Today I would like to blog about a result of Atiyah from the 1950s, from his paper “Bott periodicity and the parallelizability of the spheres.” Namely:

Theorem 1 (Atiyah) On a nine-fold suspension ${Y = \Sigma^9 X}$ of a finite complex, the Stiefel-Whitney classes of any real vector bundle vanish.

In particular, this means that any real vector bundle on a sphere $S^n, n \geq 9$ cannot be distinguished using Stiefel-Whitney classes from the trivial bundle. The argument relies on the Bott periodicity theorem and some calculations with Stiefel-Whitney classes. There is also an analog for the Chern classes of complex vector bundles on spheres; they don’t necessarily vanish but are highly divisible.

These sorts of integrality theorems often have surprising geometric consequences. In this post, I’ll discuss the classical problem of when spheres admit almost-complex structures, a problem one can solve using the second of the integrality theorems mentioned above. Atiyah was originally motivated by the question of parallelizability of the spheres.

1. Bott’s integrality theorem

To start with, I would like to mention the following theorem of Bott:

Theorem 2 The top Chern class ${c_n}$ of any (complex) vector bundle on ${S^{2n}}$ is divisible by ${(n-1)!}$.

To prove this, one shows that the Chern character of any complex vector bundle on ${S^{2n}}$ is integral (though a priori it is only in therational cohomology ring). This is done by introducing the reduced ${K}$-group ${\widetilde{K}(S^{2n})}$ of stable classes of vector bundles; the Chern character defines a map

$\displaystyle \mathrm{ch}: \widetilde{K}(S^{2n}) \rightarrow \widetilde{H}^*(S^{2n}; \mathbb{Q}).$

This map is additive and multiplicative. To check that the Chern character of any complex vector bundle on ${S^{2n}}$ is integral, one thus reduces to analyzing the image of ${\mathrm{ch}}$ as above.

This is convenient because the group ${\widetilde{K}(S^{2n})}$ is very simple. By the Bott periodicity theorem, it is generated by the ${n}$th power of ${(\eta - 1)}$ where ${\eta - 1}$ is the Hopf bundle over ${S^2}$, so that

$\displaystyle (\eta - 1)^{n} := (\eta - 1) \boxtimes (\eta - 1) \boxtimes \dots \boxtimes (\eta - 1) \in \widetilde{K}(S^{2n}) .$

So, all we need to do is to check that ${\mathrm{ch}(\eta - 1)}$ in ${H^2(S^2)}$ is integral; then, because of the multiplicative properties of the Chern character, the general result will follow.

However, ${\mathrm{ch} (\eta - 1) = c_1(\eta) }$ is clearly an integer. Thus, the claim about ${\mathrm{ch}(\widetilde{K}(S^{2n})}$ is proved.

The final step is to translate from the statement about ${\mathrm{ch}}$ to a statement about ${c_n}$. Here, one has to recall exactly how ${\mathrm{ch}}$ is defined. If ${V}$ is a vector bundle such that

$\displaystyle V = L_1 \oplus \dots \oplus L_m$

then

$\displaystyle \mathrm{ch}V = e^{ c_1(L_1))} + \dots + e^{c_1(L_m)} .$

If a subscript indicates a graded component, then this implies that

$\displaystyle (\mathrm{ch}V)_{2n} = \frac{1}{n!} \left( c_1(L_1)^n + \dots + c_1(L_m)^n \right).$

We note that the elementary symmetric polynomials ${s_i( c_1(L_1), \dots, c_1(L_m))}$ are the Chern classes ${c_i(V)}$, and that there is a certain universal expression $E( X_1, \dots, X_m)$ such that for any ${a_1, \dots, a_m}$ with ${s_1,\dots, s_m}$ the elementary symmetric polynomials as before,

$\displaystyle E(s_1(a_1, \dots, a_m), \dots, s_n(a_1, \dots, a_m)) = \frac{1}{n!} \left( a_1^n + \dots + a_n^n \right).$

We only need the first ${n}$ symmetric polynomials. Thus, we have that (where ${(\mathrm{ch} V)_{2n}}$ denotes the ${2n}$th component)

$\displaystyle (\mathrm{ch}V)_{2n} = E( c_1(V), \dots, c_n(V)).$

This is true for any vector bundle ${V}$, not necessarily a sum of line bundles, by the splitting principle; actually, it is probably how one defines ${\mathrm{ch} V}$ in the first place.

