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scelta di punto base per il caso ii.b) ovvero quando $K$ è fo per $p$ punto base direzionato su $K$. -$ - L_K (a, z) colon.eq - 1 / 2 [ - sum_(q = p, overline(p)) - ((-1)^(abs(lambda(q))+1) L[hat(K)(lambda(q))] + z sum_K (lambda(q))) - ] -$ +// $ +// L_K (a, z) colon.eq +// 1 / 2 [ +// sum_(q = p, overline(p)) +// ((-1)^(abs(lambda(q))+1) L[hat(K)(lambda(q))] + z sum_K (lambda(q))) +// ] +// $ -#lemma[ - Consideriamo i due modi di fare uno splice di un nodo banale standard al primo incrocio subito dopo un punto base direzionato. Ci sono due casi: in uno otteniamo due nodi banali in forma standard, nell'altro un solo nodo banale (non necessariamente in forma standard). -] +// #lemma[ +// Consideriamo i due modi di fare uno splice di un nodo banale standard al primo incrocio subito dopo un punto base direzionato. Ci sono due casi: in uno otteniamo due nodi banali in forma standard, nell'altro un solo nodo banale (non necessariamente in forma standard). +// ] -#proof[ - #todo[work in progress] -] +// #proof[ +// #todo[work in progress] +// ] -#lemma[ - La proprietà del lemma @std-unknot-to-curls si estende al diagramma $e_i hat(K)$. -] +// #lemma[ +// La proprietà del lemma @std-unknot-to-curls si estende al diagramma $e_i hat(K)$. +// ] -#proof[ - #todo[work in progress] -] +// #proof[ +// #todo[work in progress] +// ] -Vediamo ora alcune identità per nodi banali standard che ci mostrano cosa succede quando scorriamo il punto base di un incrocio nella direzione del punto base. +// Vediamo ora alcune identità per nodi banali standard che ci mostrano cosa succede quando scorriamo il punto base di un incrocio nella direzione del punto base. -#lemma[ - Sia $p$ un punto di partenza direzionato e $hat(K) colon.eq hat(K)(cal(U), p)$ un nodo banale standard. Sia $i$ il primo incrocio in $K$ subito dopo il punto di partenza $p$. Sia $q$ un altro punto base direzionato posto nell'arco subito dopo l'incrocio $i$ con la sua stessa direzione e poniamo $hat(K)' colon.eq hat(K)(cal(U), q)$. Allora valgono le seguenti proprietà +// #lemma[ +// Sia $p$ un punto di partenza direzionato e $hat(K) colon.eq hat(K)(cal(U), p)$ un nodo banale standard. Sia $i$ il primo incrocio in $K$ subito dopo il punto di partenza $p$. Sia $q$ un altro punto base direzionato posto nell'arco subito dopo l'incrocio $i$ con la sua stessa direzione e poniamo $hat(K)' colon.eq hat(K)(cal(U), q)$. Allora valgono le seguenti proprietà - 1. $S_i hat(K) = hat(K)'$ +// 1. $S_i hat(K) = hat(K)'$ - 2. $L[hat(K)] + L[hat(K)'] = z (L[E_i hat(K)] + L[e_i hat(K)])$ -] +// 2. $L[hat(K)] + L[hat(K)'] = z (L[E_i hat(K)] + L[e_i hat(K)])$ +// ] -#proof[ - 1. Per definizione di nodo banale standard $hat(K)(cal(U), q) = S_i hat(K)(cal(U), p)$ poiché l'incrocio $i$ in $hat(K)(cal(U), p)$ compare come sopra-incrocio mentre in $hat(K)(cal(U), q)$ è l'ultimo incrocio visitato quindi è certamente la seconda occorrenza di $i$ quindi è un sotto-incrocio. Tutti gli altri incroci compaiono nello stesso ordine