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I studying the following proposition and have a question about the very last part of its proof:

$\textbf{Proposition}$ The set $\mathcal{T}$ of topes of an oriented matroid $\mathcal{M}=(E,\mathcal{F})$ determines the set of covectors by $$ \mathcal{F}=\{X\in\{-,+,0\}^E|X\circ T\in\mathcal{T},\forall T\in\mathcal{T}\} $$

The statement can be found in the following phd-thesis on page 42, proposition 0.7.3. Let me first introduce some notation. A matroid $M$ is a pair $(E,\mathcal{A})$ of a finite set $E$ and a set $\mathcal{A}\subset 2^E$ s.t.

  1. $E\in\mathcal{A}$
  2. If $X,Y\in\mathcal{A}$ then $X\cap Y\in\mathcal{A}$
  3. For all $X,Y\in\mathcal{A},e\in E\backslash (X\cup Y)$, and $f\in X\backslash Y$ there exists $Z\in\mathcal{A}$ such that $e\in Z,f\notin Z$, and $X\cap Y\subset Z$.

An oriented matroid $\mathcal{M}$ is a pair $(E,\mathcal{F})$ of a finite set $E$ and a set $\mathcal{F}\subset\{-,+,0\}^E$ of sign vectors or covectors with some properties (can be found on page 21.). For a given oriented matroid $\mathcal{M}=(E,\mathcal{F})$ we can associate a matroid by $M=(E,\{X^0|X\in\mathcal{F}\}$, where $X^0=\{e\in E|X_e=0\}$. The composition of $X$ and $Y$, denoted by $X\circ Y$ is given by $(X\circ Y)_e=X_e$ if $X_e\not=0$ and $Y_e$ otherwise. Furthermore we write $X\le Y$ if $X_e\not= 0$ implies $X_e=Y_e$ and $X<Y$ if $X\le Y$ and $X\not= Y$. And last we set $D(X,Y):=\{e\in E|X_e=-Y_e\not= 0\}$. All these things can be found on page 21.

Now my question about the proposition above. The hard direction is $"\supset "$. They say: We will prove $X\circ Y\in\mathcal{F}$ for all $Y\in \mathcal{F}$; the proof is by induction on $|(X\circ Y)^0|$, and the claim finally will follow for $|(X\circ Y)^0|=|X^0|$, since then $X\circ Y= X\in\mathcal{F}$.

$\textbf{my first question:}$ Assuming $|(X\circ Y)^0|=|X^0|$ it is clear that $X\circ Y=X$. But why can we always find a $Y\in\mathcal{F}$ such that $|(X\circ Y)^0|=|X^0|$. Where is this shown in the proof?

$\textbf{my second question:}$ At the very end of the prove, why does the assumption on $Z$ implies that $X\circ Y= X\circ W$?

Any help would be appreciated! Thank you in advance.

math
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1 Answers1

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For question 1 note that the all zero sign vector $0$ belongs to $\mathcal F$ and $X\circ 0=X$. So take $Y=0$.

For question 2, he is using the assumption there is no $Z'$ in $\mathcal F$ with $X\circ Y\prec X\circ Z'\prec X\circ Z$ (take $Z'=W$).

Benjamin Steinberg
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  • Thank for your help. There is another question about the thesis. Maybe you can also answer this one, since it seems that you are familiar with the topic. I'm very thankful for your help. If you are interested, here is the link for the other question: http://math.stackexchange.com/questions/523482/why-is-it-true-operatornamerank-underline-mathcalme0-0 – math Oct 12 '13 at 12:27