On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction $$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ of a differential graded algebra $(A,d_A)$. The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$. As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$?

Thanks to anyone who can shed some light on the two-sided bar construction.

On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction $$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ of a differential graded algebra $(A,d_A)$ (over a field). The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$. As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$?

Thanks to anyone who can shed some light on the two-sided bar construction.

On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction $$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ of a differential graded algebra $(A,d_A)$. The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$. As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by 1?

Thanks to anyone who can shed some light on the two-sided bar construction.

On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction $$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ of a differential graded algebra $(A,d_A)$. The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$. As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module.

What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$?

Thanks to anyone who can shed some light on the two-sided bar construction.