10 Miniproject: Fujisaki’s Lemma
10.1 The goal
There is an idelic compactness statement which encapsulates both finiteness of the class group of a number field and Dirichlet’s units theorem about the rank of the unit group. In fact there is even a noncommutative version of this statement. In John Voight’s book [ 16 ] this is Main Theorem 27.6.14(a) and Voight calls it Fujisaki’s lemma. I know nothing of the history but I’m happy to adopt this name. In the quaternion algebra miniproject we will use this compactness result to prove finite-dimensionality of a space of quaternionic modular forms.
10.2 Initial definitions
Let \(K\) be a field. A central simple \(K\)-algebra is a \(K\)-algebra \(B\) (not necessarily commutative) with centre \(K\) such that \(B\) has exactly two two-sided ideals, namely \({0}\) and \(B\) (or \(\bot \) and \(\top \), as Lean would call them). We will be concerned only with central simple \(K\)-algebras which are finite-dimensional as \(K\)-vector spaces, and when \(K\) is clear we will just refer to them as central simple algebras. We remark that a 4-dimensional central simple algebra is called a quaternion algebra; we will have more to say about these later on.
Matrix algebras \(M_n(K)\) are examples of finite-dimensional central simple \(K\)-algebras. If \(K=\mathbb {C}\) (or more generally if \(K\) is algebraically closed) then matrix algebras are the only finite-dimensional examples up to isomorphism. There are other examples over the reals: for example Hamilton’s quaternions \(\mathbb {H}:=\mathbb {R}\oplus \mathbb {R}i\oplus \mathbb {R}j\oplus \mathbb {R}k\) with the usual rules \(i^2=j^2=k^2=-1\), \(ij=-ji=k\) are an example of a central simple \(\mathbb {R}\)-algebra (and a quaternion algebra), and matrix algebras over \(\mathbb {H}\) are other central simple \(\mathbb {R}\)-algebras. For a general field \(K\) one can make an analogue of Hamilton’s quaternions \(K\oplus Ki\oplus Kj\oplus Kk\) with the same multiplication rules \(i^2=-1\) etc to describe the multiplication, and if the characteristic of \(K\) isn’t 2 then this is a quaternion algebra (which may or may not be isomorphic to \(M_2(K)\) in this generality).
Some central simple algebras \(B\) are division algebras, meaning that they are division rings, or equivalently that every nonzero \(b\in B\) has a two-sided inverse. For example Hamilton’s quaternions are a division algebra over \(\mathbb {R}\), but \(2\times 2\) matrices are not (even over \(\mathbb {C}\)) because a nonzero matrix with determinant zero such as \(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) has no inverse.
10.3 Enter the adeles
The adeles of a number field are discussed in far more detail in the adele miniproject 8. We just recall here that if \(K\) is a number field then there are two huge commutative topological \(K\)-algebras called the finite adeles \(\mathbb {A}_K^\infty \) and the adeles \(\mathbb {A}_K\) of \(K\), and that they’re both locally compact as topological spaces. We also know from theorem 8.10 that \(\mathbb {A}_K\cong \mathbb {A}_{\mathbb {Q}}\otimes _{\mathbb {Q}}K\) (both topologically and algebraically), meaning that if \(R\) is a \(K\)-algebra then \(R_{\mathbb {A}} := R\otimes _K\mathbb {A}_K\) is naturally isomorphic to \(R\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}\). One can furthermore check that if \(R\) is a finite \(K\)-algebra then the \(\mathbb {A}_K\)-module topologies and \(\mathbb {A}_{\mathbb {Q}}\)-module topologies on \(R_{\mathbb {A}}\) coincide (why?).
Let \(K\) be a number field and let \(D/K\) be a finite-dimensional central simple \(K\)-algebra (later on \(D\) will be a division algebra (hence the name) but we do not need this yet). Then \(D_{\mathbb {A}}:=D\otimes _K\mathbb {A}_K\) is an \(\mathbb {A}_K\)-algebra which is free of finite rank, and if we give \(D_{\mathbb {A}}\) the \(\mathbb {A}_K\)-module topology then it is topological ring (this result is already formalized in the FLT repo and is on the way to mathlib). By the theory of Haar characters (see Chapter 9) there is a canonical character \(\delta _{D_{\mathbb {A}}}:D_{\mathbb {A}}^\times \to \mathbb {R}_{{\gt}0}\) measuring how left multiplication by an element of \(D_{\mathbb {A}}^\times \) changes the additive Haar measure on \(D_{\mathbb {A}}\). Let \(D_{\mathbb {A}}^{(1)}\) denote the kernel of \(\delta _{D_{\mathbb {A}}}\). Theorem 9.18 from the Haar character miniproject shows that \(D^\times \) (regarded as a subgroup of \(D_{\mathbb {A}}^\times \) via the map \(d\mapsto d\otimes 1\)) is in \(D_{\mathbb {A}}^{(1)}\), thus the below theorem typechecks.
