I would like to know if there exist submersions $f\colon \mathbb{S}^4\to \mathbb{S}^2$ and $g\colon\mathbb{S}^6\to \mathbb{S}^2.$ It is simply a question of curiosity.

Any suggestion or ideas are appreciated.

  • 4
    $\begingroup$ Note that the Hopf fibration $\Bbb S^3 \to \Bbb S^2$ is a submersion, so it would suffice to find submersions into $\Bbb S^3$. $\endgroup$ May 22, 2015 at 13:54
  • $\begingroup$ thank you, but i still find it difficult. $\endgroup$
    – Đức Anh
    May 22, 2015 at 14:54
  • $\begingroup$ @QiaochuYuan Thanks for this, the argument in your answer is quite nice! $\endgroup$ May 23, 2015 at 8:56
  • $\begingroup$ @Travis: I went ahead and edited in the generalization. $\endgroup$ May 23, 2015 at 20:10

2 Answers 2


In fact there are no submersions from an even sphere $S^{2n}$ into any smooth manifold $M$ of strictly smaller dimension.

The pullback of the tangent bundle of $M$ along such a submersion would be a nontrivial quotient bundle of the tangent bundle of $S^{2n}$, but the tangent bundle of $S^{2n}$ has no nontrivial quotient bundles. This is because its Euler class is nonzero but the Euler class of any orientable bundle of dimension smaller than $2n$ on $S^{2n}$ is necessarily zero, since it lives in a cohomology group which vanishes. (And every bundle on $S^{2n}$ is orientable.)

This argument doesn't apply in the odd case, where sometimes the tangent bundle is even trivializable, and in fact in the odd case there are smooth fiber bundles

$$S^1 \to S^{2n+1} \to \mathbb{CP}^n$$

for all $n$, as well as smooth fiber bundles

$$S^3 \to S^{4n+3} \to \mathbb{HP}^n$$

for all $n$. Then there is an exceptional smooth fiber bundle

$$S^7 \to S^{15} \to \mathbb{OP}^1 \cong S^8$$

but this doesn't generalize.

Edit #2: Here's a different strategy, closer to the one employed in Mike's answer, that handles the odd case. Below all cohomology is with integer coefficients.

Suppose that $S^{n+k} \to S^k$ is a submersion, $n \ge 1$. As in Mike's answer, Ehresmann's theorem shows that such a submersion gives rise to a smooth fiber bundle $F \to S^{n+k} \to S^k$ where $F$ is a closed smooth manifold of dimension $n$. If $k = 1$, the long exact sequence in homotopy shows that $\pi_0(F) \cong \mathbb{Z}$, but this contradicts $F$ compact.

Now suppose $k \ge 2$. By taking homotopy fibers twice, such a smooth fiber bundle extends to a fiber sequence

$$\Omega S^{n+k} \to \Omega S^k \to F \to S^{n+k} \to S^k$$

showing that there is a map $\Omega S^k \to F$ whose homotopy fiber is $n+k-2$-connected, hence which induces an isomorphism on cohomology in degrees $i \le n+k-2$. The ring structure on $H^{\bullet}(\Omega S^k)$ is known (via the cohomological Serre spectral sequence):

  • if $k$ is odd, it's a divided power algebra on a generator $a$ of degree $k-1$.
  • if $k$ is even, it's the tensor product of an exterior algebra on a generator $a$ of degree $k-1$ and a divided power algebra on a generator $b$ of degree $2k-2$.

Since $n+k-2 \ge n$, this determines $H^{\bullet}(F)$ as a ring.

Now, $F$ is frameable, and in particular is orientable, so we must have $H^n(F) \cong \mathbb{Z}$. Since $H^{\bullet}(\Omega S^k)$ is concentrated in degrees that are a multiple of $k-1$, it follows that $n = d(k-1)$ for some $d$.

If either $k$ is odd and $d \ge 2$ or $k$ is even and $d \ge 4$, then the ring structure above is incompatible with Poincaré duality due to the divided power structure. This is a bit annoying to spell out because there are cases depending on the parity of both $d$ and $k$, but for example if $k$ is odd then the product of a generator of $H^{k-1}(F)$ and a generator of $H^{(d-1)(k-1)}(F)$ fails to be a generator of $H^n(F)$.

