Proof to Theorem 2
To prove Theorem 2, we shall need to discuss Adams operations in K-theory. An operation in K-theory is a natural transformation \(F: K \to K\) where K is regarded as a set-valued functor. We shall first define some operations in K-theory
dirnensional vector spaces Uy li(V) = V n vn.,.n\, (i ci.:L'sl. They define natural transformacions \i : \,ect - \'ec! wtlere Vect : Top ' Ens. is the contravariant functor given by
X + set of isom. classes of conplex vectot bundles over X
f: X-Y-f-, the pullback, We L,ish to extend these to operations tr1 : K * K. To do rhis we use a neat trick. Define ,\. [V] e K(X)ttr1l, IVl I Vecr(X), to be the power series t' -
\[\sum_{i=0}^{\infty} t^{i}[\lambda^{i}[V]].\]
Since \[\lambda^{i}(V \oplus W) = \sum_{j+k=i} \lambda^{j}(V) \bigotimes \lambda^{k}(W)\],
\[\lambda_{t}[V \bigoplus W] = \lambda_{t}[V]\lambda_{t}[W].\]
AIso note that each ^t[v] is a unit ln K(x)tttll since it has constant lending term 1, Thus we have a homonorphisn
\[\lambda_{t} : Vect(X) \rightarrow 1 + K(X)[[t]]^{+}\] of the addj.tive serligroup Vect (X) into the xnultiplication group of power serles over K(X) with constant term 1. By the universal property of K(X), this extends uoiquely to a homomorphi$n
\[\lambda_{t} : K(X) \rightarrow 1 + K(X)[[t]]^{+}\]
taking the coefficients of tl, have defined the operations
\[\lambda^{i} : K(X) \to K(X)\].
Note that if I is a line bundle
\[\frac{1}{2}[L] = [1] + t[L].\]
So \[\lambda^{\alpha}[L] = [1], \lambda^{\hat{I}}[L] = [L], \text{ and } \lambda^{\hat{I}}[L] = [0], i + i.\]
Now we can define the Adams operations \(F^i: E \times K\). Let \(F^0(x) = \operatorname{rank} x\) (component-wise the trivial bundle with dimension equal to that of a fiber). Define
\[\Psi_{\varepsilon}(x) \in K(X)[[t]], x \in K(X), by\]
\[\Psi_{t}(x) = \Psi^{0}(x) - t \frac{d}{dt}(\log \lambda_{-t}(x)).\]
Then define the Adams operations \(\Psi^{i}\) as the coefficients of the \(t^{i}\); i.e.,
\[\Psi_{t}(x) = \sum_{i=0}^{\infty} t^{i} \Psi^{i}(x).\]
Lemma. The Adams operations \(\psi^{i}: K \rightarrow K\) satisfy the following properties:
- (1) \(\Psi^{i}(x + y) = \Psi^{i}(x) + \Psi^{i}(y) x, y \in K(X)\).
- (2) If L is a line bundle, \(\Psi^{i}(L) = L^{i}\). Properties (1) and (2) uniquely characterize the operations \(\Psi^{i}\). In addition,
- (3) \(\Psi^{i}(xy) = \Psi^{i}(x) \Psi^{i}(y), x,y \in K(X).\)
- (4) \(\Psi^{\mathbf{k}}\Psi^{\mathbf{j}}(\mathbf{x}) = \Psi^{\mathbf{k}\mathbf{j}}(\mathbf{x}), \mathbf{x} \in K(\mathbf{X}).\)
- (5) If p is prime, \(\Psi^{p}(x) \equiv x^{p} \mod p\).
- (6) If \(u \in \widetilde{K}(S^{2n})\), \(\Psi^{i}(u) = i^{n}u\).
