Author: Eiko

Tags: affinoid, p-adic geometry, affinoid algebra, projective line, residue theorem, holomorphic functions

Time: 2024-09-12 15:06:50 - 2024-09-12 15:10:22 (UTC)

Holomorphic functions on affinoid subsets

Definition 1.

  1. Let \(F\) be an affinoid subset of \(\mathbb{P}\), we define the ring of Holomorphic functions on \(F\), \(\mathcal{O}(F)\) be the completion of the \(K\)-algebra of rational functions regular on \(F\), taken with respect to the supremum norm \[\|f\| = \sup_{a\in F} |f(a)|.\] Thus \(f\) is holomorphic on \(F\) iff it is a uniform limit of rational functions regular on \(F\).

  2. One also defines \[\mathcal{O}(F)^\circ=\{f\in \mathcal{O}(F) : \|f\|\le 1\},\] (similar to ring of integers) \[\mathcal{O}(F)^{\circ\circ} = \{f\in \mathcal{O}(F) : \|f\|<1\}.\] (similar to the maximal ideal) \(\mathcal{O}(F)^\circ\) is a \(K^\circ\)-algebra and \(\mathcal{O}(F)^{\circ\circ}\) is an ideal. The quotient ring \[\overline{\mathcal{O}(F)} = \mathcal{O}(F)^\circ/\mathcal{O}(F)^{\circ\circ}\] gives us a \(\overline{K}\)-algebra.

Example 1. For \(F=\{a\in \mathbb{P}^1 : |a|\le 1\}\), we have the norm defined above happens to be the Gauss norm \[\left\|\sum a_n z^n\right\| = \sup |a_n|.\] The ring of holomorphic functions on \(F\) is the completion of the rational functions regular on \(F\) with respect to the Gauss norm, which is the Tate algebra \(T_1=K\langle z\rangle\). \[\mathcal{O}(F)=K\langle z\rangle, \quad \mathcal{O}(F)^\circ = K^\circ\langle z\rangle, \quad \mathcal{O}(F)^{\circ\circ} = K^{\circ\circ}\langle z\rangle.\] And the quotient \[\overline{\mathcal{O}(F)} = K^\circ\langle z\rangle/K^{\circ\circ}\langle z\rangle = \overline{K}\langle z\rangle.\]

Proposition 1. Let \(F\) be the closed disk \(\{a\in \mathbb{P}^1 : |a|\le 1\}\) and \(g\in \mathcal{O}(F)\) be any function. Define \(d\) to be the degree of the normalized polynomial \(\overline{cg}\in \overline{K}[z]\) where \(c\in K^*\) and \(\|cg\|=1\). Then we can write for any \(f\in \mathcal{O}(F)\) \[f=gq+r\] where \(q\in \mathcal{O}(F)\) and \(r\in \mathcal{O}(F)\) with \(\deg(r)<d\), and \(\|f\|=\max(\|q\|\cdot\|g\|,\|r\|)\).

Lemma 1. When \(g\) is a monic polynomial with \(\|g\|=1\), and \(f\in \mathcal{O}(F)\), there exists \[f=gq+r, \quad \deg(r)<\deg(g), \quad \|f\|=\max(\|q\|,\|r\|).\]

Proof.

  1. First we investigate the case \((f,g)=(cz^n,g)\) with \(\|g\|=1\) and \[g=\sum_{i=0}^d c_i z^i\] a monic (\(|c_d|=1\), wlog we can assume \(c_d=1\), \(|c_i|\le 1\)) degree \(d\) polynomial. Then, we note that the first quotient \(q_{n}=cz^{n-d}\) \[f=gq_{n}+r_n, \quad r_n = f-gq_n\] satisfies \(\|q_n\|\le |c|\) and \(\|r_n\|\le \|g\|\cdot \|q_n\| \le |c|\). Repeating the division process should give us \(f=gq+r\) with \(\|q\|\le |c|\) and \(\|r\|\le |c|\). In fact, we have a stronger statement \[|c|=\|f\| = \max(\|q\|,\|r\|).\]

