Hopf invariant one, part II: cohomology operations and vector fields on spheres

Let’s begin by talking about cohomology operations.

Cohomology Operations

I will state a few definitions and results now, and explain their significance in a bit.

Definition: Let $E$ and $F$ be cohomology theories. A cohomology operation of “type $k$ and degree $n$ ” is a natural transformation $E^k\to F^{k+n}$ . A stable cohomology operation of degree $n$ is a collection of homomorphisms $\phi_k:E^k\to E^{k+n}$ of type $k$ and degree $n$ such that the following diagram commutes.

Because of Brown representability for ordinary cohomology and the Yoneda lemma, we observe that cohomology operations $H^m(-;G)\to H^{m+n}(-; H)$ are in bijection with elements of $H^{m+n}(K(G,m);H)$ . (To expand on this: if $F(X)=[X,Y]$ , then natural transformations $F\to G$ are in bijection with elements of $G(Y)$ . To see this, consider $\alpha:F\to G$ . Let the corresponding element of $G(Y)$ be $\alpha(\mathrm{id}_Y)$ . Given an element $x$ of $G(Y)$ , we define $\alpha:F\to G$ via $\alpha(f)=f^\ast(x)$ .)

Theorem: Let $n\geq 0$ . There are stable cohomology operations $\mathrm{Sq}^n: H^k(X;\mathbf{Z}/2\mathbf{Z})\to H^{k+n}(X;\mathbf{Z}/2\mathbf{Z})$ such that:

$\mathrm{Sq}^0=1$

$\mathrm{Sq}^{\deg x}(x)=x^2$ and $\mathrm{Sq}^n(x)=0$ if $n>\deg x$ .

The Cartan formula holds:

These are called the Steenrod operations.

The Adem relations hold as can be checked by the reader:

adem-1

if $a<2b$ . Let $I=(i_1,i_2,\cdots,i_n)$ be a sequence of positive integers. Define $\mathrm{Sq}^I=\mathrm{Sq}^{i_1}\cdots\mathrm{Sq}^{i_r}$ . Inspired by the Adem relations, say that $I$ is admissible if $i_k\geq 2i_{k+1}$ for all $k$ . In particular, this means that no Adem relations can be applied.

Theorem: The algebra of stable cohomology operations in $\mathbf{Z}/2\mathbf{Z}$ -cohomology is the quotient of the free associative $\mathbf{F}_2$ -algebra $\mathbf{F}_2\{\mathrm{Sq}^i\}$ by imposing the Adem relations. Equivalently, it is the $\mathbf{F}_2$ -algebra generated by $\mathrm{Sq}^I$ for admissible $I$ . This is called the Steenrod algebra, and is denoted $\mathcal{A}$ .

Say that $x\in \mathcal{A}$ is indecomposable if it cannot be expressed as $\sum_i a_i b_i$ for $a_i,b_i$ having degree lower than that of $x$ . What are the indecomposable elements of the mod $2$ Steenrod algebra?

Example: Suppose $i=2^n$ . Then $\mathrm{Sq}^i$ is indecomposable. Let’s write $\mathrm{Sq}=\sum_i\mathrm{Sq}^i$ . The Cartan formulas tell us that $\mathrm{Sq}$ is a homomorphism of graded rings. In particular, $\mathrm{Sq}(\alpha^n)=\mathrm{Sq}(\alpha)^n$ . We know that $H^\ast(\mathbf{RP}^\infty;\mathbf{Z}/2\mathbf{Z})=\mathbf{Z}/2\mathbf{Z}[\alpha]$ where $\deg\alpha=1$ . We compute that $\mathrm{Sq}(\alpha)=\mathrm{Sq}^0(\alpha)+\mathrm{Sq}^1(\alpha)$ , because all the higher terms vanish. This is $\alpha+\alpha^2$ , by the characterization of Steenrod squares. In particular, $\mathrm{Sq}(\alpha^i)=\mathrm{Sq}(\alpha)^i=(\alpha+\alpha^2)^i$ . Because $i$ is a power of $2$ , this is $\alpha^i+\alpha^{2i}$ . What this means is that $\mathrm{Sq}^k(\alpha)=0$ if $0<k<i$ , and $\mathrm{Sq}^i(\alpha^i)\neq 0$ . Consequently, $\mathrm{Sq}^i$ can’t be written as a composition of lower degree terms, and thus $\mathrm{Sq}^{2^n}$ is indecomposable.

It turns out:

Proposition: The $\mathrm{Sq}^{2^n}$ are the only indecomposables in $\mathcal{A}$ .

