Left and right multiplication

If \(R\) is a locally compact topological ring, and if multiplication on \(R\) is not commutative, then left and right multiplication by an element of \(R\) can scale Haar measure in different ways. For example if \(R\) is the upper-triangular \(2\times 2\) matrices with real entries, then left multiplication by \(\begin{pmatrix}a&0\\0&1\end{pmatrix}\) sends \(\begin{pmatrix}x&y\\0&z\end{pmatrix}\) to \(\begin{pmatrix}ax&ay\\0&z\end{pmatrix}\) and thus scales \(R\)'s additive Haar measure by \(|a|^2\), but right multiplication by \(\begin{pmatrix}a&0\\0&1\end{pmatrix}\) sends \(\begin{pmatrix}x&y\\0&z\end{pmatrix}\) to \(\begin{pmatrix}ax&y\\0&z\end{pmatrix}\) and thus scales \(R\)'s additive Haar measure by a factor of \(|a|\).

What's going on here is that if we regard left and right multiplication as \(\mathbb{R}\)-linear maps from \(R\) to \(R\), then their associated matrices with respect to the obvious basis are \(diag(a,a,1)\) and \(diag(a,1,1)\), which have different determinants.

However, if \(k\) is now any field and if \(B\) is a finite-dimensional central simple algebra over \(k\) (for example a quaternion algebra, the case we'll care about later), and if \(u\in B^\times\) then \(x\mapsto ux\) and \(x\mapsto xu\) are both \(k\)-linear endomorphisms of \(B\), and I claim that they have the same determinant.

LemmaIsSimpleRing.mulLeft_det_eq_mulRight_det #

Say \(B\) is a finite-dimensional central simple algebra over a field \(k\), and \(u\in B^\times\). Let \(\ell_u:B\to B\) be the \(k\)-linear mapping \(x\) to \(ux\) and let \(r_u:B\to B\) be the \(k\)-linear map sending \(x\) to \(xu\). Then \(\det(\ell_u)=\det(r_u)\).

Proof

Determinants are unchanged by base extension, so WLOG \(k\) is algebraically closed. Then it's known that \(B\) must be a matrix algebra, say \(M_n(k)\). Now \(u\) can be thought of as a matrix which has its own intrinsic determinant \(d\), and \(B\) as a left \(B\)-module becomes a direct sum of \(n\) copies of \(V\), the standard \(n\)-dimensional representation of \(B\). Thus \(\det(\ell_u)=d^n\). Similarly \(\det(r_u)=d^n\) and in particular they are equal.

Lean

Formalized as IsSimpleRing.mulLeft_det_eq_mulRight_det.

Sources

Fermat's Last Theorem (Imperial) blueprint: chapter/HaarCharacterProject.tex, line 416

LeanIsSimpleRing.mulLeft_det_eq_mulRight_det

Discussionhttps://github.com/ImperialCollegeLondon/FLT/issues/518

Used by 1

Used byIsSimpleRing.ringHaarChar_eq_addEquivAddHaarChar_mulRight

CorollaryIsSimpleRing.ringHaarChar_eq_addEquivAddHaarChar_mulRight #

If \(B\) is a central simple algebra over a locally compact field \(F\), and if \(u\in B^\times\), then \(d_B(r_u)=\delta_B(u)\) (recall that the latter is defined to mean \(d_B(\ell_u)\)).

Proof

If \(\ell_u\) and \(r_u\) denote left and right multiplication by \(u\) on \(B\), then we have seen in lemma MeasureTheory.addEquivAddHaarChar_eq_ringHaarChar_det that \(d_B(r_u)=\delta_F(\det(r_u))\). Lemma IsSimpleRing.mulLeft_det_eq_mulRight_det tells us that this is \(\delta_F(\det(\ell_u))\) and this is \(\delta_B(u)\) again by corollary MeasureTheory.addEquivAddHaarChar_eq_ringHaarChar_det.