I am looking for a complete classification in terms of characteristic classes and "computable" (preferably geometric) invariants. There is this work where the authors classify oriented vector bundles for manifolds with $w_2(M)\neq0$ and a condition on $H^4(M;\mathbb Z)$. But this is a purely homotopy theoretic approach which does not take into account, that the underlying space is a manifold.

Is there anything else in the literature about this topic?

Edit: Found something.

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    $\begingroup$ I don't know what such a classification would look like. If $M = S^5$, then there are precisely two isomorphism classes of real rank five bundles, namely the trivial bundle and the tangent bundle. Both of these have the same characteristic classes. Do you have any idea what type of invariant you would use to distinguish these two bundles? $\endgroup$ Mar 29, 2018 at 12:22
  • $\begingroup$ Yes, for spin bundles with $w_4=0$ there is an invariant $k$ which is zero if there is a trivial rank 2 subbundle and 1 else. I can prove the following: Let $E$ and $F$ be two rank 5 vector bundles which are stably isomorphic. Then $E$ and $F$ are isomorphic if and only if their $k$ invariants agree. $\endgroup$ Mar 29, 2018 at 12:42
  • $\begingroup$ Do you have a reference for this invariant? $\endgroup$ Mar 30, 2018 at 1:44
  • $\begingroup$ For the tangent bundle it is the Semi-Kervaire characteristic mod 2, see Atiyah's article "Vector fields on manifolds". This construction can be generalized to arbitrary vector bundles. $\endgroup$ Mar 30, 2018 at 10:44
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    $\begingroup$ Yes, this is the paper I am mentioning in the post. $\endgroup$ Mar 31, 2018 at 23:27

1 Answer 1


I don't know how to answer this question completely, but I'll make some observations.

The oriented vector bundles over $M$ are classified by homotopy classes of maps $f:M\to \tilde{G}_5$, the oriented Grassmannian, denoted $\pi_0 ([M, \tilde{G}_5])$, where $[X,Y]$ denotes the space of mappings from $X$ to $Y$. Let $E\to M$ be a 5-dimensional vector bundle over $M$. Then the Euler class $e(E)=0$, and hence there is a global section. So $E=F\oplus \varepsilon_1$, where $\varepsilon_1=M\times \mathbb{R}$ is the trivial bundle, and $F\to M$ is a 4-dimensional oriented vector bundle. Hence $\pi_0 [M,\tilde{G}_4]\to \pi_0 [M,\tilde{G}_5]$ is onto. I'm not sure if this helps, but it is related to the observation made in the paper that you linked to that in the case of spin 5-manifolds, this bundle is quaternionic.

A second observation is that $M$ admits a cell structure with one $5$-cell. The 4-skeleton $M^4\subset M$ is a cofibration (note that up to homotopy, $M^4 \simeq M - pt.$, so is well defined up to homotopy). Hence the map $[M, \tilde{G}_5]\to [M^4,\tilde{G}_5]$ is a fibration, with fiber $[(S^5,\ast),(\tilde{G}_5,\ast)]$. The fiber has two components, in bijection with the components of $[S^5,\tilde{G}_5]$ from the homotopy exact sequence and the fact that $\tilde{G}_5$ is simply-connected for the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [S^5,\tilde{G}_5]\to \tilde{G}_5$.

A third observation is that by cellular approximation, $\pi_0[M^4,\tilde{G}_5] = \pi_0[M^4, \tilde{G}_5^5]$, where $\tilde{G}_5^5$ is the 5-skeleton. This is because any map $f:M^4 \to \tilde{G}_5$ may be homotoped to $f: M^4\to \tilde{G}_5^4$. Moreover, homotopies between such maps may be homotoped to $\tilde{G}_5^5$, since $M\times [0,1]$ is 5-dimensional.

A fourth observation is that $\tilde{G}_5^5 \cong \tilde{G}_\infty^5$, where $\tilde{G}_\infty$ is the infinite oriented Grassmannian obtained as the nested union $\tilde{G}_1\subset \tilde{G}_2 \subset \cdots$, where each inclusion is induced by stabilization. This can be proved using the cell structure on Grassmannians given by Schubert cells, but I suspect there may be a more direct approach. Then $\pi_0 [M^4,\tilde{G}_5]=\pi_0[M^4,\tilde{G}_5^5]=\pi_0[M^4,\tilde{G}_\infty^5] = \pi_0[M^4,\tilde{G}_\infty] =\pi_0 [M^4,BSO]\leq \tilde{KO}(M^4)$, the real reduced K-theory of $M^4$. The image of $\pi_0 [M^4,\tilde{G}_\infty] \to \pi_0[M^4,G_\infty]=\tilde{KO}(M^4)$ should correspond to the components where $im \{\pi_1(M^4)\to \pi_1(G_\infty)\}=0$, where $\pi_1(G_\infty)=\mathbb{Z}/2$. In principle, this should be computable from the Atiyah-Hirzebruch spectral sequence and the cohomology of $M^4$, which is determined by the cohomology of $M$.

A fifth observation is that with the fibration $[(S^5,\ast),(\tilde{G}_5,\ast)]\to [M, \tilde{G}_5] \to [M^4,\tilde{G}_5]$, the components in the image will correspond to maps where $im\{\pi_0 [M^4,\tilde{G}_5]\to \pi_0 [S^4,\tilde{G}_5]\} =0$, where $S^5\subset M^4$ is the attaching map of $\partial B^5$, the 5-handle. Again, this should be determined by the map $\tilde{KO}(M^4) \to \tilde{KO}(S^4)$.

Now for the above fibration, the fiber has two components, so the number of components should be at most twice the number of components of the base (determined by the above K-theory computations). I'm not sure how to determine the image $\pi_1 [M^4,\tilde{G}_5]\to \pi_0 [(S^5,\ast),(\tilde{G}_5,\ast)]$.

The upshot is that for each 5-dimensional vector bundle over $M$, one may modify it by "inserting" a copy of the tangent bundle to $S^5$ (this can be descibed in terms of clutching functions of the attaching map of the 5-cell). Then the above question reduces to whether this changes the vector bundle up to isomorphism or keeps it the same? I think the above description classifies them up to this operation.

  • $\begingroup$ Thank you for your detailed comments! What I did in my preprint, if I understand this right, is almost the same as you said, however I considered the infinite quaternionic projective spaces instead the Grassmanian. The short exact sequence, which I am using is described in your fifth observation. What you asking, namely to determine the image of $\pi_0[M^4,\tilde G_5] \to \pi_0[S^5,\tilde G_5]$ is probably "my" $\kappa$-invariant in case one substitutes $\tilde G_5$ with $\mathbb H P^\infty$. $\endgroup$ Dec 21, 2018 at 10:08
  • $\begingroup$ Moreover, you are right with your last comment, there are at most two isomorphism classes if one fixes a stable class of a vector bundle. In my preprint I show that it depends if $w_4$ if there the stable class contains only one or two isomorphism classes. $\endgroup$ Dec 21, 2018 at 10:12
  • $\begingroup$ Geometrically, on a vector bundle $V$ of rank $n$ over an $n$-manifold $M$, you always can "add" a vector bundle $E\to S^n$ to $V$ in the following sense: Pinch of a small neighboorhood of a point in $M$ and you obtain a map $M \to M\vee S^n$. You can postcompone this map with $\phi\vee \psi$ where $\phi$ is the classifying map of $V$ and $\psi$ that of $E$. For $n=5$ there is only the tangent bundle of the sphere to add. $\endgroup$ Dec 21, 2018 at 10:19

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