  2. By the linearity of residue division, we know that when \(g\) is monic with \(\|g\|=1\), with \[f=\sum_{i=0}^\infty a_i z^i, \quad |a_i|\to 0\] we should have \(q=\sum q_i, r=\sum r_i\) with \((q_i,r_i)=\mathrm{Resdiv}(a_iz^i,g)\), so \(|a_i|=\max(\|q_i\|,\|r_i\|)\). Therefore \[\begin{aligned} \|f\| &= \max |a_i|\\ &= \max_i \max(\|q_i\|,\|r_i\|) \\ &= \max\left(\max_i \|q_i\|,\max_i \|r_i\|\right)\\ &= \max\left(\|q\|,\max_i \|r_i\|\right)\\ &\ge \max(\|q\|,\|r\|). \end{aligned}\] On the other hand clearly \(\|f\|=\|gq+r\|\le \max(\|q\|,\|r\|)\).

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Now we are ready to prove the proposition.

Proof.

  1. Without loss of generality we can assume \(\|g\|=1\), and write out \[g=g_0+g_1\] where \(g_0=\sum_{i=0}^d c_i z^i\) satisfy \(|c_d|=1\) and \(|c_i|\le 1\), \(\|g_1\|<1\). Since \(g_0\) is monic, we can write \[f=q_0 g_0 + r_0=q_0g + r_0 - q_0g_1\] with \(\deg(r_0)<d\) and \(\|f\|=\max(\|q_0\|,\|r_0\|)\). We also note that \(\|q_0g_1\|=\|q_0\|\cdot \|g_1\|\) is strictly decreasing. Continuing division along \(-q_0g_1\) we can get \[\begin{aligned} f &= q_0g + r_0 + q_1g+r_1 + \dots\\ &= g(q_0+q_1+\dots) + (r_0+r_1+\dots) \end{aligned}\] with \(\|q_i\|\le \|f\|\cdot \|g_1\|^i\) and \(\|r_i\|\le \|f\|\cdot \|g_1\|^i\). The series converges since \(\|g_1\|<1\). In fact \(\max(\|q_1\|,\|r_1\|)=\|q_0g_1\|\le \|f\|\|g_1\|\), similarly \(\max(\|q_i\|,\|r_i\|)\le \|f\|\|g_1\|^i\). This tells us \[\|f\|=\max(\|q_0\|,\|r_0\|)\ge \max(\|q\|,\|r\|),\] and the other direction is derived from the equality \(f=gq+r\) by applying the ultrametric inequality.

  2. In the general case where \(\|g\|=|c_g|\neq 1\), we can reduce to the case by applying the proposition with \(g/c_g\) and write \[f = (g/c_g)(c_gq) + r\] where we have \(\|f\|=\max(\|c_gq\|,\|r\|)=\max(\|g\|\cdot \|q\|,\|r\|)\).

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Corollary 1. Every non-zero \(f\in\mathcal{O}(F)\) admits a factorization of the form \[f=u\cdot (z-a_1)\dots (z-a_d), \quad a_i\in F\] where \(u=c_f(1+s)\) is a ’unit’ in \(\mathcal{O}(F)\), \(c\in K^*\) and \(s\in \mathcal{O}(F)^{\circ\circ}\).

Proof. Apply the proposition to \((z^d, c_f^{-1}f)\) where \(c_f\in K^*\) is the maximal coefficient of \(f\), i.e. making \(d=\deg_{\overline{K}}\left(\overline{f/c_f}\right), \left\| f/c_f\right\|=1\) and \(f/c_f\) monic. We should get \[z^d = q\cdot f/c_f + r\] where taking reduction \(q\) reduces to \(1\), so \(q=1+s\). Clearly \(z^d-r\) have all roots in \(F\) (recall that \(K\) is algebraically closed). ◻

Proposition 2. Let \(\infty \in F\subset \mathbb{P}^1\) be a connected affinoid. Write \[F^c=\bigsqcup_{i=1}^n B(a_i, |\pi_i|),\] define \(\mathcal{O}(F)_+:=\{f\in \mathcal{O}(F): f(\infty)=0\}\) and \(F_i=B_i^c=\left\{\left|\frac{\pi}{z-a_i}\right|\le 1\right\}\). Then

  1. We have \[\mathcal{O}(F_i)=\left\{\sum_{n\ge 0} b_n \left(\frac{\pi_i}{z-a_i}\right)^n: b_n\to 0 \right\}\] and \(\left\|\sum b_n \left(\frac{\pi_i}{z-a_i}\right)^n\right\|_{F_i} = \max |b_n|\).

