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:

  1. \mathrm{Sq}^0=1
  2. \mathrm{Sq}^{\deg x}(x)=x^2 and \mathrm{Sq}^n(x)=0 if n>\deg x.
  3. The Cartan formula holds:

These are called the Steenrod operations.

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


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:


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


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.

This means that if n\neq 2^k, then S^{n-1} isn’t parallelizable.

Theorem (Adams): If \pi_{2n-1}(S^{n-1}) has an element of Hopf invariant 1, then n=1,2,4, or 8.

Proof. K-theory, which we will proceed to study.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s