Steenrod operations are an example of what's known as a power operation. Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity.
A cohomology class on $X$ amounts to a map $a: Xto R$, where $R = prod_{ngeq0} K(F_2,n)$. So the cup product of $a$ and $b$ is given by
$$Xtimes X to Rtimes R xrightarrow{mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes addition of cohomology classes.)
You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/Sigma_n$, where $Sigma_n$ is the symmetric group, i.e.,
$$X^n xrightarrow{a^n} R^n rightarrow R$$
should factor through the quotient $X^n/Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $Sigma_n$, and a product map:
$$mu_n' : E(n)times R^nto R$$
which is $Sigma_n$ invariant, so it factors through $(E(n)times R^n)/Sigma_n$. Thus, given $a: Xto R$, you get
$$P'(a): (E(n)times X^n)/Sigma_n to (E(n)times R^n)/Sigma_n to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/Sigma_n times Xto R.$$
If $n=2$, then $E(2)/Sigma_2$ is what Hatcher seems to call $L^infty$; it is the infinite real proj. space $RP^infty$. So $P(a)$ represents an element
in $H^* RP^infty times X approx H^*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$.
Other cohomology theories have power operations (for K-theory, these are the Adams operations).
You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.
No comments:
Post a Comment