Are there nontrivial continuous maps between complex projective spaces?

Question

Are there maps $f: \Bbb{CP}^n \rightarrow \Bbb{CP}^m$, with $n>m$, that are not null-homotopic? In particular, is there some non-null-homotopic map $\Bbb{CP}^n \rightarrow S^2$ for $n>1$? Can we say anything more generally about the pairs $(n,m)$ that admit such maps?

Note that, in the real case, the answer to the first quesiton is yes and the second no: the Hopf map descends to a non-null-homotopic map $\Bbb{RP}^3 \rightarrow \Bbb{RP}^2$, and there's a natural bijection between $[\Bbb{RP}^n, S^1]$ and $H^1(\Bbb{RP}^n; \Bbb Z) = 0$.

This question asks about the $n \leq m$ situation.

Edit: Jason DeVito's post below answers both questions in the affirmative; there remains, though, the following interesting case:

Is there a map $f: \Bbb{CP}^2 \rightarrow \Bbb{CP}^1$ that is not null-homotopic?

Answer 1

I believe there is a nontrivial map $\mathbb{C}P^k\rightarrow \mathbb{C}P^{k-1}$ when $k > 2$ is odd. I don't know what happens when $k$ is even.

Consider the following composition $$S^{2k+1}\rightarrow \mathbb{C}P^k\rightarrow S^{2k}\rightarrow S^{2k-1}\rightarrow \mathbb{C}P^{k-1}$$ with $k$ odd, where I'll now describe all the maps.

The maps from the odd dimensional spheres to complex projective spaces are the Hopf maps, that is, they are the projections from the family of Hopf fibrations $$S^1\rightarrow S^{2n-1}\rightarrow \mathbb{C}P^n.$$ The map from $\mathbb{C}P^k$ to $S^{2k}$ is obtained by collapsing the $2k-2$ skeleton of $\mathbb{C}P^k$ to a point, and the map from $S^{2k}$ to $S^{2k-1}$ generates $\pi_{2k}(S^{2k-1})$, which is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ because $k > 1$.

I will prove the the composition $S^{2k+1}\rightarrow \mathbb{C}P^{k-1}$ is homotopically nontrivial, which will then imply the map from $\mathbb{C}P^{k}\rightarrow \mathbb{C}P^{k-1}$ is as well.

First, the long exact sequence in homotopy groups associated to the Hopf fibration above implies the projection map from $S^{2k-1}\rightarrow \mathbb{C}P^{k-1}$ is an isomorphism, except on $\pi_2$. Hence, if the map $S^{2k+1}\rightarrow S^{2k-1}$ is nonzero as an element of $\pi_{2k+1}(S^{2k-1})$, the overall composition is nontrivial.

From a previous question of mine, it follows that, when $k$ is odd, the composition $S^{2k+1}\rightarrow S^{2k}$ is the generator of $\pi_{2k+1}(S^{2k})$.

So, the map $S^{2k+1}\rightarrow S^{2k-1}$ is obtained by composing a generator of $\pi_{2k+1}(S^{2k})\cong \mathbb{Z}/2\mathbb{Z}$ with a generator of $\pi_{2k}(S^{2k-1})\cong\mathbb{Z}/2\mathbb{Z}$.

But, according to http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres, in the "Ring structure" portion, this composition is nontrivial.

Answer 2

I can prove that there at most $2$ homotopy classes of maps from $\mathbb{C}P^2$ to $\mathbb{C}P^1$. However, I can't prove or disprove that there are exactly $2$. Perhaps someone can answer the final question and finish this off? (Or provide a completely different proof!)

I'll use the notation $[X,Y]$ to refer to homotopy classes of maps from $X$ to $Y$.

Claim 1: Fix a space $X$ and suppose every map $f:X\rightarrow S^2$ induces a trivial map on $H^2$. Then, the Hopf projection map $\pi:S^3\rightarrow S^2$ induces a bijection from $[X,S^3]$ to $[X,S^2]$.

Proof: First, note that $f$ induces $0$ on $H^2$ iff there is a map $\tilde{f}:X\rightarrow S^3$ with $\pi\circ \tilde{f} = f$. (I can supply a proof if desired). Then the map which assigns to $f\in[X,S^2]$ the map $\tilde{f}\in [X,S^3]$ is the inverse of the map induced by $\pi$. $\square$

Edit A proof that $f$ induces the $0$ map on $H^2$ iff it lifts was desired, and resulted in this question and answer. End edit

Claim 2: If $X = \mathbb{C}P^n$ with $n > 1$, then every map $f\in [X,S^2]$ is trivial on $H^2$.

