Grading in Eilenberg-Moore spectral sequence

Question

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.

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 $e_i \in E_2^{-1,2i}$ ?

Answer 1

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.