Image of identity under the adjunction of upper shriek and lower shriek for Euclidean spaces?

Question

Let $p: \mathbb{R}^n \longrightarrow \{\ast\}$ be the projection to a point. We have the adjunction $(R p_!,p^!)$, i.e., $$ \operatorname{Hom}_{D^b(\mathbb{Z})}(R p_!F,N) \cong \operatorname{Hom}_{D^b(\mathbb{R}^n)}(F,p^!N)$$ for any $F \in D^b(\mathbb{Z})$ and $ N \in D^b(\mathbb{R}^n)$. Since $R \Gamma_c(\mathbb{R}^n,\mathbb{Z}_{\mathbb{R}^n}) \cong \mathbb{Z}[-n] $, one can deduce that $p^! N \cong N [n]$ where $[n]$ means shifting by $n$. In the case when $F$ is the skyscraper sheaf $ \mathbb{Z}_x$ for some $x \in \mathbb{R}^n$ and $N = \mathbb{Z}$, the isomorphism becomes $$ \operatorname{Hom}_{D^b(\mathbb{Z})}( \mathbb{Z},\mathbb{Z}) \cong \operatorname{Hom}_{D^b(\mathbb{R}^n)} (\mathbb{Z}_x,\mathbb{Z}_{\mathbb{R}^n}[n]).$$ My question is what's an explicit representative for the morphism corresponding to $\operatorname{id}_{\mathbb{Z}}$ on the right hand side?

Answer 1

This is a good question, without an easy answer, even in the case $n=1$.

I will treat this case first. So we want a morphism $\mathbb{Z}_x\rightarrow\mathbb{Z}[1]$ in the derived category of $\mathbb{R}$. Morphism of this kind correspond to extensions $$0\longrightarrow \mathbb{Z}\longrightarrow\mathcal{F}\longrightarrow\mathbb{Z}_x\longrightarrow 0$$ The sheaf $\mathcal{F}$ can be described as follow : $$\mathcal{F}(]a,b[)=\left\{\begin{array}{ll}\mathbb{Z} & \text{if $b<0$ or $a>0$}\\\mathbb{Z}\oplus\mathbb{Z} & \text{if $a<0<b$}\end{array}\right.$$ And with restrictions $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ being the following :

• if $b<0$ then $d<0$ and $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ is the identity of $\mathbb{Z}$.
• if $a>0$ then $c>0$ and $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ is again the identity of $\mathbb{Z}$.
• if $a<0<b$ and $c<0<d$ then $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ is the identity of $\mathbb{Z}\oplus\mathbb{Z}$.
• if $a<0<b$ and $d<0$ then $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ is the projection onto the first factor.
• if $a<0<b$ and $c>0$ then $\mathcal{F}(]a,b[)\rightarrow\mathcal{F}(]c,d[)$ is the map $(a,b)\rightarrow a+nb$.

I think of the first factor as a locally constant function (and this is indeed the image of $\mathbb{Z}$). The second factor exists only around $0$. It is just a number, and when we move away from 0, if we go towards the negative numbers, we forget it, if we go towards the positive number, we add it (with a factor $n$) to the locally constant function.

For each different value of $n$, you get an extension. This gives the isomorphism $\mathbb{Z}\simeq\operatorname{Hom}_{D(\mathbb{R})}(\mathbb{Z}_x,\mathbb{Z}[1])$. The map corresponding to $\operatorname{id}$ is the case $n=1$.

Note that I used the order of $\mathbb{R}$ to define this extension. This is not surprising, the isomorphism $p^!N\simeq N[1]$ depends on the choice of an orientation of $\mathbb{R}$.

I let you check that this is indeed a sheaf (it requires a bit of works for open subset that are union of intervals...). You can also check (very easy) that if $n=0$, then the extension splits as expected.

So the map $\mathbb{Z}_x\rightarrow\mathbb{Z}[1]$ is the boundary of the triangle $\mathbb{Z}\rightarrow\mathcal{F}\rightarrow\mathbb{Z}_x\rightarrow\mathbb{Z}[1]$.

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Now this construction can in fact be used for any codimension 1 submanifold $Y\subset X$ such that the normal bundle of $Y$ is free. Try this for $\mathbb{R}\subset\mathbb{R}^2$. Of course, it is more complicated since an open subset in $\mathbb{R}^2$ can have some weird shape.

The construction for $\{x\}\subset \mathbb{R}^n$ can be done inductively using affine subspaces of increasing dimension : $\{x\}=A_0\subset A_1\subset A_2\subset ....\subset A_n=\mathbb{R}^n$ where $A_i$ is an affine subspace of dimension $i$, and a choice of relative orientation for $A_i\subset A_{i+1}$. These choices gives an orientation of $\mathbb{R}^n$.

Using the previous constructions (with $n=1$), we have short exact sequences : $$0\longrightarrow \mathbb{Z}_{A_i}\longrightarrow\mathcal{F}_{i-1}\longrightarrow\mathbb{Z}_{A_{i-i}}\longrightarrow 0$$

Putting all this together you get a complex $$\mathcal{F}_.=0\rightarrow\mathcal{F}_{n-1}\rightarrow\mathcal{F}_{n-1}\rightarrow...\rightarrow\mathcal{F}_0\rightarrow 0$$ with $\mathcal{F}_0$ in degree $0$. Up to homotopy the complex $\mathcal{F}_.$ only depends on the orientation of $\mathbb{R}^n$.

The inclusion $\mathbb{Z}\rightarrow\mathcal{F}_{n-1}$ and the projection $\mathcal{F}_0\rightarrow\mathbb{Z}_x$ fit into a distinguished triangle $$ \mathbb{Z}[n-1]\rightarrow\mathcal{F}_.\rightarrow\mathbb{Z}_x\rightarrow\mathbb{Z}[n]$$ where the boundary $\mathbb{Z}_x\rightarrow\mathbb{Z}[n]$ is the map you seek.