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I've been reading about Hopf fibrations. In the Wikipedia page, they state that the Hopf construction generalises to higher-dimensional projective spaces. More specifically, they write that

The Hopf construction gives circle bundles $p:S^{2n+1}\to\mathbb{CP}^n$ over complex projective space [sic]. This is actually the restriction of the tautological line bundle over $\mathbb{CP}^n$ to the unit sphere in $\mathbb{C}^{n+1}$."

I understand the "standard" $n=1$ case as follows: define $p:S^3\to S^2$ as: $$p(x_1,y_1,x_2,y_2) = (2(x_1 x_2+y_1 y_2), 2(x_1 y_2-x_2 y_1),x_1^2+y_1^2-x_2^2-y_2^2),$$ so that $p(x_1,y_1,x_2,y_2)\in S^2$ whenever $(x_1,y_1,x_2,y_2)\in S^3$. We then use the standard diffeomorphism $S^2\simeq\mathbb{CP}^1$ to conclude that there is a projection $\tilde p:S^3\to\mathbb{CP}^1$ which, for all $(x_1,y_1,x_2,y_2)\in S^3$ and $\theta\in\mathbb R$, satisfies $$\tilde p(\cos\theta\, x_1,\sin\theta\, y_1,\cos\theta\, x_2,\sin\theta\, y_2) = \tilde p(x_1,y_1,x_2,y_2),$$ meaning that $\tilde p^{-1}(x)\simeq S^1$ for all $x\in\mathbb{CP}^1$.

While that is fine, it hinges on the Hopf map $S^3\to S^2$, which seems rather ad hoc. How does this mapping generalise to $\mathbb{CP}^n$?

glS
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    I think it’s just by looking at the holomorphic projection $\mathbb C^{n+1} \setminus {0} \to \mathbb P^n$ and noting that it factors through the unit sphere. That is a circle bundle over projective space that is the restriction of the tautological line bundle to the unit sphere. – Gunnar Þór Magnússon Jun 24 '21 at 19:13
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    To add to Gunnar's point, the stated mapping $p:S^3\to S^2$ is "better" viewed as taking values in the complex projective line rather than the unit sphere in real Cartesian $3$-space. – Andrew D. Hwang Jun 24 '21 at 20:23
  • @AndrewD.Hwang in what sense? I thought I could just consider the mapping $S^3\to \mathbb{CP}^1$ by composition of the one $S^3\to S^2$ and the identification $\mathbb{CP}^1\simeq S^2$? Isn't this just a more explicit version of what you are proposing? – glS Jun 24 '21 at 22:33

2 Answers2

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You are right, the Hopf map $p : S^3 \to S^2$ is a very special construction. It is based on two ingredients:

  1. $S^3$ can be regarded as the set $\{(z_0,z_1) \in \mathbb C^2 \mid \lvert z_0 \rvert^2 + \lvert z_1 \rvert^2 = 1\}$.

  2. $S^2$ can be regarded as the set $\{(w,t) \in \mathbb C \times \mathbb R \mid \lvert w \rvert^2 + t^2 = 1\}$.

The Hopf map is then defined by using complex multiplication: $$p(z_0,z_1) = (2\bar z_0 z_1,\lvert z_0 \rvert^2 - \lvert z_1 \rvert^2 ). \tag{1}$$

However, $S^2$ can be identified naturally with $\mathbb{CP}^1$ and then the Hopf map corresponds to the quotient map $$q : S^3 \to \mathbb{CP}^1, q(z_0,z_1) = [z_0:z_1]. \tag{2}$$ For details see Prove that Hopf maps on $S^3, S^2$ and $\mathbb{C} \mathbb{P}^1$ are smooth submersions and Proving that the Hopf Fibration is a fiber bundle.

But $(2)$ shows that the quotient map $$q : S^{2n+1} \to \mathbb{CP}^n, q(z_0,\ldots,z_n) = [z_0:\ldots:z_n]. \tag{3}$$ is the desired generalization of the Hopf map.

Paul Frost
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    so, if I understand correctly, is the point that we should really think of the Hopf map as $S^{2n+1}\to\mathbb{CP}^n$, with its specific form $S^3\to S^2$ being an "accident" of the identification $\mathbb{CP}^1\simeq S^2$? – glS Jun 25 '21 at 11:44
  • @glS Yes, that is correct! – Paul Frost Jun 25 '21 at 12:57
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Let $X$ be a unital topological monoid. The Hopf construction converts the product $m:X\times X\rightarrow X$ into a quasifibration $\theta_m:X\ast X\rightarrow \Sigma X$, where $X\ast X$ denotes the join of $X$ with itself.

The generalised construction requires two ingredients: an action $\rho:Y\times X\rightarrow Y$ of $X$ on a space $Y$ together with a map $f:Y\rightarrow Z$ satisfying $f(y\cdot x)=f(y)$ for all $y\in Y$ and $x\in X$. The generalised Hopf construction returns from this data a quasifibration $\theta_{\rho,f}:Y\ast X\rightarrow C_f$, where $C_f$ is the mapping cone of $f$. The construction is due to Dold and Lashof.

Note that you can view the original Hopf construction as the special case of the generalised construction in which $X=Y$, with $X$ acting on itself via its product, and where $f:X\rightarrow\ast$.

The case you are interested in is realised as follows. Starting with the topological group $S^1$ you have $S^1\ast S^1\cong S^3$ and $\Sigma S^1\cong S^2$. The Hopf construction returns the Hopf map $\eta:S^3\rightarrow S^2$. There is an induced action of $S^1$ on $S^1\ast S^1\cong S^3$ and the generalised construction yields $S^1\ast S^1\ast S^1\cong S^5\rightarrow \mathbb{C}P^2$, where the complex projective plane is the cofibre of the Hopf map $\eta$. Iterating the construtcion we obtain maps $S^{2n+1}\rightarrow\mathbb{C}P^n$ for all $n$.

Actually the Hopf constructions accept more general input. For instance $X$ need only be a suitable H-space, and the action $\rho$ need only satisfy some of its requirements up to homotopy. Due to the fact that $S^1$ is a compact Lie group it is normally more convenient to choose smooth models and identifications and realise the Hopf maps as smooth fibre bundles.

I don't know of any analogue of the Hopf construction in the smooth category (by which I mean that the wiki article is misleading with its statement). Paul discusses the essentially ad hoc construction of the Hopf bundles.

Tyrone
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