Proof. Consider $\mathrm{Sq}^{m+2^n}$ for $0<m<2^n$ , so that, in particular, $m<2^{n+1}$ . We want to write $\mathrm{Sq}^{m+2^n}$ as the sum of lower degree terms. This follows from the Adem relations, because:

eqn1

Now, $\binom{2^k-1}{a}\equiv 1\bmod 2$ , so:

eqn2

And hence $\mathrm{Sq}^{m+2^n}$ is decomposable. QED.

This means that $\mathrm{Sq}^{2^k}$ generates $\mathcal{A}$ as an algebra. I now want to give an application of Steenrod squares to a classical question in algebraic topology.

Vector fields on spheres

Recall the following classical result.

Proposition: Any vector field on $S^{n-1}$ for odd $n$ vanishes at some point. Moreover, there is a nowhere vanishing vector field on $S^{n-1}$ for any even $n$ .

Proof. Consider a point $x\in S^{n-1}$ and a tangent vector $v(x)$ . The unit circle in the plane spanned by $x$ and $v(x)$ is defined by $x\cos \pi t + v(x)\sin \pi t$ where $0\leq t\leq 1$ . As $t$ goes from $0$ to $1$ , we get a homotopy between the antipodal map and the identity map. Looking at the degree tells us that $(-1)^n=1$ , i.e., $n$ is even and $S^{n-1}$ is an odd-dimensional sphere.

Now suppose $n$ is even, so that $S^{n-1}$ is odd-dimensional. Consider $x\in S^{n-1}$ , and write $x=(x_1,x_2,\cdots,x_{n-1},x_n)$ . We need to construct a vector $v(x)$ such that $v(x)\cdot x=0$ . An obvious choice is just $v(x):=(-x_2,x_1,\cdots,-x_n,x_{n-1})$ because then $v(x)\cdot x=x_1x_2-x_2x_1+\cdots+x_{n-1}x_n-x_nx_{n-1}=0$ . This is clearly continuous, and gives the desired vector field. QED.

A question that is now natural to ask, and has motivated a lot of algebraic topology, is the following.

Question: How many linearly indepedent vector fields exist on $S^{n-1}$ ? Namely, what is the maximum number of vector fields $X_1,\cdots,X_k$ on $S^{n-1}$ such that $X_1(x),\cdots,X_k(x)$ are all linearly independent for every $x\in S^{n-1}$ ?

By Gram-Schmidt orthonormalization, the vectors $X_1(x),\cdots,X_k(x)$ can be replaced by $X^\prime_1(x),\cdots,X^\prime_k(x)$ that span the same subspace of $\mathbf{R}^n$ . Gram-Schmidt orthonormalization is continuous, and thus we get vector fields $X^\prime_1,\cdots,X^\prime_k$ on $S^{n-1}$ .

Consider the Stiefel manifold $V_{n,k}$ of orthonormal $k$ -frames (ordered collection of $k$ orthonormal vectors) in $\mathbf{R}^n$ . This can be topologized as a subspace of $(S^{n-1})^k$ . This is equivalently the space of $n\times k$ -matrices with orthonormal columns. For example, $V_{n,n}=O(n)$ . There is a map $\pi:V_{n,k+1}\to S^{n-1}$ given by projection onto the first factor.

Suppose there are $X_1,\cdots,X_k$ linearly independent vector fields on $S^{n-1}$ . There is a map $S^{n-1}\to V_{n,k+1}$ sending $x\mapsto(x,X_1(x),\cdots,X_k(x))$ , that is clearly a section of $\pi$ . If one has a map $\sigma:S^{n-1}\to V_{n,k+1}$ that is a section of $\pi$ , then there are vector fields $X_1,\cdots,X_k:S^{n-1}\to\mathbf{R}^n$ defined via $\sigma(x)=(x,X_1(x),\cdots,X_k(x))$ . We want to therefore find some $m$ such that for every $\ell>m$ , there is no continuous map $S^{n-1}\to V_{n,\ell+1}$ that is a section of $\pi$ .

Let’s look at a special case of this problem.

Definition: A manifold $M$ is parallelizable if its tangent bundle is the trivial bundle.

In other words, if $\dim M=n$ , $M$ is parallelizable if there are vector fields $X_1,\cdots,X_n$ such that $X_1(p),\cdots,X_n(p)$ form a basis for the tangent space $T_p(M)$ at every point $p$ of $M$ .

Recall that $O(n)$ acts on $S^{n-1}$ by rotations. Suppose $S^{n-1}$ is parallelizable. Then there is a section $\sigma$ of $\pi:V_{n,n}=O(n)\to S^{n-1}$ . We can now consider the composition $S^{n-1}\times S^{n-1}\xrightarrow{\sigma\times 1} V_{n,n}\times S^{n-1}=O(n)\times S^{n-1}\xrightarrow{\text{action}}S^{n-1}$ , denoted $\mu$ .

