A closed, oriented $d$-manifold $M$ is said to dominate another such manifold $N$ if there exists a map $M \to N$ of non-zero degree. (This notion should not be confused with the unrelated concept of homotopical domination that Wall's finiteness obstruction relates to.)

This relation turns $d$-manifolds into a poset (not quite; the relation is not antisymmetric even if we identify diffeomorphic manifolds, as there are examples of manifolds that dominate each other and are not homotopy equivalent; but this subtlety is rather irrelevant) with $S^d$ as its smallest element. It has been of interest to geometric topologists at least since the times of Hopf to study this poset. For a survey on this topic, see http://www.map.mpim-bonn.mpg.de/images/c/cf/Brouwer_degree.pdf.

My question is: can every manifold $N$ be dominated by another manifold $M$ that is (stably) parallelizable, i.e., has trivial stable tangent bundle?

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    $\begingroup$ As mentioned in Theorem 2.13 of the survey you linked, Gaifullin proved that every oriented closed $n$-manifold is dominated by a hyperbolic $n$-manifold for $n\le 4$. Since hyperbolic manifolds become stably parallelizable in a finite cover (by a theorem of Sullivan), this answers the question in dimensions $\le 4$. Gaifullin also proved that any oriented closed manifold is dominated by a Tomei manifold. Are Tomei manifolds stably parallelizable? $\endgroup$ Mar 5, 2020 at 16:44
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    $\begingroup$ @IgorBelegradek Correct me if I am wrong, but let $U$ be the space of symmetric tridiagonal n x n matrices with distinct eigenvalues. This is open in the space of symmetric tridiagonal matrices, which is itself a Euclidean space, hence $TU$ is trivial. Then the map $F: U \to \Bbb R^n$ sending a matrix to its eigenvalues (listed in ascending order) is a smooth map, with any $(\lambda_1, \cdots, \lambda_n)$ in the image a regular value. Thus the Tomei manifold $F^{-1}(1, 2, \cdots, n)$ has framed normal bundle and hence is stably parallelizable. $\endgroup$
    – mme
    Mar 5, 2020 at 19:41
  • $\begingroup$ Mike Miller: this sounds right. A related interesting question is when a closed manifold is homotopy equivalent to a stably parallelizable manifold. E.g. simply-connected surgery shows that a closed simply-connected $n$-manifold $N$ with stably fiber homotopy trivial tangent bundle is homotopy equivalent to a stably parallelizable manifold if $n$ is odd, or $n$ is divisible by $4$ and the signature of $N$ is zero. $\endgroup$ Mar 5, 2020 at 20:17

1 Answer 1


I think this follows from the stable Hurewicz map $\pi_n^s(N) \to H_n(N;\mathbb{Z})$ being an isomorphism modulo torsion. If we set $n = \dim(M)$ then some positive multiple of the fundamental class of $M$ is hit. Under the Pontryagin-Thom isomorphism this gives a stably framed $M$ and a continuous $f: M \to N$ such that $f_*([M])$ is a positive multiple of $[N]$.


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