Proof: Look at the cohomology ring structure. We have $H^\ast(\mathbb{C}P^n) \cong \mathbb{Z}[z]/z^{n+1}$ with $z$ in degree $2$. If $x$ generates $H^2(S^2)$, then $f^\ast x = kz$ for some $k\in \mathbb{Z}$. But then, since $x^2 = 0$, $0 = f^\ast (x^2) = (f^\ast x)^2 = k^2z^2$ which implies $k = 0$. $\square$

Corollary: There is a bijection between $[\mathbb{C}P^n, S^2]$ and $[\mathbb{C}P^n,S^3]$ for any $n > 1$.

Now, we show there are at most 2 elements in $[\mathbb{C}P^2, S^3]$ for any $n$. The main tool we use is the Thom-Pontryagin construction which provides, for any closed manifold $X$, a bijection between $[X,S^k]$ for any $k$ and codimension $k$ closed framed submanifolds of $X$, modulo framed cobordism. (A framed submanifold is a submanifold with trivial normal bundle, together with a choice of trivialization. A framed cobordism $W$ between two framed manifolds $M_1$ and $M_2$ is a manifold with boundary $\partial W = M_1\coprod M_2$, together with trivialization of the normal bundle of $W$ which restricts to the chosen trivializations on $M_1$ and $M_2$)

Since we're focusing on $k=3$ and $\mathbb{C}P^2$, of dimension $4$, we're looking at codim 3, that is, $1$-dim closed submanifolds of $\mathbb{C}P^2$. Recall that every $1$-d compact manifold is a disjoint union of a finite number of circles.

Given two embedded circles in $\mathbb{C}P^2$, there are obviously homotopic as $\pi_1(\mathbb{C}P^2) = 0$. Whitney has proved that of you have two homotopic embeddings of $M$ into $N$ with $\dim N \geq 2\dim M + 2$, then they are isotopic. The image of an isotopy is a cobordism between start and end times, so, we may assume wlog that any framed $1$-dim submanifold we are considering lives in a chart of $\mathbb{C}P^2$, that is, as a subset of $\mathbb{R}^4$.

Claim 3: There are precisely two framed $1$-d submanifolds of $\mathbb{R}^4$, up to framed cobordism.

Proof: Since any framed cobordism is a compact subset of $\mathbb{R}^4$, it's still embedded into the compactification of $\mathbb{R}^4$, that is, of $S^4$. By the Pontryagin-Thom construction, there is a bijection beteen $[S^4, S^3] = \pi_4(S^3) = \mathbb{Z}/2\mathbb{Z}$ and framed $1$-d submanifolds of $S^4$ modulo framed cobordisms. $\square$

In $\mathbb{R}^4$, or $S^4$ representatives for the two framed cobordism classes are as follows. First, one can use $S^1$ with a trivial parallel framing. More specifically, using $S^1 = \{(\cos \theta, \sin \theta, 0,0)\}$, one can use the framing $\{v_1,v_2,v_3\} = \{ (\cos\theta, \sin\theta, 0,0), (0,0,1,0), (0,0,0,1)\}$. Second, one can use $S^1$ with a once twisted framing. More specifically, one can use the framing $\{w_1,w_2,w_3\} = \{ \cos \theta v_1 + \sin\theta v_2, -\sin\theta v_1 + \cos\theta v_2, v_3\}$.

Now, for the question.

Question: Thinking of both $S^1$s with framing as subsets of $\mathbb{R}^4\subseteq \mathbb{C}P^2$, is there a framed cobordism between them in $\mathbb{C}P^2$?

If so, then all maps from $\mathbb{C}P^2$ to $S^2$ are homotopically trivial. If not, there is precisely one nontrivial map, up to homotopy. Note that if such a cobordism exists, it must, by claim 3, not be contained in any chart of $\mathbb{C}P^2$.

Answer 3

One person offering three answers to one post seems ludicrous (and this is a first for me), but I've really enjoyed this problem.

Theorem: There is a homotopically nontrivial map $\mathbb{C}P^2\rightarrow S^2$.

(By my second post, this map must be unique up to homotopy.)

We will prove this by answering the question in post 2. Namely, we show that given the two framed $S^1\subseteq \mathbb{R}^4\subseteq \mathbb{C}P^2$ (which I'll denote $S^1_1$ and $S^1_2$) are NOT framed cobordant in $\mathbb{C}P^2$. More specifically, given a framed cobordism between the two $S_i^1$s, we show we can lift the whole picture to $S^5$. However, since $\pi_5(S^4) = \mathbb{Z}/2\mathbb{Z}$, the Pontryagin-Thom construction will imply that there is no such cobordism in $S^5$.

So, assume for a contradiction that there is such a cobordism $W$ between the $S_1^2$ and $S_2^1$. Since $W$ is framed, the normal bundle is trivial. Let $i:W\rightarrow \mathbb{C}P^2$ denote the inclusion map. Since $W$ is a surface with boundary, $H^2(W) = 0$, so the induced map $i^\ast:H^2(\mathbb{C}P^2)\rightarrow H^2(W)$ is trivial for trivial reasons.

