We discuss the computation of $i^\ast R^m j_\ast\mathbb{Q}_\ell$ where $j:\mathbb{A}^n_{\overline{k}}-\{0\}\hookrightarrow \mathbb{A}^n_{\overline{k}}$ and $i:\{0\}\hookrightarrow\mathbb{A}^n_{\overline{k}}$ are the obvious maps.

Motivation

A curious case of reverse engineering

Inspired mostly by the computation performed here, I wanted to see what $i^\ast R^m j_\ast\mathbb{Q}_\ell$ looked like when embedding punctured $n$ -space into $n$ -space. What I found interesting about this computation is that there are, ostensibly, three different ways to attack this computation and, as far as my abilities are concerned, only one is viable. Specifically, Methods and Method 2 lead to other, equivalent (but independently interesting) computations which, if not for Method 3, would still be undoable to me.

Why this is interesting to me is that the computations that Method 1 and Method 2 result in are (again, only for me) undoable without recourse to Method 3 which requires retracing these computations back to their source. In other words, the only way that I could solve the independently interesting computations afforded to me by Methods 1 and 2 would be reverse engineering them to the original problem of computing $i^\ast R^m j_\ast\mathbb{Q}_\ell$ and applying Method 3. This shows the sort of capriciousness of mathematical problems that sometimes in practice one really needs a total reinterpretation to solve them.

The intuition

Of course, we still can guess at what the answer should be based on the topological picture. Namely, if we consider the obvious inclusions $j:\mathbb{C}^n-\{0\}\hookrightarrow\mathbb{C}^n$ and $i:\{0\}\hookrightarrow \mathbb{C}^n$ then we can understand $i^\ast R^m j_\ast\mathbb{Q}$ as a direct limit of cohomology groups:

$\displaystyle i^\ast R^m j_\ast\mathbb{Q}=(R^m j_\ast\mathbb{Q})_0=\varinjlim_U H^m(j^{-1}(U),\mathbb{Q})$

where $U$ travels over neighborhoods of $0$ in $\mathbb{C}^n$ . As in the one-dimensional case cofinal in this system of $U$ are open balls around $0$ , in which case $j^{-1}(U)$ is a punctured ball.

Now, if $B$ is a ball around $0$ so that $j^{-1}(U)$ is the punctured ball $B^\ast$ then one can easily compute its cohomology. Namely:

$H^m(B^\ast,\mathbb{Q})=\begin{cases} \mathbb{Q} & \mbox{if}\quad m=0\\ \mathbb{Q} & \mbox{if}\quad m=2n-1\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

One way of seeing this directly is to see (obviously) that $B^\ast$ is homotopy equivalent to $S^{2n-1}$ . Moreover, since for two balls $B$ and $B'$ , say $B\subseteq B'$ , the natural map $B^\ast\to (B')^\ast$ is a homotopy equivalence, we have that the colimit of cohomology groups just gives

$i^\ast R^m j_\ast\mathbb{Q}=\begin{cases}\mathbb{Q} & \mbox{if}\quad m=0\\ \mathbb{Q} &\mbox{if}\quad m=2n-1\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

as one might expect.

Too many indices…

From this topological computation one would expectthat

$i^\ast R^mj_\ast\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,2n-1\\ 0 & \mbox{otherwise}\end{cases}$

and that is what we endeavor to prove in this post (as well as the interesting tidbits we get by looking at the other methods). That said, purely for notational convenience we focus on the case $n=2$ which is precisely indicative of the general case (just with less indices!).

Method 1

Remark: Again, as a matter of notational convenience, we shall denote by $0$ the origin in any $\mathbb{A}^n$ . We shall conflate it, as per usual, with the obvious closed embedding $i:\text{Spec}(k)\hookrightarrow\mathbb{A}^n_k$ . Also, let us denote $\mathbb{A}^n_k-\{0\}$ by $U_n$ .

Let us attack this computation precisely as we did in the previous $n=1$ case. Namely we know that

$i^\ast R^m j_\ast\mathbb{Q}_\ell=H^m(U_2\times_{\mathbb{A}^2_k}\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}),\mathbb{Q}_\ell)$

Now, as one can easily see we have that

$U_2\times_{\mathbb{A}^2_k}\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}})=\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}})-\{\mathfrak{m}\}$

where $\mathfrak{m}=(x,y)\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}$ is the maximal idea of the strictly local ring. Thus, in essence, we see that we’re trying to compute the etale cohomology of the space $X^\ast:=\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}})-\{\mathfrak{m}\}$ (an example of a punctured spectrum).

