# Can every manifold be dominated by a parallelizable one?

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?

• 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? – Igor Belegradek Mar 5 at 16:44
• @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. – Mike Miller Mar 5 at 19:41
• 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. – Igor Belegradek Mar 5 at 20:17

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 $$[N]$$.
• Is it right to say that modulo torsion this map is an isomorphism or rather that after rationalization it is an isomorphism? There are probably easier examples, but what comes to mind is that the Hurewicz map for $MU$ in certain degrees is not an isomorphism, but rather multiplication by p. – Connor Malin May 27 at 2:34