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Recall that $\mathbb{RP}^2\#\mathbb{RP}^2$ is the Klein bottle and can be seen as a non-trivial $S^1$-bundle over $S^1$. In particular, it is the total space of the sphere bundle of $\gamma\oplus\varepsilon^1$ over $S^1$ where $\gamma$ is the Möbius line bundle and $\varepsilon^1$ is the trivial line bundle; note, $\gamma\oplus \varepsilon^1$ is the unique non-trivial rank two vector bundle on $S^1$ (because $\pi_0(O(2)) = \mathbb{Z}_2$) and it is not orientable (i.e. $w_1(\gamma\oplus\varepsilon^1) \neq 0$) as is $K$.

Also $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ can be seen as a non-trivial $S^2$-bundle over $S^2$. One way to do this is to view it as the first Hirzebruch surface $\mathbb{P}(\mathcal{O}(-1)\oplus\mathcal{O})$; here $\mathcal{O}(-1)$ and $\mathcal{O}$ are holomorphic line bundles over $\mathbb{CP}^1$. Alternatively, it is the total space of the sphere bundle of $\mathcal{O}(-1)\oplus\varepsilon^1$ over $S^2$. Note that $\mathcal{O}(-1)\oplus\varepsilon^1$ is the unique non-trivial rank three vector bundle on $S^2$ (because $\pi_1(SO(3)) = \mathbb{Z}_2$) and it is not spin (i.e. $w_2(\mathcal{O}(1)\oplus\varepsilon^1)) \neq 0$) as is $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$.

This naturally leads one to ask:

Can $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ be realised as (non-trivial) sphere bundles over $S^4$ and $S^8$ respectively?

Not all sphere bundles in these dimensions necessarily arise as sphere bundles of vector bundles. However, given the constructions above, I would expect them to in this case.

Unlike in the real and complex cases, there is not a unique non-trivial vector bundle to take the sphere bundle of because $\pi_3(SO(5)) = \mathbb{Z}$ and $\pi_7(SO(9)) = \mathbb{Z}$. All of these homotopy groups are in the stable range, so it is no coincidence that there we have two $\mathbb{Z}_2$'s and two $\mathbb{Z}$'s appearing. As $w_4(\mathbb{HP}^2\#\overline{\mathbb{HP}^2}) \neq 0$ and $w_8(\mathbb{OP}^2\#\overline{\mathbb{OP}^2}) \neq 0$, the desired vector bundles should have non-zero $w_4$ and $w_8$ respectively (in direct analogy with the cases above).

Note that both $\gamma$ and $\mathcal{O}(-1)$ can be viewed as the tautological (real/complex) line bundles over $\mathbb{RP}^1$ and $\mathbb{CP}^1$ respectively. This suggests that we should try the direct sum of the tautological (quaternionic/octionic) line bundles over $\mathbb{HP}^2$ and $\mathbb{OP}^2$ with $\varepsilon^1$. This is perfectly reasonable in the quaternionic case, but $\mathbb{OP}^2$ does not arise as the projectivisation of $\mathbb{O}^3$, so there is no such bundle; maybe there is a natural replacelement though (probably associated to one of the generators of $\pi_7(SO(8)) \cong \mathbb{Z}\oplus\mathbb{Z}$).

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$\mathbb CP^2\setminus\{pt\}$ is a disk bundle over $\mathbb CP^1$. Therefore the double of $CP^2$ is the bundle over $\mathbb CP^1$ where the fiber is the double of the disk, namely the 2-sphere.

Similarly, $\mathbb HP^2\setminus\{pt\}$ is a $4$-disk bundle over $\mathbb HP^1$. Therefore doubling it one gets a $4$-sphere bundle over $\mathbb HP^1$.

The same argument works for $\mathbb O$.

