9 Miniproject: Haar Characters
9.1 The goal
The goal of this miniproject is to develop the theory (i.e., the basic API) of Haar characters. “Haar character” is a name I’ve made up to describe a certain character of the units of a locally compact topological ring. The main result we need here is that if \(B\) is a finite-dimensional algebra over a number field \(K\), then \(B^\times \) is in the kernel of the Haar character of \(B\otimes _K\mathbb {A}_K\), where \(\mathbb {A}_K\) is the ring of adeles of \(K\). Most if not all of this should probably be in mathlib.
9.2 Initial definitions
Let \(A\) be a locally compact topological abelian group. There’s then a regular Haar measure \(\mu \) on \(A\), unique up to a positive scalar factor. If \(\phi :(A,+)\cong (A,+)\) is a homeomorphism and an additive automorphism of \(A\), then we can pull back \(\mu \) to get a second measure \(\phi ^*\mu \) on \(A\), which is translation-invariant and regular, and hence also a Haar measure on \(A\). It must thus differ from \(\mu \) by a positive scalar factor, which we call \(d_A(\phi )\). There is a choice of normalization here between \(d_A(\phi )\) and \(d_A(\phi )^{-1}\); we normalize in such that way that if \(X\) is Borel measurable then \(\mu (\phi (X))=d_A(\phi )\mu (X)\). It is easily checked that the definition of \(d_A(\phi )\) is independent of the initial choice \(\mu \) of regular Haar measure.
Note that if \(X\) is a Borel subset of \(A\) with positive finite measure then we can read off \(d_A(\phi )\) by \(d_A(\phi ):=\mu (\phi (X))/\mu (X)\). A nice special case is when \(\mu (X)=1\), in which case we have \(d_A(\phi )=\mu (\phi (X))\) for all \(\phi \).
Now let \(R\) be a locally compact topological ring. The Haar character of \(R\), or more precisely the left Haar character of \(R\), is a group homomorphism \(R^\times \to \mathbb {R}^\times \) defined in the following way. If \(u\in R^\times \) then left multiplication by \(u\), namely the map \(\phi :(R,+)\to (R,+)\) defined by \(r\mapsto ur\), is homeomorphism and an additive automorphism of \((R,+)\), and we define \(\delta _R(u)\) (or just \(\delta (u)\) when \(R\) is clear) to be \(d_R(\phi )\).
9.3 Examples
We discuss some examples of Haar characters.
If \(R=\mathbb {R}\) then \(\delta _R(u)=|u|\).
Take \(\mu \) to be Lebesgue measure and \(X=[0,1]\). We have \(\delta (u)=\mu (uX)\). If \(u{\gt}0\) then \(u[0,1]=[0,u]\) which has measure \(u=|u|\), and if \(u{\lt}0\) then \(u*[0,1]=[u,0]\) which has measure \(-u=|u|\).
If \(R=\mathbb {C}\) then \(\delta _R(u)=|u|^2\).
Multiplication by a positive real \(r\) sends a unit square to a square of area \(r^2=|r|^2\). Multiplication by \(e^{i\theta }\) is a rotation and thus does not change area. The general case follows.
If \(R=\mathbb {Q}_p\) then \(\delta _R(u)=|u|_p\), the usual \(p\)-adic norm.
Normalise Haar measure so that \(\mu (\mathbb {Z}_p)=1\). If \(u\) is a \(p\)-adic unit then \(u\mathbb {Z}_p=\mathbb {Z}_p\) so multiplication by \(u\) didn’t change Haar measure. If however \(u=p\) then \(u\mathbb {Z}_p\) has index \(p\) in \(\mathbb {Z}_p\) and, because \(\mu (i+p\mathbb {Z}_p)=\mu (p\mathbb {Z}_p)\) we have that \(\mu (\mathbb {Z}_p)=p\mu (p\mathbb {Z}_p)\) and thus \(\delta (p)=p^{-1}\). These elements generate \(\mathbb {Q}_p^\times \) and two characters which agree on generators of a group must agree on the group.
