I am trying to understand "de Cataldo, Migliorini. The perverse filtration and the Lefschetz hyperplane theorem. Annals of Mathematics, 171(2010), 2089-2113." My question is about one detail in the paper. Here is my reproduction of the situation.

Let $\Sigma$ be a stratification of $\mathbb{P}^N$ adapted to a bounded complex $K$ with $\Sigma$-constructible cohomology. Let $\Sigma$ be also adapted to the embedding $Y \subseteq \mathbb{P}^N$ of an affine variety $Y$. Let $\Lambda \subset \mathbb{P}^N$ be a hyperplane, $H = \Lambda \cap Y$ and $\bar{H} = \Lambda \cap \bar{Y}$. Set $\bar{U} = \bar{Y} \setminus \bar{H}$ and $U = Y \setminus H$. Consider the cartesian diagram

$$\begin{array}[c]{ccccc} H&{\xrightarrow{i}}&Y&{\xleftarrow{j}}&U\\ \downarrow\scriptstyle{J}&&\downarrow\scriptstyle{J} && \downarrow\scriptstyle{J}\\ \bar{H}&{\xrightarrow{i}}&\bar{Y}&{\xleftarrow{j}}&\bar{U} \end{array}$$

Now the two authors stated (page 2101, following the same diagram) "By the octahedron axiom, the map $J_!j_*j^*K \to j_*J_!j^*K$ is an isomorphism if and only if the natural base change map $i^*J_*K \to J_*i^*K$ is an isomorphism."

My question is: what is exactly used about the octahedron axiom for the statement?



I'm not sure what's going on, but here's a guess:

Firstly, the base change map goes the other way: $i^*J_*\mathcal{F}\rightarrow J_*i^*\mathcal{F}$. Say this is an isomorphism for the Verdier dual of $K$ (they end up saying that the condition they impose makes the base change morphism an isomorphism independent of $K$, so this is ok).

Then taking the Verdier dual of this isomorphism we get an isomorphism $J_!i^!K\rightarrow i^!J_!K$. Next note that the similarly defined base change map $J_!j^*K\rightarrow j^*J_!K$ gives an isomorphism, since $j$ is an open embedding and $j^*=j^!$.

The distinguished triangles to plug into the octahedron axiom are then

$J_!K\rightarrow J_!j_*j^*K\rightarrow J_!i_*i^!K[1]\rightarrow$


$J_!K\rightarrow j_*j^*J_!K\rightarrow i_*i^!J_!K[1]\rightarrow$


$J_!j_*j^*K\xrightarrow{f}j_*j^*J_!K\rightarrow cone(f)\rightarrow$

By the axiom, there is a distinguished triangle

$J_!i_*i^!K[1]\xrightarrow{g} i_*i^!J_!K[1]\rightarrow cone(f)\rightarrow$

and so if $g$ is an isomorphism, $cone(f)=0$ and $f$ is an isomorphism.

  • $\begingroup$ Thank you very much for the thorough answer! You are certainly right that my base change arrow should be reversed. $\endgroup$ – Xudong Mar 15 '17 at 12:01

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