By a small extension of Claim 1 from the second post, there is a lift $\tilde{i}:W\rightarrow S^5$, in the sense that $\pi\circ\tilde{i} = i$, where $\pi:S^5\rightarrow \mathbb{C}P^2$ is the canonical projection.

Note that the normal bundle of $\tilde{i}(W)$ is the pullback under $\pi$ of the normal bundle to $W$, so is trivial. In fact, a choice of trivialization of the normal bundle of $W$ induces one of $\tilde{i}(W)$ in the following fashion. First, since $\pi\circ\tilde{i} = i$, it follows that $\tilde{i}(W)$ is transverse to the Hopf fibers. In particular, if $V\in T_p S^5$ with $p\in \tilde{i}(W)$ denotes the vector field tangent to the Hopf vectors, then the projection of $V$ onto $T_p \tilde{i} W$, denote $\rho(V)$ is not the identity, so $V - \rho(V)$ is non-zero and normal to $\tilde{W}$.

To trivialize the rest of the normal bundle over $\tilde{i}(W)$, we start by embedding the normal bundle of $W$ into $\mathbb{C}P^2$ as a tubular neighborhood of $W$ (and I'll call this map $i$ as well). This tubular neighborhood deformation retracts onto $W$, so $\tilde{i}$ extends to this neighborhood, and gives an embedding of the tubular neighborhood into $S^5.$ Then one can use this extended $\tilde{i}$ to transport a choice of trivialization of $W$ to one of $\tilde{i}(W).$

In particular, we have a framing on $\tilde{i}(W)$ given by taking framing given by the previous paragraph together with $V - \rho(V)$ $\tilde{i}(W)$. Thus, $\tilde{i}(W)$ is a framed cobordism between $\tilde{i}(S_1^1)$ and $\tilde{i}(S_2^1)$.

But, there is no such framed cobordism. The idea is that since both $S^1_i$ are in a small chart in $\mathbb{C}P^2$, their lifts to $S^5$ are given by essentially the same formulas (in an appropriate chart on $S^5$), together with $V - \rho(V)$. According to post 2, under the Pontryagin-Thom construction, these are representatives from the two equivalence classes corresponding to the fact that $\pi_5(S^4)$ has exactly two elements.

This contradiction implies $W$ can't exist. So, we've found at least two framed submanifolds of $\mathbb{C}P^2$ which are not framed cobordant. By the Pontryagin-Thom construction, this implies $[\mathbb{C}P^2,S^2]$ has at least 2 elements. In particular, there a homotopically nontrival map from $\mathbb{C}P^2$ to $S^2$.

$$ $$

If one traces through the Pontryagin-Thom construction, it follows that the following map $\mathbb{C}P^2\rightarrow S^2$ is homotopically nontrivial. We first define it as a map into $S^3$, then compose with the Hopf map $S^3\rightarrow S^2$.

Consider the normal bundle $\nu$ of $S_2^1$, embedded into $\mathbb{C}P^2$, and once and for all, pick a point $p\in S^3$ and a basis $\{e_1,e_2, e_3\}$ of $T_p S^3$. We map the entire $0$ section of $\nu$ to $p$. Each fiber of $\nu$, isomorphic to $\mathbb{R}^3$, is mapped onto $S^3$ with all points sufficiently far from the origin being mapped to $-p$ (think stereographics projection) and so that the differential maps the basis $\{w_1,w_2,w_3\}$ from Post 2 to the basis $\{e_1,e_2,e_3\}$. Finally, map points of $\mathbb{C}P^2$ wich are not in $\nu\subseteq \mathbb{C}P^2$ to $-p$.

Answer 4

I seem to be able to argue via obstruction theory that there is no nontrivial map $\mathbb{C} P^2\to S^2$. That is, taking any map $f$ on the 2-skeleton $\mathbb{C}P^1\subset \mathbb{C}P^2$, $f$ extends to the 4-skeleton iff the cellular cochain that takes the unique 4-cell $e_4$ to the homotopy class of $f|_{\partial e_4}$ is trivial in $C^4(\mathbb{C} P^2;\pi_3(S^2))$. If $f$ is of degree $1$, then $f(e_4)=1\in\pi_3(S^2)$, where $1$ is the Hopf map, since the 4-cell of $\mathbb{C} P^2$ is attached via the Hopf map. Then if $f$ is of degree $n$, it maps the 4-cell to $n$. Thus the obstruction is insurmountable except in case $n=0$.

The problem with this argument is that it will show that there are no maps from $\mathbb{C}P^n$ to $\mathbb{C}P^1$ for any $n>1$, and Jason deVito's comment to his first answer indicates there is one for $n=5$.