Problem Set MATH 4550: Applicable Geometry. Due at 2:54pm before class starts on March 7, 2013

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1 MATH 4550: Applicable Geometry Problem Set 6 Due at 2:54pm before class starts on March 7, 203 You are allowed to work in groups, but the solutions you hand in should be written by you only. If you work in a group, you must write the names of your collaborators at the top of your assignment. Explain your reasoning to receive full credit. All problems are worth 0 points. You are strongly encouraged to type your solutions in LaTeX. In any case, please staple your psets! P We constructed d triangulations of the d-crosspolytope in class. Are these triangulations regular? Justify, preferrably without explicitly referring to the definition of regular.... P2 The next 2 pages are solutions to the first 4 psets given by a set of anonymous persons. The solutions are generally correct, though there may be omissions, redundancies, and perhaps even errors. Your assignment is to read and understand all of the solutions, and on a printed copy of them mark any omissions, redundancies, errors you find. If you have a much shorter/more elegant solution to any of the problems you should also make that available. -

2 P Problem Set Let {S i : i 2 [n]} be a finite set of convex sets, and let S = T i2[n] S i be the intersection of all the convex subsets. Now let s look at some x, y 2 S. By the definition of set intersection, we can say that 8i 2 [n] :x, y 2 S i. And by the definition of a convex set this means that 8i 2 [n], 8t 2 [0, ] : tx +( t)y 2 S i. Because this is true for all i, we know that these points must also be in the intersection of all S i, so we can say that 8x, y 2 S and 8t 2 [0, ] : tx +( t)y 2 S. Therefore, S is a convex set. The definition of the convex hull conv({p,...,p t }) is the set of all the convex combinations of p,...,p t. We define the convex hull as 8 9 < X C = : ip i : i 0, X = i = ; i2[n] i2[n] Let s look at two di erent points in C, P i2[n] ip i and P i2[n] ip i where i and i are subject to the constraints for i in the definition of C above. To check whether C is convex, consider 0 0 X i p i A +( X i2[n] i2[n] ip i A = X p i (t i +( t) i ) i2[n] for t 2 [0, ]. Note that since i 0 and i 0, and X t i +( i2[n] So P i2[n] p i(t i +( 0 t i +( t) i 0 8i 0 t) i = X i A +( X i2[n] i2[n] i A = t() + ( t) =. t) i ) 2 conv({p,...,p t }). Therefore, C is convex. Let S denote the intersection of all convex sets containing p,...,p t. Since we just proved that C = conv({p,...,p t }) is convex, we must have S C. To show that C S we will use induction on t. The base case is trivial. Let s assume that convex combinations of n of p i s are in S. Let s prove that convex combinations of n of p i s must also be in S. W.L.O.G., let y = a p a n p n + a n p n. If one of the a i = 0, then we are done by inductive hypothesis. Otherwise, we re-write y =( a n ) a n (a p a n p n ) + a n p n. Since the P n i= a i =, we know that a n = a a n, so the equation for y reduces to y =( a n )x + a n p n where x is a convex combination of n of the p i s. This now reduces to the base case where n = 2. P2 C d = (x,...,x d ) 2 R d : x i 0 x i apple,i2 [d]

