If someone has gone through the Griffiths' paper ``On the periods of certain rational integrals: I,'' could you help me to understand Lemma 8.10?
I don't get why $\eta\in Z^{q,k+1}(l-1)$; although $\eta$ is surely a closed form of type $(q,0)+\cdots+(q-k-1,k+1)$ with a pole of order $l-1$, I think we have to say more to conclude $\eta\in Z^{q,k+1}(l-1)$.


In the paper's notation, $B^{q,k}(l)$ denotes the space of differential forms $\varphi$ on $\mathbb{P}^n$ which are of type $(q,0)+\cdots+(q-k,k)$, are $C^\infty$ on $\mathbb{P}^n-V$, and which have the local property that $f^l\varphi$ and $f^{l-1}df\wedge\varphi$ are $C^\infty$ if $f=0$ is a local holomorphic defining equation for $V$.
$Z^{q,k}(l)$ is the space of closed forms in $B^{q,k}(l)$.


In Lemma 8.7, we write $\varphi\in B^{q,k}(l)$ as $\varphi=d\psi+\eta$ locally, where $\psi,\eta$ have poles of order $l-1$ along $V$, $\psi$ is of type $(q-1,0)+\cdots+(q-1-k,k)$, and $\eta$ is of type $(q,0)+\cdots+(q-k-1,k+1)$.
In Lemma 8.9, we glue them by using a partition of unity to obtain $\varphi=d\psi+\eta$ globally.
Lemma 8.10 contains a claim that if $\varphi\in Z^{q,k}(l)$ and $l>1$ then $\eta\in Z^{q,k+1}(l-1)$, and this is what I am in trouble with.
Indeed, taking a look at the construction of $\eta$ in Lemma 8.7 and 8.9, we can say that $\eta$ is of type $(q,0)+\cdots+(q-k-1,k+1)$ and has a pole of order $l-1$ ,i.e. $f^{l-1}\eta$ is $C^\infty$. In addition, since $\varphi=d\psi+\eta$ is a closed form, so is $\eta$. However, how can we show that $f^{l-2}df\wedge\eta$ is $C^\infty$? It seems a stronger condition than just having a pole of order $l-1$.


In the same reason, I don't get why the statement $\psi-\psi'\in B^{q-1,k}(l-1)$ in the proof of Lemma 8.10 holds. Excuse me if I'm asking a dumb question, for I'm not from the background of mathematics. Any comments are helpful.



A specific question on the Griffiths' paper: the reduction of the pole order
In fact, it just follows from that $\eta$ has a pole of order $l-1$ and is $\textit{closed}$: We have the identity $$f^{l-2}df\wedge\eta=\frac{1}{l-1}d(f^{l-1})\wedge\eta=\frac{1}{l-1}d(f^{l-1}\wedge\eta),$$


which is $C^{\infty}$ since $f^{l-1}\wedge\eta$ is $C^{\infty}$.


