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Let $E$ be a smooth real vector bundle of even rank, over a smooth manifold $M$. Suppose there exist an orientation-reversing vector bundle isomorphism $\Phi:E \to E$.

Is it true that $E$ has a subbundle of rank $1$?

(We don't need an orientation on $E$, since for maps from a vector space to itself, the notion of orientation-preserving or reversing is always well-defined, without the need to actually choose an orientation on the space).

Here is one approach (which was suggested to me by Amitai Yuval, who also raised this question):

We can put a metric on $E$, and take the orthogonal polar factor $Q$ of $\Phi$, which will now be an isometric orientation-reversing isomorphism. So, $Q$ will have at least one non-zero fixed point at each fiber $E_x$ (see below**).

My hope is that somehow we can extract a continuously changing family of fixed points along the different fibers (one at each fiber) that will form a subbundle of rank $1$. Of course, there can be problems of multiplicity, so perhaps some perturbation argument is needed.

Can this approach work?

**Here we use the fact $\text{rank}(E)$ is even: at each fiber $Q_x$ is essentially an orthogonal matrix with negative determinant. Since its complex eigenvalues comes in conjugate pairs, and the determinant is real negative, there must be some real negative eigenvalues. Since $\det Q_x<0$ the number of the negative eigenvalues must be odd. Since $\dim E_x$ is even, we conclude there must be a positive eigenvalue, which must be $1$, since $Q_x$ is orthogonal.

Motivation:

I am trying to find out which real vector bundles admit orientation-reversing isomorphisms. Of course, every bundle of odd rank admits one: the map $x \to -x$. Now suppose the rank is even. If $E$ admits a subbundle $F$ of rank $1$, we can define an orientation-reversing isomorphism as follows:

$$\Phi|_F=\text{Id}_F,\Phi|_{{F}^{\perp}}=-\text{Id}|_{{F}^{\perp}}$$

where ${F}^{\perp}$ is some complement of $F$.

My question is if this condition ("there exist a subbundle of rank $1$") is necessary.

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  • $\begingroup$ Interesting question. Just a comment - one can imagine a situation in which an even rank vector bundle doesn't split a line bundle but does split an odd rank bundle (if this happens, the rank of the bundle must be $\geq 6$) and then your construction still yields an orientation reversing isomorphism. If this can indeed happen, then one might replace your condition by "$E$ has an odd rank bundle". I would guess that this can happen but I don't know nor I know if this is a necessary condition $\endgroup$
    – levap
    Commented Dec 18, 2017 at 15:49
  • $\begingroup$ What about this? The tuatological bundle over CP^1 has a natural orientation given by the complex structure, as well as its dual. These bundles are real isomorphic (for example via a choice of riemannian metric). Does this isomorphism preserve the orientations? The realification of the bundle does not have a subbundle of rank 1. $\endgroup$
    – Thomas Rot
    Commented Mar 20, 2018 at 12:48

1 Answer 1

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Let us say that a vector bundle is reversible if it has an orientation-reversing automorphism. I claim the following, which provides a negative answer to the question:

Proposition: Up to isomorphism, there is a unique non-trivial real vector bundle of rank $6$ over $S^{10}$. This bundle is reversible and does not have any proper subbundle.

Proof:

As $S^{10}$ is simply-connected, we may assume that our bundles are oriented. Recall that the clutching construction determines a bijection $$ \operatorname{Vect}_r^+(S^n) \cong \pi_{n-1} SO(r), $$ where $\operatorname{Vect}_r^+(S^n)$ is the set of isomorphism classes of oriented bundles of rank $r$ over $S^n$ (where isomorphisms are required to preserve the orientation). A good reference for this is Section 1.2 of this unfinished book by Hatcher.

In the case of $S^{10}$, the first few homotopy groups are as follows:

$$ \begin{array}{c|c|c|c|c|c|} &SO(2) &SO(3) &SO(4) &SO(5) &SO(6) \\ \hline \pi_9 &0 &\mathbb{Z}_3 &\mathbb{Z}_3 \oplus \mathbb{Z}_3 &0 &\mathbb{Z}_2 \end{array} $$ I've taken these values from this nlab page. Another reference is the table on p.970 of the survey [1] by Mimura.

From the above table, we learn the following:

  1. There is a unique non-trivial oriented bundle of rank $6$, up to isomorphism. Suppose that $E$ is such a bundle. By uniqueness $E \cong \overline{E}$, i.e. $E$ is reversible.

  2. Every bundle of rank $5$ is trivial. Therefore, since $E$ is non-trivial, it cannot have a subbundle of rank $1$ or $5$.

  3. Every map $SO(3) \to SO(6)$ induces the zero map on $\pi_9$. Hence, the direct sum map $SO(3) \times SO(3) \to SO(6)$ is trivial on $\pi_9$. This implies that the direct sum of any two bundles of rank $3$ is trivial. Since $E$ is non-trivial, it does not have a subbundle of rank $3$.

  4. Every map $SO(4) \to SO(6)$ induces the zero map on $\pi_9$. Hence, the direct sum map $SO(2) \times SO(4) \to SO(6)$ is trivial on $\pi_9$. By the argument of 3, $E$ does not have a subbundle of rank $2$ or $4$.

Conclusion: $E$ is a reversible bundle that has no proper subbundle. $\square$

[1] MR1361904 Mimura, M. Homotopy theory of Lie groups. Handbook of algebraic topology, 951-991. North-Holland, Amsterdam, 1995.

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