  2. \[\mathcal{O}(F)_+ = \bigoplus_{i=1}^n \mathcal{O}(F_i)_+\] with \(\|\sum f_i\|_F = \max_i \|f_i\|_F, \|f_i\|_F=\|f_i\|_{F_i}\).

Residue Theorem

Let \(F\) be a connected affinoid set, \(\omega\) a meromorphic differential on \(F\). We would like to define an interior \(F^\circ\) and boundary \(\partial F\) of \(F\), with some kind of ’integral’ of \(\omega\) along \(\partial F\) so that there is some kind of residue theorem.

Definition 2.

  1. A meromorphic function on an affinoid \(F\) is an element of the ring of total quotients of \(\mathcal{O}(F)\). If \(F\) is connected, this is just an element of the fraction field of \(\mathcal{O}(F)\), represented as \(\frac{f}{g}\) with \(f,g\in \mathcal{O}(F)\) having no common zeros.

  2. We can define \(\mathrm{ord}_a(f)\) for \(f\in \mathcal{O}(F), a\in F\) to be the maximal \(n\) such that \((z-a)^n|f\). For meromorphic functions, we define \(\mathrm{ord}_a\left(\frac{f}{g}\right)=\mathrm{ord}_a(f)-\mathrm{ord}_a(g)\).

  3. For a meromorphic differential \(\omega =f\cdot \,\mathrm{d}{z}\), we define \[\mathrm{ord}_a(\omega)=\mathrm{ord}_a(f), \quad \mathrm{ord}_{\infty}(\omega) = \mathrm{ord}_0\left(-\frac{f(t^{-1})}{t^2}\,\mathrm{d}{t}\right).\]

  4. Let \(t\) be a local parameter at \(a\), which \(\in \mathcal{O}(D(a,r))\) with \(\mathrm{ord}_a(t)=1\). Then locally \(\omega\) can be written as \[\omega = \sum_{n\in \mathbb{Z}} a_n t^n \,\mathrm{d}{t}.\] Its residue at \(a\) is defined as \[\mathrm{Res}_a(\omega) = a_{-1}.\]

  5. A ring domain is an affinoid set \(F\subset \mathbb{P}^1\) that is projective isomorphic to \(\{|z|=1\}\). A thick ring domain is isomorphic to \(\{r_1\le |z|\le r_2\}\).

For a ring domain \(F\) and a holomorphic differential \(\omega\) and an invertible holomorphic function \(f\in\mathcal{O}(F)\), we want to define what the residues \(\mathrm{Res}_F(\omega)\) and \(\mathrm{ord}_F(f)\) are, similar to the contour integral along circle and the winding number (degree) of a function along a circle.

Example 2. Take the classical \(F=\{|z|=1\}\), we know \(\mathcal{O}(F)=K\langle z,z^{-1}\rangle\) \(\omega = \sum a_n z^n \,\mathrm{d}{z}\) where \(a_n\xrightarrow{|n|\to \infty} 0\). We would like to define (not to be confused with the residue at a point) \[\mathrm{Res}_F(\omega) := a_{-1}.\] This \(a_{-1}\) in fact will ’sum up’ the residues at all points in \(F\) due to the peculiar non-archimedean geometry, imagine an expression like \[\frac{A}{z-a}+\frac{B}{z-b} = \frac{1}{z}\left(\frac{A}{1-\frac{a}{z}}+\frac{B}{1-\frac{b}{z}}\right)=\frac{A+B}{z} + \dots\] where \(|a|,|b|<1\) and the rest terms converges. We see that the \(z^{-1}\) coefficient actually collects the residues at \(a,b\).