If \(D\) is a division algebra then the quotient \(D^\times \backslash D_{\mathbb {A}}^{(1)}\) with its quotient topology coming from \(D_{\mathbb {A}}^{(1)}\), is compact.
The rest of this miniproject is devoted to a proof of this theorem.
10.4 The proof
We prove the theorem via a series of lemmas.
There’s a compact subset \(E\) of \(D_{\mathbb {A}}\) with the property that for all \(x\in D_{\mathbb {A}}^{(1)}\), the obvious map \(xE\to D\backslash D_{\mathbb {A}}\) is not injective.
We know that if we pick a \(\mathbb {Q}\)-basis for \(D\) of size \(d\) then this identifies \(D\) with \(\mathbb {Q}^d\), \(D_{\mathbb {A}}\) with \(\mathbb {A}_{\mathbb {Q}}^d\), and \(D\backslash D_{\mathbb {A}}\) with \((\mathbb {Q}\backslash \mathbb {A}_{\mathbb {Q}})^d\). Now \(\mathbb {Q}\) is discrete in \(\mathbb {A}_{\mathbb {Q}}\) by theorem 8.13, and the quotient \(\mathbb {Q}\backslash \mathbb {A}_{\mathbb {Q}}\) is compact by theorem 8.15. Hence \(D\) is discrete in \(D_{\mathbb {A}}\) and the quotient \(D\backslash D_{\mathbb {A}}\) is compact.
Fix a Haar measure \(\mu \) on \(D_{\mathbb {A}}\) and push it forward to \(D\backslash D_{\mathbb {A}}\); by compactness this quotient has finite and positive measure, say \(m\in \mathbb {R}_{{\gt}0}\). Choose any compact \(E\subseteq D_{\mathbb {A}}\) with measure \({\gt} m\) (for example, choose a \(\mathbb {Z}\)-lattice \(L\cong \mathbb {Z}^d\) in \(D\cong \mathbb {Q}^d\), define \(E_f:=\prod _p L_p\in D\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}^\infty \), and define \(E_{\infty }\subseteq D\otimes _{\mathbb {Q}}\mathbb {R}\cong \mathbb {R}^n\) to be a huge closed ball, large enough to ensure the measure of \(E:=E_f\times E_{\infty }\) is bigger than \(m\)). Then \(\mu (xE)=\mu (E){\gt}m\) so the map can’t be injective.
Define \(X:=E-E:=\{ e-f:e,f\in E\} \subseteq D_{\mathbb {A}}\) and \(Y:=X.X:=\{ xy:x,y\in X\} \subseteq D_{\mathbb {A}}\). Then \(X\) and \(Y\) are also compact subsets of \(D_{\mathbb {A}}\) as they’re continuous images of compact sets.
If \(\beta \in D_{\mathbb {A}}^{(1)}\) then \(\beta X\cap D^\times \not=\emptyset \).
Indeed by the previous lemma, the map \(\beta E\to D\backslash D_{\mathbb {A}}\) isn’t injective, so there are distinct \(\beta e_1,\beta e_2\in \beta E\) with \(e_i\in E\) and \(\beta e_1-\beta e_2=b\in D\). Now \(b\not=0\) and \(D\) is a division algebra, so \(b\in D^\times \). And \(e_1-e_2\in X\) so \(b=\beta (e_1-e_2)\in \beta X\), so we’re done.
Similarly, if \(\beta \in D_{\mathbb {A}}^{(1)}\) then \(X\beta ^{-1}\cap D^\times \not=\emptyset \).
Indeed, \(\beta ^{-1}\in D_{\mathbb {A}}^{(1)}\), and so left multiplication by \(\beta ^{-1}\) doesn’t change Haar measure on \(D_{\mathbb {A}}\), so neither does right multiplication (by theorem 9.16). So the same argument works: \(E\beta ^{-1}\to D\backslash D_{\mathbb {A}}\) is not injective so choose \(e_1\beta ^{-1}\not=e_2\beta ^{-1}\) with difference \(b\in D\) and then \((e_1-e_2)\beta ^{-1}\in D^\times \).
\(Y\cap D^\times \) is finite.
It suffices to prove that \(Y\cap D\) is finite. But \(D\subseteq D_{\mathbb {A}}\) is a discrete additive subgroup, and hence closed. And \(Y\subseteq D_{\mathbb {A}}\) is compact. So \(D\cap Y\) is compact and discrete, so finite.
