A novice to the Adams spectral sequence, I am attempting to follow a computation in McCleary's book in the mod 3 Adams spectral sequence for $\pi_*(S)$. By working out part of a minimal resolution of $\mathbb{Z}/3\mathbb{Z}$ over the mod 3 Steenrod algebra, we compute $Ext^{s, t}_A(\mathbb{Z}/3\mathbb{Z}, \mathbb{Z}/3\mathbb{Z})$ for $t \leq 11$ and all $s$. McCleary claims that there is a lack of differentials here, but I don't see why. If we arrange the spectral sequence with $t-s$ increasing horizontally to the right and $s$ increasing vertically, then $d_r$ goes left one space and up $r$ spaces. In the $t-s=3$ column, for $s \leq 3$ there is only a generator in degree $s=2$ and there is nothing in the $s \leq 3$ range in the $t-s=2$ column. But as far as I know, there could be nontrivial things in the $t-s=2$ column for large enough $s$ for which I don't know $Ext^{s, t}$ for $t=s+2$. So conceivably, there is room for differentials here (from t-s=3, s=2 to something in the $t-s=2$ column). Why does McCleary conclude that there aren't any, or what am I misunderstanding here? Thanks!
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$\begingroup$ I thought about it on the walk home, and I have a guess as to what is going on here. In $P_3$, the third stage of our resolution, the first generator (besides the ones coming from the Bocksteins, which won't generate things that map to 0 on their own) occurs in degree greater than 11. So something that maps to zero will be in at least one degree greater than 13, so the first non-Bockstein related generator in $P_4$ won't occur until at least 14. In general, the first generator in $P_{s+3}$ won't occur until degree at least 13+s. So we actually know the Ext groups for $t-s < 13+s-(s+3)=10$. $\endgroup$– Vitaly LormanCommented Nov 16, 2011 at 7:09
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$\begingroup$ McCleary doesn't quite make this explicit. Is this correct? $\endgroup$– Vitaly LormanCommented Nov 16, 2011 at 7:09
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$\begingroup$ In the first comment above, the 11 should be a 13. $\endgroup$– Vitaly LormanCommented Nov 21, 2011 at 19:42
1 Answer
I am going to post more or less what I wrote in the comments above and maybe someone can tell me if this is correct.
I wrote that we know the resolution for $t \leq 11$ and all $s$, but something stronger is true. In the $s$th stage of the resolution, we have a generator $a_s$ in degree $s$ which ultimately comes from the Bockstein $\beta$, and $d(\beta a_s)=0$, so we put a generator in the $(s+1)$ stage of the resolution in degree $s+1$ to map to $\beta a_s$. No other product of an element of the Steenrod algebra (which is not in the ideal generated by $\beta$) and $a_s$ will map to zero, so all other things in the $s$ stage that map to zero must contain at least one other generator. But in the 3rd stage of the resolution, the first generator not related to the Bockstein is in degree 13. To find something in terms of this generator that maps to zero, we must multiply it by something in the Steenrod algebra, so the first non-Bockstein generator in the 4th stage appears in degree greater than or equal to 14. Continuing in this way, the first non-Bockstein generator in the $s+3$ stage does not occur until degree at least $13+s$, and below that the only generator is $a_{s+3}$. So we know the resolution for $t-s< 13+s-(s+3)=10$. When we look at the $E_2$ page of the Adams spectral sequence, this means that we know the first 9 columns, and since diferentials go one to the left and $r$ units upward, from our computations we see that there are no differentials in this part of the spectral sequence.