quindi restano invariati. +// #proof[ +// 1. Per definizione di nodo banale standard $hat(K)(cal(U), q) = S_i hat(K)(cal(U), p)$ poiché l'incrocio $i$ in $hat(K)(cal(U), p)$ compare come sopra-incrocio mentre in $hat(K)(cal(U), q)$ è l'ultimo incrocio visitato quindi è certamente la seconda occorrenza di $i$ quindi è un sotto-incrocio. Tutti gli altri incroci compaiono nello stesso ordine quindi restano invariati. - 2. Per questo secondo punto usiamo che $L_hat(K) = a^w(hat(K))$, assumiamo ora che tra $E_i hat(K)$ e $e_i hat(K)$ il primo sia il caso con due componenti composte da nodi banali standard ed il secondo quello con una sola componente di un nodo banale. +// 2. Per questo secondo punto usiamo che $L_hat(K) = a^w(hat(K))$, assumiamo ora che tra $E_i hat(K)$ e $e_i hat(K)$ il primo sia il caso con due componenti composte da nodi banali standard ed il secondo quello con una sola componente di un nodo banale. - Supponiamo che $w(e_i hat(K)) = w$, allora in quanto abbiamo solo rimosso un incrocio necessariamente +// Supponiamo che $w(e_i hat(K)) = w$, allora in quanto abbiamo solo rimosso un incrocio necessariamente - $ - {w(hat(K)), w(S_i hat(K))} = {w-1, w+1} - $ +// $ +// {w(hat(K)), w(S_i hat(K))} = {w-1, w+1} +// $ - ovvero i writhe di $hat(K)$ e $S_i hat(K)$ differiscono di $2$, assumiamo di essere nel caso $w(hat(K)) = w - 1$. +// ovvero i writhe di $hat(K)$ e $S_i hat(K)$ differiscono di $2$, assumiamo di essere nel caso $w(hat(K)) = w - 1$. - Analogamente $E_i hat(K)$ avrà due componenti dunque $K = K_1 hash K_2$, ognuna con writhe rispettivamente $w_1$ e $w_2$. Per il teorema della curva di Jordan, la somma dei segni degli incroci tra componenti diverse ha somma zero poiché una delle due _sovrasta_ l'altra dunque $w_1 + w_2 = w$. +// Analogamente $E_i hat(K)$ avrà due componenti dunque $K = K_1 hash K_2$, ognuna con writhe rispettivamente $w_1$ e $w_2$. Per il teorema della curva di Jordan, la somma dei segni degli incroci tra componenti diverse ha somma zero poiché una delle due _sovrasta_ l'altra dunque $w_1 + w_2 = w$. - $ - L[hat(K)] &= a^(w+1) \ - L[S_i hat(K)] &= a^(w-1) \ - L[e_i hat(K)] &= a^w \ - L[E_i hat(K)] &= d a^(w_1) a^(w_2) = d a^(w_1 + w_2) = d a^w = (a^(w+1) + a^(w-1)) slash z - a^w - $ +// $ +// L[hat(K)] &= a^(w+1) \ +// L[S_i hat(K)] &= a^(w-1) \ +// L[e_i hat(K)] &= a^w \ +// L[E_i hat(K)] &= d a^(w_1) a^(w_2) = d a^(w_1 + w_2) = d a^w = (a^(w+1) + a^(w-1)) slash z - a^w +// $ - che dunque verifica l'identità: +// che dunque verifica l'identità: - $ - a^(w+1) + a^(w-1) &= cancel(z) ((a^(w+1) + a^(w-1))/cancel(z) - cancel(a^w) + cancel(a^w)) \ - => L[hat(K)] + L[hat(K)'] &= z (L[E_i hat(K)] + L[e_i hat(K)]) \ - $ +// $ +// a^(w+1) + a^(w-1) &= cancel(z) ((a^(w+1) + a^(w-1))/cancel(z) - cancel(a^w) + cancel(a^w)) \ +// => L[hat(K)] + L[hat(K)'] &= z (L[E_i hat(K)] + L[e_i hat(K)]) \ +// $ - E questo conclude la dimostrazione delle due identità. -] +// E questo conclude la dimostrazione delle due identità. +// ] -Possiamo ora mostrare l'invarianza per scelta di punto