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I am reading John McCleary's A User's Guide to Spectral Sequence and was quite confused about one result: On page 15 of the version I was reading, it says that if $E^{\star,\star}_2$ is the bigraded vector space in Example 1.E, then $P(E^{\star,\star}_2,t)=(1+t^{11})(1+t^4+t^8+t^{12})(1+t^3)$. I am quite confused on how to obtain this result from Example 1.E. It seems to me that $P(E^{\star,\star}_t)$ has a term $t^{11+12+3}=t^{26}$, which by definition of $P(E^{\star,\star}_2,t)$ implies that $\text{dim}_k(\bigoplus _{p+q=26}E^{p,q})=1$. Why is that? I am not sure if I have understood Example 1.E wrongly. Any explanation will be greatly appreciated.

Update: Thank you @Neil Strickland for reminding me. The conditions of Example 1.E are: Suppose $E^{\star,\star}_2$ is given as an algebra by

$E^{\star,\star}_2\cong\mathbb{Q}[x,y,z]/(x^2=y^4=z^2=0)$,

where the bidegree of each generator is given by $\text{bideg}x=(7,1)$, $\text{bideg}y=(3,0)$ and $\text{bideg}z=(0,2)$. Furthermore, suppose $d_2(x)=y^3$ amd $d_3(z)=y$. In this case, the spectral sequence collapses at $E_4$ and, though $x$ and $y$ do not survive to $E_{\infty}$, the product $xy$ does.

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    $\begingroup$ You'll have a much better chance of getting a useful answer if you actually tell us what Example 1.E is. $\endgroup$
    – Neil Strickland
    Commented Apr 9, 2012 at 21:23
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    $\begingroup$ Also, I had found McCleary's book quite unenlightening, but maybe it's just me. $\endgroup$
    – Igor Rivin
    Commented Apr 9, 2012 at 21:38
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    $\begingroup$ My guess is that the numbers in example 1.E were changed to (7,1), (3,0) and (0,2) from (10,1), (4,0) and (0,3), perhaps to fit the page better, but the discussion a page later was not changed. $\endgroup$
    – Ben Williams
    Commented Apr 10, 2012 at 0:25
  • $\begingroup$ @Ben Williams, this change is not mentioned in the book. Does this mean that the case is that this part of the book contains some flaws, not I understood it wrongly? $\endgroup$
    – Zuriel
    Commented Apr 10, 2012 at 2:47
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    $\begingroup$ That's my guess. I think you understood it fine, and it's an erratum in the book. $\endgroup$
    – Ben Williams
    Commented Apr 10, 2012 at 4:56

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I think the Poincar\'{e} polynomial should be $P(E_2^{*,*},t)=(1+t^{8})(1+t^3+t^6+t^9)(1+t^2)$ instead of the one you mentioned so this is an erratum in the book. Its clear from the algebra structure that there can not be any term of degree 26.

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