So either $k$ is odd and $d = 1$, hence $n = k-1$, or $k$ is even and $d = 1, 2, 3$. If $d = 2$ then $F$ has Euler characteristic $1$, and this contradicts that the base $S^k$ has Euler characteristic $2$ but the total space $S^{3k-3}$ has Euler characteristic $0$; it also contradicts $F$ frameable. The conclusion is that

If there is a submersion $S^{n+k} \to S^k, n \ge 1$, then either $k \ge 3$ is odd and $n = k-1$ or $k \ge 2$ is even and $n = k-1$ or $n = 3k-3$.

When $k = 2$ this gives that either $n = 1$, which we already know occurs, or $n = 3$, which is ruled out in Mike's answer.

When $k \ge 3$ and $n = k-1$, the long exact sequence in homotopy shows that if $F \to S^{2k-1} \to S^k$ is a fiber sequence with $k \ge 3$, then $F$ is simply connected. By the computation above, $F$ is an integral homology sphere, and a simply connected integral homology sphere must be homotopy equivalent to a sphere, and so by the topological Poincaré conjecture must be homeomorphic to a sphere. Adams classified all fibrations where all three spaces are spheres, and barring the degenerate case $S^0 \to S^1 \to S^1$ these are the only possibilities for the dimensions of the spheres:

$$S^1 \to S^3 \to S^2$$ $$S^3 \to S^7 \to S^4$$ $$S^7 \to S^{15} \to S^8.$$

The only remaining case is that $k$ is even and $n = 3k-3$, which I asked a question about here.

  • $\begingroup$ I agree, this is much better. I'll leave my answer up because it deals with an interesting odd-dimensional case. $\endgroup$
    – user98602
    May 22, 2015 at 18:21
  • $\begingroup$ Thank you very much. Could you explain these details: - the pullback of $TM$ is non-trivial, - the Euler class of any orientable bundle of dimension smaller than 2n is necessarily zero? $\endgroup$
    – Đức Anh
    May 22, 2015 at 21:34
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    $\begingroup$ @Đức Anh: 1) I don't mean that it's nontrivial as a bundle, I mean that it's nontrivial as a quotient of the tangent bundle (that is, it isn't just the tangent bundle itself, or zero; this is guaranteed by the hypothesis that $M$ has strictly smaller dimension, and of course I'm assuming that $M$ has positive dimension as well). 2) The Euler class of an oriented bundle on $S^{2n}$ of dimension $k$ lives in $H^k(S^{2n}, \mathbb{Z})$, and if $1 \le k \le 2n-1$ then this cohomology group vanishes. $\endgroup$ May 22, 2015 at 21:35
  • $\begingroup$ Thank you. Yes, I understand it's nontrivial as a quotient, but I'm sorry, I have not understood yet why the quotient of the tangent bundle of $\mathbb{S}^{2n}$ is non-trivial. $\endgroup$
    – Đức Anh
    May 22, 2015 at 21:38
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    $\begingroup$ @Đức Anh: the Euler class is multiplicative in short exact sequences; that is, if $U \to V \to W$ is a short exact sequence of oriented vector bundles, then $e(V) = e(U) \cup e(W)$. Hence if either $e(U)$ or $e(W)$ vanishes, then so does $e(V)$. $\endgroup$ May 22, 2015 at 22:05

I can prove there aren't any submersions $S^4 \to S^2$ or $S^5 \to S^2$.

First, we want to apply Ehresmann's theorem. Because $S^m$ is compact any submersion $f: S^m \to S^n$ for is a proper submersion; we need to show that it's surjective. If $f: S^m \to S^n$ was not surjective, you could puncture $S^n$ at one of the points not in the image of $f$ to get a smooth submersion $f: S^m \to \Bbb R^n$. Because submersions are open maps and $S^m$ is compact, $f(S^m)$ must be both open and closed, hence all of $\Bbb R^n$; but this is silly, as $\Bbb R^n$ is not compact.

