# Nearby cycles and extension by zero

Let $$f: X\to \text{Spec}(R)$$ be a proper and smooth morphism, with $$R$$ a strictly henselian dvr. Call $$s = \overline{s}$$ the closed point and $$\eta$$ the geometric point of $$\text{Spec}(R)$$.

Call $$i_s : X_{\overline{s}}\to X$$ the closed immersion and $$h : X_{\overline{\eta}}\to X$$ the inclusion of the geometric generic fiber.

The proper base change theorem and universal local acyclicity of smooth morphisms give an isomorphism

$$(*)\ \ \ \ H^p(X_{\overline{s}},F)\xrightarrow{\simeq} H^p(X_{\overline{\eta}},F),$$

where $$F$$ is a locally constant constructible $$\ell$$-adic sheaf.

Suppose that $$f$$ is not proper anymore but affine of finite type and separated. We still assume that $$f$$ is smooth.

Do we still have an isomorphism between $$\ell$$-adic cohomologies with proper supports:

$$H^p_c(X_{\overline{s}},\mathbf{Q}_{\ell})\xrightarrow{\simeq} H^p_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})\ ?$$

ATTEMPTED SOLUTION:

The main idea I have in mind is as follows.

Choose a proper $$g : \overline{X}\to \text{Spec}(R)$$ that compactifies $$f$$.

Although $$g$$ may not be smooth, the inclusion $$j : X\to \overline{X}$$ is quasi-compact by the assumptions, and $$j_!\mathbf{Q}_{\ell}$$ is constructible on $$\overline{X}$$. Proper base change still applies to $$j_!\mathbf{Q}_{\ell}$$.

Call $$j_s$$ the base change of $$j$$ along $$s\to\text{Spec}(R)$$, and same for $$j_{\eta}$$.

Call $$i_s$$ and $$h$$ still the maps $$\overline{X}_s\to \overline{X}$$ and $$\overline{X}_{\overline{\eta}}\to \overline{X}$$ by abuse of notation.

I want to show that $$i_s^*\Psi_g(j_!\mathbf{Q}_{\ell}) = 0$$, for $$\Psi_g$$ the nearby cycles functor. If this is true, then the natural map

$$(j_s)_!\Psi_f(\mathbf{Q}_{\ell}) \to \Psi_g(j_!\mathbf{Q}_{\ell})$$ is an isomorphism, since its cofiber is $$(i_s)_*i_s^*\Psi_g(j_!\mathbf{Q}_{\ell})$$ (shifted). This is because $$j_s^*\Psi_g = \Psi_f$$ (where I use that $$j$$, an open immersion, is smooth).

In other words the question has a positive answer if the nearby cycles functor commutes with extension by zero.

Suppose that this is true. By properness of $$g$$, we have

$$R\Gamma(\overline{X}_{\overline{s}}, \Psi_g(j_!\mathbf{Q}_{\ell})) = R\Gamma(\overline{X}_{\overline{\eta}}, j_!\mathbf{Q}_{\ell}).$$

By the claim we have $$R\Gamma(\overline{X}_{\overline{s}}, \Psi_g(j_!\mathbf{Q}_{\ell})) = R\Gamma(\overline{X}_{\overline{s}}, (j_s)_!\Psi_f(\mathbf{Q}_{\ell}))$$. By smoothness of $$f$$ we have $$\Psi_f(\mathbf{Q}_{\ell}) = \mathbf{Q}_{\ell}[0]$$.

Putting everything together we have an isomorphism

$$R\Gamma(\overline{X}_{\overline{s}},j_!\mathbf{Q}_{\ell}) = R\Gamma(\overline{X}_{\overline{\eta}},j_!\mathbf{Q}_{\ell})$$ which gives what we want.

So the question really is:

do we have $$(j_s)_!\Psi_f(\mathbf{Q}_{\ell}) =\Psi_g(j_!\mathbf{Q}_{\ell})$$?

For every closed geometric point $$\overline{x}\to \overline{X}_{\overline{s}}$$, we call $$\overline{X}_{\overline{s}}(\overline{x})$$ the strict henselianization of $$\overline{X}_{\overline{s}}$$ at $$\overline{x}$$ and $$\overline{X}_{\overline{s}}(\overline{x})_{\overline{\eta}}$$ its generic fiber.

Call $$t_x : \overline{X}_{\overline{s}}(\overline{x})\to \overline{X}_{\overline{s}}$$ the obvious map.

If $$\overline{x}\to \overline{X}_{\overline{s}}$$ does not factor through $$X_{\overline{s}}$$, then $$t_x^*j_!\mathbf{Q}_{\ell} = 0$$. Right?

In particular the stalk $$\Psi_g(j_!\mathbf{Q}_{\ell})_{\overline{x}}$$ is zero and so $$i_s^*\Psi_g(j_!\mathbf{Q}_{\ell}) = 0$$ and we're done.

Am I correct?

SUMMARY OF THE ANSWER BELOW: the answer is no, and below there is an example of a smooth $$X$$ of dimension $$1$$ for which $$H^1_c(X_{\overline{s}},\mathbf{Q}_{\ell})\to H^1_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})$$ is not an isomorphism.

