8
$\begingroup$

Let $\mathbf{X}$ be a Grothendieck topos and let $A$ be an abelian group in $\mathbf{X}$. Verdier's Theorem allows one to describe $\mathrm{H}^n(\mathbf{X},A)$ in terms of hypercoverings, namely, as the colimit of $\mathrm{H}^n(U_\bullet,A)$ where $U_\bullet$ ranges over the category of hypercoverings (of the terminal object of $\mathbf{X}$).

One can further form the Cech cohomology $\check{\mathrm{H}}^n(\mathbf{X},A)$, which is the colomit of $\mathrm{H}^n(U_\bullet,A)$ as $U_\bullet$ ranges over hypercoverings with $U_\bullet=\mathrm{cosk}_0(U_\bullet)$. Let me call them Cech hypercoverings. (These are the hypercoverings with $U_n=U_0^{\times (n+1)}$ and the standard simplicial structure.)

A well-known spectral sequence relates the derived-functor cohomology with the Cech cohomology, implying in particular that $\mathrm{H}^1(\mathbf{X},A)$ is canonically isomorphic to $\check{\mathrm{H}}^1(\mathbf{X},A)$. I am not absolutely certain, but it seems correct that under the description provided by Verdier's Theorem, the isomorphism $\check{\mathrm{H}}^1(\mathbf{X},A)\to \mathrm{H}^1(\mathbf{X},A)$ is the obvious one, namely, if a Cech cohomology class is represented by a $1$-cocycle in $Z^1(U_\bullet,A)$, then its image in $\mathrm{H}^1(\mathbf{X},A)$ is the cohomology class represented by that $1$-cocycle.

The latter statement means that for a hypercovering $U_\bullet$ (not necessarily Cech) and any $1$-cocycle $\alpha\in Z^1(U_\bullet,A)$, one can find a Cech hypercovering $U'_\bullet$ and $\alpha'\in Z^1(U'_\bullet,A)$ representing the same cohomology class. I am looking for a way to construct these $U'$ and $\alpha'$ directly. The main problem is that $U_\bullet$ cannot be refined to a Cech hypercovering in general.

An alternative approach to the problem (which also applies to non-abelian $A$) is via the correspondence with $A$-torsors. It is a standard fact that $\check{H}^1(\mathbf{X},A)$ is in 1-1 correspondence with isomorphism classes of $A$-torsors. Explicitly, if $\alpha\in Z^1(U_\bullet,A)$ with $U_\bullet$ being a Cech hypercovering, then the $A$-torsor $P$ corresponding to $\alpha$ can be described by $$ P(V)=\{a\in A(U_0\times V)~:~ \alpha_V \cdot d_0^1a =d_1^1 a\} $$ where $d^1_i:A(U_0\times V)\to A(U_1\times V)=A(U_0\times U_0\times V)$ is induced by $d^1_i:U_1\to U_0$. Suppose now that $U_\bullet$ is an arbitrary hypercovering (not necessarily Cech). The question will be resolved if one can show that $P$ constructed above is still an $A$-torsor. The difficult thing to check is that $P\neq \emptyset$. (However, when $U_\bullet$ is Cech, one can check that $\alpha\in P(U_0)$.)

$\endgroup$

1 Answer 1

2
$\begingroup$

Some hints in the literature led me to an answer, which I find a bit surprising: One can take $U'_\bullet=\mathrm{cosk}_0(U_\bullet)$ and the $1$-cocycle $\alpha\in Z^1(U_\bullet,A)\subseteq A(U_1)$ descends uniquely to a $1$-cocycle in $\alpha'\in Z^1(\mathrm{cosk}_0(U_\bullet),A)\subseteq A(U_0\times U_0)$ along $(d_0,d_1):U_1\to U_0\times U_0$. In other words:

Proposition: For any hypercovering $U_\bullet$, the canonical map $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism. Consequently, the map $\mathrm{H}^1(\mathrm{cosk}_0(U_\bullet),A)\to \mathrm{H}^1(U_\bullet,A)$ is an isomorphism.

This is a nontrivial statement so let me sketch the ad-hoc proof I have.

Step 1: We may assume $U_\bullet=\mathrm{cosk}_1(U_\bullet)$.

Indeed, since $U_\bullet$ is a hypercovering, the map $U_2\to \mathrm{cosk}_1(U_\bullet)_2$ is a covering, and this easily implies that $Z^1(\mathrm{cosk}_1(U_\bullet),A)\to Z^1(U_\bullet,A)$ is an isomorphism, so replace $U_\bullet$ with its $1$-coskeleton.

One consequence of this assumption is that $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$, where $\partial \Delta^2$ is the boundary of the $2$-simplex, realized as a constant simplicial object (i.e. sheaf) in $\mathbf{X}$, and $\underline{\mathrm{Hom}}$ denote internal $\mathrm{Hom}$ in $\mathbf{X}$.

Step 2: Let $\alpha\in Z^1(U_\bullet,A)$. We claim that $a\in A(U_1)$ descends along $(d_0,d_1):U_1\to U_0\times U_0$ to some $\alpha'\in G(U_0\times U_0)$.

Let $V=U_1\times_{U_0\times U_0}U_1$ and let $\pi_1,\pi_2:V\to U_1$ denote the first and second projections. Let $S$ denote the simplicial object of $\mathbf{X}$ obtained by gluing two copies of $\Delta^1$ along their vertices. Then $V=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(S, U_\bullet)$. There is a simplicial map $\partial \Delta^2\to S$ which degenerate the edge $\{1,2\}$ into a vertex. This gives rise to a map $V\to U_2$, which in turn gives rise a map $A(U_2)\to A(V)$. Applying this map to the cocycle equation $d_0^2\alpha-d_1^2\alpha+d^2_2\alpha=0$ in $A(U_2)$ gives $0-\pi_2^*\alpha+\pi_1^*\alpha=0$ in $A(V)$, which means that $\alpha$ descends to $\alpha'\in G(U_0\times U_0)$.

Step 3: We finally claim that $\alpha'$ lies in $Z^1(\mathrm{cosk}_0(U_\bullet),A)$, which proves the surjectivity of $Z^1(\mathrm{cosk}_0(U_\bullet),A)\to Z^1(U_\bullet,A)$. The injectivity follows easily from the fact that $(d_0,d_1):U_1\to\mathrm{cosk}_0(U_\bullet)_1=U_0\times U_0$ is a covering.

To show the claim, it is enough to show that the canonical map $U_2\to \mathrm{cosk}_0(U_\bullet)_2=U_0\times U_0\times U_0$ is a covering. Indeed, if this holds, then the fact that the $1$-cocycle equation holds for $\alpha$ in $A(U_2)$ implies that it holds for $\alpha'$ in $A(U_0\times U_0\times U_0)$. Proving that $U_2\to U_0\times U_0\times U_0$ is a covering amounts to showing that any $3$ "vertices" in $U_0$ can be joined by a "triangle" in $U_2$ locally. But this follows from $U_2=\underline{\mathrm{Hom}}_{\mathrm{Simp}}(\partial \Delta^2, U_\bullet)$ and the fact that $(d_0,d_1):U_1\to U_0\times U_0$ is a covering.

I would value references for this proposition in the literature, if you know them.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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