So, now our aim is clear: we need to compute the etale cohomology of $X^\ast$ with coefficients in $\mathbb{Q}_\ell$ . And now that our way is illuminated things should be easy–right? Well, no. The issue being that, in practice, it is usually only easy to compute $H^1$ or, more to the point, it’s easy to compute $\pi_1^{\acute{e}\text{t}}$ . This case is no different:

Theorem(Purity of the branch locus): Let $X$ be a regular locally Noetherian scheme and $U\subseteq X$ an open subset such that $\text{codim}(X-U)\geqslant 2$ . Then, the inclusion $U\hookrightarrow X$ induces an isomorphism $\pi_1^{\acute{e}\text{t}}(U,\overline{u})\to \pi_1(X,\overline{u})$ for any geometric point $\overline{u}$ of $U$ .

This theorem (usually attributed to Zariski and Nagata) is certainly difficult techincally but is not so surprising intuitively. Namely, it’s intuitive content is that if $f:Y\to X$ is a cover such that the pullback $Y_U\to U$ is etale then $f$ must also be etale. That said, if $f$ is thought to be relative dimension $0$ (and perhaps flat) then the points where $f$ should fail to be etale should be something like $V(\det(J))$ where $J$ is the Jacobian matrix for $f$ , in particular it should be at least codimension $1$ subset of $X$ . Since $X$ and $U$ differ by codimension at least $2$ we see that $V(\det(J))$ , if not empty, would intersect $Y_U$ . Thus, if we know that $Y_U\to U$ is etale, then so should be $Y\to X$ .

Now, since $\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}=\mathcal{O}_{\mathbb{A}^2_k,0}^\text{sh}$ (and the fact that regular local and Noetherian are preserved under strict Henselization) we know that $\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}$ is a Noetherian regular local ring with maximal ideal $\mathfrak{m}$ . Moreover, since

$2=\dim(\mathcal{O}_{\mathbb{A}^2_k,0})=\dim(\mathcal{O}_{\mathbb{A}^2_k,0}^\text{sh})=\dim(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}})$

we know that $\text{codim}(\{\mathfrak{m}\})=2$ and thus, by the purity of the branch locus, we may conclude that the inclusion $X^\ast\hookrightarrow\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}})$ induces an ismorphism

$\pi_1^{\acute{e}\text{t}}(X^\ast,\overline{0})\xrightarrow{\approx}\pi_1^{\acute{e}\text{t}}(\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}},\overline{0})$

but the latter is obviously $0$ .

Thus, from this, we may conclude that

$i^\ast R^1 j_\ast\mathbb{Q}_\ell=H^1(X^\ast,\mathbb{Q}_\ell)=\text{Hom}_{\text{cont.}}(\pi_1^{\acute{e}\text{t}}(X^\ast,\overline{0}),\mathbb{Z}_\ell)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell=0$

but, as mentioned above, this is essentially where this line of inquiry into the problem ends. Namely, this sort of method is ill-equipped to deal with higher cohomology computations without further remark.

For example, one might hope, naively, that $X^\ast$ is an ‘algebro-geometric $K(\pi,1)$ ‘ so that one can compute $H^i(X^\ast, \mathcal{F})$ , for any LCC sheaf $\mathcal{F}$ on $X^\ast$ , as $H^i(\pi_1^{\acute{e}\text{t}}(X^\ast,\overline{0}),\mathcal{F}_{\overline{0}})$ . Unfortunately, this is not the case. Moreover, one wouldn’t expect it to be the case. This intuitively says that $B^\ast$ is a $K(\pi,1)$ for a small punctured ball around $0$ in $\mathbb{C}^2$ . But, this is obviously false since its universal cover (which is itself) is not contractible.

Thus, this is the end of the line for Method 1.

Method 2

Now that we see that a direct attack on this problem, using the same methods as in the $n=1$ case, is somewhat difficult we instead try to reduce to the $n=1$ case. This is, essentially, by a Mayer-Vietoris argument.

