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3 votes
0 answers
42 views

Exercise 2.2 In Stanley's Algebraic Combinatorics

This is Exercise $2.2$ In Stanley's Algebraic Combinatorics. I don't have much work to show because despite being stuck on this problem for a long time, I haven't got a clue how to start. $\mathcal{C}...
koifish's user avatar
  • 3,100
0 votes
0 answers
23 views

Affiness, $U_{2,4}$ and $M(K_4).$

I do not know why $M(K_4)$ is not affine over $GF(2)$ or $GF(3)$ but it is affine over all fields with more than 3 elements. I proved that $U_{2,4}$ is $\mathbb F$-representable iff $|\mathbb F| \geq ...
Intuition's user avatar
  • 3,139
0 votes
0 answers
20 views

Lift and frame matroids.

I want to read more about lift matroid and frame matroid and their flats and relations to signed graphs, do you know any basic resources for this?
Emptymind's user avatar
  • 2,087
0 votes
0 answers
90 views

Hyperplane Areangements and contraction.

I am trying to understand an idea presented in McNulty book, matriods a geometric introduction about the new hyperplane arrangement $\mathcal{A}^{''} = \{ H \cap H_x | H \in \mathcal{A}\}$ where $\...
Intuition's user avatar
  • 3,139
0 votes
1 answer
46 views

The basis of a regular matroid.

I know that a regular matroid is one that can be represented by a totally unimodular matrix. I also know that a rank r totally unimodular matrix is a matrix over $\mathbb R$ for which every submatrix ...
Intuition's user avatar
  • 3,139
1 vote
1 answer
56 views

affine geometries that are self-dual matroids.

I want to know which of the affine geometries $AG(n,q)$ are self-dual matroids? I have proved before that uniform matroids $U_{n,m}$ that are self duals are those who satisfies that $m = 2n.$ I am ...
Intuition's user avatar
  • 3,139
3 votes
1 answer
101 views

Understanding how to find the dual of a matroid.

I am trying to understand the following picture of a matroid and its dual but I found myself not understanding exactly what we are doing to find the dual: ]1 Roughly speaking, according to some ...
Hope's user avatar
  • 95
2 votes
1 answer
94 views

Show that $C^*(e,B)$ is the unique cocircuit that is disjoint from $B - e.$

I want to prove the following question: Let $B$ be a basis of a matroid $M.$ If $e \in B,$ denote $C_{M^*}(e, E(M) - B)$ by $C^*(e,B)$ and call it the fundamental cocircuit of $e$ with respect to $B.$\...
Intuition's user avatar
  • 3,139
0 votes
1 answer
68 views

if $X$ is independent and $E(M) - X$ is coindependent, show that $X$ is a basis and $E(M) - X$ is a cobasis.

here is the question I am trying to solve: In a matroid $M,$ if $X$ is independent and $E(M) - X$ is coindependent, show that $X$ is a basis and $E(M) - X$ is a cobasis. I know how to prove that a set ...
Intuition's user avatar
  • 3,139
0 votes
1 answer
36 views

Let $A\subset B$ be flats of a matroid $M$ such that $r(A)=r(B)<\infty$. Then $A = B$.

I want to prove the following lemma: Let $r$ denotes the rank. Lemma. Let $A\subset B$ be any flats of a matroid $M$ such that $r(A)=r(B)<\infty$. Then $A = B$. My thoughts are: I know that $cl(A) =...
Intuition's user avatar
  • 3,139
0 votes
1 answer
82 views

Proving that $ \beta(M) = \beta (M - e) + \beta (M /e).$

Here is the statement I am trying to prove: If $e \in E$ is neither a loop nor an isthmus, then $$ \beta(M) = \beta (M - e) + \beta (M /e).$$ Here are all the properties I know about the Crapo's beta ...
Intuition's user avatar
  • 3,139
0 votes
1 answer
71 views

Why always the Crapo beta invariant value greater than or equal zero?

Here are the definitions of the Crapo beta invariant I know: My definition of the Crapo's beta invariant of a matroid from the book "Combinatorial Geometries" from page 123 and 124 is as ...
Intuition's user avatar
  • 3,139
2 votes
1 answer
69 views

Sum of squares of chromatic roots of a bipartite graph

Given a graph $G = (V, E)$, we can calculate its chromatic polynomial $P(G, k)$, and it has $n$ (complex) roots, also known as chromatic roots. It is a well-known fact that the sum of chromatic roots ...
hedgehog0's user avatar
2 votes
0 answers
47 views

Can we characterize the “associate classes” of a unipotent quasi-commutative quasigroup as some combinatorial design?

$I_n$ is the $n \times n$ or order $n$ identity matrix, $J_n$ is the order $n$ matrix of all ones, and $n \in \mathbb{Z}^+$. We define a Latin square $\mathcal{L_n}$ to be a set of $n$ permutation ...
Naiim's user avatar
  • 317
2 votes
2 answers
90 views

In a primitive symmetric association scheme, why does $E_j$ occur in some power of $E_i$ for each $i,j$?

I am having some trouble in the proof of the Absolute Bound Condition for primitive symmetric association Schemes in the book Algebraic Combinatorics I by Bannai and Ito (Chapter 2, Section 4, Theorem ...
F.Tomas's user avatar
  • 508

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