If \(R\) is a finite extension of \(\mathbb {Q}_p\) then \(\delta _R(u)\) is the norm on \(R\) normalised in the following way: \(\delta _R(\varpi )=q^{-1}\), where \(\varpi \) is a uniformiser and \(q\) is the size of the (finite) residue field. The proof is the same as for \(\mathbb {Q}_p\), but we won’t need this.
9.4 Algebras
Say \(F\) is a locally compact topological field (for example \(\mathbb {R}\) or \(\mathbb {C}\) or \(\mathbb {Q}_p\)), \(V\) is a finite-dimensional \(F\)-vector space, and \(\phi :V\to V\) is an invertible \(F\)-linear map. Then \(V\) with its module topology (which is the product topology if one picks a basis) is a locally compact topological abelian group, and \(\phi \) is additive. One can check that linearity implies continuity (this is IsModuleTopology.continuous_of_linearMap in mathlib), so in fact \(\phi \) is a homeomorphism and our theory applies. The following lemma gives a formula for the scale factor \(d_V(\phi )\).
\(d_V(\phi )=\delta _F(\det (\phi ))\), where \(\det (\phi )\in F\) is the determinant of \(\phi \) as an \(F\)-linear map.
The proof should be inspired by Real.map_matrix_volume_pi_eq_smul_volume_pi, which crucially uses the induction principle Matrix.diagonal_transvection_induction_of_det_ne_zero. In short, one needs to check it for diagonal matrices and for matrices which are the identity except that one off-diagonal entry is non-zero. Note that we only need the result for \(F=\mathbb {R}\) and \(F=\mathbb {Q}_p\), so if it helps we can assume that \(F\) is second countable.
Now say \(F\) is a locally compact topological field, and \(R\) is a (possibly non-commutative) \(F\)-algebra, finite-dimensional as an \(F\)-vector space. Then both left and right multiplication by a fixed element of \(R\) are \(F\)-linear. We used left multiplication in our definition of the Haar character. If \(u\in R^\times \) then left multiplication by \(u\) is an \(F\)-linear automorphism of \(R\), and hence has a determinant \(\det (u)\in F^\times \). The preceding lemma then immediately implies
If \(u\in R^\times \) then \(\delta _R(u)=\delta _F(\det (u))\).
Follows immediately from the preceding lemma.
9.5 Products
Here are two facts which we will need about products.
If \((A,+)\) and \((B,+)\) are locally compact topological abelian groups, and if \(\phi :A\to A\) and \(\psi :B\to B\) are additive homeomorphisms, then \(\phi \times \psi :A\times B\to A\times B\) is an additive homeomorphism (this is obvious), and \(d_{A\times B}(\phi \times \psi )=d_A(\phi )d_B(\psi )\).
We only need this result in the case where both \(A\) and \(B\) are second-countable, in which case Prod.borelSpace can be used to show that Haar measure on \(A\times B\) is the product of Haar measures on \(A\) and \(B\), and in this case the result follows easily. Without this assumption, the product of these measures may not even be a Borel measure and one has to be more careful. The proof in this case is explained here.
If \(A_i\) are a finite collection of locally compact topological abelian groups, with \(\phi _i:A_i\to A_i\) additive homeomorphisms, then \(d_{\prod _i A_i}(\prod _i\phi _i)=\prod _i d_{A_i}(\phi _i)\).
Induction on the size of the finite set, using the previous lemma.
If \(R\) and \(S\) are locally compact topological rings, then \(\delta _{R\times S}(r,s)=\delta _R(r)\times \delta _S(s)\).
Follows immediately from lemma 9.7.
If \(R_i\) are a finite collection of locally compact topological rings, and \(u_i\in R_i^\times \) then \(\delta _{\prod _i R_i}((u_i)_i)=\prod _i\delta _{R_i}(u_i)\).
Follows immediately from lemma 9.8.
