I have been following Stanley's Enumerative combinatorics for the definition of an Eulerian poset. It is defined as follows:
Definition: A finite graded poset $P$ with $\hat{0}$ and $\hat{1}$ is Eulerian if $\mu_P(s,t)=(-1)^{l(s,t)}$ for all $s\leq t$ in $P$.
Also after checking online, I found another definition that says an Eulerian poset is a graded poset in which every non-trivial interval has the same number of elements of even ranks as of odd ranks.
Can someone guide me a little to see how these two definitions are equivalent? I might be missing something pretty basic here.
It is indeed true that $\mu(s,t)=(-1)^{\ell(s,t)}\iff $ every nontrivial interval has an equal number of elements with even and odd ranks.
$(\!\!\implies\!\!)$ Fix $s<t$. Since $\mu$ is the inverse of the zeta function, $\zeta(s,t)=1$ for all $s\le t$, we know that $$ \begin{align} 0 =\sum_{s\le r\le t}\mu(s,r)\color{gray}{\cdot 1} =\sum_{s\le r\le t}(-1)^{\ell(s,r)} =(-1)^{\ell(\hat 0,s)}\sum_{s\le r\le t}(-1)^{\ell(0,r)}. \end{align} $$ Note that $\ell(0,r)$ is the rank of $r$. Since the sum of $(-1)^{\text{rank}(r)}$ over the interval $[s,t]$ is zero, it follows that $[s,t]$ has an equal number of elements of even and odd rank.
($\Longleftarrow$) To prove, $\mu(s,t)=(-1)^{\ell(s,t)}$, define $\sigma(s,t):=(-1)^{\ell(s,t)}$. It suffices to show that $\sigma$ is an inverse to the zeta function. For any $s<t$, we compute $$ (\sigma*\zeta)(s,t) =\sum_{s\le r\le t}\sigma(s,r)\cdot \zeta(r,t) =\sum_{s\le r\le t}(-1)^{\ell(s,r)}\cdot 1 =(-1)^{\ell(\hat 0,s)}\sum_{s\le r\le t}(-1)^{\text{rank}(r)} $$ The assumption that $[s,t]$ has an equal number of even and odd rank elements then implies that $\sum_{s\le r\le t}(-1)^{\text{rank}(r)}=0$, meaning that $(\sigma * \zeta)(s,t)=0$ when $s<t$, while $(\sigma * \zeta)(s,s)=1$. This means $\sigma * \zeta$ is the identity, proving that $\sigma(s,t)=(-1)^{\ell(s,t)}=\mu$.