# Grading in Eilenberg-Moore spectral sequence

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

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 $$E_2^{s,t}\Rightarrow E_{\infty }^{t}$$.