So Ehresmann's theorem applies, and our submersion is actually a fiber bundle $F \to S^m \to S^n$. Because the map is a submersion, pick any point $x \in S^n$; $f^{-1}(x)$ is a closed manifold of dimension $m-n$. The fiber $F$ is also orientable, because $TS^m$ splits as $\text{Ker}_f \oplus f^*TS^n$. Pulling back under the inclusion we obtain $f^*TS^m = TF \oplus \textbf{n}$, where the last symbol means the trivial bundle of rank $n$. Because all the bundles here but possibly $TF$ here are orientable, $F$ is orientable.

Now we need to work with examples. Start by supposing there's a fiber bundle $\Sigma \to S^4 \to S^2$. The long exact sequence in homotopy groups says $\pi_2(S^4) \to \pi_2(S^2) \to \pi_1(\Sigma) \to \pi_1(S^4)$ is exact; that is, $\pi_1(\Sigma) \cong \Bbb Z$. No closed surface has fundamental group $\Bbb Z$, so this is impossible.

There's also no fiber bundle $F \to S^5 \to S^2$. We see again from the long exact sequence that $\pi_2(S^5) \to \pi_2(S^2) \to \pi_1(M) \to \pi_1(S^5)$ is exact, so $\pi_1(M) \cong \Bbb Z$. The only orientable closed 3-manifold with fundamental group $\Bbb Z$ is $S^2 \times S^1$, so the fiber $F = S^1 \times S^2$. So we see $\pi_k(M) \cong \pi_k(S^2)$. In particular, $\pi_5(M)$ is finite. So looking at the long exact sequence of homotopy groups again we get a contradiction from the exactness of $\pi_5(M) \to \pi_5(S^5) \to \pi_5(S^2)$.

More generally there is no fiber bundle $F \to S^{m+3} \to S^m$ unless $m=4$, and there is no fiber bundle $F \to S^{m+2} \to S^m$. Essentially the same arguments as above apply. In that specific case there is a fiber bundle: $S^3 \to S^7 \to S^4$, as mentioned in Qiaochu's answer.

As for why $S^2 \times S^1$ is the only orientable closed 3-manifold with fundamental group $\Bbb Z$: We say a 3-manifold is prime if every connected sum decomposition of $M$ is of the form $M \# S^3$ (in other words, if an embedded sphere disconnects $M$, then there's a ball on one side). There's also the notion of irreducible: every embedded sphere bounds a ball on one side. Because $\Bbb Z$ does not split as a free product, we could only possibly have $M \# S$, where $S$ is a homotopy sphere; so by the Poincare conjecture $S = S^3$, and $M$ is prime. By the sphere theorem an irreducible 3-manifold with infinite fundamental group is a $K(\pi,1)$. But no closed manifold other than $S^1$ is homotopy equivalent to $\Bbb S^1$, so $M$ is not irreducible. So embed a sphere $S \subset M$ that does not disconnect $M$. So you can pick on the sphere, and a point just to its left and right, and a curve connecting them in the complement of the sphere, and then join them up. Taking a tubular neighborhood of this circle $\cup$ sphere you get $S^2 \times [0,1]$ with a 1-handle glued on, its ends on different sides of $S^2 \times [0,1]$. You can check by hand that the boundary here is a sphere; because it disconnects $M$ the other side must be a ball. There is only one way to glue a ball on a 3-manifold up to diffeomorphism. So one checks by hand that $S^2 \times S^1$ satisfied this description, so $M$ must be $S^2 \times S^1$.

  • $\begingroup$ thank you very much. Your answer is very nice. $\endgroup$
    – Đức Anh
    May 22, 2015 at 18:04
  • $\begingroup$ @ĐứcAnh Thanks, but you should accept Qiaochu's answer. It gives a full and simple solution to your question. $\endgroup$
    – user98602
    May 22, 2015 at 18:21
  • $\begingroup$ Yes, it's better, but your text is very nice to me, since I will learn a lot of things. $\endgroup$
    – Đức Anh
    May 22, 2015 at 18:26
  • $\begingroup$ @Mike: thanks for editing in the details about $S^2 \times S^1$! I edited in a generalization of the argument in my answer. $\endgroup$ May 23, 2015 at 20:11

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