Proof. Since \(\Psi_{t}(x + y) = \Psi_{t}(x) + \Psi_{t}(y)\), (1) follows. We have
shown that \(\lambda_{-1}(L) = 1 - tL\) where L is any line bundle, hence
\[\Psi_{t}(L) = 1 - t(\frac{-L}{1-tL}) = 1 + tL(\sum_{i=0}^{\infty} (tL)^{i})\] \[= \sum_{i=0}^{\infty} t^{i}L^{i}\] \[= 0\] and (2) is proved. That properties (1) and (2) uniquely characterize the operations \(\Psi^{\hat{\mathbf{1}}}\) follows immediately from Lemma (Splitting Principle). Let \(\mathbf{E}_{\hat{\mathbf{1}}} \to \mathbf{X}\), \(1 \le \hat{\mathbf{1}} \le \mathbf{n}\), be complex vector bundles over X. Then \(\frac{1}{2}\) map \(\pi: \mathbf{Y} \to \mathbf{X}\)
- (a) \(\pi^*\): K(X) \(\rightarrow\) K(Y) is injective
- (b) each \(\pi^{\bigstar}\) (E;) is a direct sum of line bundles.
Similarly (3), (4), and (5) follow from the Splitting Principle. Finally, in \(\widetilde{K}(S^2)\), \(\Psi^i(\beta_2) = i\beta_2\), so, applied to the generator \(\beta_{2n} = \beta_2 \bigotimes \beta_2 \bigotimes \ldots \bigotimes \beta_2\) (n times) of \(\widetilde{K}(S^{2n})\), we get
\[\Psi^{i}(\beta_{2n}) = i^{n}\beta_{2n}\] and hence (6). QED.
Note that the Adams operations restrict to K. Then we are in a position to prove Theorem 2. Since we are assuming n is even, we have the short exact sequence
\[0 \to \widetilde{K}(S^{4n}) \overset{\Theta}{\to} \widetilde{K}(C_{\underline{}}) \overset{\varphi}{\to} \widetilde{K}(S^{2n}) \to 0\] where \(\Theta(\beta_{4n})=y\), \(\phi(x)=\beta_{2n}\), and \(x^2=\mathbb{N}(\overline{m})y\). We shall apply some Adams operations to x and y. Note that
\[\Psi^{2}(\mathbf{x}) = 2^{n}\mathbf{x} + \mathbf{a}\mathbf{y}, \quad \mathbf{a} \in \mathbf{Z},\] \[\Psi^{3}(\mathbf{x}) = 3^{n}\mathbf{x} + \mathbf{b}\mathbf{y}, \quad \mathbf{b} \in \mathbf{Z}\] \[\Psi^{k}(\mathbf{y}) = \mathbf{k}^{2n}\mathbf{y}\] and by property (6) of the Adams operations. But by property (5),
\[\mathbb{P}^2(\mathbf{x})\] . \(\mathbf{x}^2 \mod 2 = \mathbb{H}(\tilde{\mathbf{m}}) \text{ y mod } 2\).
So, since \(\mathbb{H}(\overline{m})\) is odd, a must be odd. Hence, by property (4),
\[\Psi^{6}(x) = \Psi^{3}\Psi^{2}(x) = \Psi^{3}(2^{n}x + ay)\] \[= 2^{n}3^{n}x + 2^{n}by + 3^{2n}ay\] and
\[\Psi^{6}(x) = \Psi^{2}\Psi^{3}(x) = \Psi^{2}(3^{n}x + by)\]
= \(3^{n}2^{n}x + 3^{n}ay + 2^{2n}by\).
Thus
\[3^{n}a(3^{n}-1) = 2^{n}b(2^{n}-1),\] and, since a is odd,
\[2^{n}/3^{n} - 1\].
By elementary number theory this can happen only if n = 1, 2, or 4. QED.
Ordinary Cohomology
The multiplication problem was originally solved using Steenrod's equivalent definition of Hopf invariant given in terms of ordinary cohomology. We shall discuss this approach. Our aim will be to obtain our previous assumption that n must be even. We shall also outline Adem's proof that n must be a power of 2.
