I am reading Segal's paper "All Autoequivalences are Spherical Twists", https://arxiv.org/abs/1603.06717, and I am trying to understand a detail in example 2.4. The situation is this:
Let $X$ be a scheme and let $\mathcal{L}$ be a line bundle, so that $(-) \otimes \mathcal{L}: D^b(Coh(X)) \to D^b(Coh(X))$ is an autoequivalence (so far so good). Consider the total space of the dual bundle $\mathcal{L}^{-1}$ and the zero section
$i: X \hookrightarrow \mathcal{L}^{-1}$. The claim is:
For any object $\mathcal{E}$ in $D^b(X)$ we have that
$i^*i_*\mathcal{E} = \mathcal{E} \oplus (\mathcal{E} \otimes \mathcal{L}[1])$
I have read further ahead and I intuitively see why it $\textit{should}$ be true, but I don't understand how to compute it. I tried to use Fourier-Mukai transforms but it always just reduced back to $i^*i_*$. I believe I am supposed to use a Koszul resolution of some kind, but the only one I can think of is
$0 \to \pi^*\mathcal{L} \to \mathcal{O}_{\mathcal{L}^{-1}} \to i_*\mathcal{O}_X \to 0$
(where $\pi$ is projection onto $X$) and I don't know for sure if that is correct. I also don't know how to adapt that for arbitrary $i_*\mathcal{E}$.
Also as an aside, is this the right level of generality? Does it really hold for any scheme?
Thanks in advance.
By projection formula $$ i_*i^*i_*E \cong i_*E \otimes i_*\mathcal{O}_X. $$ Using Koszul complex as a resolution of the second factor we obtain the distinguished triangle $$ \pi^*L \otimes i_*E \to i_*E \to i_*i^*i_*E. $$ Note also that the adjunction morphism $i^*i_*E \to E$ induces a morphism $i_*i^*i_*E \to i_*E$ such that the composition $i_*E \to i_*i^*i_*E \to i_*E$ is the identity. Therefore, the above triangle splits and gives a direct some decomposition $$ i_*i^*i_*E \cong i_*E \oplus (\pi^*L \otimes i_*E[1]) \cong i_*E \oplus i_*(L \otimes E[1]). $$ Finally, applying the functor $\pi_*$ and taking into account that $\pi_* \circ i_* = \mathrm{id}$, we deduce $$ i^*i_*E \cong E \oplus (L \otimes E[1]). $$