# Is there an explicit construction of Tate-Beilinson residue?

Background: Tate's construction of abstract residues is generalized in Beilinson's constructon as follows:

Let $E$ be a unital, cubically decomposed $k$-algebra, i.e., (a) there are two sided ideals $E_i^+, E_i^-$ such that $E_i^+ + E^-_i= E$ for $i=1,\ldots, n$. (b) Let $E_i^0:=E_i^+ \cap E_i^-$ and $E_{\text{tr}}:= \bigcap_{i=1}^{n} E^0_i$. Then there exists a $k$-linear map $\text{trace}: E_{\text{tr}}/[E_{\text{tr}}, E] \to k$.

Now let $K$ be a commutative $k$-subalgebra (with unity) of $E$. Then we can obtain a $k$-linear map $\text{res}: \Omega_{K/k}^n \to k$. The construction of $\text{res}$ by Beilinson makes use of lie algebra homology and some spectral sequences. Now for the case $n=1$, Tate's construction gives an explicit description of $\text{res}$ as follows:

Let $f,g \in K$ and $f = f_1 + f_2$ and $g= g_1 + g_2$, where $f_1, g_1 \in E^+$ and $f_2,g_2 \in E^-$. Then $\text{res} (f \text{d}g) = \text{trace}(f_1g_1 - g_1f_1)$. (It can be shown that this is independent of the choice of $f_1, f_2, g_1, g_2$.)

Question: Is there an explicit way to write down the map $\text{res}$ in Beilinson's construction for any $n$ (maybe with some further assumptions)?

• Your formula for the residue seems to have some wrong subscripts. May 30, 2016 at 13:02
• Maybe, but I don't see it. Would you please explain what is wrong? May 30, 2016 at 13:09
• If I am not mistaken, this is just Tate's construction with $E^+ = E_1$ and $E^- = E_2$. May 30, 2016 at 13:41