Let $L$ be a finite lattice with least element $\hat{0}$, greatest element $\hat{1}$, and Möbius function $\mu$.

Question 1: What class of lattices the following property characterizes? $$\mu(\hat{0},a)=\mu(\hat{0},\hat{1})\mu(a,\hat{1}), \ \forall a \in L$$ It follows that $\mu(\hat{0},\hat{1}) = \pm1$.

Remark: It is satisfied by any boolean lattice, more generally by the face lattice of any convex polytope (as suggested by John Shareshian), and more generally by any Eulerian lattice.
Proof: An Eulerian lattice is a graded lattice $L$ such that for any $a,b \in L$ with $a \le b$, we have $\mu(a,b)= (-1)^{|b|-|a|} $, with $a \mapsto |a|$ the rank function. The result is immediate. $\square$

Question 2: Is there a non-Eulerian lattice with the above property on the Möbius function?
Yes, see the answer of John Machacek.

As suggested by Sam Hopkins:
Question 3: Is there a non-Eulerian atomistic lattice with the above property on the Möbius function?
No for $|L| \le 13$, as checked by the following Sage program (using these lists of Martin Malandro):

from itertools import product
def relationtest(L,n):
    for l in L:
        b = P.bottom()
        t = P.top()
        if all(P.moebius_function(b,x) == P.moebius_function(b,t)*P.moebius_function(x,t) for x in P):
            if L.is_atomic():
                if not L.is_graded():
                if L.is_graded():
                    for x, y in product(P, P):
                        if P.compare_elements(x,y)==-1:
                            if not P.moebius_function(x,y)==(-1)^(P.rank(y)-P.rank(x)):

Are the small atomistic lattices listed somewhere?

  • 4
    $\begingroup$ Hi Sebastien, how about the face lattice of an arbitrary polytope? $\endgroup$ – John Shareshian Sep 11 '17 at 23:14
  • $\begingroup$ Hi John, you are right, it works for some non-simplex polytopes I tried. I don't know if it works for any arbitrary polytope, or if we need convexity. In fact, I'm interested to know all the lattices satisfying this property, so I hope you will not mind if I improve the post. $\endgroup$ – Sebastien Palcoux Sep 12 '17 at 0:09
  • 4
    $\begingroup$ You might want to read about Eulerian posets in Chapter 3 of Richard Stanley's "Enumerative Combinatorics". $\endgroup$ – John Shareshian Sep 12 '17 at 0:24

I found the following example with SageMath. Below is the Hasse diagram and verification in SageMath that the poset is a lattice which is not Eulerian (in fact not graded) that satisfies the desired Möbius function condition. I have nothing more enlightening to say currently, but I'll post this example since it answers the second question about a non-Eulerian example and will perhaps helps toward an answer to the first question about characterizing such lattices.


sage: P = Poset(([0,1,2,3,4,5,6,7,8],[[0, 1], [0, 2], [0, 5], [1, 4], [1, 6], [2, 3], [2, 7], [3, 4], [4, 8], [5, 6], [5, 7], [6, 8], [7, 8]]))
sage: b = P.bottom()
sage: t = P.top()
sage: P.is_lattice()
sage: P.is_graded()
sage: all(P.moebius_function(b,x) == P.moebius_function(b,t)*P.moebius_function(x,t) for x in P)
  • 1
    $\begingroup$ Yes! From your example I deduce the following process generating new lattices with the expected property: take a lattice $L$ with this property (for example an Eulerian lattice), and take two elements $a, b \in L$ with $$\hat{0} < a \lessdot b < \hat{1}.$$ Create a new lattice $\tilde{L}$ by adding an extra element $c$ between $a$ and $b$. By construction $$\mu_{\tilde{L}}(\hat{0},c) = \mu_{\tilde{L}}(c,\hat{1}) = 0,$$ and the rest is unchanged, so $\tilde{L}$ works also. Your example is this process applied to $B_3$. $\endgroup$ – Sebastien Palcoux Sep 12 '17 at 19:24
  • 1
    $\begingroup$ Question: Is there an example which does not come the application finitely many times of this process to an Eulerian lattice? For example, a lattice $L$ having an element $a$ such that $\mu(\hat{0},a) \not \in \{ -1,0,1 \}$. $\endgroup$ – Sebastien Palcoux Sep 12 '17 at 19:29
  • 2
    $\begingroup$ @SebastienPalcoux: for starters, you could ask if every atomistic lattice satisfying your condition is Eulerian. $\endgroup$ – Sam Hopkins Sep 12 '17 at 20:04
  • $\begingroup$ SebastienPalcoux, nice construction. I just ran a computer search but did not have much time to actually think. Glad you were able to make something of it. @SamHopkins, yes I think it would be interesting to find conditions which force Eulerian (or don't). $\endgroup$ – John Machacek Sep 13 '17 at 1:18
  • 1
    $\begingroup$ @SebastienPalcoux not that I know of. I only know Posets(n) will give a list of posets on n elements up to isomorphism. That could be a start, but then you would need to check for properties you want. $\endgroup$ – John Machacek Oct 10 '17 at 20:23

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