\(v_1\)-periodic 2-exponents of \(SU(2^e)\) and \(SU(2^e+1)\)
Let \(n\) be a natural number and let \(\nu(n)\) denote the exponent of \(2\) in \(n\). Let NEWLINE\[NEWLINET_j(k)=\sum_{\mathrm{odd}\;i}{j\choose i}i^kNEWLINE\]NEWLINE denote a family of partial Stirling numbers. Suppose \(\mathbf{e}(k,n)=\min\{ \nu(T_j(k): j\geq n\}\) and \(\overline{\mathbf{e}}(n)= \max\{\mathbf{e}(k,n): k\in \mathbb{Z}\}\). The author mentions that \(\overline{\mathbf{e}}(n)\) is what we might call the \(\upsilon_1\)-periodic \(2\)-exponent of \(SU(n)\), where \(SU(n)\) is the special unitary group of degree \(n\).NEWLINENEWLINEThe main theorem of the paper under review which verifies a conjecture of the author's paper [Integers 8, No. 1, Article A56, 25 p., electronic only (2008; Zbl 1232.11034)] is as follows.NEWLINENEWLINETheorem. {\parindent=6mm\begin{itemize}\item[(a)] If \(e\geq 3\), then \(\mathbf{e}(k,2^e)\leq 2^e+2^{e-1}-1\) with equality occurring iff \(k\equiv 2^e-1\;\mathrm{mod}\;2^{2^{e-1}+e-1}\). \item[(b)] If \(e\geq 2\), then \(\mathbf{e}(k,2^e+1)\leq 2^e+2^{e-1}\) with equality occurring iff \(k\equiv 2^e+2^{2^{e-1}+e-1}\;\mathrm{mod}\;2^{2^{e-1}+e}\). NEWLINENEWLINE\end{itemize}} The homotopy \(2\)-exponent of a topological space \(X\), denoted \(\exp_2(X)\), is the largest \(k\) such that \(\pi_*(X)\) contains an element of order \(2^k\). An immediate consequence of the main theorem of the paper is as follows.NEWLINENEWLINECorollary. For \(\epsilon=\{0,1\}\) and \(2^e+\epsilon\geq 5\), NEWLINE\[NEWLINE \exp_2(SU(2^e+\epsilon))\geq 2^e+2^{e-1}-1+\epsilon .NEWLINE\]
scientific article; zbMATH DE number 6094116
scientific article