3 This is the intersection of 2d halfspaces. P3 Let {A i : i 2 [n]} be a set of a ne subspaces in R d, and let A = T i2[n] A i be the intersection of those a ne subspaces. If A = ;, we are done. Otherwise, 9a 2 A. So there exist linear subspaces W i such that A i = W i + a. From linear algebra, we know that W = T i2[n] W i is a linear subspace and so A = W + a is an a ne subspace of R d. P4 =): In this direction, we know that x,...,x n are a nely independent and we will show that this means that ˆx,...,ˆx n are linearly independent in R d+. We showed in class that an a ne subspace can be translated to its corresponding linear subspace by translating times a vector in that a ne subspace. For this proof, we will choose to translate the a ne subspace created by x,...,x n by x, so we get the corresponding vectors in the linear subspace (x x ), (x 2 x ),...,(x n x ). We know in this direction that x,...,x n are a nely independent so we know that the corresponding linear subspace has dimension n. We can see this because the first vector in the linear subspace as listed above is the zero vector. If we drop that, then we have n vectors that we know must be linearly independent. If we write the equation for linearity c 2 (x 2 x ) c n (x n x )=0 c 2 x c n x n (c c n )x =0 c x + c 2 x c n x n =0 Where c = (c c n ). But we know that because the vectors (x 2 x ),... are linearly independent, the only values that work for c i are c i = 0. If we look at that last line, another way to put the constraints of linear independence and on c is to write them as a set of linearly independent vectors: apple apple ˆx =,...,ˆx x n = x n And we re done with this direction. (=: In this direction, we know that ˆx,...,ˆx n are linearly independent in R d+ and we will show that x,...,x n are a nely independent. If ˆx,...,ˆx n are linearly independent, then for the equation c ˆx c nˆx n =0 the only values of c,...,c n that will work are 0. Breaking the vectors into parts we get two equations: c + c c n =0 c x + c 2 x c n x n = ~0 2

4 Solving for c gets us: (c c n )=c c 2 x c n x n (c c n )x = ~0 c 2 (x 2 x ) c n (x n x )=~0 So from this we can see that we have a set of n linearly independent vectors (x 2 x ),...,(x n x ). These vectors form an n dimensional space. If we translate the vectors by x we get n vectors (x,x 2,...,x n ) that define an a ne space with dimension n, so they are a nely independent. 3

5 Applicable Geometry Problem Set 2 7February203 P. P = conv ({(0, 0), (, 0), (0, )}). Geometrically, we can describe P as a triangle in R 2. Let us consider its faces F by dimension, using the definition F = P {x R cx = c 0 } where cx c 0 is a valid inequality for P. Elements y P must be such that a 0 (0, 0) + a (, 0) + a 2 (0, ) = (a,a 2 ), such that a 0 + a + a 2 =anda 0,a,a 2 are non-negative. Dimension -: The inequality 0x is a valid inequality for P (as it is for any x R 2, or in any vector space) so P {x 0x =} is a face; the image of 0x ==Ø,thusP Ø = Ø is a face. Dimension 0: These faces are the three vertices: Consider the inequality (, ) T x 0. An element in P, y =(a,a 2 )must have non-negative coordinates. Thus, (, ) T x 0 is a valid inequality; (, ) T y = a{ a 2 = (a + a 2 ), which } must be non-positive, and zero when a 0 =. P x R (, ) T x =0, then, is the set of (x,x 2 ) R 2 such that x,x 2 = a,a 2 as above and such that x 2 = x, thus, the only possible solution is that x = x 2 =0. Thus,(0, 0) is a face. Consider the inequality (, ) T x. As above, an element in P, y = (a,a 2 ), so (, ) T y = a 2 a, which is at most when a 2 =;a 0,a =0, and at least whena =;a 0,a 2 =0. Thus,(, ) T x is a valid inequality for P. P {x 2 x =} as we just showed consists of the point that occurs only when x 2 =andx =0,thus(0, ) is a face. Consider the inequality (, ) T x. (, ) T y = a a 2 is at most, when a =;a 0,a 2 = 0, and at least, when a 2 =;a 0,a =0. Thus, (, ) T x is a valid inequality for P. P {x x 2 =} consists of the point that occurs only when x =,andx 2 =0,thus,(, 0) is a face. Dimension : These faces are the three edges: Consider the inequality (0, ) T x 0. (0, ) T y = a 2, which is at most 0 and at least. Thus, this is a valid inequality for P. P { x 2 =0} thus restricts y P to where a 2 =0;thus,a 0 + a =, and so is the set of points y =(a, 0) such that a [0, ]. Thus, the line segment from (0, 0) to (, 0) is a face. Consider the inequality (, 0) T x 0. (, 0) T y = a, which, similarly to above, is at most 0 and at least. Thus, this is a valid inequality for P. P { x =0} restricts y P to where a = 0, thus is the set of points y =(0,a 2 )suchthata 2 [0, ]. Thus, the line segment from (0, 0) to (0, ) is a face. Consider the inequality (, ) T x. (, ) T y = a + a 2, which is at most whena 0 = 0 and at least 0 when a 0 =. Thus, this is a valid inequality for P. P {x + x 2 =} restricts y P to where a 0 =0,hencea + a 2 = and this is the set of points y =(a, a ), such that a [0, ]. Therefore, the line segment from (0, ) to (, 0) is a face. Dimension 2: The inequality 0x 0 is a valid inequality for P (as it is for any x R 2, or in any vector space) so P {x 0x =0} is a face; the image of 0x =0 is all of R 2 and we know P R 2,thusP {x 0x =0} = P is a face.

