Is this algebra isomorphic to an incidence algebra? This question is motivated by trying to establish a converse to Theorem 7.8 of our paper.
I have a finite poset $P$ with the following properties:


*

*$P$ has binary meets (and hence a least element).

*$P$ is ranked (or graded). This means every maximal chain from the least element of $P$ to a given element has the same length.

*$P$ is strongly connected: 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 $I(P,K)$ of $P$ over $K$ is quotient of the path algebra $KQ$ 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.
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:


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?


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 $p-q=0$. 
 A: Sorry about the long delay. The answer is no. Take a regular CW subdivision of a connected non-orientable surface $X$, such that the intersection of any two faces is a face of both. For example, $\mathbb{RP}^2$ can be triangulated with $6$ vertices, $15$ edges and $10$ triangles by taking an icosahedron and quotienting by the antipodal map. Let $P$ be the poset of faces of this subdivision, plus an additional minimal element $\hat{0}$ and maximal element $\hat{1}$. I claim that $P$ has all required properties, but the quotients of $KP$ by $\langle p-q \rangle$ and $\langle p+q \rangle$ (with $p$ and $q$ the two sides of a diamond) are not isomorphic when $K$ is any field of characteristic not $2$.
$P$ is a lattice: The intersection of any two faces is another face, or empty. The corresponding meet is that intersection, or $\hat{0}$. And, of course, the meet of any $x$ with $\hat{1}$ is $x$.
$P$ is graded: By dimension of the face.
$P$ is strongly connected (sketch): Given any two chains $(v,e,f)$ and $(v',e',f')$ of vertex in edge in face, we want to show that we can change one to the other by changing one element of the chain at a time. Choose a path $v_0$, $v_1$, ..., $v_N$ from $v$ to $v'$, and let $e_i$ be the edge between $v_i$ and $v_{i+1}$. We can swivel $(v,e,f)$ around the vertex $v=v_0$ to be $(v_0, e_0, f_0)$ for some $f_0$ containing $e_0$. Then we can switch to $(v_1, e_0, f_0)$. Then we can swivel around to $(v_1, e_1, f_1)$ for some $f_1$ etcetera. Finally, we reach $(v_N, e_{N-1}, f_{N-1})$, and swiveling around $v_N=v'$ brings us home to $(v',e',f')$.
$P$ is thin: Consider any length $2$ interval $[x,y]$. If $x = \hat{0}$, then this is the fact that an edge has two endpoints. If $x$ is a vertex and $y$ a face, this is the fact that the vertex of a polygon is in two edges. If $y=\hat{1}$, this is the fact that every edge is in two faces.
It is impossible to assign the edges of $P$ weights of $\pm 1$ so that product of the weights around every diamond is $-1$: Suppose we had such an assignment. Consider any edge $e$, contained in faces $f_1$ and $f_2$, and containing vertices $v_1$ and $v_2$. In the diamond $\hat{0} < v_1, v_2 < e$, one of the sides has weight $1$ and the other has weight $-1$: Say $(\hat{0}, v_1, e)$ has weight $1$. Similarly, suppose that $(e, f_1, \hat{1})$ has weight $1$ and $(e, f_2, \hat{1})$ has weight $-1$. Draw tangent vectors to $X$ at the midpoint of $e$, with the first vector pointing in direction of $v_1$ and the second pointing into $f_1$. Their wedge gives an orientation of $X$ at the midpoint of $e$. The condition on diamonds of the form $v < e_1, e_2 < f$ implies that this orientation is consistent across all of $X$. This contradicts that $X$ is nonorientable.
The more general nonisomrophism What about a stranger isomorphism between the algebras? The OP confirms below that he already knows that such an isomorphism must send arrows $x \overset{a}{\longrightarrow} y$ to $\epsilon(x \to y) \cdot a$ for some $\epsilon(x \to y) \in K^{\ast}$. I claim this is also impossible. To simplify exposition, I'll only do this for the specific triangulation of $\mathbb{RP}^2$ mentioned above. 
Consider a Mobius strip embedded in $X$. To be concrete, take the equatorial belt of the icosahedron. It's image in $\mathbb{RP}^2$ is a simplicial complex with $5$ vertices $v_1$, $v_2$, $v_3$, $v_4$, $v_5$. (All subscripts will be periodic modulo $5$.) There are $10$ edges: Let $e_i$ connect $v_i$ to $v_{i+1}$ and $e'_i$ connect $v_i$ to $v_{i+2}$. There are $5$ triangles: Let $t_i$ have vertices $(v_i, v_{i+1}, v_{i+2})$. Now, look at the product
$$\prod_{i=1}^5 \frac{\epsilon(\hat{0} \to v_i) \epsilon(v_i \to e_i)}{\epsilon(\hat{0} \to v_{i+1}) \epsilon(v_{i+1} \to e_i)} \prod_{i=1}^5 \frac{\epsilon(v_i \to e_{i-1}) \epsilon(e_{i-1} \to f_{i-1})}{\epsilon(v_i \to e_i) \epsilon(e_i \to f_{i-1})}\prod_{i=1}^5 \frac{\epsilon(e_i \to f_{i-1}) \epsilon(f_{i-1} \to \hat{1})}{\epsilon(e_i \to f_{i}) \epsilon(f_{i} \to \hat{1})}.$$
The whole product telescopes to $1$. But each fraction is the ratio of two sides of a diamond, and there are $15$ terms, so the product is also $(-1)^{15} = -1$. 
I have a much messier version of this formula which works for any cellular division of a Mobius strip, but one is enough for the counterexample.
A: 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.
