I am reading a paper by Terry Lawson, and there's something that I don't quite understand:

Let $T_k$ denote the 2-disk bundle over ${\bf S}^2$ with Euler number $k$; $T_2$ is just the tangent bundle of ${\bf S}^2$. Let $N_k$ be the nonorientable 2-disk bundle over ${\bf RP}^2$ with Euler number $k$; $N_1$ is the tangent bundle of ${\bf RP}^2$.

There are lots of things happening in those two sentences:

  • What does he mean by the 2-disk bundle?
  • How does he define the Euler number of such a bundle?
  • Why is it, in each case, the bundle with prescribed number? (there seems to be a uniqueness result involved.)
  • How come $T_2$ and $N_1$ be tangent bundles?

I know, that's more that one question; I will self-moderate my question if you think it's against the rules, but let me try to explain my point first.

I tried looking at related questions on this forum (the ones you end up on when typing “Euler number vector bundle”), but none seemed to clarify the notion. By 2-disk bundle, does he mean a bundle whose fibers are 2-disks, as in here? (thus, the 4th question.)

Even the Euler class doesn't seem to do it, because of that nonorientable ${\bf RP}^2$ part. (Milnor's Characteristic classes didn't help much.)

Maybe my third question will become a terribly easy exercise once I understand the rest?

Thus, I am looking for a reference dealing with that notion: would anyone know where to look for?


1 Answer 1


$1.$ A $2$-disk bundle is a fiber bundle with fiber $D^2$, the two-dimensional disk. More generally, an $F$-bundle over $B$ is a fiber bundle $F \to E \to B$.

$2.$ Every $2$-disk bundle arises as the unit disk bundle of a rank two real vector bundle; this follows from a theorem of Smale that $\operatorname{Diff}(D^2)$ deformation retracts onto $O(2)$. The real rank two bundle is orientable if and only if the $2$-disk bundle is orientable. The Euler class of the $2$-disk bundle is the Euler class of the corresponding real rank two bundle.

There is a non-orientable analogue of the Euler class. If $V \to B$ is a non-orientable real rank $k$ vector bundle, then $e(V)$ is an element of the twisted cohomology group $H^k(B; \mathbb{Z}_w)$. A reference for the construction of this class is chapter $39$ of Steenrod's The Topology of Fibre Bundles.

If $B$ is a closed surface, then the Euler number of $V$ is given by $\langle e(V), [B]\rangle$, where if $B$ is non-orientable, one must use twisted cohomology and homology.

$3.$ Orientable real rank two bundles over $B$ are determined up to isomorphism by their Euler class. In fact, $\operatorname{Vect}_{\mathbb{R}}^{2,+}(B) \to H^2(B; \mathbb{Z})$, $V \mapsto e(V)$ is an isomorphism of groups where $\operatorname{Vect}_{\mathbb{R}}^{2,+}(B)$ denotes isomorphism classes of orientable real rank two bundles with binary operation direct sum. See Propostion $3.10$ of Hatcher's Vector Bundles and K-Theory for example (note that an orientable real rank two vector bundle can be viewed as a complex line bundle, and the Euler class coincides with the first Chern class).

I don't know if there is an analogous statement for non-orientable bundles, but I suspect uniqueness holds for $\mathbb{RP}^2$ at least. (There is; see Moishe Kohan's comment and this answer.)

$4.$ The Euler number of $V = TB$ for a closed surface $B$ is $\langle e(TB), [B]\rangle = \chi(B)$. In particular, the Euler number of $TS^2$ is $\chi(S^2) = 2$, so $TS^2 \cong T_2$ by uniqueness. The Euler number of $T\mathbb{RP}^2$ is $\chi(\mathbb{RP}^2) = 1$, so $T\mathbb{RP}^2 \cong N_1$ by uniqueness.

  • 1
    $\begingroup$ Right, but item 3 needs a clarification in the case of non-orientable bundles. You have to fix a character $\chi: \pi_1(B)\to {\mathbb Z}_2$ of the non-orientable bundle. (Corresponding to $w_1$ of the bundle.) Then one can talk about cohomology with coefficients twisted by $\chi$ and the corresponding twisted Euler class $e$. Then the bundle is determined by the pair $(\chi, e)$ (the proof is by the same argument as in the orientable case, but the only reference I know is in greater generality, for Seifert fibrations, e.g. in Peter Scott's article "Geometry of 3-manifolds"). $\endgroup$ Sep 7, 2021 at 14:21
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    $\begingroup$ @MoisheKohan: Thanks. I suspected something along these lines. In this case $B = \mathbb{RP}^2$, so there is only one non-zero homomorphism $\pi_1(B) \to \mathbb{Z}_2$. In particular, the definition of $N_k$ by Terry Lawson is unambiguous. $\endgroup$ Sep 7, 2021 at 15:48
  • $\begingroup$ I must thank you very much for that great answer! $\endgroup$
    – Anthony
    Sep 7, 2021 at 18:36
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    $\begingroup$ @MoisheKohan: This answer points to the appropriate framework for such a proof. $\endgroup$ Sep 7, 2021 at 20:08

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