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This question is motivated solely by idle curiousity.

There is a natural map $p:S^{2n+1}\rightarrow \mathbb{C}P^n$ mapping a point on $S^{2n+1}\subseteq \mathbb{C}^{n+1}$ to the unique complex line it spans in $\mathbb{C}^{n+1}$.

It is well know that $p$ gives $S^{2n+1}$ the structure of an $S^1$ bundle over $\mathbb{C}P^n$. From this, one gets a long exact sequence in homotopy groups. Since $\pi_k(S^1) = 0$ for $k > 1$, one sees from this that $p_\ast :\pi_{2n+1}(S^{2n+1})\rightarrow \pi_{2n+1}(\mathbb{C}P^n)$ is an isomorphism. In particular, $p$ is homotopically nontrivial.

Likewise, notice that by collapsing the $2n-2$ skeleton of $\mathbb{C}P^n$ to a point, one gets a map $q:\mathbb{C}P^n\rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong S^{2n}$, the latter homeomorphism coming from the usual cellular picture of $\mathbb{C}P^n$. It is easy to see that this map induces an isomorphism on $H_{2n}$, so is also homotopically nontrivial.

My question is about the composition $q\circ p : S^{2n+1}\rightarrow S^{2n}$.

Is the composition $q\circ p:S^{2n+1}\rightarrow S^{2n}$ also homotopically nontrivial?

Here are some data points:

  1. The composition of two homotopically nontrivial maps need not be homotopically nontrivial. The simplest examples are probably the following: If $X$ and $Y$ are both noncontractible, then any of the "natural" inclusions $i:X\rightarrow X\times Y$ is homotopically nontrivial, the projection $X\times Y\rightarrow Y$ is homotopically nontrivial, but the composition is homotopically trivial.

  2. In the case $n=1$, $q$ doesn't actually collapse anything - it's a homeomorphism. Thus, in this case $q \circ p$ is homotopically nontrivial. (In fact, it's known to generate $\pi_3(S^2)$).

  3. For higher $n$, $\pi_{2n+1}(S^{2n})\cong \mathbb{Z}/2$, so the square of $q\circ p$ in $\pi_{2n+1}(S^{2n})$ is homotopically trivial.

  4. If we use $\mathbb{R}P^n$ instead of $\mathbb{C}P^n$, we end up getting a map $S^n\rightarrow S^n$. This map is homotopically nontrivial iff $n$ is odd, where it acts by multiplication by 2 on the top homology group.

Thanks, and please feel free to retag as appropriate!

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A minor point, I think $q$ should be the quotient map $q:\mathbb{C}P^n\rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}$, not the homeomorphism $\mathbb{C}P^n/\mathbb{C}P^{n-1}\rightarrow S^{2n}$. Otherwise the composition $q\circ p$ is not defined. – Moss Feb 19 '13 at 19:41
@Sebastian: You're right - I didn't correctly write what I was thinking :-). I'll edit in one moment... – Jason DeVito Feb 19 '13 at 19:42
up vote 6 down vote accepted

As far as I see, the cone of the map $S^{2n+1}→S^{2n}$ is precisely $CP^{n+1}/CP^{n−1}$. So we have to distinguish the latter from $S^{2n+2}\vee S^{2n}$ (at least when $n\geq 2$, since then we are already in the stable range).

Let $s: CP^{n+1}\rightarrow CP^{n+1}/CP^{n-1}$ be the projection. We have $H^*(CP^{n+1}; Z/2) = Z/2[x]/(x^{n+2})$ and there are $u$, $v\in H^*(CP^{n+1}/CP^{n-1}; Z/2)$ such that $s^*(u) = x^n$, $s^*(v) = x^{n+1}$. Then: $$s^*(Sq^2(u)) = Sq^2(x^n) = nx^{n+1} = ns^*(v).$$ Hence, $Sq^2(u) = n v$. This implies that the map is essential if $n$ is odd.