6 2 P2. Please see the attached drawings for visual representations of the face posets as Hasse diagrams. Drawn carefully such that the faces of the octahedron correspond to the dual cube inside it (and vice-versa), if we turn the one poset graph upside down (in other words, Ø P or Q, v f, and vice-versa) we obtain exactly the poset graph of the other. This thus defines a one-to-one correspondence between the cube and the octahedron, and therefore their face posets are isomorphic. P3. (Please see the attached drawing.) The chains are the totally ordered subsets of the face poset. The maximally ordered chains are those of length 3 (i.e. they have 4 elements) in this poset; they all consist of {Ø,v i,e j,r} for some i {, 2, 3, 4} and some j such that v i e j. If we were to add another element from the poset, it would have to be either another vertex or edge, and they are not comparable with the vertex or edge already present (respectively) and thus we do not have total order. Because all of the maximally ordered chains are the same length, this poset is graded. The maximally ordered chains of this poset are: {Ø,v,e,R}, {Ø,v,e 2,R}, {Ø,v 2,e,R}, {Ø,v 2,e 3,R}, {Ø,v 3,e 2,R}, {Ø,v 3,e 4,R}, {Ø,v 4,e 3,R}, {Ø,v 4,e 4,R}. The chains of length 2 are the subsets of these deleting exactly one element: {Ø,v,e },{Ø,v,e 2 }, {Ø,v 2,e }, {Ø,v 2,e 3 }, {Ø,v 3,e 2 }, {Ø,v 3,e 4 }, {Ø,v 4,e 3 }, {Ø,v 4,e 4 },{v,e,r}, {v,e 2,R}, {v 2,e,R}, {v 2,e 3,R}, {v 3,e 2,R}, {v 3,e 4,R}, {v 4,e 3,R}, {v 4,e 4,R},{Ø,v,R}, {Ø,v 2,R}, {Ø,v 3,R}, {Ø,v 4,R}, {Ø,e,R}, {Ø,e 2,R}, {Ø,e 3,R}, {Ø,e 4,R}. Those subsets with jumps (i.e. a vertex or edge deleted from the superset) are still totally ordered because of transitivity: v i R, for example. The chains of length are the subsets of these deleting exactly one element: {Ø,v }, {Ø,v 2 }, {Ø,v 3 }, {Ø,v 4 }, {Ø,e }, {Ø,e 2 }, {Ø,e 3 }, {Ø,e 4 }, {Ø,R}, {v,r}, {v 2,R}, {v 3,R}, {v 4,R}, {e,r}, {e 2,R}, {e 3,R}, {e 4,R}, {v,e }, {v,e 2 }, {v 2,e }, {v 2,e 3 }, {v 3,e 2 }, {v 3,e 4 }, {v 4,e 3 }, {v 4,e 4 }. The chains of length 0 are the sets containing exactly one element from the poset (by reflexivity, x x for all elements x, so these are indeed totally ordered and thus chains.) This poset is bounded because there is a maximal element, R, such that no element x is such that R x (and all elements y are such that y R) and a similarly minimal element, Ø. This poset is a lattice: every pair of elements has a unique minimal upper bound and a unique maximal lower bound. The minimal upper bound of any element with R or any two edges e j,e j2 is R, the maximal lower bound of any element with Ø or any two vertices v i,v i2 is Ø. The maximal lower bound of any vertex and edge v i,e j is v i, and the minimal upper bound is e j. The minimal upper bounds for: v,v 2 is e, v,v 3 is e 2, v v 4 is R, v 2,v 3 is R, and v 3,v 4 is e 4. The maximal lower bounds: e,e 2 is v, e,e 3 is v 2, e,e 4 is Ø, e 2,e 3 is Ø, e 2,e 4 is v 3,ande 3,e 4 is v 4.