This also gives an explicit example for which formation of the nearby cycles complex does not commute with extension by zero. My argument above breaks down when I say "$$t_x^*j_!\mathbf{Q}_{\ell} = 0$$".

However, if $$X$$ is of dimension $$d$$, smooth and with geometrically connected fibers, the specialization map is indeed an isomorphism in degree $$0$$ and $$2d$$ even though $$X$$ is not proper.

The statement is false without proper assumption.

Consider any "degeneration of a smooth elliptic curve to a nodal curve" and delete a singular point in a special fibre. This will give you a counterexample for the dimension reasons.

Details: Start with any proper morphism $$f: \mathcal E' \to \operatorname{Spec}R$$ such that its generic fibre is an elliptic curve (smooth geometrically connected curve of genus $$1$$ with a fixed section) and its special fibre is a nodal curve. For example, take $$R=\overline{\mathbf F_p}[[t]]$$ and consider a curve $$\mathcal E'\subset \mathbf P^2_{R}$$ given by the equation $$Y^2Z-X^3-X^2Z-t^3=0.$$

This curve has exactly one singular point in a special fibre given by $$p=[0:0:1] \subset \mathbf P^2_{R}(\overline{\mathbf F_p})$$. Define $$\mathcal E:=\mathcal E' \setminus {p}.$$ By the construction $$\mathcal E'$$ is a smooth $$R$$-scheme of relative dimension one, but it is not proper. Let us compute cohomology with compact support of the geometric fibres of this curve.

Geometric generic fibre: We know $$\mathcal E_{\overline{\eta}}$$ is a connected proper smooth curve of genus 1, so $$\mathrm{H}^1_{c}(\mathcal E_{\overline{\eta}}, \mathbf Q_l)=\mathrm{H}^1(\mathcal E_{\overline{\eta}}, \mathbf Q_l)=\mathbf Q_l^2.$$

Geometric special fibre: Here we see that $$\mathcal E_{\overline{s}}$$ is a nodal cubic $$Y^2Z-X^3-X^2Z=0$$ minus one point $$[0:0:1]$$. It is a standard computation to show that this scheme is actually isomorphic to $$\mathbf G_m$$. We know that $$\mathrm{H}^1(\mathbf G_m, \mathbf Q_l)=\mathbf Q_l.$$

And finally using Poincare duality between usual cohomology and cohomology with compact support (using that $$\mathbf G_m$$ is smooth) we conclude that $$\mathrm{H}^1_{c}(\mathcal E_{\overline{s}}, \mathbf Q_l) = \mathrm{H}^1_{c}(\mathbf G_m, \mathbf Q_l)=\mathrm{H}^1(\mathbf G_m, \mathbf Q_l)^{\vee}=\mathbf Q_l.$$

So, just for dimension reasons we can't have an isomorphism

$$\mathrm{H}^1_{c}(\mathcal E_{\overline{s}}, \mathbf Q_l) \to \mathrm{H}^1_{c}(\mathcal E_{\overline{\eta}}, \mathbf Q_l).$$

Remark 1: I assume everywhere that $$\ell$$ is coprime with $$p$$.

Remark 2: This example may be a little bit misleading. Actually the failure of a specialization map to be an isomorphism has nothing to do with singularities of the "compactified family". You can produce a lot of examples by deleting points from a special fibre of smooth proper families of curves as well. But I think that this example is easier to visualize.

• OK excellent. Thanks. I’m wondering if, however, if my $X$ is smooth as in the question, and in addition connected of pure dimension $d$, the specialization map in top degree is indeed an isomorphism. It should be, since we have trace maps $H^{2d}_c(X_{\overline{s}},\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}$ and $H^{2d}_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}$, and both trace maps are isomorphisms. Also, traces are compatible with base change so they fit into a commutative square with the specialization map on top and the identity on $\mathbf{Q}_{\ell}$ at the bottom – Ari Feb 20 '19 at 2:21
• Am I right at least in this special case when $p = 2d$? Thanks a lot – Ari Feb 20 '19 at 2:22
• It seems reasonable to me. But there is an issue with twists, so things should work well once you assume that $\mu_{l^{\infty}}$ lie inside $R$. Part of the Poincare duality says that Trace map $R^{2d}f_!\mathbf Q_l(d) \to \mathbf Q_l$ is an isomorphism for smooth $R$-schemes with geometrically connected fibres of dimension $d$. Since formation of $R^i f_!$ commutes with base change by the Proper Base Change Theorem you get your statement (since it holds for $\mathbf Q_l$). – gdb Feb 20 '19 at 2:34
• Ok, sorry, my comment about twists was rather dumb. It works well for $\mathbf Q_l$ just because $R^{2d}f_! \mathbf Q_l \cong \mathbf Q_l(-d)$. So there is no need to make any assumption about roots of unity. – gdb Feb 20 '19 at 2:45
• Yes I agree. Excellent. Great answer too. – Ari Feb 20 '19 at 3:33