have the exact sequence $$0 \longrightarrow R[i-1] \stackrel{\times \theta_i}{\longrightarrow} R[i-1] \longrightarrow R[i] \longrightarrow 0$$ for $i=1,2,\dots,d-1$. The exact sequences show that $H_{R[i]}(t)=(1-t)H_{R[i-1]}(t)$ for $i=1,2,\dots,d-1$. Then, since $H_{K[P]}(t)=(h_0+h_1t+\cdots+h_dt^d)/(1-t)^d$, we have $H_{R[d-1]}(t)=(h_0+h_1t+\cdots+h_dt^d)/(1-t)$. This fact says $$\dim_K \big(R[d-1]\big)_k = h_k + h_{k-1} + \cdots + h_1 +h_0$$ for $k=0,1,\dots,d.$ Also, since $\theta_1,\dots,\theta_{d-1},\theta_d^n$ is an h.s.o.p. of $K[P]$, the sequence $$0 \longrightarrow R[d-1] \stackrel{\times \theta_d^n}{\longrightarrow} R[d-1] \longrightarrow A/\big(I_P+(\theta_1,\dots,\theta_{d-1},\theta_d^n)\big) \longrightarrow     0$$ is exact. Hence $$\begin{aligned}
\dim_K \big(A/\big(I_P+(\theta_1,\dots,\theta_{d-1},\theta_d^n)\big)\big)_k
&=& \dim_K \big(R[d-1] \big)_k - \dim_K \big( R[d-1] \big)_{k-n}\\
&=& h_k+h_{k-1} + \cdots + h_{k-n+1}\end{aligned}$$ for $k=n,n+1,\dots,d$, as desired.

\[necessary1\] Let $P$ be a simplicial cell $(d-1)$-ball and ${\mathbf{h}}=h(P)=(h_0,h_1,\dots,h_d)$. If $\partial h_n=0$ for some $1 \leq n \leq d-2$ then $$h_k + h_{k-1} + \cdots +h_{k-n+1} \geq \partial h_k
\ \ \mbox{ for } k=n,n+1,\dots,d-1.$$

Let $A=K[x_\sigma: \sigma \in P \setminus \{ \hat 0\}]$ and $J=I_P + (x_\sigma: \sigma \not \in \partial P) \subset A$. Then $A/J \cong K[\partial P]$ by Lemma \[subposet\]. For a general choice of linear forms $\theta_1,\dots,\theta_d \in A_1$, we have (e.g. use a criterion of an l.s.o.p. for face rings [@Ko Lemma 6])

-   $\theta_1,\dots,\theta_{d-1}$ is an l.s.o.p. of $A/J$.

-   $\theta_1,\dots,\theta_{d-1},\theta_d$ is an l.s.o.p. of $A/I_P$.

Let $\overline \Theta = \theta_1,\dots,\theta_{d-1}$. Since 