base di $Omega_K (p)$. +// Possiamo ora mostrare l'invarianza per scelta di punto base di $Omega_K (p)$. -#lemma[ - Sia $K$ un diagramma di un nodo. Allora @kauffman-rec-single-component non dipende dalla scelta di punto base. Inoltre se $p$ è un punto base direzionato su $K$ e $lambda(p)$ la sequenza di scambi determinata da $p$ allora +// #lemma[ +// Sia $K$ un diagramma di un nodo. Allora @kauffman-rec-single-component non dipende dalla scelta di punto base. Inoltre se $p$ è un punto base direzionato su $K$ e $lambda(p)$ la sequenza di scambi determinata da $p$ allora - $ - Omega_K (p) = (-1)^(abs(lambda(p)) + 1) L_(hat(K)(p)) + z sum_K (lambda(p)) - $ +// $ +// Omega_K (p) = (-1)^(abs(lambda(p)) + 1) L_(hat(K)(p)) + z sum_K (lambda(p)) +// $ - non dipende dalla scelta di punto base. -] +// non dipende dalla scelta di punto base. +// ] -#proof[ - Possiamo assumere che la sequenza di scambi determinata da $p$ sia etichetta $lambda = (n, dots, 0)$, in questo modo +// #proof[ +// Possiamo assumere che la sequenza di scambi determinata da $p$ sia etichetta $lambda = (n, dots, 0)$, in questo modo - $ - Omega_K (p) = (-1)^(n+1) L_(hat(K)(p)) + z sum_K (lambda(p)) - $ +// $ +// Omega_K (p) = (-1)^(n+1) L_(hat(K)(p)) + z sum_K (lambda(p)) +// $ - Ricordiamo come era definito $L_K$ nel caso di una sola componente @kauffman-rec-single-component e notiamo quanto segue +// Ricordiamo come era definito $L_K$ nel caso di una sola componente @kauffman-rec-single-component e notiamo quanto segue - $ - L_K (a, z) &colon.eq 1 / 2 [ - sum_(q = p, overline(p)) - ((-1)^(abs(lambda(q))+1) kL(hat(K)(lambda(q))) + z sum_K (lambda(q))) - ] \ - &= 1/2 (Omega_K (p) + Omega_K (overline(p))) - $ +// $ +// L_K (a, z) &colon.eq 1 / 2 [ +// sum_(q = p, overline(p)) +// ((-1)^(abs(lambda(q))+1) kL(hat(K)(lambda(q))) + z sum_K (lambda(q))) +// ] \ +// &= 1/2 (Omega_K (p) + Omega_K (overline(p))) +// $ - Dunque ci basta mostrare che $Omega_K (p)$ non dipenda dalla scelta di punto base. +// Dunque ci basta mostrare che $Omega_K (p)$ non dipenda dalla scelta di punto base. - Come in precedenza possiamo mostrarlo facendo scorrere $p$ nell'arco successivo al primo incrocio dopo $p$ (questo ci induce l'invarianza per permutazioni cicliche e ci permette di concludere). +// Come in precedenza possiamo mostrarlo facendo scorrere $p$ nell'arco successivo al primo incrocio dopo $p$ (questo ci induce l'invarianza per permutazioni cicliche e ci permette di concludere). - Sia $i$ l'etichetta del primo incrocio dopo $p$. In $hat(K)(p)$ questo incrocio sarà sicuramente un sopra-incrocio in quanto $i$ è visitato per la prima volta essendo il primo incrocio dopo $p$. Invece in $K(p)$ può essere sia un sopra-incrocio che un sotto-incrocio, abbiamo quindi due casi: +// Sia $i$ l'etichetta del primo incrocio dopo $p$. In $hat(K)(p)$ questo incrocio sarà sicuramente un sopra-incrocio in quanto $i$ è visitato per la prima volta essendo il primo incrocio dopo $p$. Invece in $K(p)$ può essere sia un sopra-incrocio che un sotto-incrocio, abbiamo quindi