# Vigre 2009: algebraic geometry

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Lemma 3.1 [link] Formula 1.1.2 Given a continued fraction $\left[{r}_{1},{r}_{2},{r}_{3},...\right],{h}_{v}$ and k v satisfy the following recursions:

${h}_{v}={r}_{v}{h}_{v-1}+{h}_{v-2},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{h}_{-1}=1,{h}_{-2}=0$
${k}_{v}={r}_{v}{k}_{v-1}+{k}_{v-2},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{k}_{-1}=0,{k}_{-2}=1.$

Lemma 3.2 [link] Formula 1.2.1 Given a continued fraction as above, A v satisfies the following relations:

$\begin{array}{cc}\hfill {A}_{\delta }& \ge {A}_{v}\phantom{\rule{1.em}{0ex}}v\phantom{\rule{0.166667em}{0ex}}odd\hfill \\ \hfill {A}_{\delta }& \le {A}_{v}\phantom{\rule{1.em}{0ex}}v\phantom{\rule{0.166667em}{0ex}}even.\hfill \end{array}$

Moreover, this means that ${A}_{v}\le {A}_{v+1}$ for odd v.

The following is a claim about the n 's from the Euclidean Algorithm.

Lemma 3.3 In the above notation, ${n}_{v}={\left(-1\right)}^{v}\left(p{h}_{v}-{k}_{v}q\right)$ .

This will be done by induction with two base cases. If $v=1$ , then

${n}_{1}=q-p{r}_{1}=-\left(p{r}_{1}-q\right)=-\left(p{h}_{1}-{k}_{1}q\right).$

If $v=2$ , then

${n}_{2}=p-{r}_{2}{n}_{1}=p-{r}_{2}\left(q-{r}_{1}p\right)=p\left({r}_{1}{r}_{2}+1\right)-{r}_{2}q={\left(-1\right)}^{2}\left(p{h}_{2}-{k}_{2}q\right).$

Now we will assume the claim is true for v and $v+1$ and we will prove that it is true for $v+2$ . From the Euclidean algorithm, we have that

${n}_{v+2}={n}_{v}-{r}_{v+2}{n}_{v+1}$
$={\left(-1\right)}^{v}\left(p{h}_{v}-{k}_{v}q\right)-{\left(-1\right)}^{v+1}{r}_{v+2}\left(p{h}_{v+1}-{k}_{v+1}q\right).$

For v even, using Lemma [link] we obtain

${n}_{v+2}=p\left({h}_{v}+{r}_{v+2}{h}_{v+1}\right)-\left({k}_{v}+{r}_{v+2}{k}_{v+1}\right)q=p{h}_{v+2}-{k}_{v+2}q.$

If v is odd, using Lemma [link] we have

${n}_{v+2}=-p\left({h}_{v}+{r}_{v+2}{h}_{v+1}\right)+\left({k}_{v}+{r}_{v+2}{k}_{v+1}\right)q=-\left(p{h}_{v+2}-{k}_{v+2}q\right).$

Since v and $v+2$ will be either both even or both odd, this proves the claim. The following lemmas relate recursions that occur in the proof of Theorem [link] to continued fraction expressions.

Lemma 3.4 Suppose $\left\{{C}_{i}\right\}$ is a sequence of integers satisfying the following recurrence relation

${C}_{i}=\left({C}_{i-1}+{n}_{i-1}\right){r}_{i}+{C}_{i-2},\phantom{\rule{1.em}{0ex}}{C}_{1}={r}_{1}p,\phantom{\rule{0.166667em}{0ex}}{C}_{2}={r}_{2}q.$

Then

${C}_{v}=\left\{\begin{array}{cc}p{h}_{v}\hfill & v\phantom{\rule{0.166667em}{0ex}}odd\hfill \\ q{k}_{v}\hfill & v\phantom{\rule{0.166667em}{0ex}}even.\hfill \end{array}\right)$

