Tate's definition of residues

Question

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?

Answer 1

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.

Answer 2

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.