Namely, let us denote by $V_1\subseteq\mathbb{A}^2_k-\{0\}$ the space $\mathbb{G}_m\times_k\mathbb{A}^1_k$ and similarly let $V_2$ denote $\mathbb{A}^1_k\times_k\mathbb{G}_m$ . Note then that $V_1\cup V_2=\mathbb{A}^2_k-\{0\}$ . Thus, we see that we might try and understand $R^mj_\ast\mathbb{Q}_\ell$ in terms of the Mayer-Vietoris sequence for this cover.

To wit, we have the following exact sequence of sheaves on the etale site of $\mathbb{A}^2_k$ :

$\cdots\to R^m j_\ast\mathbb{Q}_\ell \to R^m j_{1\ast}\mathbb{Q}_\ell\oplus R^mj_{2\ast}\mathbb{Q}_\ell\to R^m j_{12\ast}\mathbb{Q}_\ell\to\cdots$

where $j_1:V_1\hookrightarrow \mathbb{A}^2_k$ and $j_2:V_2\hookrightarrow\mathbb{A}^2_k$ are the obvious inclusions and $j_{12}:V_{12}\hookrightarrow\mathbb{A}^2_k$ is the inclusion with $V_{12}:=V_1\cap V_2$ . By passing to stalks we see that we’ve reduced ourselves to computing $(R^m j_{r\ast}\mathbb{Q}_\ell)_{\overline{0}}$ for $r=1,2$ and $(R^m j_{12\ast}\mathbb{Q}_\ell)_{\overline{0}}$ .

So, what now? Well, now that we are dealing with different pushforwards let us, again, try the trick reducing stalks of pushforwards to cohomologies of strict Henselizations. Namely, we’re now trying to compute the following three quanities:

$\displaystyle (R^m j_{1\ast}\mathbb{Q}_\ell)_{\overline{0}}=H^m(V_1\times_{\mathbb{A}^2_k}\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}),\mathbb{Q}_\ell)=H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{x}\right]\right),\mathbb{Q}_\ell\right)$

$\displaystyle (R^m j_{2\ast}\mathbb{Q}_\ell)_{\overline{0}}=H^m(V_2\times_{\mathbb{A}^2_k}\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}),\mathbb{Q}_\ell)=H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{y}\right]\right),\mathbb{Q}_\ell\right)$

$\displaystyle (R^m j_{12\ast}\mathbb{Q}_\ell)_{\overline{0}}=H^m(V_{12}\times_{\mathbb{A}^2_k}\text{Spec}(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}),\mathbb{Q}_\ell)=H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{xy}\right]\right),\mathbb{Q}_\ell\right)$

unfortunately we’re back in the same sort of situation we were in above but worse. Namely, we need to compute these higher cohomology groups and, as far as I know, the only thing which is easy to directly compute is the first cohomology groups (again because we can get a handle on the fundamental group).

Here we need to use something even more sophisticated than the purity of the branch locus (in fact purity is used in the proof of this result):

Theorem(Generalized Abhyankar’s lemma): Let $(A,\mathfrak{m},\kappa)$ be a strictly Henselian regular local ring of dimension $d$ . Let $t_1,\ldots,t_r$ , for $r\leqslant d$ , be regular local parameters for $A$ . Then,

$\displaystyle \pi_1^{\acute{e}\text{t}}\left(\text{Spec}\left(A\left[\frac{1}{t_1\cdots t_r}\right]\right),\overline{x}\right)^{(p)}\cong \mu^p(\kappa)^r$

where, here, $p$ is characteristic of $\kappa$ , the superscript $(p)$ means the maximal prime-to- $p$ quotient, and $\mu^p(\kappa)$ means $\displaystyle \varinjlim_{N,\, (N,p)=1}\mu_N(\kappa)$ . Thus, non-canonically we have that

$\displaystyle \pi_1^{\acute{e}\text{t}}\left(\text{Spec}\left(A\left[\frac{1}{t_1\cdots t_r}\right]\right),\overline{x}\right)^{(p)}\cong \left(\widehat{\mathbb{Z}}^{(p)}\right)^r$

where, as per usual, $\displaystyle \widehat{\mathbb{Z}}^{(p)}=\prod_{\ell\ne p}\mathbb{Z}_\ell$ .

