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:

  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: 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 $p-q=0$.

  • $\begingroup$ I think I'm missing something. Since $P$ is graded, can't you just flip the signs on all edges between grades $2i$ and $2i+1$, while leaving the signs the same on the edges between grades $2i-1$ and $2i$? $\endgroup$ Sep 13 '15 at 12:46
  • $\begingroup$ Since each diamond covers three consecutive ranks, this should change each $p+q$ into a $p-q$. $\endgroup$ Sep 13 '15 at 12:49
  • $\begingroup$ @DavidSpeyer, wouldn't this change p+q to -p-q? $\endgroup$ Sep 13 '15 at 12:55
  • $\begingroup$ for example, if P is a diamond I would want to change just one edge to its negative and not 2. My worry is that diamonds overlap so how do I do it consistently? $\endgroup$ Sep 13 '15 at 12:56
  • $\begingroup$ I just read your paper, which clarifies. If you have a diamond $a < b,c < d$, you want the relation $(a \to b \to d) + (a \to c \to d)$. The phrase "parallel" made me think of $(a\to b) + (c \to d)$, which made me confused. So what I wrote is nonsense. $\endgroup$ Sep 13 '15 at 14:05

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.

  • $\begingroup$ Actually Brown proves a triangulation gives the interval from bottom to top is a Chamber complex. I misspoke. Since most of your other intervals are Boolean lattices, the only question is intervals to the top. Anyway, there certainly are triangulation of non-orientable manifolds like the one you gave before $\endgroup$ Sep 18 '15 at 15:15
  • $\begingroup$ Your welcome! I only understood it this morning. Until then I was thinking "sure is convenient that I got orientations for surfaces; I wonder what strange structure I get for higher dimensional manifolds?". $\endgroup$ Sep 18 '15 at 15:29

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.

  • $\begingroup$ I will need time to digest this. It is well known that any triangulation of a manifold without boundary is strongly connected. This is an exercise in Brown's building book. I think it follows from basic quiver theory that if they are isomorphic then there is one taking an arrow to a multiple of an arrow. $\endgroup$ Sep 16 '15 at 12:56
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    $\begingroup$ Conjugating by a unit if necesssary we can assume the isomorphism sends the standard primitive idempotents of my algebra to the standard primitive idempotents of the incidence algebra. Then since the pierce components between consecutive vertices are one dimensional we must send edges to multiples of edges. $\endgroup$ Sep 16 '15 at 13:02
  • $\begingroup$ I meant any triangulation of a closed manifold gives a chamber complex. But the small rank intervals are trivial in your setup $\endgroup$ Sep 16 '15 at 13:15
  • $\begingroup$ This looks convincing. I'll null it over just a little more before accepting. $\endgroup$ Sep 16 '15 at 13:35
  • $\begingroup$ many thanks for this answer. I probably would have wasted more time on trying to find a proof. It still leaves open whether I can get something like the face lattice of a non-orientable surface from the algebras we are considering in our paper, but it is good to know the general case is false. $\endgroup$ Sep 16 '15 at 21:30

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