# Showing when the spectral sequence associated to a filtered complex $C$ collapses using a similar complex $C\otimes\mathbb{F}[t]$

Let $\mathbb{F}$ be a field, and let $C=\bigoplus_{i,j\in\mathbb{Z}} C^{i,j}$ be a bigraded $\mathbb{F}$-vector space of finite total dimension. Suppose there are two differentials $d$ and $\widetilde{d}$ on $C$ such that

• $d$ is of bidegree $(1,0)$, i.e. $d:C^{i,j}\to C^{i+1,j}$,
• $\widetilde{d}$ is of bidegree $(1,1)$, i.e. $d:C^{i,j}\to C^{i+1,j+1}$, and
• $d$ and $\widetilde{d}$ anti-commute, i.e. $\widetilde{d}\circ d = -d \circ\widetilde{d}$.

Then $d+\widetilde{d}$ is a differential on $C$. There is a filtration $\mathcal{F}^p C = \bigoplus_{i\in\mathbb{Z}, j\geq p} C^{i,j}$ which induces a spectral sequence on the complex $(C,d+\widetilde{d})$. See Wikipedia reference. Since $C$ is of finite total dimension, the spectral sequence is guaranteed to collapse at some point.

I am interested in relating the homology of $(C,d+\widetilde{d})$ and the page at which the above spectral sequence collapses to the following construction.

Consider the complex $C\otimes \mathbb{F}[t]$ where $t$ has bidegree $(0,-1)$. If $x_0\in C^{i,j}$, then the bidegree of $x_0 t^k$ in $C\otimes \mathbb{F}[t]$ is $(i,j-k)$. The differential in the complex is $d + t\widetilde{d}$. Since multiplication by $t$ is of bidegree $(0,-1)$ and $\widetilde{d}$ is of bidegree $(1,1)$, the map $t\widetilde{d}$ is of bidegree $(1,0)$. Thus the map $d+t\widetilde{d}$ preserves the second grading on $C\otimes \mathbb{F}[t]$.

Let's denote the homology of $(C,d+\widetilde{d})$ by $H=\bigoplus_{i\in\mathbb{Z}} H^i$, and let's denote the homology of $(C\otimes\mathbb{F}[t],d+t\widetilde{d})$ by $\mathcal{H} = \bigoplus_{i,j\in\mathbb{Z}} \mathcal{H}^{i,j}$. Suppose that the homology of $(C,d+\widetilde{d})$ is $H\cong\mathbb{F}^N=\underbrace{\mathbb{F}\oplus\cdots\oplus\mathbb{F}}_{N}$.

Let $\operatorname{T}(\mathcal{H})$ consist of those elements $\alpha\in\mathcal{H}$ such that $t^n\;\alpha=0$ for some nonnegative integer $n$. If $\alpha\in\operatorname{T}(\mathcal{H})$ and $n$ is the smallest integer in $\mathbb{Z}_+$ such that $t^n\; \alpha=0$, then we say the order of $\alpha$ is $n$. The order of the zero element in $\mathcal{H}$ is declared to be zero.

1. Prove that $\mathcal{H}\cong \operatorname{T}(\mathcal{H})\oplus\mathbb{F}[t]^N = \operatorname{T}(\mathcal{H})\oplus\underbrace{\mathbb{F}[t]\oplus\cdots\mathbb{F}[t]}_{N}$.

2. Prove that the page where the spectral sequence for $(C,d+\widetilde{d})$ collapses is one more than the maximum order of any element in $\operatorname{T}(\mathcal{H})$.

I am especially interested in Problem 2 above.