even.

We sketch a proof of the above theorem. The vectors ${\mathbf{h}}$ and $\partial {\mathbf{h}}$ are non-negative since $h$-vectors of Cohen-Macaulay simplicial posets are non-negative [@St Theorem 3.10]. Also, since $h_d(P)$ is equal to the reduced Euler characteristic of $P$ times $(-1)^{d-1}$, $h_d(P)=0$ if $P$ is a simplicial cell ball. Finally, (3) follows from Theorem \[1.1\] as follows: If $d$ is odd then $$\partial h_{\frac {d-1} 2}= (h_0+\cdots +h_{\frac {d-1} 2}) - (h_{\frac {d+1} 2}+ \cdots +h_d)$$ is equal to $\sum_{k=0}^d h_k$ mod $2$. Since $\sum_{k=0}^{d-1} \partial h_k$ is even by Theorem \[1.1\](3), by the symmetry $\partial h_i=\partial h_{d-1-i}$ of $\partial {\mathbf{h}}$, $\partial h_{\frac {d-1} 2}$ must be even.

In the rest of this section, we prove that conditions (4), (5), (6) and (7) in Theorem \[1.2\] are necessary conditions of $h$-vectors of simplicial cell balls.

Proof of (4)
------------

\[trivial\] Let $P$ be a $(d-1)$-dimensional simplicial poset and $h(P)=(h_0,\dots,h_d)$. Let $K[P]=A/I_P$ be the face ring of $P$ and $\theta_1,\dots,\theta_d \in A_1$ an l.s.o.p. of $K[P]$. If $K[P]$ is Cohen-Macaulay then, for all integers $n >0$ and $n \leq k \leq d$, one has $$\dim_K \big(A/\big(I_P+(\theta_1,\dots,\theta_{d-1},\theta_d^n)\big)\big)_k
=h_k+h_{k-1} + \cdots + h_{k-n+1}.$$

Let $R[i]=A/(I_P+(\theta_1,\dots,\theta_{i}))$ for $i=0,1,\dots,d-1$. Since $R[0]=K[P]$ is Cohen-Macaulay, we 