due casi: - - Se $i$ è un sotto-incrocio in $K(p)$ allora $i$ sarà il primo incrocio nella sequenza di scambi ed avremo $i = n$ +// - Se $i$ è un sotto-incrocio in $K(p)$ allora $i$ sarà il primo incrocio nella sequenza di scambi ed avremo $i = n$ - #todo[disegnino] +// #todo[disegnino] - Consideriamo ora la situazione di $K(q)$ +// Consideriamo ora la situazione di $K(q)$ - #todo[disegnino] +// #todo[disegnino] - come già detto prima se l'ultimo incrocio è un sotto-incrocio allora è già in forma di nodo banale standard. Segue che $i$ non appartiene alla sequenza di scambi in questo caso ed avremo $(n-1, dots, 0)$ e $hat(K)(q) = S_(n-1) dotss S_0 K$. Vorremo vedere che $Omega_K (p) = Omega_K (q)$, ovvero: +// come già detto prima se l'ultimo incrocio è un sotto-incrocio allora è già in forma di nodo banale standard. Segue che $i$ non appartiene alla sequenza di scambi in questo caso ed avremo $(n-1, dots, 0)$ e $hat(K)(q) = S_(n-1) dotss S_0 K$. Vorremo vedere che $Omega_K (p) = Omega_K (q)$, ovvero: - $ - Omega_K (p) &= (-1)^(n+1) L[hat(K)(p)] + z sum_K (lambda(p)) \ - Omega_K (q) &= (-1)^((n-1)+1) L[hat(K)(q)] + z sum_K (lambda(q)) - $ +// $ +// Omega_K (p) &= (-1)^(n+1) L[hat(K)(p)] + z sum_K (lambda(p)) \ +// Omega_K (q) &= (-1)^((n-1)+1) L[hat(K)(q)] + z sum_K (lambda(q)) +// $ - come in precedenza studiamo la differenza: +// come in precedenza studiamo la differenza: - $ - Omega_K (p) - Omega_K (q) - =& (-1)^(n+1) L[hat(K)(p)] + z sum_K (lambda(p)) \ - &+ underbrace((-1)^(n+1), = -(-1)^n) L[hat(K)(q)] - z sum_K (lambda(q)) \ - =& (-1)^(n+1) (L[hat(K)(p)] + L[hat(K)(q)]) \ - &+ z (-1)^n (L[A_n^lambda K] + L[B_n^lambda K]) - $ +// $ +// Omega_K (p) - Omega_K (q) +// =& (-1)^(n+1) L[hat(K)(p)] + z sum_K (lambda(p)) \ +// &+ underbrace((-1)^(n+1), = -(-1)^n) L[hat(K)(q)] - z sum_K (lambda(q)) \ +// =& (-1)^(n+1) (L[hat(K)(p)] + L[hat(K)(q)]) \ +// &+ z (-1)^n (L[A_n^lambda K] + L[B_n^lambda K]) +// $ - dove abbiamo usato il fatto che tutti i termini in $sum_K (lambda(p)) - sum_K (lambda(q))$ si cancellano tra loro tranne l'ultimo di $sum_K (lambda(p))$. +// dove abbiamo usato il fatto che tutti i termini in $sum_K (lambda(p)) - sum_K (lambda(q))$ si cancellano tra loro tranne l'ultimo di $sum_K (lambda(p))$. - Ora notiamo che +// Ora notiamo che - $ - A_n^lambda K &= E_n S_(n-1) dotss S_0 K = E_n hat(K)(p) \ - B_n^lambda K &= e_n S_(n-1) dotss S_0 K = e_n hat(K)(p) \ - hat(K)(p) &= S_n space.med hat(K)(p) - $ +// $ +// A_n^lambda K &= E_n S_(n-1) dotss S_0 K = E_n hat(K)(p) \ +// B_n^lambda K &= e_n S_(n-1) dotss S_0 K = e_n hat(K)(p) \ +// hat(K)(p) &= S_n space.med hat(K)(p) +// $ - inoltre per il @lemma-slide-identities abbiamo anche che +// inoltre per il @lemma-slide-identities abbiamo anche che - $ - L[hat(K)(p)] + L[hat(K)(q)] = z (L[E_n hat(K)(p)] + L[e_n hat(K)(p)]) - $ +// $ +// L[hat(K)(p)] + L[hat(K)(q)] = z (L[E_n hat(K)(p)] + L[e_n hat(K)(p)]) +// $ - Per concludere questo caso basta sostituire in queste ultime identità +// Per concludere questo caso basta sostituire in queste