Cohomology of quaternions on an abelian variety - MathOverflow

Cohomology of quaternions on an abelian variety

2010-01-14T20:43:52Z

Given two non-isogenous elliptic curves $E_1$ and $E_2$ over $\mathbb{C}$. Set $A:=E_1 \times E_2$. Given a nontrivial sheaf of quaternion algebras $D$ over $A$, what is the dimension of the vector space $H^1(A,D)$?

If one thinks of $D$ as an element in the Brauer group $Br(A)$, then it is $2$-torsion, hence belongs to $Br(A)[2]$. Since the curves are non-isogenous there is an isomorphism $Pic(E_1)[2] \otimes Pic(E_2)[2] \to Br(A)[2]$. So there should be a connection between such quaternions and $2$-torsion line bundles on the curves, but i cannot find an explicit description for this isomorphism. If there is one, i thought one could use the Künneth formula to compute $H^1(A,D)$ in terms of the cohomology of the line bundles on the curves.

For now i could only work out the bound $d=dim(H^1(A,D)) \geq 2$: using Hirzebruch-Riemann-Roch and simplifying terms one gets $d=c_2(D)+2$. After a result of M.Lieblich one has $c_2(D)\geq 0$.

Does anyone see/have an explicit description of the isomorphism mentioned above? Is the idea using Künneth a promising approach to this problem at all? Or does anyone have another approach? Are there some calculations regarding this in the literature (i couldn't find one)?

Another question in this context is: what is the image of such an algebra under the map $Br(A) \rightarrow Br(\mathbb{C}(A))$. This should be nontrivial $\mathbb{C}(A)$-quaternions, since the map "looking at the genric point $\eta$" is injective, i.e. $D_{\eta}$ is generated by elements $i,j$ with $i^2=a, j^2=b and ij=-ji$. But what are a resp. b? I think they should have something to do with functions h such that 2*Y=div(h), where Y defines one of the line bundles. Is this true?

Answer by Oren Ben-Bassat for Cohomology of quaternions on an abelian variety

2010-01-27T18:27:48Z

To compute the cohomology of $D$, you could consider its pullback via some finite to one map $A\to A$ to for which $D$ trivializes. Then there is a Lyndon-Serre spectral sequence using the group cohomology of $G$ the group you quotient. The $E_{2}$ term looks like $H^{i}(G,H^{j}(A,p^{-1}D))$ and it converges to $H^{i+j}(A,D)$. See the book by Mumford Abelian Varieties. There are 3 other helpful references that you can find in http://arxiv.org/abs/0811.2746. See the articles cited in that paper by Hoobler, Elencwajg and Narasimhan, and Berkovich.

The map from the tensor product of the Picard Groups to the Brauer group can also probably be made explicit using group or Cech Etale cohomology.

--Oren

Answer by Tyler Lawson for Cohomology of quaternions on an abelian variety

2010-01-27T18:47:05Z

For the description of the quaternion algebra associated to a pair of torsion line bundles, try the following. Take line bundles ${\cal L}_i$ on $E_i$ equipped with isomorphisms ${\cal L}_i^{\otimes 2} \to {\cal O}$ , and pull these back to $A$. Define $$D = {\cal O} \oplus {\cal L}_1 \oplus {\cal L}_2 \oplus {\cal L}_1 \otimes {\cal L}_2$$ with multiplication induced by the maps ${\cal L}_i^{\otimes 2} \to {\cal O}$ , ${\cal O}$ being the unit, and the elements of ${\cal L}_1$ and ${\cal L}_2$ anticommuting.

ADDED: I wish I had a more conceptual explanation for why this represents the cup-product in the Brauer group, but here is a cocycle description along the lines of what Oren suggested.

Suppose $X$ is given with 2-torsion line bundles $\cal L$ and $\cal M$. Choose cocycles representing these, in the form of a cover (either open in the analytic case, or etale in the algebraic case) $U_\alpha$ of $X$ together with sections $s_\alpha$ of $\cal L$ and $t_\alpha$ of $\cal M$ on $U_\alpha$ such that $s_\beta / s_\alpha = u_{\alpha \beta} \in \{\pm 1\}$ and similarly $t_\beta / t_\alpha = v_{\alpha \beta}$ ; these latter two are the representing cocycles.

Then D has basis $\{1,s_\alpha, t_\alpha, s_\alpha t_\alpha\}$ on $U_\alpha$ , where $s_\alpha^2 = t_\alpha^2 = 1$ , and you can explicitly make this isomorphic to a matrix algebra. The change-of-basis sends $s_\alpha$ to $s_\beta = u_{\alpha \beta} s_\alpha$ and similarly for $t$. This can be achieved by conjugation by the element $t_\alpha^{(1 - u_{\alpha \beta})/2} s_\alpha^{(1 - v_{\alpha \beta})/2} = g_{\alpha\beta} \in D \cong M_2(\mathbb{C})$ . These change-of-basis matrices reduce to a cocycle in $PGL_2(\mathbb{C})$ representing the algebra, and so the image in the Brauer group is represented by the coboundary $(\delta g)_{\alpha \beta \gamma} \in \{\pm 1\}$ .

EDIT: fixed up following description of the coboundary so that it correctly described where the cup product lands.

Explicit computation finds $(\delta g)_{\alpha \beta \gamma}$ is equivalent to the cocycle $v_{\alpha \beta} \otimes u_{\beta \gamma}$ with coefficients in $\{\pm 1\} \otimes \{\pm 1\} \cong \{\pm 1\}$ (I may have mixed the indices, if I did please let me know and I'll correct it), which is precisely the formula for the cup product of the cocycles $u$ and $v$.

So based on your description of the cup product inducing an isomorphism between 2-torsion in the Brauer group and cup products of 2-torsion elements in the Picard group, this genuinely should provide you with the bundles you're looking for.