# Euler number of 2-disk bundles

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?

$$1.$$ 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.

• 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"). Sep 7, 2021 at 14:21
• @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. Sep 7, 2021 at 15:48
• I must thank you very much for that great answer! Sep 7, 2021 at 18:36
• @MoisheKohan: This answer points to the appropriate framework for such a proof. Sep 7, 2021 at 20:08