Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 24 / 2

For based maps $\alpha :S^p\to Y$ and $\beta :S^q\to Y$, use the standard cell decompositionThe attaching map $w:S^{p+q-1}\to S^p\vee S^q$ is obtained from the boundary of $D^p\times D^q$, collapsing the appropriate coordinate boundaries. The geometric Whitehead product is the homotopy classOrient the product cell consistently with the ordinary product orientation. Its class depends only on the two homotopy classes, and it is natural: $f_{\ast }[\alpha ,\beta {]}_W=[f_{\ast }\alpha ,f_{\ast }\beta {]}_W$.

For $p,q\ge 2$ it is additive in both variables:Thus it is bilinear over $\mathbb{Z}$, and after rationalization of a topological space over $\mathbb{Q}$. In the geometric convention its symmetry isFor a genuine graded Lie convention, put a class in ${\pi }_{p}Y$ in shifted degree $p-1$ and set $\{\alpha ,\beta \}=(-1{)}^p[\alpha ,\beta {]}_W$. Equivalently transport the Samelson product through reduced suspension and loop space adjunction. Writing $|\alpha |=p-1$, this bracket has degree zero, satisfiesand has the graded Jacobi identityThese are two equivalent sign presentations of the same Whitehead product. In particular the square of an odd-dimensional geometric class is two-torsion, whereas an even-dimensional sphere class can have a nonzero rational square. Reduced suspension kills Whitehead products, since the suspended attaching map of the product cell is null-homotopic. No proofs of these remaining algebraic properties are needed here.

Bilinearity has a specific fundamental-group exception. If both inputs lie in ${\pi }_{1}Y$, the attaching loop of the torus is the commutator, so the operation is $gh{g}^{-1}{h}^{-1}$, with the possible inverse corresponding to reversal of its boundary orientation. This is not a homomorphism in either input in a general nonabelian group.

If one input is $g\in {\pi }_{1}Y$ and the other is $\beta \in {\pi }_{q}Y$, $q\ge 2$, the product records the action difference $g\cdot \beta -\beta$, up to the fixed overall boundary sign. Choose the action convention matching that boundary orientation and write ${\delta }_g\beta =g\cdot \beta -\beta$. ThenThe second formula is a crossed-additivity law, not ordinary additivity in $g$. It follows by expanding $gh\cdot \beta -\beta =g\cdot (h\cdot \beta -\beta )+(g\cdot \beta -\beta )$. Consequently the higher-homotopy variable remains additive, while the fundamental-group variable is generally not. If the action is trivial the mixed product is zero. This explains exactly where the usual bilinearity statement fails.

Put $m=2n$ and let $\alpha$ be the identity class of $S^m$. With the product-cell orientation used in part (i), form the folding map $\nabla :S^m\vee S^m\to S^m$, which is identity on each summand. The top cell of $S^m\times S^m$ is attached by the product of the two inclusions, so folding gives the attaching map $[\alpha ,\alpha {]}_W$. Thus there is a cellular mapwhich is the fold on the lower skeleton and degree one on the top relative cell.

Let $a,b\in H^m(S^m\times S^m;\mathbb{Z})$ be the classes from the two factors, and let $x\in H^m(C;\mathbb{Z})$ restrict to the generator of the bottom sphere. Orient the top class $z\in H^{2m}(C;\mathbb{Z})$ so that $F^{\ast }z=ab$. Then $F^{\ast }x=a+b$. Since $a^2=b^2=0$ and $m$ is even, graded commutative algebra givesThe map $F^{\ast }$ is an isomorphism in degree $2m$, because of its degree-one top-cell map. Therefore $x^2=2z$. By the definition of the Hopf invariant this provesThe shifted Lie convention in part (i) multiplies the geometric bracket by $(-1{)}^m=1$, so it gives the same square and the same invariant. Reversing the orientation chosen for the attached top cell would reverse both its generator and the numerical sign convention; with the standard orientation the value is the stated positive two.