Tate's definition of residues In http://www.numdam.org/article/ASENS_1968_4_1_1_149_0.pdf, Tate defines residues on a curve over an arbitrary field as a trace of some commutator. What is the intuition for the definition? If I knew the definition of a residue in complex analysis what might lead me to think about traces of commutators?
 A: Residues satisfy $$\operatorname{res}_x(fdg) + \operatorname{res}_x(gdf) = \operatorname{res}_x(d (fg)) =0$$
which tells you that they are antisymmetric, and many antisymmetric pairings in math arise from commutators.
Usually I find the best way to get intuition for these things is to calculate them explicitly for $z^n , n \in \mathbb Z$. I don't think it's too hard to calculate this trace definition for $z^n$, and see how it gets you the correct answer. Then linearity does the rest. But I'm not sure how much that will help explain how someone could think of it.
A: I don't have a conceptual answer, but here is what you get by tracing through Tate's definitions.
Let $V$ be the vector space of Laurent series in $t$ and $A$ the subspace of power series.
We will work with linear operators $\phi: V \to V$, which we think of as $\infty \times \infty$ matrices with rows and columns labeled by $\mathbb{Z}$.
We will often divide these infinte $4$ blocks. For example, the condition that $\phi(A) \subseteq A$ says that $\phi$ has block form $\left( \begin{smallmatrix} \ast & 0 \\ \ast&\ast \end{smallmatrix} \right)$. 
We can represent Tate's ring $E$ with ideals $E_1$,$E_2$ and $E_0$ visually as follows:
$$
E = \begin{pmatrix} \ast & 0' \\ \ast & \ast \end{pmatrix} \qquad
E_1 = \begin{pmatrix}  0' &  0' \\ \ast & \ast \end{pmatrix} $$
$$E_2 = \begin{pmatrix} \ast &  0' \\ \ast &  0' \end{pmatrix} \qquad
E_0 = E_1 \cap E_2 = \begin{pmatrix}  0' &  0' \\ \ast &  0' \end{pmatrix} $$
where $0'$ means ``confined to finitely many rows".
Let $f(t)$ and $g(t)$ be Laurent series and write $f$ and $g$ for the operations of multiplication by $f(t)$ and $g(t)$. Let $f_1$ and $g_1$ be approximations with $f_1$, $g_1 \in E_1$ and $f \equiv f_1$, $g \equiv g_1 \bmod E_2$. Tate's claim is that $\mathrm{Tr}\ [f_1, g_1] = \mathrm{Res} (fdg)$.
Let's check for $f(t)=t^a$, $g(t)=t^b$, and one particular choice of $f_1$, $g_1$. Multiplication by $t^c$ is given by the matrix
$$M(c)_{ij} =  \begin{cases} 1 & i=j+c \\ 0 & \mbox{otherwise} \end{cases}.$$
We choose the approximation
$$S(c)_{ij} = \begin{cases} 1 & 0 \leq i=j+c \\ 0 & \mbox{otherwise} \end{cases}.$$
In other words, we take $M(c)$ and change all elements in the top blocks to $0$.
We note that $S(a)S(b) - S(b) S(a)$ has finitely many nonzero entries, so it makes sense to take its trace. Those entries lie on the diagonal $i=j+a+b$, so the trace is zero if $a+b \neq 0$. In the case $a+b=0$, I compute $\mathrm{Tr}\ [ S(-b), S(b) ] = b$. So we have
$$\mathrm{Tr}\ [S(a),S(b)] = \left\{ \begin{matrix} b & a+b=0 \\ 0 & \mbox{otherwise} \end{matrix} \right\} = \mathrm{Res} {\Big(} x^a d(x^b) {\Big)}$$
as desired.
