Trace of a matrix associated to posets

Question

Let $P$ be a finite connected poset with $n$ elements. Let $C=(c_{x,y})$ be the $n \times n$ matrix with entry 1 in case $x \leq y$ and 0 else. The Coxeter matrix of $P$ is defined as the matrix $M_P=-C^{-1}C^T$.

Let $u_P$ be defined as the trace of $M_P^2$.

Question 1: Is it true that $u_P$ is an odd integer in case $P$ is a lattice? Does it have a nice interpretation in this case?

I can prove question 1 for distributive lattices.

Question 2: Is $u_P$ an odd integer in case $P$ is just a bounded poset?

Questions 1 and 2 have a positive answer for posets with at most 9 points. In case this is true for general bounded posets, there surely is a nice reason.

For a general connected poset $P$ $u_p$ can be zero.

Here is how to get a poset (not bounded) with $u_p=0$ using Sage:

n=6 

posets=[P for P in Posets(n) if P.is_connected()]

U=[P for P in posets if ((P.coxeter_transformation())^2).trace()==0]

P=U[0]

display(P)

Answer 1

Here is a nice proof for posets $P$ with a global maximum $M$ (it works dually for posets with a global minimum, but not for general posets as the example in the question shows) suggested by comments of Darij Grinberg and Fedor Petrov.

Note first that the entries of the coxeter matrix $Co=Co_P$ (formerly known as $M_P$) are given by $Co_{x,y}=- \sum\limits_{z \in P: z \geq x}^{}{\mu(y,z)}$, where $\mu$ is the Moebius function of $P$.

Now for any square matrix $A$ over the ring of integers, we have $\operatorname{tr}(A^2) \equiv \operatorname{tr}(A)^2 \equiv \operatorname{tr}(A) \mod 2$. Thus when we show that $\operatorname{tr}(A) \equiv 1 \mod 2$, we are done.

Now by Hall's theorem on chains we have $\mu(x,y)=-c_1+c_2-c_3+c_4-....$ when $c_i$ denotes the number of length $i$ chains starting at $x$ and ending at $y$. Mod 2 this is equal to the number of chains from $x$ to $y$ (since there is difference between - and + in mod 2).

Thus $\operatorname{tr}(Co)=- \sum\limits_{x \in P} \sum\limits_{z \in P: z \geq x}^{}{\mu(x,z)}$ is equal mod 2 to the total number of non-empty chains in the poset.

Now for any chain $K$ we can add or remove the maximum $M$ of the poset $P$, so we can partition all non-empty chains except $\{ M \}$ into pairs. Thus $\operatorname{tr}(Co) \equiv 1 \mod 2$.