This is done by induction again. For $v=1$ and $v=2$ , the initial conditions of the recursion of C give ${C}_{1}=p{h}_{1}$ and ${C}_{2}=q{k}_{2}$ . Now we assume that the claim is true for v and $v+1$ and we will prove that it is true for $v+2$ . First, we have

${C}_{v+2}=\left({C}_{v+1}+{n}_{v+1}\right){r}_{v+2}-{C}_{v}=\left[{C}_{v+1}+{\left(-1\right)}^{v+1}\left(p{h}_{v+1}-{k}_{v+1}q\right)\right]{r}_{v+2}-{C}_{v}$

by the original recursion of C i and Lemma [link] . There are two cases. First we will consider the case where v is odd. Then

${C}_{v+2}=\left(q{k}_{v+1}+p{h}_{v+1}-{k}_{v+1}q\right){r}_{v+2}+p{h}_{v}$

by the inductive hypothesis. It follows from Lemma [link] that

${C}_{v+2}=p\left({h}_{v+1}{r}_{v+2}+{h}_{v}\right)=p{h}_{v+2}.$

Now consider the case where v is even. Then

${C}_{v+2}=\left(p{h}_{v+1}-p{h}_{v+1}+{k}_{v+1}q\right){r}_{v+2}+q{k}_{v}$

by the inductive hypothesis. It follows from Lemma [link] that

${C}_{v+2}=q\left({k}_{v+1}{r}_{v+2}+{k}_{v}\right)=q{k}_{v+2}.$

This proves the claim.

Lemma 3.5 Suppose that $\left\{{K}_{i}\right\}$ is a sequence of integers satisfying the following recurrence relation

${K}_{i}=\left({K}_{i-1}+1\right){r}_{i}+{K}_{i-2},\phantom{\rule{1.em}{0ex}}{K}_{1}={r}_{1},{K}_{2}={r}_{2}\left({r}_{1}+1\right).$

Then ${K}_{v}+1={h}_{v}+{k}_{v}$ .

We again use induction with two base cases. First, if $v=1$ , then

${K}_{1}+1={r}_{1}+1={h}_{1}+{k}_{1}.$

Second, if $v=2$ , then

${K}_{2}+1={r}_{2}\left({r}_{1}+1\right)+1={r}_{2}{r}_{1}+{r}_{2}+1={h}_{2}+{k}_{2}.$

Now we will assume that the claim is true for v and $v+1$ and we will prove that it is true for $v+2$ . By the original recursion of K i ,

${K}_{v+2}+1=\left({K}_{v+1}+1\right){r}_{v+2}+{K}_{v}+1.$

Then by the inductive hypothesis,

$\begin{array}{cc}\hfill {K}_{v+2}+1& =\left({h}_{v+1}+{k}_{v+1}\right){r}_{v+2}+{h}_{v}+{k}_{v}\hfill \\ & =\left({h}_{v+1}{r}_{v+2}+{h}_{v}\right)+\left({k}_{v+1}{r}_{v+2}+{k}_{v}\right)\hfill \\ & ={h}_{v+2}+{k}_{v+2}\hfill \end{array}$

by the recursion of k v and h v from Lemma [link] . This proves the claim.

## Proof of theorem

Without loss of generality, assume that $q\ge p$ . For this proof, we will use a “reduced” notion of blow-ups in which consecutive blow-ups that use the same variable will be reduced to a single blow-up. For example, if we have three typical blow-up steps with $y=sx$ , $s=tx$ , and $t=ax$ , we will combine these to be $y=a{x}^{3}$ . Note that the multiplicities of these blow-ups are the same. We will soon justify the use of these blow up “chains”.