ultime identità - $ - Omega_K (p) - Omega_K (q) =& (-1)^(n+1) (L[hat(K)(p)] + L[hat(K)(q)]) \ - &+ z (-1)^n (L[A_n^lambda K] + L[B_n^lambda K]) \ - = & cancel(z (-1)^(n+1) (L[E_n hat(K)(p)] + L[e_n hat(K)(p)])) \ - &+ cancel(z (-1)^n (L[E_n hat(K)(p)] + L[e_n hat(K)(p)])) \ - = & 0 - $ +// $ +// Omega_K (p) - Omega_K (q) =& (-1)^(n+1) (L[hat(K)(p)] + L[hat(K)(q)]) \ +// &+ z (-1)^n (L[A_n^lambda K] + L[B_n^lambda K]) \ +// = & cancel(z (-1)^(n+1) (L[E_n hat(K)(p)] + L[e_n hat(K)(p)])) \ +// &+ cancel(z (-1)^n (L[E_n hat(K)(p)] + L[e_n hat(K)(p)])) \ +// = & 0 +// $ - e quindi $Omega_K (p) = Omega_K (q)$. +// e quindi $Omega_K (p) = Omega_K (q)$. - - Se $i$ è un sopra-incrocio in $K(p)$ siamo nella seguente situazione +// - Se $i$ è un sopra-incrocio in $K(p)$ siamo nella seguente situazione - #todo[disegnino] +// #todo[disegnino] - In questo caso, $i$ non fa parte della sequenza di scambi per $K(q)$. Però dopo aver spostato il punto base avremo la seguente situazione +// In questo caso, $i$ non fa parte della sequenza di scambi per $K(q)$. Però dopo aver spostato il punto base avremo la seguente situazione - #todo[disegnino] +// #todo[disegnino] - Questa volta $i$ fa parte della sequenza di scambi per $K(q)$ e quando compare come ultimo incrocio è un sotto-incrocio. Avremo quindi le seguenti sequenze di scambi +// Questa volta $i$ fa parte della sequenza di scambi per $K(q)$ e quando compare come ultimo incrocio è un sotto-incrocio. Avremo quindi le seguenti sequenze di scambi - $ - lambda(q) &= (n, n-1, dots, i, dots, 1, 0) \ - lambda(p) &= (n, n-1, dots, i+1, i-1, dots, 1, 0) - $ +// $ +// lambda(q) &= (n, n-1, dots, i, dots, 1, 0) \ +// lambda(p) &= (n, n-1, dots, i+1, i-1, dots, 1, 0) +// $ - a meno di riordinare gli scambi è facile vedere che $S_i hat(K)(q) = hat(K)(p)$, e che il @lemma-slide-identities si applica alla coppia $hat(K)(p), hat(K)(q)$ per l'indice $i$. Per il @lemma-sum-switches-rotation abbiamo anche che $sum_K (p), sum_K (q)$ sono invarianti a meno di permutazioni cicliche. Possiamo quindi sostituire $lambda(p)$ e $lambda(q)$ con le seguenti sequenze di scambi applicando le giuste rotazioni. +// a meno di riordinare gli scambi è facile vedere che $S_i hat(K)(q) = hat(K)(p)$, e che il @lemma-slide-identities si applica alla coppia $hat(K)(p), hat(K)(q)$ per l'indice $i$. Per il @lemma-sum-switches-rotation abbiamo anche che $sum_K (p), sum_K (q)$ sono invarianti a meno di permutazioni cicliche. Possiamo quindi sostituire $lambda(p)$ e $lambda(q)$ con le seguenti sequenze di scambi applicando le giuste rotazioni. - $ - lambda'(p) &= (i-1, dots, 1, 0, n, n-1, dots, i+1) \ - lambda'(q) &= (i, i-1, dots, 1, 0, n, n-1, dots, i+1) - $ +// $ +// lambda'(p) &= (i-1, dots, 1, 0, n, n-1, dots, i+1) \ +// lambda'(q) &= (i, i-1, dots, 1, 0, n, n-1, dots, i+1) +// $ - A questo punto possiamo applicare un argomento simile a quello del punto precedente. +// A questo punto possiamo applicare un argomento simile a quello del punto precedente. - E questo completa la dimostrazione dell'indipendenza