In any event, what we have proved is that

$\displaystyle E(c_1(V), \dots, c_n(V)) \in \mathbb{Z}$

for any vector bundle ${V}$ on ${S^{2n}}$. In this case, of course, ${c_1, \dots, c_{n-1} = 0}$, so we only need the coefficient of ${s_n}$ in ${E}$ to conclude. Here we can use Newton’s identities to conclude that

$\displaystyle E( c_1, \dots, c_n) = \pm \frac{n c_n}{n!} = \pm \frac{c_n}{(n-1)!}.$

This implies the integrality theorem, since we have seen that the left side is integral.

In fact, it seems that this integrality result can be deduced from the Atiyah-Singer index theorem by intepreting ${c_n(V)/(n-1)!}$ as the index of a “twisted signature operator”; I don’t understand this well enough to comment yet.

There are a lot of surprising integrality results on the characteristic classes of the tangent bundle to a manifold; for instance, the Hirzebruch signature formula implies that ${\frac{p_1}{3}}$ of the tangent bundle is always an integer for an oriented four-manifold. Another example is that the Todd genus of a complex manifold is always an integer, by the Hirzebruch-Riemann-Roch theorem. However, there don’t seem to be that many that apply to every vector bundle on a space.

2. Almost complex structures on spheres

Recall that an almost complex structure on a manifold is a complex structure on its tangent bundle. Any complex manifold is an almost complex manifold. Among the spheres, it is a classical theorem that only ${S^2 = \mathbb{P}^1(\mathbb{C})}$ and ${S^6}$ admit almost complex structures.

We can deduce most of this result very quickly from Bott’s integrality theorem above.

Theorem 3 No sphere other than ${S^2}$ and ${S^6}$ can admit an almost complex structure.

In fact, suppose the tangent bundle ${TS^{2n}}$ could be given the structure of a complex vector bundle. Then it would have Chern classes, and in particular, it would have a top Chern class ${c_n(TS^{2n})}$. This is necessarily the Euler class ${\chi(TS^{2n}) = 2}$, since the Euler class is the Euler characteristic. However, we have also seen that ${c_n(TS^{2n})}$ is divisible by ${(n-1)!}$, and for ${n \geq 4}$ this cannot thus be ${2}$ (or rather, ${2}$ times a generator of ${H^{2n}}$).

The argument does not show that ${S^{4}}$ does not admit an almost complex structure. But one can give a direct argument as follows. Suppose ${TS^4}$ admitted the structure of a complex vector bundle. We know that ${p_1(TS^4) = 0}$ since ${TS^4 }$ is stably trivial. However, if we assume that ${TS^4}$ is a complex vector bundle then we have the following identity for the total Pontryagin class:

$\displaystyle p(TS^4) = c(TS^4 \otimes_{\mathbb{R}} \mathbb{C}) = c( TS^4 \oplus \overline{TS^4}) = ( 1 + c_2(TS^4))(1 + c_2(TS^4)) = 1 + 2c_2(TS^4).$

Necessarily ${c_2(TS^4) }$ is nonzero (as it is the Euler class, which is twice the generator of ${H^4(S^4)}$). So this is a contradiction.

In fact, using the signature theorem, one can moreover conclude that any compact four-manifold with ${H^2 = 0}$ and nonzero Euler characteristic does not admit an almost complex structure. The argument is the same, once one notes that ${p_1 = 0}$, at least after tensoring with ${\mathbb{Q}}$.

3. Atiyah’s theorem

Atiyah’s theorem will be proved in a similar way as Bott’s integrality theorem, except that we will get something identically zero, since we are working in a torsion ring. The necessary tool is the (harder) form of Bott periodicity for ${KO}$-theory.

Namely, the periodicity theorem states that:

Theorem 4 There is an eight-dimensional bundle ${\rho}$ over ${S^8}$, whose Stiefel-Whitney in ${H^8(S^8; \mathbb{Z}/2)}$ is nonzero, such that multiplication by ${(\rho- 8)}$ determines an isomorphism

$\displaystyle \beta = (\rho - 8) : \widetilde{KO}(X) \simeq \widetilde{KO}(S^8 \wedge X)$

for any CW complex ${X}$.

Recall that our goal is to show that the Stiefel-Whitney classes of any vector bundle over a nine-fold suspension vanish. To do so, we’ll use a relation between the Stiefel-Whitney classes of an element ${x \in \widetilde{KO}(X)}$ and those of ${\beta(x) = (\rho - 8) \otimes x }$. This is the analog of the calculation of the Chern classes of the generator of ${\widetilde{K}(S^{2n})}$.

Let ${g \in H^8(S^8; \mathbb{Z}/2)}$ be the nonzero element. There is an isomorphism ${\widetilde{H}^*(X; \mathbb{Z}/2) \simeq \widetilde{H}^{*+8}(S^8 \wedge X; \mathbb{Z}/2)}$ given by multiplication by ${g}$.

