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I am trying to compute $ [\mathbb{HZ}/4,\mathbb{HZ}/4 ]$ the mod 4 Steenrod Algebra. For some reason, I need to work it out till dimension 6 or so. My approach is to use the cofiber sequence

$\mathbb{HZ}/4 \to \mathbb{HZ}/2 \xrightarrow{Sq^{1}}\Sigma \mathbb{HZ}/2 $

twice.

I did the computation till first five degrees. A description would be as follows, Let $g$ be the generator of $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{0}$, for convenience let us denote $g' = 2g$. So whenever there is an element with $g'$ would mean that it has $2$ torsion.

Degree | elements

$ 1: \beta g$

$ 2: Sq^{2} g' $ (this would mean that $2Sq^{2} g' =0$)

$ 3 : Sq^{3}g$, (but satisfies $2Sq^{3}g =0$ ), $Sq^{2} \beta g'$

$ 4 : Sq^{3} \beta g$( $2 Sq^{3} \beta g = 0$), $Sq^{4}g'$

$ 5 : Sq^{5}g$( $2 Sq^{5} g = 0$), $Sq^{4} \beta g' = 0$

I do not know if they are write, but if somebody has done it please verify. I suspect the all the groups $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{n}$ are two torsion except $n=0,1$. Is this some sort of known result? Also another thing that I am worrying about is the equivalent Cartan formula in Mod 4 Steenrod algebra? Is there a way to detect the Cartan formula?

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I asked a similar question here mathoverflow.net/questions/106602/… and got some helpful comments but no answers. –  Mark Grant Jan 22 '13 at 22:13
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1 Answer

up vote 11 down vote accepted

Let me write $H$ for $H\mathbb{Z}/2$ and $X$ for $H\mathbb{Z}/4$. In these terms we can compute $H^*(X) = [H\mathbb{Z}/4,H\mathbb{Z}/2]$; it is isomorphic to $A/Sq^1 \oplus \Sigma A/Sq^1$, where $A$ is the Steenrod algebra.

Let's take the sledgehammer to the walnut and hit this with the Adams spectral sequence, and leave convergence for some poor other sod to clean up. It computes the groups you're interested in and starts with $$ Ext_A(A/Sq^1 \oplus \Sigma A/Sq^1, A/Sq^1 \oplus \Sigma A/Sq^1). $$ This allows a change of rings to Ext over an exterior algebra: $$ Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2 \oplus \Sigma \mathbb{Z}/2, A/Sq^1 \oplus \Sigma A/Sq^1). $$ This decomposes into a sum of four terms, all coming from $Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2, A/Sq^1)$.

Now we need some structure about the Steenrod algebra: namely, the basis in terms of admissible monomials. A sum of distinct admissible monomials is in the image of right multiplication by $Sq^1$ if and only if it the monomials all visibly end with $Sq^1$, and in the image of left multiplication by $Sq^1$ if and only if terms all end with $Sq^{odd}$.

The net result of this is that, other than $\mathbb{Z}/2$ in degree 0, the module $A/Sq^1$ is a direct sum of free modules with respect to the left action by $Sq^1$. This means that almost all of the Ext groups vanish in positive degrees. The only exceptions are four copies of $Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2, \mathbb{Z}/2)$ in total degrees $-1, 0, 0, 1$. These are spawning the two copies of $\mathbb{Z}/4$ that you mention in (cohomological) degrees 0 and 1. The rest of the terms are concentrated in Adams filtration zero, and so multiplication by 2 annihilates them.

Generalization from $4$ is an exercise for the reader. Sorry for my cryptic comments on the question of Mark Grant's, which also appear to be a description of the self-transformations of $H\mathbb{Z}_{(2)}$. I also don't have a good answer for you about the Cartan formula.

(This could probably be phrased in a more pedestrian fashion using the cofiber sequence $H\mathbb{Z} \to H\mathbb{Z} \to H\mathbb{Z}/4$.)

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Thanks Tyler, this is a great computation. Glad that you posted it here. This solves a lot of my worries. –  Prasit Jan 23 '13 at 6:57
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