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I'll begin with a broad question: if $M$ is a smooth manifold and $E \to M$ is a stably trivial bundle, can one determine lower bounds on the rank $k$ of the trivial bundle needed such that $E \oplus \underline{\mathbb{R}}^k$ is trivial?

An obvious example is for the tangent bundle of spheres: $TS^n \to S^n$. Here, $k=1$. If I'm not mistaken, if $\Sigma$ is a homotopy $n$-sphere and $f:S^n \to \Sigma$ is a homotopy equivalence, then $f^*T\Sigma \cong TS^n$ which means that $T\Sigma$ is also stably trivial and $k=1$. Kervaire-Milnor showed that $\mathbb{Z}$-homology spheres also have stably trivializable tangent bundles. I do not really understand their proof but it doesn't seem to give any quantitative results on how large the rank needs to be in order to stabilize.

Refined question: In the broad question above, let's require $M$ to be a homology sphere and $E \to M$ to be the tangent bundle, and then ask the same question about lower bounds.

Related question: For a given positive integer $k$, does there exist a smooth manifold with stably trivial tangent bundle that requires at least a rank $k$ trivial bundle to stabilize?

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  • $\begingroup$ Why is $f^*T\Sigma \cong TS^n$? In general, if $f : M \to N$ is a homotopy equivalence, then we do not necessarily have $f^*TN \cong TM$. For example, there are homeomorphic smooth manifolds such that their tangent bundles are not even stably isomorphic, see this answer. $\endgroup$ Commented Mar 10, 2022 at 1:37
  • $\begingroup$ As for the stably parallelisability of integral homology spheres, you might find Oscar Randall-Williams' answer here to be enlightening. $\endgroup$ Commented Mar 10, 2022 at 1:38
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    $\begingroup$ @MichaelAlbanese For the homotopy sphere comment, Johannes Ebert's answer in the link below seems to confirm this. This came to my attention when a paper of Abouzaid suggested something to the effect of: tangent bundles of homotopy spheres do not detect exotic smooth structure but the symplectic topology of cotangent bundles can. As for your second comment, it was that answer which inspired me to ask this question! mathoverflow.net/questions/58131/… $\endgroup$
    – inkievoyd
    Commented Mar 10, 2022 at 1:55

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If $E \to X$ is a rank $r$ real vector bundle, then it is classified by a map $X \to BO(r)$. The existence of an isomorphism $E \cong E_0\oplus\underline{\mathbb{R}}$ (equivalently, the existence of a nowhere-zero section of $E$), corresponds to lifting $X \to BO(r)$ through the map $BO(r-1) \to BO(r)$ induced by the inclusion $O(r-1)\hookrightarrow O(r)$. The obstructions to such a lift lie in $H^n(X; \pi_{n-1}(S^{r-1}))$ as the homotopy fiber of $BO(r-1) \to BO(r)$ is $O(r)/O(r-1) = S^{r-1}$. Moreover, the obstructions to the uniqueness of such a lift, which corresponds to the uniqueness of $E_0$ up to isomorphism, lie in $H^n(X; \pi_n(S^{r-1}))$. This allows us to conclude the following:

  • If $\operatorname{rank}E > \dim X$ (the cohomological dimension of $X$), then all the obstructions to existence vanish, so $E \cong E_0\oplus\underline{\mathbb{R}}$ for some $E_0$ with $\operatorname{rank}E_0 = \operatorname{rank}E - 1$.
  • If $\operatorname{rank}E > \dim X + 1$, then all the obstructions to uniqueness also vanish, so $E \cong E_0\oplus\underline{\mathbb{R}}$ and $E_0$ is unique up to isomorphism.

In particular, if $M$ is an $n$-dimensional smooth manifold with stably trivial tangent bundle, then $TM\oplus\underline{\mathbb{R}}^m \cong \underline{\mathbb{R}}^{n+m}$ for some $m \geq 0$. If $m \geq 2$, then $\operatorname{rank}(TM\oplus\underline{\mathbb{R}}^m) > \dim M + 1$, so there is a unique vector bundle $E_0$ up to isomorphism with $TM\oplus\underline{\mathbb{R}}^m \cong E_0\oplus\underline{\mathbb{R}}$. Now note that $TM\oplus\underline{\mathbb{R}}^m = (TM\oplus\underline{\mathbb{R}}^{m-1})\oplus\underline{\mathbb{R}}$ and $TM\oplus\underline{\mathbb{R}}^m \cong \underline{\mathbb{R}}^{n+m} \cong \underline{\mathbb{R}}^{n+m-1}\oplus\underline{\mathbb{R}}$, so by uniqueness, we have $TM\oplus\underline{\mathbb{R}}^{m-1} \cong \underline{\mathbb{R}}^{n+m-1}$. After finitely many applications of this argument, we see that $TM\oplus\underline{\mathbb{R}}$ is trivial - that is, $k = 1$ unless $M$ is parallelisable, in which case $k = 0$. More generally, if $E \to X$ is a stably trivial bundle with $\operatorname{rank}E = \dim X$, then $k = 1$ unless $E$ is trivial, in which case $k = 0$.

If $E \to X$ is stably trivial and $\operatorname{rank}E < \dim X$ then it is possible that a larger value of $k$ is needed. Such examples can be constructed as in this answer. First note that $TS^n$ is non-trivial for $n \neq 1, 3, 7$. On the other hand, if $n$ is odd, then $TS^n \cong E_0\oplus\underline{\mathbb{R}}^{\rho(n+1)-1}$ where $\rho(n+1)$ denotes that $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$. Therefore $E_0\oplus\underline{\mathbb{R}}^{\rho(n+1)} \cong TS^n\oplus\underline{\mathbb{R}} \cong \underline{\mathbb{R}}^{n+1}$; i.e. for the vector bundle $E_0$, the value of $k$ is $\rho(n+1)$.

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  • $\begingroup$ In the first paragraph, should the $n$'s be $r$'s? $\endgroup$
    – inkievoyd
    Commented Mar 11, 2022 at 0:01
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    $\begingroup$ @inkievoyd: No. For every $n$ there could be an obstruction. Of course, for $n < r$ the groups $H^n(X; \pi_{n-1}(S^{r-1}))$ are trivial. If the bundle is orientable (which is the case if it is stably trivial), then the obstruction in $H^r(X; \pi_{r-1}(S^{r-1})) = H^r(X; \mathbb{Z})$ is the Euler class. $\endgroup$ Commented Mar 11, 2022 at 1:24

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