Applying this with $A=\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}$ , $t_1=x$ and $t_2=y$ shows that

$\displaystyle \pi_1^{\acute{e}\text{t}}\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{x}\right]\right),\overline{0}\right)^{(p)}=\pi_1^{\acute{e}\text{t}}\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{y}\right]\right),\overline{0}\right)^{(p)}=\mu^p(k)$

and

$\displaystyle \pi_1^{\acute{e}\text{t}}\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{xy}\right]\right),\overline{0}\right)^{(p)}=\left(\mu^p(k)\right)^2$

from where it easily follows that

$\displaystyle H^1\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{x}\right]\right),\mathbb{Q}_\ell\right)=H^1\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{y}\right]\right),\mathbb{Q}_\ell\right)=\mathbb{Q}_\ell$

and

$\displaystyle H^1\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{xy}\right]\right),\mathbb{Q}_\ell\right)=\mathbb{Q}_\ell^2$

But, once again, we’re stuck. Where do we go from now? It’s not even obvious if we can use these computations to compute $i^\ast R^1j_\ast\mathbb{Q}_\ell$ since we’re still lacking the other terms in the Mayer-Vietoris sequence! So, we must resort to a third, different, method.

Method 3

This method starts as the previous one by attempting to find $i^\ast R^m j_\ast\mathbb{Q}_\ell$ using the Mayer-Vietoris sequence. But, what it does differently is to compute $i^\ast R^m j_{?\ast}\mathbb{Q}_\ell$ (with $?\in\{1,2,12\}$ ) by realizing an extra structure present in these pushforwards. Namely, the maps $j_?$ are all products of maps between one-dimensional objects (in a way to soon be made precise) and thus are amenable to attack via a Kunneth type formula.

So, let us note that we can decompose our $j$ -maps as follows:

$j_1=u_1\times\text{id}, \quad j_2=\text{id}\times u_2,\quad j_{12}=u_1\times u_2$

where $\text{id}$ denotes the identity map on $\mathbb{A}^1_k$ and $u_i$ (for $i=1,2$ ) is the inclusion $\mathbb{G}_m\hookrightarrow\mathbb{A}^1_k$ . Thus, we see that, in all cases, we’re trying to compute something of the form $R^m(f_1\times f_2)_\ast$ and this can be done fruitfully using the Kunneth formula:

Theorem(Kunneth formula): Let $k$ be a separably closed field, $X_i$ and $Y_i$ for $i=1,2$ finite type $k$ -schemes, and $f_i:X_i\to Y_i$ morphisms over $k$ . Let $p_i:X_1\times_k X_2\to X_i$ and $q_i:Y_1\times Y_2\to Y_i$ be the obvious projections. Then, for any $K_i^\bullet\in D^b_c(X_i,\mathbb{Q}_\ell)$ we have that

$R(f_1\times f_2)_\ast\left(p_1^\ast K_1^\bullet \otimes^L p_2^\ast K_2^\bullet\right)=\left[q_1^\ast(R f_{1\ast}K_1^\bullet)\right]\otimes^L \left[q_2^\ast(Rf_{2\ast}K_2^\bullet)\right]$

Thus, in the case $j_1$ , this reads

$Rj_{1\ast}\mathbb{Q}_\ell=R(u_1\times\text{id})(p_1^\ast\mathbb{Q}_\ell\otimes^L p_2^\ast\mathbb{Q}_\ell)=(Ru_{1\ast}\mathbb{Q}_\ell)\otimes^L (R\text{id}_\ast\mathbb{Q}_\ell)$

Passing to stalks in the Kunneth formula (and noting that we don’t have to worry about any $\text{Tor}$ -terms since we’re dealing with $\mathbb{Q}_\ell$ -spaces) we see that

$\displaystyle \left(R^m(f_1\times f_2)_\ast\mathbb{Q}_\ell\right)_{\overline{(y_1,y_2)}}=\bigoplus_{a+b=m}(R^1 f_{1\ast}\mathbb{Q}_\ell)_{\overline{y_1}}\otimes_{\mathbb{Q}_\ell}(R^b f_{2\ast}\mathbb{Q}_\ell)_{\overline{y_2}}$

Thus, to compute the higher pushforward’s stalks we’re interested in it suffices to understand the following two more basic computations:

$(R^a u_{i\ast}\mathbb{Q}_\ell)_{\overline{0}}=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad a=0,1\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

for $i=1,2$ and

$\left(R^b \text{id}_\ast\mathbb{Q}_\ell\right)_{\overline{0}}=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad b=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

The first of these is literally the content of the $n=1$ case, and the second is a triviality.