Using the complex $J$-homomorphism, one can also show that the map is stably trivial if $n$ is even. For background see for instance Hatcher's VB&K-Theory. Given a map $f\colon S^1\rightarrow U(n)$, one can construct a new map

$$ Jf\colon S^1* S^{2n-1}\rightarrow D^{2n}/S^{2n-1}, (z, t, (z_1, \ldots, z_n))\mapsto \sqrt{1-t^2}f(z)(z_1, \ldots z_n).$$ Actually, usually $t$ is used instead of $\sqrt{1-t^2}$ in the last formula but this doesn't matter. Here, $X*Y$ is the topological join. The important point is that this yields a group homomorphism $J\colon \pi_1(U(n))\rightarrow \pi_1^{st}\cong \mathbb{Z}/2$, which is surjective, since one can check that $J([S^1\subset U(n)])$ is the Hopf element. Note that $$ D^{2n} \rightarrow \mathbb{C}P^n, z=(z_1, \ldots, z_n)\mapsto [z_1: \ldots : z_{n} : \sqrt{1-|z|^2}] $$ is a homeomorphism away from $S^{2n-1}$ and induces a homeomorphism $$D^{2n}/S^{2n-1}\cong \mathbb{C}P^n/\mathbb{C}P^{n-1}.$$

Under this identification, one checks that the map $S^{2n+1} \rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong D^{2n}/S^{2n-1}$ in question is given by $(z_1, \ldots, z_{n+1}) \mapsto |z_{n+1}|/z_{n+1} (z_1, \ldots, z_n)$, if $z_{n+1}\neq 0$ and $(z_1, \ldots, z_{n+1})\mapsto [S^{2n-1}]$ otherwise. Finally, we have a homeomorphism $$S^1*S^{2n-1}\rightarrow S^{2n+1}, (z, t, (z_1, \ldots, z_n))\mapsto (\sqrt{1-t^2} z_1, \ldots, \sqrt{1-t^2} z_n, tz).$$ Composition with the former map, yields $$S^1*S^{2n-1}\rightarrow D^{2n}/S^{2n-1}, (z, t, (z_1, \ldots, z_n))\mapsto \bar{z}\sqrt{1-t^2}(z_1, \ldots, z_n).$$

This is precisely $J(\bar{i}_n)$, where $\bar{i}_n\colon S^1\rightarrow U(n)$ maps $z$ to the matrix representing multiplication by $\bar{z}$. Then $[\bar{i}_n] = -n*[S^1\subset U(n)]\in \pi_1(U(n))$, consequently, $J$ being a homomorphism, $J([\bar{i}_n]) = n[\eta]$ which is zero if $n$ is even and non-zero if $n$ is odd.

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This is going to take me awhile to digest. I'm familiar with Steenrod operations, but not too familiar with K-theory. To begin with, however, I'm confused at the very beginning: If the cone is $\mathbb{C}P^{n+1}/\mathbb{C}P^n$, doesn't this already give that the map is essential? This follows since $\mathbb{C}P^{n+1}/\mathbb{C}P^n$ is homeomorphic to $S^{2n+2}$ (map the collapsed $\mathbb{C}P^{n}$ to the south pole of $S^{2n+2}$ and map the attaching disc to all other points of $S^{2n+2}$), so, in particular, not homotopy equivalent to $S^{2n+2}\vee S^{2n}$. – Jason DeVito Feb 20 '13 at 13:31
(I won't be able to look at this again for another 4 or 5 hours or so. Sorry!) – Jason DeVito Feb 20 '13 at 13:35
The cone isn't $CP^{n+1}/CP^n$ but $CP^{n+1}/CP^{n-1}$. – ruediger Feb 20 '13 at 14:55
Thanks for the clarification. I agree that that is the mapping cone and I understand your argument in the case of $n$ odd with Steenrod squares. Unfortunately, I literally have only heard of Adam's relations and I know very little about K-theory in general, so that part's out of reach for now. If you don't mind, I'm going to hold of accepting this answer for a day or so in case someone comes up with a simpler argument that $\mathbb{C}P^{n+1}/\mathbb{C}P^{n-1}$ and $S^{2n+2}\vee S^{2n}$ are homotopically distinct. But thank you again! – Jason DeVito Feb 20 '13 at 18:45
I am sorry for being suspicious. I think the fact that the map is null-homotopic for $n$ even is easier. In any case you have this map $f: S^{2n+1} \to S^{2n}$ and if $n \geq 3$ this is a stable map, hence either zero or $\eta$. Hence the Cone of this map is either $\Sigma^n(\mathbb{C}P^2)$ or $S^{2n+2}\vee S^{2n}$. In both cases you can detect the homotopy type of that space by $Sq^2$ which is either zero or not. So I think for $n$ even the map you describe is null-homotopic for these simple reasons. – mland Feb 22 '13 at 10:28

Here's another proof coming from Pontrjagin's framed cobordisms.

