Let $S\_1 = S\_2 = S^d$ be two copies of the $d$-dimensional sphere. Let $p\_i : S\_1 \times S\_2 \to S\_i$ be the projection, $j : U \to S\_1 \times S\_2$ the inclusion of the complement of the diagonal and $q\_i : U \to S\_i$ the restriction of $p\_i$.


I want to know how to get the following formula for a constructible sheaf $\mathcal F \in D^b(Sh(S^d))$ : $${q\_2}\_! q\_1^\*(\mathcal F) \cong {p\_2}\_\* (p\_1^\* \mathcal F \otimes j\_! \mathbb Q\_U[1])$$ It looks like some kind of base change but I don't understand where the $[1]$-shift is coming from.



Fourier transform for constructible sheaves on spheres
We have $q\_1 = p\_1 \circ j$ and $q\_2 = p\_2 \circ j$ so


$$ {q\_2}\_! q\_1^\*(\mathcal F) \cong {p\_2}\_! j\_! j^\* p\_1^\*(\mathcal F) \cong {p\_2}\_! j\_! ( j^\* p\_1^\*(\mathcal F) \otimes \mathbb Q\_U) \ \cong {p\_2}\_! ( p\_1^\*(\mathcal F) \otimes j\_! \mathbb Q\_U) $$ where the last isomorphism is the projection formula (indeed a variant of proper base change).


I do not know where the $[1]$ shift is coming from either. It seems to me that ${p\_2}\_! ( p\_1^\*(\mathcal F) \otimes j\_! \mathbb Q\_U[1])$ is the correct formula for $SR$ so the earlier formula should be $T\_1(\mathcal F) = q\_{2!} (q\_1^\* \mathcal F [1])$.


Also the shift appearing in the formula for $T\_1'$ should be $[d-1]$, with the $[d]$ coming from the shift in $L$ (which I think should be $[d]$ instead of $[-d]$).


With these shifts, I can calculate that that they are quasi-inverses following the suggested approach. So one possible interpretation is that the claims are morally right but the shifts are all wrong. (Or maybe my claims are morally right but the shifts are all wrong.)