Now say \(A=\prod '_i A_i\) is the restricted product of a collection of topological abelian additive groups \(A_i\) equipped with compact open subgroups \(C_i\). Say \(\phi _i:A_i\to A_i\) are additive homeomorphisms with the property that \(\phi _i(C_i)=C_i\) for all but finitely many \(i\). Then the restricted product \(\prod '_i\phi _i\) is an additive homeomorphism \(\phi :A\to A\). Note also that \(d_{A_i}(\phi _i)=1\) for all the \(i\) such that \(\phi _i(C_i)=C_i\), as \(d_{A_i}(\phi _i)\) can be computed as \(\mu (\phi _i(C_i))/\mu (C_i)\) and \(\mu (C_i)\) is guaranteed to have positive finite measure as it is open and compact. Thus the product \(\prod _i\delta (\phi _i)\) is a finite product (and in particular cannot diverge to \(0\)). Moreover, we have
With \(A\), \(A_i\), \(C_i\), \(\phi _i\), \(\phi \) defined as above, we have \(\delta _A(\phi )=\prod _i\delta _{A_i}(\phi _i)\).
Assume \(\phi _i(C_i)=C_i\) for all \(i\not\in S\), a finite set, and work in the open subgroup \(\prod _{i\in S}A_i\times \prod _{i\not\in S}C_i\). Then \(\phi \) is an automorphism of this open subgroup of \(A\), and in particular we can compute \(\delta (\phi )\) using this group instead. Because \(\prod _{i\not\in S}\phi _i\) is an automorphism of a compact group we must have \(\delta (\phi _{i\not\in S}\phi _i)=1\). Finally \(\delta (\prod _{i\in S}\phi _i)=\prod _{i\in S}\delta (\phi _i)\) by the previous result.
As a special case, if \(R\) is the restricted product of a collection of topological rings \(R_i\) (not necessarily commutative) each equipped with a compact open subring \(C_i\), then we have
If \(u=(u_i)_i\in R^\times \) then \(\delta _R(u)=\prod _i\delta _{R_i}(u_i)\).
By definition of restricted product we have \(u_i\in C_i\) for all but finitely many \(i\). Note also that \(u\) has an inverse \(v=(v_i)_i\) with \(v_i\in C_i\) for all but finitely many \(i\). The fact that \(u_iv_i=v_iu_i=1\) means that \(u_i,v_i\in C_i^\times \) for all but finitely many \(i\). Thus the previous lemma applies.
9.6 Adeles
We have seen in lemmas 9.1 and 9.3 that if \(v\) is a place of \(\mathbb {Q}\) (i.e., a prime number or \(+\infty \)) then \(\delta _{\mathbb {Q}_v}=|\cdot |_v\). We can deduce from this, lemma 9.12 (at the finite places) and lemma 9.9 (to include the infinite place) that if \((x_v)_v\in \mathbb {A}_{\mathbb {Q}}^\times \) then \(\delta _{\mathbb {A}_{\mathbb {Q}}}((x_v)_v)=\prod _v|x_v|_v\), where the product runs over all places of \(\mathbb {Q}\). The product formula (on the way to mathlib) says that if \(x\in \mathbb {Q}^\times \subseteq \mathbb {A}_{\mathbb {Q}}^\times \) then \(\prod _v|x|_v=1.\) A quick proof: if \(x=\pm \prod _pp^{e_p}\) then \(\prod _p|x|_p=\prod _pp^{-e_p}\) and \(|x|_\infty =\prod _pp^{e_p}\) so they cancel. We conclude that \(\delta _{\mathbb {A}_{\mathbb {Q}}}(\mathbb {Q}^\times )=\{ 1\} \).
Next we generalize this to finite-dimensional \(\mathbb {Q}\)-vector spaces. So say \(V\) is an \(N\)-dimensional \(\mathbb {Q}\)-vector space, and define \(V_{\mathbb {A}}:= V\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}\) with its \(\mathbb {A}_{\mathbb {Q}}\)-module topology. If we choose an isomorphism \(V\cong \mathbb {Q}^N\) then \(V_{\mathbb {A}}\cong \mathbb {A}_{\mathbb {Q}}^N\) as an additive topological abelian group. In particular, \(V_{\mathbb {A}}\) is locally compact.