Let $\mathbf{e}_1$ denote the standard basis vector $(1,0,\cdots,0)$ . Then $(x,\mathbf{e}_1)\mapsto(\sigma(x),\mathbf{e}_1)\mapsto x$ because $\sigma(x)$ acts on $\mathbf{e}_i$ by sending it to $X_{i-1}(x)$ where $X_0(x):=x$ and the $X_i$ are obtained from $\sigma$ as above. Also, by definition, $(\mathbf{e}_1,x)\mapsto(\sigma(\mathbf{e}_1),x)$ under $\sigma\times 1$ . Since $S^{n-1}$ is parallelizable, we can consider $(n-1)$ vector (linearly indepedent) fields $X_1,\cdots,X_{n-1}$ defined by $\sigma$ . We can assume that $X_1(\mathbf{e}_1),\cdots,X_{n-1}(\mathbf{e}_1)$ are the other standard basis vectors $\mathbf{e}_2,\cdots,\mathbf{e}_n$ by changing signs if necessary, and “straightening” things out near $\mathbf{e}_1$ . Thus $\mu$ is a multiplication on $S^{n-1}$ with $\mathbf{e}_1$ as the identity. This is a H-space structure on $S^{n-1}$ .

Lemma: $\mathbf{R}^n$ is a division algebra if and only if $S^{n-1}$ is a H-space.

Proof. Define a multiplication $\mu:S^{n-1}\times S^{n-1}\to S^{n-1}$ via $\mu(x,y)=\frac{xy}{||xy||}$ . QED.

Studying H-spaces through the Hopf invariant

Construction: Suppose $f:S^{2n-1}\to S^n$ , so that $[f]\in\pi_{2n-1}(S^n)$ . Its cofiber is the pushout $C_f:=S^n\cup_f D^{2n}$ . Suppose that $n\geq 2$ . Then it is easy to see that $H^i(C_f;\mathbf{Z})=\begin{cases}\mathbf{Z} & i=0,n,2n\\ 0 & \text{else}\end{cases}$ . Suppose $\alpha$ generates $H^n(C_f;\mathbf{Z})$ and $\beta$ generates $H^{2n}(C_f;\mathbf{Z})$ . Then $\alpha^2\in H^{2n}(C_f;\mathbf{Z})$ , so $\alpha^2=H(f)\beta$ for some integer $H(f)$ . This is called the Hopf invariant of $f$ .

Example: Suppose $n$ is odd. Because of graded commutativity, it follows that $\alpha^2=0$ , so that $H(f)=0$ .

Proposition [Prop 4B.1 in Hatcher]: The Hopf invariant gives a homomorphism $\pi_{2n-1}(S^n)\to \mathbf{Z}$ .

It’s rather easy to see that the Hopf invariant of $S^{2n-1}\xrightarrow{f}S^{2n-1}\xrightarrow{g}S^n$ is $H(g)\deg f$ , and that the Hopf invariant of $S^{2n-1}\xrightarrow{g}S^n\xrightarrow{f}S^n$ is $H(g)(\deg f)^2$ . More interesting is the statement that we can find a map $S^{2n-1}\to S^n$ whose Hopf invariant is $2$ for even $n$ . Consider $S^n\times S^n/(x_0,\ast)\sim(\ast,x_0)$ where $x_0$ is a chosen basepoint. This has one cell in dimensions $0,n,2n$ . Thus it’s of the form $C_f$ for some $f$ , and Proposition 3.22 of Hatcher shows that $H(f)=2$ .

Suppose $\mu:S^{n-1}\times S^{n-1}\to S^{n-1}$ is a H-space structure on $S^{n-1}$ . We will define a map $f:S^{2n-1}\to S^{n-1}$ as follows. We know that $S^{2n-1}=\partial D^{2n}$ , which is $(D^n\times \partial D^n)\cup(\partial D^n\times D^n)$ . On the first factor, define $f(x,y)=||x||\mu\left(\frac{x}{||x||},y\right)$ , and on the second factor, define $f(x,y)=||y||\mu\left(x,\frac{y}{||y||}\right)$ .

Lemma: The Hopf invariant of $f$ constructed above is $1$ .

Proof. Claim 2.6 in http://math.uchicago.edu/~may/REU2015/REUPapers/Kirsche.pdf. QED.

Now we can prove the following important result.

Theorem: If $f:S^{2n-1}\to S^n$ has Hopf invariant $1$ , then $n$ is a power of $2$ .

Proof. Recall that $\alpha^2=H(f)\beta$ . But $\alpha^2=\mathrm{Sq}^n\alpha$ . If $\mathrm{Sq}^n$ was decomposable, then $\alpha^2=0$ , so $H(f)=0$ . Thus by our characterization of indecomposables in $\mathcal{A}$ , we know that $n=2^k$ for some $k$ . QED.