understanding the orientation sheaf

Question

If I understand correctly for a proper map, the upper shriek functor is the same as the pullback functor. Now, for some complex compact manifold, I read that $f^{!}(\mathbb{C}_{X})$, where $\mathbb{C}_{X}$ is the constant sheaf and f map X to a point, is the orientation sheaf. How can it be ? I am thinking that $f^{!}(\mathbb{C}_{X})$ is the constant sheaf on X. Where am I going wrong ?

Answer 1

Unless I am misunderstanding something, there are a few things that are not quite correct.

Let $f:X\rightarrow Y$ be any proper map between topological spaces. It is not true that $f^!=f^*$, but only that $f_!=f_*$ (direct image with proper support and direct image).

In fact, $f^!$ does not even exist in general. However, there are derived versions of these functors : $f^*,Rf_*,Rf_!$. We still have $Rf_*=Rf_!$ for proper map, and now $Rf_!$ has a right adjoint $Rf^!$ (at least for reasonable spaces). Note that $f^*$ is the left adjoint of $Rf_*$ so even if $f$ is proper, $f^*$ and $Rf^!$ may be very different.

Now let $f:X\rightarrow pt$ be the projection onto a point where $X$ is a manifold of (real) dimension $n$. Then it is not quite true that $Rf^!\mathbb{C}$ is the orientation sheaf, but only true up to a shift : $Rf^!\mathbb{C}=\mathfrak{or}_X\otimes\mathbb{C}[n]$.

Note however that complex manifold are oriented, so the orientation sheaf is trivial anyway, and we have $f^!\mathbb{C}=\mathbb{C}_X[2n]$ if $X$ is a complex manifold of complex dimension $n$.