a power of 2. \(_{\star}\) So, let H \(^{\prime}\) be reduced singular cohomology with coefficients in G. The map
\[m : S^{n-1} \times S^{n-1} \rightarrow S^{n-1}, n > 1,\]
gives rise to its Hopf construction
\[\bar{m}: s^{2n-1} \rightarrow s^n\]
Looking at the exact sequence of the pair \((C_-, S^n)\), we see that
\[H^{\mathbf{r}}(C_{\underline{}}) = \begin{cases} G & r = n, 2n \\ 0 & \text{otherwise.} \end{cases}\]
Let G be either Z or Z<sub>2</sub> and let x be the generator of \(\operatorname{H}^n(C_{\overline{m}})\) and y the generator of \(\operatorname{H}^{2n}(C_{\overline{m}})\). Then taking the cup product
\[x^2 = H(\bar{m})y\] for some integer \(H(\overline{m})\) defined to be the Hopf invariant. For this ordinary cohomology definition of Hopf invariant, Theorem 1 holds V \(n \ge 1\) (odd or even).
Theorem 1. Let \(m: S^{n-1} \times S^{n-1} \to S^{n-1}\) be a map of bidegree \((d_1, d_2)\). Then \(H(\overline{m}) = d_1 d_2\).
Proof. Completely analogous to the K-theory proof, only using the exact sequence
\[0 \to H^{n}(C_{\underline{n}}) \xrightarrow{\simeq} H^{n}(S^{n}) \to 0. \quad QED.\]
Note by the commutativity properties of the cup product that
\[x^2 = (-1)^{n^2} x^2\].
So letting G=Z, if n is odd, \(2x^2=0\) and \(R(\tilde{m})=0\). We have thus reduced the multiplication problem to a consideration of n even.
We can further reduce the values of n under consideration to powers of 2; i.e. \(n=2^k\). This result by Adem uses Steenth squares. We shall outline the proof. From now on assume \(C=Z_{\frac{n}{2}}\).
Theorem A. There exist unique Steenrod square operations \(Sq^{1}\): \(H^{r}(X,A) \rightarrow H^{r+i}(H,A)\), \(i \geq 0\), which are homomorphisms and have the following properties
- (1) \(Sq^0 = id\). (2) If dim x = i, \(Sq^i(x) = x^2\).
- (3) If i > dim x, \(Sq^{i}(x) = 0\).
- (4) Cartan formula:
\[\operatorname{Sq}^{i}(xy) = \sum_{j+k \approx i} \operatorname{Sq}^{j}(x) \cdot \operatorname{Sq}^{k}(y)\].
(5) \(Sq^{1}\) is the Bochstein homomorphism \(\beta\) of the exact coefficient sequence
\[0 \rightarrow Z_2 \rightarrow Z_4 \rightarrow Z_2 \rightarrow 0.\]
(6) Adem relations: if 0 < a < 2b, then
\[Sq^{a}.Sq^{b} = \sum_{j=0}^{\lfloor a/2 \rfloor} {b-1-j \choose a-2j} Sq^{a+b-j}.Sq^{j}.\]
Define R(2), the Steenrod algebra mod 2, to be the graded associative algebra generated by the \(Sq^{i}\). In detail, let M be the graded \(Z_{2}\)-module with \(M_{i} = Z_{2}\). Denote the generator of \(M_{i}\)as \(Sq^{1}\). R(2) is the quotient of the tensor algebra \(\Gamma(M)\) by relations of the form
\[Sq^{a}.Sq^{b} = \sum_{j=0}^{\lfloor a/2 \rfloor} {b-1-j \choose a-2j} Sq^{a+b-j} Sq^{j}, 0 \le a \le 2b.\]
Theorem B. The elements \(\operatorname{Sq}^{2^{k}}\) generate R(2) as an algebra.
Let \(m: S^{n-1} \times S^{n-1} \to S^{n-1}\) be a continuous multiplication with unit, n > 1. Then \(n = 2^k\).
Proof. We have the relation \(x^2 = H(\bar{m}).y\) in \(H^*(C_{-})\). By Theorem 1. \(H(\bar{m}) = 1\). Hence
\[\operatorname{Sq}^{n}(x) \approx x^{2} \neq 0.\]
Using Theorem B, \(\operatorname{Sq}^{2^{k}}(x) \neq 0\) for some k > 0. But looking at \(\operatorname{H}^{*}(C_{-})\), we must have \(2^{k} = n\). QED.