This question is motivated by trying to establish a converse to Theorem 7.8 of our [paper](http://arxiv.org/abs/1508.05446).


I have a finite poset $P$ with the following properties:


1. $P$ has binary meets (and hence a least element).
2. $P$ is ranked (or graded). This means every maximal chain from the least element of $P$ to a given element has the same length.
3. $P$ is [strongly connected](https://en.wikipedia.org/wiki/Abstract_polytope#Connectedness): given $x<y$ in $P$, one can transform any maximal chain from $x$ to $y$ to any other by successively changing one element of the chain to a different element (that is, the order complex of the interval $[x,y]$ is a chamber complex).


Fix a field $K$ and let $Q$ be the Hasse diagram of $P$, viewed as a quiver (=digraph) by orienting each edge upward in the partial order. The [incidence algebra](https://en.wikipedia.org/wiki/Incidence_algebra) $I(P,K)$ of $P$ over $K$ is quotient of the [path algebra](https://en.wikipedia.org/wiki/Quiver_%28mathematics%29) $KQ$ of $Q$ by the ideal generated by differences $p-q$ of parallel paths in $Q$. Under assumptions 1-3 above, it is not hard to show that $I(P,K)$ is the quotient of $KQ$ by the ideal generated by all differences $p-q$ of parallel paths in $Q$ of length $2$. See [Lemma 7.4](http://arxiv.org/pdf/1508.05446.pdf).


A ranked poset is called *thin* if each interval of length $2$ is a diamond (has 4 elements). If our poset $P$ satisfies 1-3 above and is thin, then $I(P,K)$ is the quotient of $KQ$ by the ideal spanned generated by differences $p-q$ where $p,q$ are the two sides of a diamond.


Now for the question:



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> > **Question.** If $P$ is a thin poset satisfying 1-3 above and $K$ is a field, is the incidence algebra $I(P,K)$ of $P$ isomorphic to $KQ/I$ where $Q$ is the Hasse diagram of $P$ and $I$ is the ideal generated by all sums $p+q$ with $p,q$ parallel paths of length $2$, that is, $p$ and $q$ are the two sides of a diamond?
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Of course, if the characteristic of $K$ is $2$, this is the case. 


**My idea.** Given any subset $X$ of edges of $Q$, there is an automorphism of $KQ$ sending each edge of $X$ to its negative. My thought is one should inductively be able to choose an edge from each diamond to make negative in order to change the relation $p+q=0$ to $p-q=0$. 



Is this algebra isomorphic to an incidence algebra?
Here is a much better exposition. Let $X$ be any connected compact manifold and let $\Delta$ be a simplicial complex realizing $X$. Let $P$ be the face lattice $\Delta$, with added minimal and maximal elements $\hat{0}$ and $\hat{1}$. (I suspect we can replace "simplicial complex" with "regular CW complex", but I don't want to think carefully enough.) Then $P$ is a thin, graded, strongly connected (the OP says this is in Browns book), lattice. I claim that the isomorphism of algebras holds if and only if $X$ is orientable.


As discussed above, an isomorphism of the algebras is equivalent to a choice of weights $\epsilon(a)$ for each arrow $a$ of $P$ satisfying $\epsilon(x \to y\_1) \epsilon(y\_1 \to z) = - \epsilon(x \to y\_2) \epsilon(y\_2 \to z)$ for each diamond $x < y\_1, y\_2 < z$. We adopt the convention that, for any path $\gamma$ through the quiver, $\epsilon(\gamma)$ means $\prod\_{a \subset \gamma} \epsilon(a)$. If $\gamma\_1$ and $\gamma\_2$ are two paths with the same endpoints, then $\epsilon(\gamma\_1) = \pm \epsilon(\gamma\_2)$.


In particular, there are two values in $K^{\ast}$, negatives of each other, such that $\epsilon(\gamma)$ takes one of these values for any path $\hat{0} \to \hat{1}$. We choose one of them to call $u$ and one to call $-u$


A path $\hat{0} \to \hat{1}$ looks like $0$, $\{ v\_0 \}$, $\{ v\_0, v\_1 \}$, $\{v\_0, v\_1, v\_2 \}$, ... , $\{ v\_0, v\_1, \ldots, v\_n \}$, $\hat{1}$ for some maximal simplex $\{ v\_0, \ldots, v\_n \}$ of $\Delta$. So, given an ordering $(v\_0, \ldots, v\_n)$ of the vertices of some simplex of $\Delta$, we get a sign.


Using diamonds at heights other then the top, we see that this gives a chosen orientation of each simplex of $\Delta$. Then the diamonds at the top show that the orientation on adjacent simplices is compatible.


So such a set of weights exists if and only $X$ is orientable.