Now let \(T:=Y\cap D^\times \) be this finite subset of \(D_{\mathbb {A}}\), and define \(K:= (T^{-1}.X) \times X\subset D_{\mathbb {A}}\times D_{\mathbb {A}}\), noting that \(K\) is compact because \(X\) is compact and \(T\) is finite.
For every \(\beta \in D_{\mathbb {A}}^{(1)}\), there exists \(b\in D^\times \) and \(\nu \in D_{\mathbb {A}}^{(1)}\) such that \(\beta =b\nu \) and \((\nu ,\nu ^{-1})\in K.\)
By an earlier lemma, \(\beta X\cap D^\times \not=\emptyset \), and by another earlier lemma, \(X\beta ^{-1}\cap D^\times \not=\emptyset \), so we can write \(\beta x_1=b_1\) and \(x_2\beta ^{-1}=b_2\) with obvious notation.
Multiplying, \(x_2x_1=b_2b_1\in Y\cap D^\times =T\) (recall that \(Y=X*X\) and \(T=Y\cap D^\times \) is finite); call this element \(t\). Note that \(T\subset D^\times \) so \(t\) is a unit, and thus \(x_1,x_2\) are units (a left or right divisor of a unit is a unit; this is a general fact about subrings of matrix rings and may be true more generally).
Then \(x_1^{-1}=t^{-1}x_2\in T^{-1}*X\), and \(x_1\in X\), so if we set \(\nu =x_1^{-1}\) and \(b=b_1\) then we have \(\beta =b\nu \) and \((\nu ,\nu ^{-1})\in K := (T^{-1}*X)\times X\). We are done!
We can now prove Fujisaki’s theorem.
\(D^\times \backslash D_{\mathbb {A}}^{(1)}\) is compact.
Indeed, if \(M\) is the preimage of \(K\) under the map \(D_{\mathbb {A}}^{(1)} \to D_{\mathbb {A}}\times D_{\mathbb {A}}\) sending \(\nu \) to \((\nu ,\nu ^{-1})\), then \(M\) is a closed subspace of a compact space so it’s compact (note that \(\delta _{D_{\mathbb {A}}}\) is continuous, by theorem 9.5). The previous lemma shows that \(M\) surjects onto \(D^\times \backslash D_{\mathbb {A}}^{(1)}\) which is thus also compact.
We note here a useful consequence.
\(D^\times \backslash (D\otimes _K\mathbb {A}_K^\infty )^\times \) is compact.
In this generality the quotient might not be Hausdorff.
There’s a natural map \(\alpha \) from \(D^\times \backslash D_{\mathbb {A}}^{(1)}\) to \(D^\times \backslash (D\otimes _K \mathbb {A}_K^\infty )^\times \). We claim that it’s surjective. Granted this claim, we are home, because if we put the quotient topology on \(D^\times \backslash (D\otimes _K \mathbb {A}_K^\infty )^\times \) coming from \((D\otimes _K \mathbb {A}_K^\infty )^\times \) then it’s readily verified that \(\alpha \) is continuous, and the continuous image of a compact space is compact.
As for surjectivity: say \(x\in (D\otimes _K \mathbb {A}_K^\infty )^\times \). We need to extend \(x\) to an element \((x,y)\in (D\otimes _K \mathbb {A}_K^\infty )^\times \times (D\otimes _K K_\infty )^\times \) which is in the kernel of \(\delta _{D_{\mathbb {A}}}\). Because \(\delta _{D_{\mathbb {A}}}(x,1)\) is some positive real number, it will suffice to show that if \(r\) is any positive real number then we can find \(y\in (D\otimes _K \mathbb {A}_K^\infty )^\times =(D\otimes _{\mathbb {Q}}\mathbb {R})^\times \) with \(\delta _{D_{\mathbb {A}}}(1,y)=r\), or equivalently (setting \(D_{\mathbb {R}}=D\otimes _{\mathbb {Q}}\mathbb {R}\)) that \(\delta _{D_{\mathbb {R}}}(y)=r\). But \(D\not=0\) as it is a division algebra,and hence \(\mathbb {Q}\subseteq D\), meaning \(\mathbb {R}\subseteq D_{\mathbb {R}}\), and if \(x\in \mathbb {R}^\times \subseteq D_{\mathbb {R}}^\times \) then \(\delta (x)=|x|^d\) with \(d=\dim _{\mathbb {Q}}(D)\), as multiplication by \(x\) is just scaling by a factor of \(x\) on \(D_{\mathbb {R}}\cong \mathbb {R}^d\). In particular we can set \(x=y^{1/d}\).