We will use the equations from the Euclidean algorithm as in the previous section. The number of r i 's used will be the number of blow-ups needed, and the value of r i will be the degree of the variable at that particular blow-up (e.g. ${r}_{i}=3$ in the previous example). Note that the r i values can also be expressed as the continued fraction of $\frac{q}{p}$ as $\left[{r}_{1},{r}_{2},...,{r}_{\delta }\right]$ . Let C i be the multiplicity of the ${i}^{th}$ exceptional divisor and let K i be the log canonical multiplicity of the differential form of the ${i}^{th}$ blow-up. We claim that the multiplicities K i and C i satisfy the following recursions:

${K}_{i}=\left({K}_{i-1}+1\right){r}_{i}+{K}_{i-2},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{K}_{1}={r}_{1},{K}_{2}={r}_{2}\left({r}_{1}+1\right),$
${C}_{i}=\left({C}_{i-1}+{n}_{i-1}\right){r}_{i}+{C}_{i-2},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{C}_{1}={r}_{1}p,{C}_{2}={r}_{2}q.$

First, it is easy to check that ${K}_{1}={r}_{1}$ and ${K}_{2}={r}_{2}\left({r}_{1}+1\right)$ . Then, after $i-1$ blow-ups, our differential form ihas the form

${{t}_{i-2}}^{{K}_{i-2}}{{t}_{i-1}}^{{K}_{i-1}}d{t}_{i-1}d{t}_{i-2}.$

By definition of r i , the next iterated blow up includes the substitution ${t}_{i-1}={t}_{i}{{t}_{i-2}}^{{r}_{i}}$ . We obtain the differential form

${{t}_{i-2}}^{{K}_{i-2}}{\left({t}_{i}{{t}_{i-2}}^{{r}_{i}}\right)}^{{K}_{i-1}}{{t}_{i-2}}^{{r}_{i}}d{t}_{i-2}d{t}_{i}={{t}_{i}}^{{K}_{i-1}}{{t}_{i-2}}^{\left({K}_{i-1}+1\right){r}_{i}+{K}_{i-2}}d{t}_{i-2}d{t}_{i}.$

Thus, ${K}_{i}=\left({K}_{i-1}+1\right){r}_{i}+{K}_{i-2}$ .

It is also easy to check that ${C}_{1}=p{r}_{1}$ and ${C}_{2}=q{r}_{2}$ . After $i-1$ blow-ups the equation is

${{t}_{i-2}}^{{C}_{i-2}}{{t}_{i-1}}^{{C}_{i-1}}\left({{t}_{i-2}}^{{n}_{i-2}}-{{t}_{i-1}}^{{n}_{i-1}}\right).$

By definition of r i , the next iterated blow up includes the substitution of ${t}_{i-1}={t}_{i}{{t}_{i-2}}^{{r}_{i}}$ . We obtain the new equation

${{t}_{i}}^{{C}_{i-1}}{{t}_{i-2}}^{{C}_{i-2}+{r}_{i}{C}_{i-1}}\left({{t}_{i-2}}^{{n}_{i-2}}-{{t}_{i}}^{{n}_{i-1}}{{t}_{i-2}}^{{r}_{i}{n}_{i-1}}\right)$
$={{t}_{i}}^{{C}_{i-1}}{{t}_{i-2}}^{\left({C}_{i-1}+{n}_{i-1}\right){r}_{i}+{C}_{i-2}}\left({{t}_{i-2}}^{{n}_{i}}-{{t}_{i}}^{{n}_{i-1}}\right).$

Thus, ${C}_{i}=\left({C}_{i-1}+{n}_{i-1}\right){r}_{i}+{C}_{i-2}$ .

The recursion relations for C i and K i allow us to apply Lemma [link] to C i and Lemma [link] to K i . Now we can justify the notion of blow-ups that groups together consecutive substitutions of the same variable into “chains.” It is not entirely clear that substitutions within a given chain do not yield smaller contributions to the LCT. To show that the smallest contribution to the LCT from a particular chain comes from its final substitution, we need to show that each successive substitution within such a chain decreases the contribution of the chain to the LCT. Let j be the number of blow-ups within the ${i}^{th}$ chain. To prove the above statement, it would suffice to show that

