I am puzzled over something I read in Quillen's *[On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field](https://www.jstor.org/stable/1970825)*.


On page 557, when computing the $E\_2$ page of a case of the Eilenberg-Moore spectral sequence, he first shows that 


$$E\_2 \cong \text{Tor}^{A\_1}(k,k) \otimes k \otimes k \otimes A\_4$$


In this formula, 


* $\otimes = \otimes\_k$
* $A\_1 = k[x\_1, x\_2, \dots]$ with $x\_i$ in degree $2i$
* $A\_4 = k[y\_1, y\_2, \dots]$ with $y\_i$ in degree $2i$.
* The isomorphism is an isomorphism of algebras.


Based on the isomorphism above he claims that 
$$E\_2 \cong P[y\_1, y\_2, \dots] \otimes \Lambda\_k[e\_1, e\_2, \dots]$$


where 


* $\Lambda\_R$ means "exterior algebra over $R$"
* $y\_i \in E\_2^{0,2i}$
* $e\_i \in E\_2^{-1,2i-1}$.


***Question:*** I am confused about the last point, that $e\_i \in E\_2^{-1,2i-1}$. My understanding of the "bigrading" on $\text{Tor}^{A\_1}(k,k)$ is that I can take a graded projective resolution $P^{\bullet} \rightarrow k$, and then the second index comes from the fact that the (co)homology objects of $P^{\bullet} \otimes\_{A\_1} k$ inherit a grading. 


Of course, I'm going to use the Koszul resolution. But then, 


1. Shouldn't the generators $e\_i$ live in degree $2i$ just like the $x\_i$'s they're mapped to under $\Lambda\_{A\_1}[e\_i] \rightarrow A\_1$?
2. If so, doesn't that imply that $e\_i \in E\_2^{-1,2i}$ ?



Grading in Eilenberg-Moore spectral sequence
Different people use different notation on gradings, for example I would have called the bigrdading of $e\_i$ $E\_2^{1,2i}$. Supposing that this is not a typo, Quillen meant
by $k$ in $E\_s^{j,k}$ the total degree, not the internal degree of the element.  

In other words, he has a spectral sequence $E\_2^{s,t}\Rightarrow E\_{\infty }^{t}$.


So you are right in 1, but in 2, the above definition of indices gives the shift.


