4
$\begingroup$

For a seperated scheme of finite type $X$ over $\mathbf{C}$, let $H_*(X)$ denote its Borel-Moore homology, which is defined by $$ H_k(X) = R^{-k}\Gamma(X, \omega_X) $$ where $\omega\in D_c(X, \mathbf{C})$ is a dualising object in the derived category of constructible $\mathbf{C}$-sheaves on $X$. It is immediate from the definition and the six operations for constructible sheaves that $H_*$ is covariant with respect to proper morphisms and contravariant with respect to smooth morphisms. Let $f_*$ and $f^*$ denote respectively the direct and inverse image maps.

Question

If we are given a cartesian diagramme of schemes over $\mathbf{C}$ $\require{AMScd}$ \begin{CD} Y' @>\tilde g>> Y \\ @VVf'V @VVfV \\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth, do we have an equality of the maps between homology groups $f'_*\tilde g^* = g^*f_*: H_*(Y)\to H_*(X')$?

In the book [N. Chriss & V. Ginzburg, Representation theory and complex geometry, 8.3.34] , this is shown in the case where $g$ is locally a trivial fibration, by reducing it to the commutativity of the following diagramme of derived functors

\begin{CD} f_*f^! @>>> \mathrm{id}_X @>>> g_*g^* \\ @VVV @. @AAA \\ f_*\tilde g_*\tilde g^*f^! @. = @. g_* f'_*{f'}^! g^* \end{CD}

in which all the morphisms are adjunction morphisms except for the base change morphism $\tilde g^*f^! \cong {f'}^! g^*$ in the bottom line, whose definition can be found in [SGA4, Exposé 18, 3.1.14.2]. One gets the result by apply the diagramme to $\omega_X$ and then take global sections.

The commutativity of the diagramme above can be easily checked in the case where $g: X'\to X$ is locally a trivial fibration. While I don't know how to do it if $g$ is only assumed to be smooth, I still suspect the commutativity to remain true.

Let me also add that there is such an equality $f'_*\tilde g^* = g^*f_*: A_*(Y)\to A_*(X')$ for Chow groups and for $f$ proper and $g$ flat, see for example [Fulton, Intersection theory, Prop. 1.7].

$\endgroup$
0
$\begingroup$

$\require{AMScd}$ After the struggle of a whole day due to my unfamiliarity with the category theory, I found the answer.

The commutativity is due essentially to the following facts:

  1. Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f^*, f_*, \epsilon_f, \eta_f)$: $$ f^*: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}),\; f_*: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}) $$ $$ \epsilon_f : \mathrm{id} \to f_*f^*, \; \eta_f : f^*f_* \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $(g_*\epsilon_fg^*)\epsilon_g = \epsilon_{gf}$, as indicated below \begin{CD} \mathrm{id}@>\epsilon_{gf}>> (gf)_*(gf)^* \\ @V\epsilon_{g}VV @| \\ g_*g^*@>g_*\epsilon_fg^*>> g_*f_*f^*g^* \end{CD}
  2. Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f_!, f^!, \sigma_f, \tau_f)$: $$ f_!: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}),\; f^!: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}) $$ $$ \sigma_f : \mathrm{id} \to f^!f_!, \; \tau_f : f_!f^! \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $\tau_g(g_!\tau_fg^!) = \tau_{gf}$, as indicated below \begin{CD} g_!f_!f^!g^!@>g_!\tau_{f}g^!>> f_!f^! \\ @| @V\tau_{f}VV \\ (gf)_!(gf)^!@>\tau_{gf}>> \mathrm{id} \end{CD}

The first one is obvious, while the second one is a sheaf-theoretic version of the Fubini theorem.

Now we turn to the diagramme in question \begin{CD} Y' @>\tilde g>> Y \\ @Vf'VV @VfVV\\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth and equidimensional.

Suppose we are given two objects $A, B\in D_c(X, \mathbf{C})$ and a morphism $\varphi\in \mathrm{Hom}(f^*A, f^!B)\cong \mathrm{Hom}(A, f_*f^!B)$. Form the following diagramme \begin{CD} g^* A @>g^*\epsilon_f>> g^*f_*f^* A @>g^*f_*\varphi>> g^*f_*f^! B @>g^*\tau_f>> g^* B \\ @V\epsilon_{f'}g^*VV @V\mathrm{BC}VV @V\mathrm{BC}VV @A\tau_{f'}g^*AA \\ f'_*{f'}^*g^*A @= f'_*\tilde g^*f^*A @>f'_*\tilde g^*\varphi>> f'_*\tilde g^*f^!B @= f'_*{f'}^!g^*B \end{CD} This diagramme is commutative:

  • The square in the middle is commutative, by applying the base change isomorphism (BC) $g^*f_*\cong f'_*\tilde g^*$ to $\varphi$.
  • The square in the left is equivalent to the following adjoint \begin{CD} A @>\epsilon_f>> f_*f^* A \\ @V(g_*\epsilon_{f'}g^*)\epsilon_gVV @Vf_*\epsilon_{\tilde g}f^*VV \\ g_*f'_*{f'}^*g^*A @= f_*\tilde g_*\tilde g^*f^*A \end{CD} This is commutative, as we have mentionned in the above fact 1. that $(g_*\epsilon_{f'}g^*)\epsilon_g = \epsilon_{gf'}= \epsilon_{f\tilde g} = (f_*\epsilon_{\tilde g}f^*)\epsilon_f$.
  • Similarly, the square in the right is commutative in regard of the fact 2 and the canonical isomorphisms $g^* \cong g^![2d], \; \tilde g^* \cong \tilde g^![2d]$, where $d = \dim X'/X$.

Now, if we set $A = \mathbf{C}$ and $B = \omega_X[-k]$ in $D_c(X, \mathbf{C})$ for $k\in \mathbf{Z}$, then every homology class $\alpha\in H_k(Y)\cong \mathrm{Hom}_{D_c(Y)}(f^*A, f^! B)$ corresponds to a morphism $\varphi_{\alpha}: f^*A \to f^! B$. Then the composite $$(g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f): g^*A \to g^*B$$ corresponds exactly to the class $g^*f_*\alpha\in H_{k+2d}(X')$, whereas the composite $$ (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*): g^*A \to g^*B $$ corresponds to the class $f'_*\tilde g^*\alpha\in H_{k+2d}(X')$.

The commutativity of the above 8-term diagramme applying to $\varphi = \varphi_{\alpha}$ then tells us that $$ (g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f) = (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*) $$ and consequently $g^*f_*\alpha = f'_*\tilde g^*\alpha$. Thus $g^*f_* = f'_*\tilde g^*:H_*(Y)\to H_*(X')$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.