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As is discussed in this answer, the connected sum of two copies of $\mathbb{RP}^3$ can be realised as the compact affine grassmannian of $2$-planes in $\mathbb{R}^3$, denoted by $\operatorname{cGraff}_2(\mathbb{R}^3)$, and there is a fiber bundle

$$S^1 \to \operatorname{cGraff}_2(\mathbb{R}^3) \to \mathbb{RP}^2.$$

Is there a fiber bundle whose total space is the connected sum of three copies of $\mathbb{RP}^3$?

Michael Albanese
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Dimpi Paul
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    Hi Dimpi Paul. If you edit the question to provide more context so that it can be reopened, I'd be happy to answer it. – Michael Albanese Aug 14 '23 at 11:46
  • What does cGraff_2(\mathbb{R}^3) mean? It's not just the affine grassmannian of 2-planes in $\mathbb{R}^3$. – Michael Albanese Aug 15 '23 at 00:51
  • cGraff_2(R^3) denotes compact affine Grassmannian of 2 planes in R^3. – Dimpi Paul Aug 15 '23 at 09:57
  • see https://math.stackexchange.com/questions/4365938/connected-sum-of-two-copies-of-rp3?rq=1 – Dimpi Paul Aug 15 '23 at 09:59
  • I have edited your question to include this information. I replaced vector bundle by fiber bundle as that is the correct name. If you are unclear of the difference, please let me know. The question will potentially be reopened, and if it is, it can be answered. – Michael Albanese Aug 15 '23 at 16:07

1 Answers1

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Let $3\mathbb{RP}^3$ denote the connected sum of three copies of $\mathbb{RP}^3$. If $F \to 3\mathbb{RP}^3 \xrightarrow{\pi} B$ is a fiber bundle with manifold fiber and base, then $F$ and $B$ are closed (compact and without boundary), and $B$ is connected. Note that $\dim F, \dim B \in \{0, 1, 2, 3\}$ and $\dim F + \dim B = \dim 3\mathbb{RP}^3 = 3$.

$\bullet$ If $\dim F = 3$ and $\dim B = 0$, then we have only one possibility: $3\mathbb{RP}^3 \to 3\mathbb{RP}^3 \to \ast$ where $\ast$ denotes a singleton.

$\bullet$ If $\dim F = 2$ and $\dim B = 1$, then we have a fiber bundle of the form $F \to 3\mathbb{RP}^3 \to S^1$. Note that a part of the associated long exact sequence in homotopy is

$$\dots \to \pi_1(3\mathbb{RP}^3) \to \pi_1(S^1) \to \pi_0(F) \to \dots$$

Since $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ has a generating set which are torsion elements and $\pi_1(S^1) \cong \mathbb{Z}$ is torsion-free, the map $\pi_1(3\mathbb{RP}^3) \to \pi_1(S^1)$ is trivial, so $\pi_1(S^1)$ injects into $\pi_0(F)$. But this would imply that $F$ has infinitely many components, contradicting the compactness of $F$.

$\bullet$ If $\dim F = 1$ and $\dim B = 2$, then we have a fiber bundle $F \to 3\mathbb{RP}^3 \to B$ where $F$ is a disjoint union of finitely many copies of $S^1$, and $B$ is a closed surface. Again, from the long exact sequence in homotopy we have

$$\dots \to \pi_1(3\mathbb{RP}^3) \to \pi_1(B) \to \pi_0(F) \to \dots$$

If $B \neq S^2$ or $\mathbb{RP}^2$, then $\pi_1(B)$ is torsion-free and infinite, so by the same argument as in the previous case, $F$ would have to have infinitely many components which is impossible.

If $B = S^2$, then from the long exact sequence in homotopy we have

$$\dots \to \pi_1(F) \to \pi_1(3\mathbb{RP}^3) \to \pi_1(S^2) \to \dots$$

Since $S^2$ is simply connected, we see that $\pi_1(F) \cong \mathbb{Z}$ surjects onto $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ which is impossible as the only infinite quotient of $\mathbb{Z}$ is $\mathbb{Z}$ itself and $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2\not\cong\mathbb{Z}$.

If $B = \mathbb{RP}^2$, we instead obtain

$$\dots \to \pi_1(F) \to \pi_1(3\mathbb{RP}^3) \xrightarrow{\alpha} \pi_1(\mathbb{RP}^2) \to \dots$$

where now $\pi_1(\mathbb{RP}^2) \cong \mathbb{Z}_2 \not\cong 0$. If the map $\alpha$ is trivial, then the same argument as in the $S^2$ case applies. Suppose then that $\alpha$ is non-trivial (and hence surjective). Then we see that $\pi_1(F) \cong \mathbb{Z}$ surjects onto the kernel of $\alpha$. Since $\pi_1(3\mathbb{RP}^3) \cong \mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ is infinite and $\mathbb{Z}_2$ is finite, we see that $\ker\alpha$ is infinite. Again, the only infinite quotient of $\mathbb{Z}$ is $\mathbb{Z}$ itself, so we must have $\mathbb{Z} \cong \ker\alpha$. Let $a, b, c$ denote the standard generators of $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$. Since these elements are torsion, they do not belong to the kernel of $\alpha$ as $\ker\alpha\cong\mathbb{Z}$ is torsion-free. It follows that $ab$ and $ac$ do belong to the kernel which is a contradiction as one must be a power of the other since $\ker\alpha$ is infinite cyclic, but the only relations in $\mathbb{Z}_2\ast\mathbb{Z}_2\ast\mathbb{Z}_2$ are $a^2 = b^2 = c^2 = 1$.

$\bullet$ If $\dim F = 0$ and $\dim B = 3$, then $3\mathbb{RP}^3$ is a finite covering space of a connected closed three-manifold. Of course, there is the trivial covering $\ast \to 3\mathbb{RP}^3 \to 3\mathbb{RP}^3$, but $3\mathbb{RP}^3$ can also be realised as the total space of a non-trivial covering. Let $L(4; 1, 1)$ be the lens space $S^3/\mathbb{Z}_4$ where $\mathbb{Z}_4$ acts by $(z_1, z_2) \mapsto (e^{\pi i/2}z_1, e^{\pi i/2}z_2)$. This is double covered by $\mathbb{RP}^3$, see here. Therefore, $\mathbb{RP}^3\# 2\mathbb{RP}^3 = 3\mathbb{RP}^3$ double covers $L(4; 1, 1)\#\mathbb{RP}^3$ by the useful trick mentioned in this answer. Similarly, since $S^3$ is a triple cover of $L(3; 1, 1)$, we see that $S^3\# 3\mathbb{RP}^3 = 3\mathbb{RP}^3$ is a triple cover of $L(3; 1, 1)\#\mathbb{RP}^3$.

Conclusion: $3\mathbb{RP}^3$ cannot be realised as the total space of a fiber bundle with positive-dimensional manifold fiber and base, but it can be realised as the total space of a non-trivial covering map.

Michael Albanese
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