The manifold of isospectral symmetric tridiagonal matrices and realization of cycles by aspherical manifolds
Given \(n\), let \(\lambda\) be a finite sequence \(\lambda_1>\lambda_2 > \cdots > \lambda_{n+1}\) of real numbers. Define \(M_{\lambda}\) as the set of real symmetric \((n+1)\times(n+1)\)-matrices \((a_{ij})\) with the spectrum \(\lambda\) and such that \(a_{ij}=0\) for \(|i-j|>1\). It turns out that \(M_{\lambda}\) is a closed aspherical manifold and that \(M_{\lambda}\) and \(M_{\lambda'}\) are diffeomorphic for any pair \(\lambda,\lambda'\), cf. \textit{C. Tomei} [Duke Math. J. 51, 981--996 (1984; Zbl 0558.57006)]. So, we have a unique (up to a diffeomorphism) manifold \(M\). Theorem 1.4. For each integral homology class \(z\in H_n(X)\), there exists a finite-fold covering \(\widehat M\) of \(M\) and a continuous map \(f: \widehat M \to X\) such that \(f_*[\widehat M]=qz\) for some nonzero integer \(q\). Furthermore, if \(X\) is path connected then \(\widehat M\) can be chosen to be connected. To my mind, it is remarkable that, in fact, a unique (up to finite covering) manifold realizes, rationally, all homology classes of given dimension. As a corollary, the author notices that, for any homology class \(z\in H_*(M)\) there exists an aspherical closed smooth manifold \(V\) and a map \(f: V \to M\) such that \(f_*[V]=qz\) for some nonzero integer \(q\), (Theorem 1.6). The last corollary is not new, see \textit{M. Davis} and \textit{T. Januszkiewicz}, [J. Differ. Geom. 34, No.~2, 347--386 (1991; Zbl 0723.57017)]. Note, however, that the proof of Theorem 1.4 is purely combinatorical. So, we get a new proof of Theorem 1.6 that avoids the Thom transversality theorem as well as the hyperbolization of Davis--Januszkiewicz.
scientific article; zbMATH DE number 5619789
scientific article