Remark 50.14.2. In the situation of Proposition 50.14.1 we get moreover that the map

$\tilde\xi : \bigoplus \nolimits _{t = 0, \ldots , r - 1} \Omega ^\bullet _{X/S}[-2t] \longrightarrow Rp_*\Omega ^\bullet _{P/S}$

is an isomorphism in $D(X, (X \to S)^{-1}\mathcal{O}_ X)$ as follows immediately from the application of Proposition 50.13.3. Note that the arrow for $t = 0$ is simply the canonical map $c_{P/X} : \Omega ^\bullet _{X/S} \to Rp_*\Omega ^\bullet _{P/S}$ of Section 50.2. In fact, we can pin down this map further in this particular case. Namely, consider the canonical map

$\xi ' : \Omega ^\bullet _{P/S} \to \Omega ^\bullet _{P/S}$

of Remark 50.4.3 corresponding to $c_1^{dR}(\mathcal{O}_ P(1))$. Then

$\xi '[2(t - 1)] \circ \ldots \circ \xi ' \circ \xi ' : \Omega ^\bullet _{P/S} \to \Omega ^\bullet _{P/S}[2t]$

is the map of Remark 50.4.3 corresponding to $c_1^{dR}(\mathcal{O}_ P(1))^ t$. Tracing through the choices made in the proof of Proposition 50.13.3 we find the value

$\tilde\xi |_{\Omega ^\bullet _{X/S}[-2t]} = Rp_*\xi '[-2] \circ \ldots \circ Rp_*\xi '[-2(t - 1)] \circ Rp_*\xi '[-2t] \circ c_{P/X}[-2t]$

for the restriction of our isomorphism to the summand $\Omega ^\bullet _{X/S}[-2t]$. This has the following simple consequence we will use below: let

$M = \bigoplus \nolimits _{t = 1, \ldots , r - 1} \Omega ^\bullet _{X/S}[-2t] \quad \text{and}\quad K = \bigoplus \nolimits _{t = 0, \ldots , r - 2} \Omega ^\bullet _{X/S}[-2t]$

viewed as subcomplexes of the source of the arrow $\tilde\xi$. It follows formally from the discussion above that

$c_{P/X} \oplus \tilde\xi |_ M : \Omega ^\bullet _{X/S} \oplus M \longrightarrow Rp_*\Omega ^\bullet _{P/S}$

is an isomorphism and that the diagram

$\xymatrix{ K \ar[d]_{\tilde\xi |_ K} \ar[r]_{\text{id}} & M \ar[d]^{(\tilde\xi |_ M)} \\ Rp_*\Omega ^\bullet _{P/S} \ar[r]^{Rp_*\xi '} & Rp_*\Omega ^\bullet _{P/S} }$

commutes where $\text{id} : K \to M$ identifies the summand corresponding to $t$ in the deomposition of $K$ to the summand corresponding to $t + 1$ in the decomposition of $M$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).