Complex structure on $S^4$

Question

I have heard that there is a proof of non-existence of complex structure on the 4-sphere $S^{4}$ using only the topological K-theory (complex $KU$ and real $KO$). Moreover this argument can not be extended to $S^6$ for some reason that I am curious to understand.

Is there a reference for that? A sketch of proof will be even better!

EDIT: The proof I'm looking for, works for all $S^{2n}$ except $n=3$. More precisely (it seems that) there exists a uniform proof using $KU$ and $KO$ which proves that $S^{2n}$ does not have a complex structure for $n>1$ and $n\neq 3$.

Second Edit: I think I have found a reference for the statement but it is an exercise 8.15 (page 268) in Karoubi's K-theory: An introduction.

Edit 08/30 The combined answers given by mme and Michael Albanese give a complete answer to my question (Thank you very much!). I can't choose which one I should accept... I will try to follow step by step the exercise given in the book.

Answer 1

(a) Because $TS^4$ is stably trivial we have $p_1(S^4) = 0$.

(b) For any complex vector bundle we have $p_1(E) = c_1(E)^2 - 2c_2(E)$.

(c) For a complex vector bundle of complex dimension 2 we have $c_2(E) = e(E)$ the Euler class, and for $E = TM$ we have $e(E) = \chi(M) [M]$.

All of these assertions can surely be found in Milnor and Stasheff's book, but I don't have it on hand to give precise references.

Now suppose towards a contradiction that $TS^4$ may be given the structure of a complex vector bundle. That is, suppose $S^4$ is almost complex. Because the second cohomology of $S^4$ is trivial, any expression of the form $c_1(E)^2$ is zero, and $\chi(S^4) = 2$, so combining (b) and (c) we obtain $p_1(S^4) = -4[S^4]$. But this contradicts (a).

A variant of this argument works to show that $S^{4n}$ is not almost complex for any $n > 0$. Such an argument can only possibly work for spheres $S^{4n}$ because Pontryagin classes lie in degrees divisible by 4, and $S^k$ only has nontrivial cohomology in degree $k$.

Answer 2

The argument via K-theory proceeds as follows.

There is a map $K(X) \to H^*(X; \mathbb{Q})$ given by the Chern character. If $X = S^{2n}$, then it follows from Bott periodicity that the image of the Chern character lies in $H^*(S^{2n}; \mathbb{Z})$. For $[E] \in K(S^{2n})$, a direct computation shows that

$$\operatorname{ch}([E]) = \operatorname{rank}E + \tfrac{(-1)^{n+1}}{(n-1)!}c_n(E).$$

In particular, if $S^{2n}$ admits an almost complex structure, then $\frac{(-1)^{n+1}}{(n-1)!}c_n(TS^{2n}) \in H^{2n}(S^{2n}; \mathbb{Z})$. Now note that

$$\tfrac{(-1)^{n+1}}{(n-1)!}\langle c_n(TS^{2n}), [S^{2n}]\rangle = \tfrac{(-1)^{n+1}}{(n-1)!}\langle e(TS^{2n}), [S^{2n}]\rangle = \tfrac{(-1)^{n+1}}{(n-1)!}\chi(S^{2n}) = \tfrac{2(-1)^{n+1}}{(n-1)!} \in \mathbb{Z}$$

so $(n-1)! \mid 2$ and therefore $n \leq 3$. As mme points out, a separate argument is needed for $n = 2$.

A good reference for this argument is Konstantis and Parton's Almost Complex Structures on Spheres. Another reference is section $24.4$ of May's A Concise Course in Algebraic Topology.

Answer 3

It is possible to use the Chern classes modulo an odd prime $p$ to give a quick argument in all cases. Let $T^n \subset U(n)$ denote the maximal torus, then we know $$H^*(BU(n); \Bbb F_p) \cong H^*(BT^n; \Bbb F_p)^{S_n} \cong H^*((\Bbb{CP}^\infty)^n; \Bbb F_p)^{S_n} \cong \Bbb F_p[e_1, \cdots, e_n]$$ where $e_i = e_i(z_1, \cdots, z_n)$ are the elementary symmetric polynomials in the generators ("Chern roots") $z_1, \cdots, z_n$ of the $i$-th factor of $H^*(\Bbb{CP}^\infty; \Bbb F_p)$ appearing above. Note that from Chern-Weil theory, $e_i$ are the mod $p$ reduction of the universal Chern classes. Observe $H^*(BU(n); \Bbb F_p)$ is an $\mathcal{A}_p$-module where $\mathcal{A}_p$ is the Steenrod $p$-th power algebra. We compute a portion of this module structure below.

The first Steenrod power of $e_m$ can be computed as follows:

$$\begin{align*}P^1(e_m) = P^1(\mathrm{Sym}(z_1 z_2 \cdots z_m)) &= \mathrm{Sym}(z_1^p z_2 \cdots z_m) \\ &= \mathrm{Sym}(z_1^{p-1}) \mathrm{Sym}(z_1 z_2 \cdots z_m) - \mathrm{Sym}(z_1^{p-1} z_2 \cdots z_{m+1})\end{align*}$$

where $\mathrm{Sym}$ denotes symmetrization under the action of $S_n$ on the indices, where we normalize by dividing out by $1/d!$ where $d$ is the number of variables in the monomial. We can induct donwards until we reach

$$\begin{align*}\mathrm{Sym}(z_1^2 z_2 \cdots z_{m+p-2}) = \ & \mathrm{Sym}(z_1) \mathrm{Sym}(z_1 z_2 \cdots z_{m+p-2}) \\ &- (m+p-1) \mathrm{Sym}(z_1 z_2 \cdots z_{m+p-1})\end{align*}$$

Combined with Newton's identities, this shows $P^1(e_m)$ is a polynomial in $e_1, \cdots, e_{m+p-1}$ and in fact from above we see $e_{m+p-1}$ has degree $1$ in this identity with associated coefficient $m+p-1$, so turning that around we get $(m+p-1)e_{m+p-1}$ is a polynomial in $e_1, \cdots, e_{m+p-2}, P^1(e_m)$.

Suppose $S^{2n}$ has an almost complex structure. Suppose $n \geq 4$ so that there is an odd prime $p < n$ which does not divide $n$. Letting $m = n-p+1 > 0$ above, and using invertibility of $m+p-1 = n$ modulo $p$, we obtain that $c_n$ is a polynomial in $c_1, \cdots, c_{n-1}, P^1(c_{n-p+1})$ modulo $p$. This is a contradiction because $c_n(S^{2n}) = \chi(S^{2n}) = 2$ is nonzero modulo an odd prime, whereas all the lower-dimensional Chern classes vanish integrally as $S^{2n}$ has no cohomology below dimension $2n$.

This leaves only finitely many possibilities, $n = 1, 2, 3$, and $n = 2$ can be ruled out by the Pontryagin class argument as explained by mme above.