Fix a \(\mathbb {Q}\)-linear automorphism \(\phi :V\to V\). By base extension \(\phi \) induces an \(\mathbb {A}_{\mathbb {Q}}\)-linear automorphism \(\phi _{\mathbb {A}}\) of \(V_{\mathbb {A}}\) which is also a homeomorphism of \(V_{\mathbb {A}}\) if \(V_{\mathbb {A}}\) is given the module topology as an \(\mathbb {A}_{\mathbb {Q}}\)-module. Our goal is
\(d_{V_{\mathbb {A}}}(\phi _{\mathbb {A}})=1.\)
Notation: write \(\mathbb {A}_{\mathbb {Q}}=\mathbb {A}_{\mathbb {Q}}^\infty \times \mathbb {R}\), where \(\mathbb {A}_{\mathbb {Q}}^\infty \) is the finite adeles of \(\mathbb {Q}\). Define \(V_f:= V\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}^\infty \) and for \(v\) a place of \(\mathbb {Q}\) (that is either \(v=p\) prime or \(v=\infty \)) write \(V_v:=V\otimes _{\mathbb {Q}}\mathbb {Q}_v\), so for example \(V_\infty =V\otimes _{\mathbb {Q}}\mathbb {R}\). Then \(V_{\mathbb {A}}=V_f\times V_\infty \). Fix once and for all a \(\mathbb {Z}\)-lattice \(L\subseteq V\) (that is, a spanning \(\mathbb {Z}^N\) in the \(\mathbb {Q}^N\)). For \(p\) a prime, define \(C_p:=L\otimes _{\mathbb {Z}}\mathbb {Z}_p\subseteq V_p\). It can be checked that as a topological abelian group \(V_f\) is the restricted product \(\prod '_p V_p\) over the subgroups \(C_p\).
We have \(\phi :V\to V\). For \(v\) a place of \(\mathbb {Q}\), let \(\phi _v:V_v\to V_v\) denote the \(\mathbb {Q}_v\)-linear base extension of \(\phi \) to \(V_v\). We have \(d_{V_{\mathbb {A}}}(\phi _{\mathbb {A}})=d_{V_f}(\prod '_p \phi _p)\times d_{V_\infty }(\phi _\infty )\), where \(\prod '_p \phi _p\) is the restricted product of the \(\phi _p\) as an endomorphism of \(V_f\). Thus \(d_{V_{\mathbb {A}}}(\phi _{\mathbb {A}})=\prod _p d_{V_p}(\phi _p)\times d_{V_\infty }(\phi _\infty )=\prod _v d_{V_v}(\phi _v)\), where we note that all but finitely many of the \(d_{V_p}(\phi _p)\) are 1.
By Lemma 9.5 we have that \(d_{V_v}(\phi _v)=\delta _{\mathbb {Q}_v}(\det (\phi _v))\), hence \(d_{V_{\mathbb {A}}}(\phi _{\mathbb {A}})=\prod _v\delta _{\mathbb {Q}_v}(\det (\phi _v))\). But \(\det (\phi _v)\) is equal to the determinant of \(\phi \) on \(V\) as \(\mathbb {Q}\)-vector space (because base change does not change determinant), which is some nonzero rational number \(q\). Thus \(d_{V_{\mathbb {A}}}(\phi _{\mathbb {A}})=\prod _v\delta _{\mathbb {Q}_v}(q)=1\) by the product formula for \(\mathbb {Q}\).
If \(B\) is a finite-dimensional \(\mathbb {Q}\)-algebra (for example a number field, or a quaternion algebra over a number field), if \(B_{\mathbb {A}}\) denotes the ring \(B\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}\), and if \(b\in B^\times \), then \(\delta _{B_{\mathbb {A}}}(b)=1\).
Follows immediately from the previous theorem.
If \(B\) is a finite-dimensional \(\mathbb {Q}\)-algebra and if \(b\in B^\times \) then right multiplication by \(b\) does not change Haar measure on \(B_{\mathbb {A}}\).
Follows immediately from the previous theorem.