di $Omega_K (p)$ dalla scelta di punto base. -] +// E questo completa la dimostrazione dell'indipendenza di $Omega_K (p)$ dalla scelta di punto base. +// ] -A questo punto possiamo vedere che $L_K$ verifica gli assiomi della @kauffman-poly-def. +// A questo punto possiamo vedere che $L_K$ verifica gli assiomi della @kauffman-poly-def. -#lemma[ - Sia $i$ un incrocio di un diagramma di link $K$. Allora $L_K$ verifica le identità: +// #lemma[ +// Sia $i$ un incrocio di un diagramma di link $K$. Allora $L_K$ verifica le identità: - 1. $L[K] + L[S_i K] = z (L[E_i K] + L[e_i K])$ +// 1. $L[K] + L[S_i K] = z (L[E_i K] + L[e_i K])$ - 2. $L[#skein.over-twist-medium] = a L [#skein.strand-medium]$, $L[#skein.under-twist-medium] = a^(-1) L [#skein.strand-medium]$ -] +// 2. $L[#skein.over-twist-medium] = a L [#skein.strand-medium]$, $L[#skein.under-twist-medium] = a^(-1) L [#skein.strand-medium]$ +// ] -#proof[ - Mostriamo per induzione sul numero di incroci nel diagramma. +// #proof[ +// Mostriamo per induzione sul numero di incroci nel diagramma. - 1. Consideriamo i seguenti casi: #margin-note[ricontrollare tutta questa] +// 1. Consideriamo i seguenti casi: #margin-note[ricontrollare tutta questa] - - Se $K$ ha una sola componente ed un incrocio $i$: consideriamo $S_i K$, $E_i K$ e $e_i K$, scegliendo bene $p$ possiamo fare in modo che $i$ sia il primo incrocio nella sequenza di scambi. Se ora consideriamo @kauffman-rec-single-component ovvero +// - Se $K$ ha una sola componente ed un incrocio $i$: consideriamo $S_i K$, $E_i K$ e $e_i K$, scegliendo bene $p$ possiamo fare in modo che $i$ sia il primo incrocio nella sequenza di scambi. Se ora consideriamo @kauffman-rec-single-component ovvero - $ - L_K (a, z) &colon.eq - 1 / 2 [ - sum_(q = p, overline(p)) - ((-1)^(abs(lambda(q))+1) L[hat(K)(lambda(q))] + z sum_K (lambda(q))) - ] \ - &= 1 / 2 [ - sum_(q = p, overline(p)) - ((-1)^(n+1) L[hat(K)(q)] + z sum_K (lambda(q))) - ] - $ +// $ +// L_K (a, z) &colon.eq +// 1 / 2 [ +// sum_(q = p, overline(p)) +// ((-1)^(abs(lambda(q))+1) L[hat(K)(lambda(q))] + z sum_K (lambda(q))) +// ] \ +// &= 1 / 2 [ +// sum_(q = p, overline(p)) +// ((-1)^(n+1) L[hat(K)(q)] + z sum_K (lambda(q))) +// ] +// $ - A questo punto otteniamo la tesi considerando la differenza nelle espansioni di $Omega_K (p)$ e $Omega_(S_i K) (p)$, in particolare tutti i termini si cancellano tranne il primo: +// A questo punto otteniamo la tesi considerando la differenza nelle espansioni di $Omega_K (p)$ e $Omega_(S_i K) (p)$, in particolare tutti i termini si cancellano tranne il primo: - $ - => Omega_K (p) + Omega_(S_i K) (p) = z(L[E_i K] + L[e_i K]) \ - $ +// $ +// => Omega_K (p) + Omega_(S_i K) (p) = z(L[E_i K] + L[e_i K]) \ +// $ - - Se $K$ ha più di una componente e $i$ è un incrocio di una componente con se stessa: in questo caso utilizziamo per induzione @kauffman-rec-multi-component ma osserviamo che $i$ non compare in nessuna sequenza di sollevamento (in quanto le sequenze di sollevamento scambiano solo incroci tra componenti diverse) dunque