Show that the inclusion \(T^n \hookrightarrow U(n)\) of the \(n\)-torus of diagonal matrices is homotopic to the map \(T^n \to U(1) \hookrightarrow U(n)\) sending an \(n\)-tuple of unit complex numbers \((z_1, \ldots, z_n)\) to the \(1 \times 1\) matrix \((z_1 \cdots z_n)\). Do the same for the diagonal subgroup of \(Sp(n)\).
Solution
We first prove the unitary statement. For two diagonal entries, \[ \operatorname{diag}(z,w) =\operatorname{diag}(zw,1)\operatorname{diag}(w^{-1},w). \] The second factor is a loop in \(SU(2)\cong S^3\), hence is nullhomotopic. Multiplying a chosen nullhomotopy by \(\operatorname{diag}(zw,1)\) gives a homotopy, natural in \((z,w)\), from \[ \operatorname{diag}(z,w) \] to \(\operatorname{diag}(zw,1)\). Applying this successively to adjacent diagonal entries yields a homotopy \[ \operatorname{diag}(z_1,\ldots,z_n) \simeq \operatorname{diag}(z_1\cdots z_n,1,\ldots,1). \] Thus the diagonal inclusion \(T^n\hookrightarrow U(n)\) is homotopic to \[ T^n\xrightarrow{(z_i)\mapsto\prod z_i}U(1)\hookrightarrow U(n). \]
For \(Sp(n)\), the diagonal subgroup is \(Sp(1)^n=(S^3)^n\). The same reduction works because \[ \operatorname{diag}(q_1,q_2) =\operatorname{diag}(q_1q_2,1)\operatorname{diag}(q_2^{-1},q_2). \] The map \[ S^3\to Sp(2),\qquad q\mapsto\operatorname{diag}(q^{-1},q), \] is nullhomotopic: under \(\pi_3(Sp(1))\to\pi_3(Sp(2))\cong\mathbb Z\), the two diagonal blocks represent \(-1\) and \(+1\), so their sum is zero. Iterating gives \[ \boxed{\operatorname{diag}(q_1,\ldots,q_n) \simeq\operatorname{diag}(q_1\cdots q_n,1,\ldots,1).} \]