A K is ±a ij ; more precisely, det(a K ) = ( ) t a ij where t = 0 if there are no pivot columns j 0 between i 0 and j, and t = otherwise. Then the Plücker coordinates of the reduced row-echelon form A uniquely determine A and hence P, thus ρ is injective. This is the Plücker embedding; the entries of ρ(p ) are called the Plücker coordinates of P. 0