We'll use the following result due to Pontrjagin, which can be found, for example, in Milnor's Topology from the Differentiable viewpoint:

For any closed $n$-manifold $M$, the homotopy classes of maps into $S^k$, denoted $[M,S^k]$ is in 1-1 correspondence with framed (n-k)-submanifolds of $M$ (submanifolds with trivial normal bundle and a particular choice of trivialization) modulo the equivalence of framed cobordism (where the cobordism is required to be a submanifold of $M$ as well).

The correspondence is as follows: given a homotopy class of map $[f]:M\rightarrow S^k$, choose a smooth representative $f$. Choose any regular value $p$. Then the associated submanifold is $f^{-1}(p)$ and the framing is pulled back from a choice of oriented basis for $T_p S^k$.

In our particular application, we want to understand a map from $S^{2n+1}$ to $S^{2n}$. By dimension count, the inverse image will be a disjoint union of circles. By drawing pictures in $\mathbb{R}^3$, one can easily convince oneself that any number of circles is frame cobordant to a single circle.

Our particular map is already smooth, at least, away from an equitorial $S^{2n-1}$ (and it can be homotoped to be smooth there as well). Since the map from $\mathbb{C}P^n$ to $S^{2n}$ is generically $1-1$, this reduces the problem to considering the induced framing on any Hopf circle.

The normal bundle of the Hopf circle $(e^{i\theta}, 0,....,0)\in S^{2n+1} (\subseteq\mathbb{C}^{2n+2})$ is a $2n$-dimensional vector bundle. Hence, it is determined by a choice of $S^1\rightarrow SO(2n)$. Note that if two choices are homotopic, the corresponding framings are frame cobordant: One can take a cylinder over $S^1$ where the $"t"$ coordinate of the cylinder corresponds to the time parameter in the homotopy. This is clearly embeddable in $S^{2n+1}$.

We conclude that the map $S^{2n+1}\rightarrow S^{2n}$ is essential iff the corresponding map $S^1\rightarrow SO(2n)$ is. It remains to compute that map.

To that end, let $v_i$ denote the vector in $\mathbb{C}^n$ whose only nonzero coordinate (in the $i$th slot), is $1$ and consider the basis $$\{v_2, iv_2, ..., v_n, iv_n$$ of $(i,0,...,0)^\bot \subseteq T_{(1,0...,0)} S^{2n+1}$. It's easy to see that the framing at the point $(e^{i\theta},0,...,0)$ in the Hopf circle (which is pulled back from $\mathbb{C}P^n$ via the Hopf map) is given by the basis $$\{e^{i\theta}\,v_2, e^{i\theta}\,iv_2,...,e^{i\theta}\,v_n, e^{i\theta}\,iv_n\}.$$

In particular, in terms of this basis, the map $S^1\rightarrow SO(2n)$ maps $e^{i\theta}$ to the block diagonal matrix $$\operatorname{diag}(R(\theta), R(\theta), ..., R(\theta))$$ where $$R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$ is the standard rotation matrix.

It is relatively well known that we can homotope such a map to one of the form $$\operatorname{diag}(R(n\theta), 1, ..., 1)$$ which is very well known to be homotopically essential iff $n$ is odd.

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This is also very nice. In the end, this is simply the inverse to the real $J$-homomorphism $J\colon\pi_1(SO)\rightarrow \pi_1^{st}$, of course. – ruediger Feb 21 '13 at 17:15
I had a sneaking suspicion that would end up being the case, but I wasn't sure. – Jason DeVito Feb 21 '13 at 18:59

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