Similarly one defines \(\mathrm{ord}_F f\) as the unique \(n\) such that \(f = cz^n(1+s)\)with \(c\in K^*, s\in \mathcal{O}(F)^{\circ\circ}\) i.e. \(\|s\|<1\) (it says that in this case \(\mathcal{O}(F)\) is a DVR).

Theorem 1 (Simple Residue Theorem). For \(F=\{|z|=1\}\), and \(\omega = f\,\mathrm{d}{z}\) a meromorphic differential holomorphic on \(F\), we have \[\sum_{|a|<1} \mathrm{Res}_a(\omega)= \mathrm{Res}_F(\omega),\] and for \(f\in \mathcal{O}(F)^\times\), we have \[\sum_{|a|<1} \mathrm{ord}_a(f) = \mathrm{ord}_F(f).\]

Theorem 2 (Residue Theorem). Let \(F=\left(\bigsqcup_i B_i\right)^c\subset \mathbb{P}\) be a connected affinoid. Let \(q\in F\) and suppose the boundaries \(\partial B_i\) with respect to \(q\) are disjoint and given suitable orientation, counting outside \(F\) and inside \(B_i\).

  1. Let \(\omega\) be a meromorphic differential on \(F\) which is holomorphic on \(\partial B_i\). Then \[\sum_i \mathrm{Res}_{\partial B_i}(\omega) + \sum_{a\in F^\circ} \mathrm{Res}_a(\omega) = 0.\]

  2. Let \(f\) be a meromorphic function on \(F\) such that \(f\) is invertible on all boundaries \(\partial B_i\) . Then \[\sum_i \mathrm{ord}_{\partial B_i}(f) + \sum_{a\in F^\circ} \mathrm{ord}_a(f) = 0.\]

Example 3. Let \(F\) be a thick ring domain with boundaries \(\partial B_1,\partial B_2\), \(f\in \mathcal{O}(F)\). Then the number of zeros of \(f\) in \(F\) is equal to the number of poles of \(f\) outside \(F\), \[N_F(f) = -\mathrm{ord}_{\partial B_1}(f) - \mathrm{ord}_{\partial B_2}(f).\] where you want the \(\partial B_i\) to be oriented to count the zeros and poles outside \(F\), inside \(B_i\). Let’s expand this formula in the simplest example of \[F = \{|\alpha|\le |z|\le |\beta|\}, \quad f = \sum_{n\in \mathbb{Z}} c_n z^n\] with \(|\alpha| < |\beta|\Leftrightarrow v(1/\alpha)<v(1/\beta)\). Then we should have, by adjusting to the correct choice of orientation, \[N_F(f) = -\mathrm{ord}_{\partial B(0, |\alpha|)}(f) + \mathrm{ord}_{\partial B(0,|\beta|)}(f).\] Then the ring of regular functions are \[\mathcal{O}(F)=\left\{\sum c_n z^n : |c_k|=o(|\beta|^{-k}),\, |c_{-k}|=o(|\alpha|^{k})\text{ as }k\to +\infty \right\}.\] or \[\mathcal{O}(F)=\left\{\sum c_n z^n : {v(c_k)-kv(1/\beta)\to \infty, v(c_{-k})-(-k)v(1/\alpha)\to \infty\, (k\to +\infty)} \right\}.\] You can actually see \(\mathrm{ord}_{B(|\beta|)}(f)-\mathrm{ord}_{B(|\alpha|)}(f)\) as the length of the part of the Newton polygon of \(\sum_{n\in \mathbb{Z}} a_n z^n\) that has slope in \((-v(1/\alpha),-v(1/\beta))\), and \(\mathcal{O}(F)\) are functions that converge above a V-shape formed by slopes \(-v(1/\alpha),-v(1/\beta)\).

Excercise 1. Let \(F=\{|z|=|z-1|=1\}\), develop residue theorem for this case and a formula for number of zeros \(N_F(f)\).