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  • $\begingroup$ Is it obvious that the pairs of boundary spheres can be glued to each other matchingly simultaneously in all fibres? $\endgroup$ – მამუკა ჯიბლაძე May 31 '18 at 14:51
  • $\begingroup$ That's what it means to take the double. The two halves are mirror images of one another so the fibrations are compatible by definition. $\endgroup$ – Mikhail Katz May 31 '18 at 14:55
  • $\begingroup$ Well yes if you already know the double is fibred compatibly to the fibrations of its halves. Do you? $\endgroup$ – მამუკა ჯიბლაძე May 31 '18 at 14:57
  • $\begingroup$ @მამუკაჯიბლაძე The clearest picture in my mind represents $\Bbb{KP}^n$ minus a ball as a solid spherical shell in $\Bbb K^n$ with the equivalence relation on the outer boundary given by identifying points equivalent under the action of $S(\Bbb K)$ (unit norm elements). The discs are the $\Bbb K$-multiples of a point in that 'boundary' $\Bbb{KP}^{n-1}$; their boundary is a copy of $S(\Bbb K)$ in the "inner boundary component". Now one doubles this by adding another spherical shell glued to the inner boundary, and an equivalence relation on the new innermost boundary... $\endgroup$ – Mike Miller Eismeier May 31 '18 at 16:54
  • $\begingroup$ the fibers are still the $\Bbb K$-multiples of a point in the innermost (or, equivalently, outermost) `boundary component' = copy of $\Bbb{KP}^{n-1}$. This picture might be clearest when applied to $\Bbb K = \Bbb R$ and $n = 3$ (note that $\Bbb{RP}^3 \# \Bbb{RP}^3$ is not the 3D Klein bottle, which should be what one calls the mapping torus of reflection $S^2 \to S^2$.) $\endgroup$ – Mike Miller Eismeier May 31 '18 at 16:55
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Another way of describing the content of the previous answers is as follows. Suppose one starts with one of the Hopf bundles $p:S^{2n-1}\to S^n$ for $n=2,4,8$ coming from $\mathbb C$, $\mathbb H$, or $\mathbb O$, with fibers $S^{n-1}$. The mapping cone of $p$ is then the associated projective plane $\mathbb {CP}^2$, $\mathbb {HP}^2$, or $\mathbb {OP}^2$. The mapping cylinder is obtained from the mapping cone by deleting an open cone on the domain $S^{2n-1}$ of $p$ so this amounts to deleting the interior of a ball $B^{2n}$. The mapping cylinder of a fiber bundle with fibers $S^{n-1}$ is a fiber bundle with fibers $B^n$ over the same base space, $S^n$ in this case. The double of this mapping cylinder is then a bundle over the same base with fibers $S^n$, the double of $B^n$. On the other hand, the double is also the connected sum of two copies of the projective plane since we are taking taking two copies of the projective plane, deleting the interior of a ball $B^{2n}$ from each, then identifying the two resulting boundary spheres $S^{2n-1}$.

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Here is a way of viewing the construction in Mikhail Katz' answer.

Define $\Bbb{KP}^n \# \overline{\Bbb{KP}}^n$ in the following way: you can identify $\Bbb{KP}^n$ with the disc of elements of norm at most 2 in $\Bbb K^n$, modulo the identification $v \sim \lambda v$ for $\lambda \in S(\mathbb K)$ and $\|v \| = 2$. We then consider $\Bbb{KP}^n \setminus D^{n \cdot \dim \Bbb K}$ as the spherical shell of vectors whose norms lie between 1 and 2, inclusive, modulo the identifications on the norm 2 elements. Taking the 'mirror image' in the norm direction, we see that we can write $\overline{\Bbb{KP}}^n$ as a quotient of the spherical shell of elements with norm between $1/2$ and $1$ inclusive (quotient on the $1/2$ side), and the connected sum therefore is described as the space

$$\Bbb{KP}^n \# \overline{\Bbb{KP}}^n \cong X := \{v \in \Bbb K^n \;\bigg|\; \frac 12 \leq \|v\| \leq 2\}/(v \sim \lambda v \text{ iff } \|v\| = 1/2 \text{ or } 2).$$ Write the spherical shell that $X$ is a quotient of as $S$.

There is then a natural projection map $S \to \Bbb{KP}^{n-1}$ given by sending $v \mapsto [v]$, thinking of $S$ as a subset of $\Bbb K^n \setminus \{0\}$. This clearly factors through $X$. A fiber in $X$ above a point in $\Bbb{KP}^{n-1}$ is the intersection of a $\Bbb K$-line through the origin with $S$, after which we pass to the quotient. That takes the norm 1/2 to 2 elements in $\Bbb K$ (so $S(\Bbb K) \times [1/2,2]$) and quotients by the action of $S(\Bbb K)$ on the boundary, which of course is simply transitive on each boundary component, so that the quotient is a sphere.

To see that this is locally trivial, pick a manifold chart of $\Bbb{KP}^n$, giving us a fiberwise norm-preserving map $\rho: U \times \Bbb K \to \Bbb K^n$; then the restriction of our bundle to $U$ is the quotient of $U \times (S(\Bbb K) \times [1/2, 2])$ by the action of $S(\Bbb K)$ on the boundary (that is, the quotient of the fibering of the spherical shell by $S(\Bbb K)$-annuli). Of course, the cone on $S(\Bbb K)$ is a sphere of one dimension larger, and this gives us a local trivialization of our sphere bundle.

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  • $\begingroup$ This would only apply in the octonion case if $\Bbb{OP}^2$ can be described as in the first sentence (the one-point compactification of the tautological line bundle over $\Bbb{OP}^1$). I don't know enough about octonions to know if this is the case. This (and of course Mikhail Katz' answer) verify the claim that the sphere bundle is the unit bundle of the sum of the trivial $\Bbb R$-bundle and the tautological bundle. $\endgroup$ – Mike Miller Eismeier May 31 '18 at 19:06

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