possiamo applicare l'induzione senza problemi. +// - Se $K$ ha più di una componente e $i$ è un incrocio di una componente con se stessa: in questo caso utilizziamo per induzione @kauffman-rec-multi-component ma osserviamo che $i$ non compare in nessuna sequenza di sollevamento (in quanto le sequenze di sollevamento scambiano solo incroci tra componenti diverse) dunque possiamo applicare l'induzione senza problemi. - - Se $K$ ha più di una componente e $i$ è un incrocio tra due componenti diverse: possiamo considerare la @kauffman-rec-multi-component e mostrare che ogni addendo verifica l'identità e quindi l'identità seguirà per la loro media: +// - Se $K$ ha più di una componente e $i$ è un incrocio tra due componenti diverse: possiamo considerare la @kauffman-rec-multi-component e mostrare che ogni addendo verifica l'identità e quindi l'identità seguirà per la loro media: - #{ - set text(size: small-size) - $ - L_K (a, z) colon.eq - 1 / (2n) [ - sum_(i=1)^n sum_(q=p_i, overline(p)_i) - ((-1)^(abs(lambda(q))+1) d kL_(K_i) kL_(K - K_i) + z sum_K (lambda(q))) - ] - $ - } +// #{ +// set text(size: small-size) +// $ +// L_K (a, z) colon.eq +// 1 / (2n) [ +// sum_(i=1)^n sum_(q=p_i, overline(p)_i) +// ((-1)^(abs(lambda(q))+1) d kL_(K_i) kL_(K - K_i) + z sum_K (lambda(q))) +// ] +// $ +// } - Per quanto riguarda gli addenti che non intersecano nessuna delle due componenti interessate dall'incrocio scelto non ci sono problemi e possiamo applicare l'induzione. Per le componenti che le intersecano ci basta scegliere come prima un punto base in modo appropriato in modo che l'identità sia verificata e procedere come prima. +// Per quanto riguarda gli addenti che non intersecano nessuna delle due componenti interessate dall'incrocio scelto non ci sono problemi e possiamo applicare l'induzione. Per le componenti che le intersecano ci basta scegliere come prima un punto base in modo appropriato in modo che l'identità sia verificata e procedere come prima. - // gli addendi che non intersecano nessuna delle due componenti possiamo procedere per induzione. Altrimenti possiamo scegliere un punto base appropriato e procedere per induzione. +// // gli addendi che non intersecano nessuna delle due componenti possiamo procedere per induzione. Altrimenti possiamo scegliere un punto base appropriato e procedere per induzione. - 2. La dimostrazione è analoga, basta osservare che possiamo sempre scegliere un punto base che non faccia comparire l'incrocio del ricciolo nella sequenza di scambi. +// 2. La dimostrazione è analoga, basta osservare che possiamo sempre scegliere un punto base che non faccia comparire l'incrocio del ricciolo nella sequenza di scambi. - E questo conclude la dimostrazione del lemma. -] +// E questo conclude la dimostrazione del lemma. +// ] -Come ultimo risultato vediamo che $L_K$ è un invariante di isotopia regolare. +// Come ultimo risultato vediamo che $L_K$ è un invariante di isotopia regolare. -#lemma[ - Siano $K, K'$ diagrammi di link equivalenti a meno di isotopia regolare allora $L_K (a, z) = L_(K') (a, z)$. -] +// #lemma[ +// Siano $K, K'$ diagrammi di link equivalenti a meno di isotopia regolare allora $L_K (a, z) = L_(K') (a, z)$. +// ] -#proof[ - Procediamo sempre per induzione sul numero di incroci e mostriamo l'invarianza per mosse che non aumentano il numero di incroci (l'invarianza per la mossa II che aggiunge due incroci segue come conseguenza dell'induzione#margin-note[da aggiustare]). +// #proof[ +// Procediamo sempre per induzione sul numero di incroci e mostriamo l'invarianza per mosse che non aumentano il numero di incroci (l'invarianza per la mossa II che aggiunge due incroci segue come conseguenza dell'induzione#margin-note[da aggiustare]). + +// - Mossa II: Ci sono più casi in base a se la mossa riguarda una o più componenti - - Mossa II: Ci sono più casi in base a se la mossa riguarda una o più componenti +// - Se la mossa è su una sola componente allora ci basta scegliere il punto base come segue - - Se la mossa è su una sola componente allora ci basta scegliere il punto base come segue +// #todo[disegnino] - #todo[disegnino] +// in questo modo i due incroci non compariranno nella sequenza di scambi. Tutti i termini di ii.a) saranno invarianti per mosse di tipo II e lo sarà anche $L_K$. - in questo modo i due incroci non compariranno nella sequenza di scambi. Tutti i termini di ii.a) saranno invarianti per mosse di tipo II e lo sarà anche $L_K$. +// - Se la mossa riguarda più componenti allora il caso peggiore è il seguente - - Se la mossa riguarda più componenti allora il caso peggiore è il seguente +// #todo[disegnino] - #todo[disegnino] +// Vediamo che $L[K] = L[S_2 S_1 K]$... - Vediamo che $L[K] = L[S_2 S_1 K]$... +// - Mossa III: - - Mossa III: +// - Se la mossa è su fili di una sola componente allora possiamo scegliere il punto base sul filo che passa sopra, in questo modo avremo la situazione seguente: - - Se la mossa è su fili di una sola componente allora possiamo scegliere il punto base sul filo che passa sopra, in questo modo avremo la situazione seguente: +// #todo[disegnino] - #todo[disegnino] +// In un caso possiamo procedere per induzione, se invece ci sono degli splice possiamo utilizzare le seguenti equivalenze di diagrammi e poi procedere per induzione - In un caso possiamo procedere per induzione, se invece ci sono degli splice possiamo utilizzare le seguenti equivalenze di diagrammi e poi procedere per induzione +// #todo[disegnino] - #todo[disegnino] +// - Nel caso in cui la mossa riguardi fili appartenenti a più componenti possiamo utilizzare l'identità della mossa precedente $L[K] = L[S_2 S_1 K]$ per semplificare la situazione e procedere nuovamente per induzione. - - Nel caso in cui la mossa riguardi fili appartenenti a più componenti possiamo utilizzare l'identità della mossa precedente $L[K] = L[S_2 S_1 K]$ per semplificare la situazione e procedere nuovamente per induzione. +// Questo completa la dimostrazione. +// ] - Questo completa la dimostrazione. -] \ No newline at end of file +#todo[work in progress] \ No newline at end of file