Let \(X\) be a path-connected H-space with \(H^*(X; R)\) free and finitely generated in each dimension. For maps \(f, g: X \to X\), the product \(fg: X \to X\) is defined by \((fg)(x) = f(x)g(x)\), using the H-space product.
Show that \((fg)^*(\alpha) = f^*(\alpha) + g^*(\alpha)\) for primitive elements \(\alpha \in H^*(X; R)\).
Deduce that the \(k\)-th power map \(x \mapsto x^k\) induces the map \(\alpha \mapsto k\alpha\) on primitive elements \(\alpha\). In particular the quaternionic \(k\)-th power map \(S^3 \to S^3\) has degree \(k\).
Show that every polynomial \(a_n x^n b_n + \dotsb + a_1 x b_1 + a_0\) of nonzero degree with coefficients in \(\mathbb{H}\) has a root in \(\mathbb{H}\). [See Theorem 1.8.]
Solution
Let \(\mu:X\times X\to X\) be the H-space multiplication and let \[ \delta_{f,g}:X\to X\times X, \qquad \delta_{f,g}(x)=(f(x),g(x)). \] Then \(fg=\mu\circ\delta_{f,g}\).
1. If \(\alpha\) is primitive, then\[ (fg)^*(\alpha)=f^*(\alpha)+g^*(\alpha). \]
Primitivity means \[ \mu^*(\alpha)=\alpha\times1+1\times\alpha. \] Therefore \[ (fg)^*(\alpha) =\delta_{f,g}^*\mu^*(\alpha) =f^*(\alpha)+g^*(\alpha). \]
2. The \(k\)th-power map sends every primitive class \(\alpha\) to \(k\alpha\).
Induct on \(k\) using part 1. For the identity map \(P_1\), \(P_1^*(\alpha)=\alpha\). Since \(P_{k+1}=P_k\cdot\operatorname{id}\), \[ P_{k+1}^*(\alpha)=P_k^*(\alpha)+\alpha=(k+1)\alpha. \] On \(S^3\), a generator of \(H^3(S^3;\mathbb Z)\) is primitive for dimensional reasons. Hence the quaternionic power map \(q\mapsto q^k\) has degree \(k\).
3. Every quaternionic polynomial\[ P(x)=a_nx^nb_n+\cdots+a_1xb_1+a_0, \qquad a_n,b_n\ne0, \]has a root.
View \(\mathbb H\) as \(\mathbb R^4\). Since the leading term dominates as \(|x|\to\infty\), \(P\) is proper and extends to a map of one-point compactifications \[ \widehat P:S^4\to S^4. \] The proper homotopy \[ P_t(x)=a_nx^nb_n+t(a_{n-1}x^{n-1}b_{n-1}+\cdots+a_0), \qquad 0\le t\le1, \] shows that \(\widehat P\) has the same degree as the leading map \(x\mapsto a_nx^nb_n\).
Left and right multiplication by a nonzero quaternion are orientation-preserving linear automorphisms of \(\mathbb R^4\), so they have degree \(+1\). Radially, the map \(x\mapsto x^n\) has boundary map \(S^3\to S^3\), \(q\mapsto q^n\), whose degree is \(n\) by part 2. Hence \[ \deg\widehat P=n\ne0. \] If \(P\) had no root, then \(0\notin P(\mathbb H)\); a proper map missing a point has degree \(0\), contradiction. Thus \(P(x)=0\) for some \(x\in\mathbb H\).