From these, and the Kunneth formula, we easily deduce

$i^\ast R^m j_{i\ast}\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,1\\ 0 & \mbox{if}\quad\text{otherwise}\end{cases}$

and

$i^\ast R^m j_{12\ast}\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,2\\ \mathbb{Q}_\ell^2 & \mbox{if}\quad m=1\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Thus, from the Mayer-Vietoris sequence we deduce that

$i^\ast R^m j_\ast\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,3\\ 0 & \mbox{if}\quad\text{otherwise}\end{cases}$

as claimed (note there is a tiny hiccup in the Mayer-Vietoris sequence where one really only gets the claimed equalities immediately for $m\geqslant 3$ –one really only sees that the dimensions of $m=1$ and $m=2$ term are equal, but we already resolved that the $m=1$ term is trivial).

The uintended computations

Of course by looking at Method 1 and Method 2 we see that the above also proves the following computations of etale cohomology groups:

$\displaystyle H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{x}\right]\right),\mathbb{Q}_\ell\right)=H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_{k},\overline{0}}\left[\frac{1}{y}\right]\right),\mathbb{Q}_\ell\right)=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,1\\ 0 & \mbox{if}\quad\text{otherwise}\end{cases}$

and

$\displaystyle H^m\left(\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{xy}\right]\right),\mathbb{Q}_\ell\right)=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,2\\ \mathbb{Q}_\ell^2 & \mbox{if}\quad m=1\\ 0 & \mbox{if}\quad\text{otherwise}\end{cases}$

and

$\displaystyle H^m\left(X^\ast,\mathbb{Q}_\ell\right)=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0,3\\ 0& \mbox{if}\quad \text{otherwise}\end{cases}$

but these computations would have been highly non-trivial to do out of context. Namely, one sees that, in all cases, we used ‘global techniques’, namely we realized these cohomologies as being stalks of global maps which actually decomposed. So, for example, one might try and imagine that

$\text{Spec}\left(\mathcal{O}_{\mathbb{A}^2_k,\overline{0}}\left[\frac{1}{x}\right]\right)$

is something like ‘a contractible piece times a topologically interesting piece’–something like “it’s combination of $\mathcal{O}_{\mathbb{A}^1_k,\overline{0}}$ and $\mathcal{O}_{\mathbb{A}^1_k,\overline{0}}\left[x^{-1}\right]$ and the former is ‘contractible’ so we should only really need to consider the latter”. Unfortunately, I don’t know how to make this rigorous beyond our observation that on the level of global maps this really happens (i.e. it’s basically a stalk of $\mathbb{A}^1$ and $\mathbb{G}_m$ and $\mathbb{A}^1$ has trivial relative cohomology over itself).

I would be interested if one could compute these cohomology groups without reinterpreting/re-engineering them as stalks of global objects which actually decompose.

Odds and ends

Here are just two comments about questions that one might have if one read the $n=1$ case.

First, in the $n=1$ case we didn’t just compute the dimensions of the stalks, but also computed them as $G_k$ -representations. But, this is simple. Namely, as the above $n=2$ case showed, the non-trivial cohomology group comes from, essentially, an $n$ -fold tensor product of the $1$ -dimensional case. Since we get $\mathbb{Q}_\ell(-1)$ there one can easily deduce that in the $n=2$ case we get

$i^\ast R^m j_\ast\mathbb{Q}_\ell=\begin{cases}\mathbb{Q}_\ell & \mbox{if}\quad m=0\\ \mathbb{Q}_\ell(-2) & \mbox{if}\quad m=3\\ 0 & \mbox{if}\quad\text{otherwise}\end{cases}$

and this extends in the obvious way to higher dimensions.

The other thing we talked about in the $n=1$ case was how one could relate to this to the cohomology of a formal disk (which made me happier because it’s easier for me to think of a formal disk as a disk than these punctured strictly local rings). So, does something similar happen here? I think the answer is yes, but I’m not positive. Namely a local elder pointed me to Prop. 5.4.53 of Gabber-Romero and this looks like it gives precisely what we want, but I haven’t checked all the details.

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Hard Arithmetic

A computation a day: a (harder) pullback pushforward

Motivation

A curious case of reverse engineering

The intuition

Too many indices…

Method 1

Method 2

Method 3

The uintended computations

Odds and ends

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Motivation

A curious case of reverse engineering

The intuition

Too many indices…

Method 1

Method 2

Method 3

The uintended computations

Odds and ends

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