$\frac{\left({K}_{i-1}+1\right)j+{K}_{i-2}+1}{\left({C}_{i-1}+{n}_{i-1}\right)j+{C}_{i-2}}\ge \frac{\left({K}_{i-1}+1\right)\left(j+1\right)+{K}_{i-2}+1}{\left({C}_{i-1}+{n}_{i-1}\right)\left(j+1\right)+{C}_{i-2}},$

which is true if and only if

$\left({C}_{i-1}+{n}_{i-1}\right)\left({K}_{i-2}+1\right)\ge {C}_{i-2}\left({K}_{i-1}+1\right).$

We will show that this is true using two cases. First, assume that i is even. Then the above equation becomes

$\left(q{k}_{i-1}\right)\left({h}_{i-2}+{k}_{i-2}\right)\ge \left(q{k}_{i-2}\right)\left({h}_{i-1}+{k}_{i-1}\right)$

$⇕$

${A}_{i-2}+1\ge {A}_{i-1}+1,$

which is true for even i by Lemma [link] . Now we will assume i is odd. Then the previous expression becomes

$\left(p{h}_{i-1}\right)\left({h}_{i-2}+{k}_{i-2}\right)\ge \left(p{h}_{i-2}\right)\left({h}_{i-1}+{k}_{i-1}\right)$

$⇕$

$\frac{1}{{A}_{i-2}}+1\ge \frac{1}{{A}_{i-1}}+1$

$⇕$

${A}_{i-2}\le {A}_{i-1},$

which is true for odd i by Lemma [link] . This shows that the LCT will be determined by these chains.

We will now show that the final chain determines the LCT. Thus, we want to show that ${\alpha }_{\delta }\le {\alpha }_{v}$ for all v . First, we will introduce the notation for ${A}_{v}=\frac{{h}_{v}}{{k}_{v}}$ to be the expression of the ${v}^{th}$ convergent. We must consider two cases. First, if v is odd, then

${\alpha }_{\delta }\le {\alpha }_{v}$

$⇕$

$\frac{{h}_{\delta }+{k}_{\delta }}{p{h}_{\delta }}\le \frac{{h}_{v}+{k}_{v}}{p{h}_{v}}$

$⇕$

$1+\frac{1}{{A}_{\delta }}\le 1+\frac{1}{{A}_{v}}$

$⇕$

$\frac{1}{{A}_{\delta }}\le \frac{1}{{A}_{v}}$

$⇕$

${A}_{\delta }\ge {A}_{v}.$

This last statement is true by Lemma [link] , so for v odd, ${\alpha }_{\delta }\le {\alpha }_{v}$ . If v is even, then

${\alpha }_{\delta }\le {\alpha }_{v}$

$⇕$

$\frac{{h}_{\delta }+{k}_{\delta }}{q{k}_{\delta }}\le \frac{{h}_{v}+{k}_{v}}{q{k}_{v}}$

$⇕$

$1+{A}_{\delta }\le 1+{A}_{v}$

$⇕$

${A}_{\delta }\le {A}_{v}$

$⇕$

${A}_{\delta }\le {A}_{v}.$

This last statement is also true by Lemma [link] , so for v even, ${\alpha }_{\delta }\le {\alpha }_{v}$ . Thus, ${\alpha }_{\delta }\le {\alpha }_{v}$ for all $v<\delta$ .

It only remains to show that ${\alpha }_{\delta }=\frac{1}{p}+\frac{1}{q}$ . We know that $\frac{q}{p}=\frac{{h}_{\delta }}{{k}_{\delta }}$ , which is to say $q{k}_{\delta }=p{h}_{\delta }$ , so by Lemma [link] ${C}_{\delta }=q{k}_{\delta }$ for both odd and even δ . Thus,

${\alpha }_{\delta }=\frac{{K}_{\delta }+1}{{C}_{\delta }}=\frac{{h}_{\delta }+{k}_{\delta }}{q{k}_{\delta }}=\frac{\frac{{h}_{\delta }}{{k}_{\delta }}+1}{q}=\frac{\frac{q}{p}+1}{q}=\frac{1}{p}+\frac{1}{q}.$

This concludes the proof of the theorem.

#### Questions & Answers

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