10 P d P 4 P 4 d P > 0 (, 0, 0,...,0) 2 R d (, 0, 0,...,0) 2 R d P a =(a,...,a d ) P 4 a (, 0, 0,...,0) apple, a (, 0, 0,...,0) apple, apple a apple. P 4 2 > 0 (0, 2, 0,...,0) 2 R d (0, 2, 0,...,0) 2 R d P a (0, 2, 0,...,0) apple, a (0, 2, 0,...,0) apple, 2 apple a 2 apple 2. P 4 (0, 0, 3, 0,...0) (0, 0, 3, 0,...0) (0, 0, 0, 4,...,0) (0, 0, 0, 4,...,0)..., (0, 0, 0, 0,..., d ) (0, 0, 0, 0,..., d ) P 3 > 0 4 > 0,..., d > 0 3 apple a 3 apple 3, 4 apple a 4 apple 4,... d apple a d apple d. P 4 r = max n,..., d o r p d P 4 P 4 P 4 R d P = conv({v,...,v n })

11 v,...,v n P 4 = {a a v apple 8v 2{v,...,v n }} {v,...,v n } P 4 P 4 (, 0), (, 0), (, ), (, ) P 4 = {a a v apple 8v 2{(, 0), (, 0), (, ), (, )}} a =(a,a 2 ) P 4 a + a 2 apple a apple,a, a + a 2 apple a = a = a 2 P 4 x z y y apple z apple x P d P 4 P P P 4 P P 4 P P 4 p q P p apple q p p 0 q q 0 p 0,q 0 P 4 q 0 p 0 q 0 apple p 0 x y y x y apple x P P d P P 4 d P v P f,...,f d P v v apple f,v apple f 2,...,v apple f d P v f 0 f 0 P 4 f i i [,d] vi 0 vi 0

12 P 4 P 4 v 0 apple f 0,v 2 apple f 0..., vd 0 apple f 0 f 0 v d + f 0 d v P P 4 P 4 P 4 d P d P f 0 P 4 v 0,...,v0 d P 4 f 0 v 0 apple f 0,v 0 2 apple f 0,...,v 0 d apple f 0 v P v f 0 f,...,f d P v 0,...,v0 d v apple f,v apple f 2,...,v apple f d v f 0 d P 4 P P d P

13 C d (n) R d d + d d d C d (n) t <...<t n C 3 (t,t 2,t 3,t 4,t 5 ) {t,t 2,t 3,t 4,t 5 } {t,t 2,t 3 }, {t,t 2,t 5 }, {t,t 3,t 4 }, {t,t 4,t 5 }, {t 2,t 3,t 5 }, {t 3,t 4,t 5 } {t,t 2 }, {t,t 3 }, {t,t 4 }, {t,t 5 }, {t 2,t 3 }, {t 2,t 5 }, {t 3,t 4 }, {t 3,t 5 }, {t 4,t 5 } t,t 2,t 3,t 4,t 5 ;

Do not open this exam until you are told to begin. You will have 75 minutes for the exam.

Math 2603 Midterm 1 Spring 2018 Your Name Student ID # Section Do not open this exam until you are told to begin. You will have 75 minutes for the exam. Check that you have a complete exam. There are 5