Lemma 5 If ${X}$ is a CW complex and ${x \in \widetilde{KO}(X)}$, then

$\displaystyle w(\beta(x)) = 1 + g \otimes \sum_{k=1}^\infty P_{8k}( w_1(x), \dots, w_{8k}(x))$

where ${P_t}$ is the polynomial which expresses the sum ${a_1^t + \dots + a_t^t}$ in terms of the elementary symmetric functions of ${a_1, \dots, a_t}$.

Let us prove this. In fact, ${x}$ can be represented by ${V - n}$ for ${V}$ a real vector bundle of dimension ${n}$ over ${X}$. Then ${\beta (x)}$ is represented by

$\displaystyle (V - n) \otimes (\eta - 8) = V \otimes \eta - n \eta - 8 V$

(modulo stable equivalence). So we need to compute the Stiefel-Whitney classes of this. As usual, we may pretend that ${V}$ is a sum of line bundles, ${V = L_1 \oplus \dots \oplus L_n}$ with Stiefel-Whitney classes ${\alpha_1, \dots, \alpha_n}$. We can also pretend that ${\eta}$ is a sum of line bundles ${M_1 \oplus \dots \oplus M_8}$ with Stiefel-Whitney classes ${\gamma_1, \dots, \gamma_8}$; although this is not literally true any manipulations we make will be symmetric in the ${\gamma_i}$ and thus will be justified by the splitting principle.

So,

$\displaystyle w(V) = \prod_{i=1}^n (1 + \alpha_i), \quad w(\eta) = \prod_{j=1}^8( 1 + \gamma_j) = 1+ g.$

In particular, since ${V \otimes \eta = \bigoplus_{i,j} L_i \otimes \gamma_j}$, and ${w_1}$ sends tensor products of line bundles to sums of cohomology classes, we also have

$\displaystyle w(V \otimes \eta) = \prod_{i,j} (1 + \alpha_i + \gamma_j).$

However, since all but one of the elementary symmetric functions of ${\gamma_j}$ is zero, and their product is ${g}$, we find:

$\displaystyle w(V \otimes \eta) = \prod_i ( (1 + \alpha_i)^8 + g). \ \ \ \ \ (1)$

Next, the Stiefel-Whitney classes of ${n \eta}$ are

$\displaystyle w(n \eta) =(1 + g)^n. \ \ \ \ \ (2)$

Finally, we have

$\displaystyle w(8 V) = \prod_i (1 + \alpha_i)^8. \ \ \ \ \ (3)$

If we put all these together, and use the fact that ${(1+g)^{-n} = 1+ng}$ (in characteristic 2!), we find that

$\displaystyle w(\beta(x)) = \left( \prod_i ( (1 + \alpha_i)^8 + g) \right) (1 - ng) \prod_i (1 + \alpha_i)^{-8}.$

Now, again we can simplify if we note that ${g^2 = 0}$. We can combine the two products to get another expression for ${w(\beta(x))}$:

$\displaystyle (1 + ng) \prod_i \left( 1 + \frac{g}{(1 + \alpha_i)^8} \right) = (1 + ng) \prod_i ( 1 + g( 1 + \alpha_i^8 + \alpha_i^{16} + \dots ) ).$

Here because ${g^2 =0}$ everything simplifies to

$\displaystyle 1 + g \sum_i \left( \alpha_i^8 + \alpha_i^{16} + \dots \right)$

which gives the expression we desired, since the symmetric polynomials of the ${\alpha_i}$ are the Stiefel-Whitney classes of ${V}$.

4. Proof of Atiyah’s theorem

With the formula for ${w(\beta(x))}$ established as in the lemma, it will be fairly quick to prove Atiyah’s theorem. The idea is the following. Consider an element ${x \in \widetilde{KO}(S^9 \wedge X)}$. Then in particular, there is ${y \in \widetilde{KO}(S^1 \wedge X)}$ such that ${x = \beta(y)}$; this means that

$\displaystyle w(x) = 1 + g \otimes \sum_{k=1}^\infty P_{8k}(w_1(y), \dots, w_{8k}(y)).$

Here the ${w_i (y) }$ live instead ${H^*(S^1 \wedge X)}$; the advantage, however, is that this ring is very simple: all cup-products in a suspension vanish. Since the top term of ${P_{8k}}$ is ${\pm 8k w_{8k}}$, and all the other terms involve nontrivial products (again, by Newton’s identities), we find that ${w(x)}$ is ${1 + 8 (\mathrm{something}) = 1}$, since we are in characteristic two.