Let’s begin by talking about cohomology operations.
I will state a few definitions and results now, and explain their significance in a bit.
Definition: Let and be cohomology theories. A cohomology operation of “type and degree ” is a natural transformation . A stable cohomology operation of degree is a collection of homomorphisms of type and degree such that the following diagram commutes.
Because of Brown representability for ordinary cohomology and the Yoneda lemma, we observe that cohomology operations are in bijection with elements of . (To expand on this: if , then natural transformations are in bijection with elements of . To see this, consider . Let the corresponding element of be . Given an element of , we define via .)
Theorem: Let . There are stable cohomology operations such that:
- and if .
- The Cartan formula holds:
These are called the Steenrod operations.
The Adem relations hold as can be checked by the reader:
if . Let be a sequence of positive integers. Define . Inspired by the Adem relations, say that is admissible if for all . In particular, this means that no Adem relations can be applied.
Theorem: The algebra of stable cohomology operations in -cohomology is the quotient of the free associative -algebra by imposing the Adem relations. Equivalently, it is the -algebra generated by for admissible . This is called the Steenrod algebra, and is denoted .
Say that is indecomposable if it cannot be expressed as for having degree lower than that of . What are the indecomposable elements of the mod Steenrod algebra?
Example: Suppose . Then is indecomposable. Let’s write . The Cartan formulas tell us that is a homomorphism of graded rings. In particular, . We know that where . We compute that , because all the higher terms vanish. This is , by the characterization of Steenrod squares. In particular, . Because is a power of , this is . What this means is that if , and . Consequently, can’t be written as a composition of lower degree terms, and thus is indecomposable.
It turns out:
Proposition: The are the only indecomposables in .
Proof. Consider for , so that, in particular, . We want to write as the sum of lower degree terms. This follows from the Adem relations, because:
Now, , so:
And hence is decomposable. QED.
This means that generates 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 for odd vanishes at some point. Moreover, there is a nowhere vanishing vector field on for any even .
Proof. Consider a point and a tangent vector . The unit circle in the plane spanned by and is defined by where . As goes from to , we get a homotopy between the antipodal map and the identity map. Looking at the degree tells us that , i.e., is even and is an odd-dimensional sphere.
Now suppose is even, so that is odd-dimensional. Consider , and write . We need to construct a vector such that . An obvious choice is just because then . 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 ? Namely, what is the maximum number of vector fields on such that are all linearly independent for every ?
By Gram-Schmidt orthonormalization, the vectors can be replaced by that span the same subspace of . Gram-Schmidt orthonormalization is continuous, and thus we get vector fields on .
Consider the Stiefel manifold of orthonormal -frames (ordered collection of orthonormal vectors) in . This can be topologized as a subspace of . This is equivalently the space of -matrices with orthonormal columns. For example, . There is a map given by projection onto the first factor.
Suppose there are linearly independent vector fields on . There is a map sending , that is clearly a section of . If one has a map that is a section of , then there are vector fields defined via . We want to therefore find some such that for every , there is no continuous map that is a section of .
Let’s look at a special case of this problem.
Definition: A manifold is parallelizable if its tangent bundle is the trivial bundle.
In other words, if , is parallelizable if there are vector fields such that form a basis for the tangent space at every point of .
Recall that acts on by rotations. Suppose is parallelizable. Then there is a section of . We can now consider the composition , denoted .
Let denote the standard basis vector . Then because acts on by sending it to where and the are obtained from as above. Also, by definition, under . Since is parallelizable, we can consider vector (linearly indepedent) fields defined by . We can assume that are the other standard basis vectors by changing signs if necessary, and “straightening” things out near . Thus is a multiplication on with as the identity. This is a H-space structure on .
Lemma: is a division algebra if and only if is a H-space.
Proof. Define a multiplication via . QED.
Studying H-spaces through the Hopf invariant
Construction: Suppose , so that . Its cofiber is the pushout . Suppose that . Then it is easy to see that . Suppose generates and generates . Then , so for some integer . This is called the Hopf invariant of .
Example: Suppose is odd. Because of graded commutativity, it follows that , so that .
Proposition [Prop 4B.1 in Hatcher]: The Hopf invariant gives a homomorphism .
It’s rather easy to see that the Hopf invariant of is , and that the Hopf invariant of is . More interesting is the statement that we can find a map whose Hopf invariant is for even . Consider where is a chosen basepoint. This has one cell in dimensions . Thus it’s of the form for some , and Proposition 3.22 of Hatcher shows that .
Suppose is a H-space structure on . We will define a map as follows. We know that , which is . On the first factor, define , and on the second factor, define .
Lemma: The Hopf invariant of constructed above is .
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 has Hopf invariant , then is a power of .
Proof. Recall that . But . If was decomposable, then , so . Thus by our characterization of indecomposables in , we know that for some . QED.
This means that if , then isn’t parallelizable.
Theorem (Adams): If has an element of Hopf invariant , then , or .
Proof. K-theory, which we will proceed to study.