abstract-algebra--nf--2 corrections

abstract-algebra--nf--1.tex

@@ -30,8 +30,8 @@ \section*{Problem  Set 1}
   $0$ is an identity under $\star$ since $x \star 0 = r(x + 0) = r(x) = x$ for all $x \in [0, 1)$.
 
   {\bf Existence of inverses:}\\
-  $0^\1 = 0$ since $0 \star 0 = r(0 + 0) = 0$.
-  For $x \in (0, 1)$ we have $x^{-1} = 1 - x$, since $x \star (1 - x) = r(x + 1 - x) = r(1) = 0$.
+  $0^\1 = 0$ since it's the identity.
+  For $x \in G - \{0\}$ we have $x^{-1} = 1 - x$, since $x \star (1 - x) = r(x + 1 - x) = r(1) = 0$.
 
   {\bf Associativity:}\\
   Let $x, y, z \in [0, 1)$. We have

abstract-algebra--nf--2.tex

@@ -186,6 +186,10 @@ \section*{Problem  Set 2}
   with $\sigma_i^M$.
 \end{proof}
 
+you can pick any element that is not taken to itself by $\sigma_i^M$ (which must exist because
+$\sigma_i^M$ isn't the identity) and then argue that since the $\sigma_j^M$'s are disjoint, this element is
+also not taken to itself by their product.
+
 
 
 ~\\~\\
@@ -243,35 +247,127 @@ \section*{Problem  Set 2}
   and $\sigma$ was arbitrary, $S_n$ is generated by its transpositions.
 \end{proof}
 
+Alternative proof:
+
+\begin{proof}
+  Note that any cycle can be written as a product of transpositions:
+  \begin{align*}
+     (a_1, a_2, \ldots, a_k) = (a_1, a_k)(a_1, a_{k-1}) \cdots (a_1, a_2).
+  \end{align*}
+  Therefore any $\sigma \in S_n$ can be written as a product of transpositions by first forming its cycle
+  decomposition and then replacing each cycle in the decomposition with a product of transpositions
+  formed as above.
+\end{proof}
+
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/algebra--nf--2-d8ce.png}
 \end{mdframed}
 
 
 The identity is even since it has zero inversions.
 
-A transposition is odd since it has one inversion.
+A transposition is odd. To see this, consider the transposition $(i, j)$. Clearly it has
+the $\{i, j\}$ inversion. And we also get $2(|i - j| - 1)$ additional inversions since, for every
+inversion induced by $i$ "crossing over" an intervening position, there is a matching inversion
+induced by $j$ "crossing over" in the opposite direction. So the total number is odd.
+
+For example, the transposition $(1, 4)$ has $5$ inversions:
+\begin{mdframed}
+\includegraphics[width=400pt]{img/abstract-algebra--nf--2-e92f.png}
+\end{mdframed}
+
+
 
 
 \begin{mdframed}
 \includegraphics[width=400pt]{img/algebra--nf--2-179c.png}
 \end{mdframed}
 
+\begin{mdframed}
+\includegraphics[width=400pt]{img/abstract-algebra--nf--2-4dc0.png}
+\end{mdframed}
+
+
 \begin{proof}
-  Let $\sigma \in S_n$ be any permutation, let $\tau \in S_n$ be a transposition, and let $I_\rho$ be the set of
-  inversions of any permutation $\rho \in S_n$.
+  Let $\sigma \in S_n$ be any permutation, let $\tau \in S_n$ be the transposition of $i$ and $j$, and for
+  any $\rho \in S_n$ define
+  \begin{align*}
+    \delta^{(\rho)}_{a,b} =
+    \begin{cases}
+      1 &~~~\text{if}~\{a, b\}~\text{is an inversion of}~\rho\\
+      0 &~~~\text{otherwise},
+    \end{cases}
+  \end{align*}
+  and
+  \begin{align*}
+    I^-(\rho) &=  \sum_{\substack{l > k\\k, l \in \{1, 2, \ldots, n\} - \{i, j\}}}\delta^{(\rho)}_{kl}\\
+    I^+(\rho) &=  \sum_{\substack{l > k\\k \in \{i, j\} ~\text{or}~ l \in \{i, j\}}}\delta^{(\rho)}_{kl},
+  \end{align*}
+  where indices $k$ and $l$ range over $1, 2, \ldots, n$. Thus $I(\rho) = I^-(\rho) + I^+(\rho)$ is the number of
+  inversions of $\rho$. Assume WLOG that $j > i$ and note that the summation in the definition
+  of $I^+(\rho)$ ranges over an odd number $2(j - i -1) + 1$ of elements.
+
+  Next, note that every inversion of $\tau$ involves either $i$ or $j$. Therefore
+  \begin{align*}
+    \delta^{(\tau\sigma)}_{kl} =
+    \begin{cases}
+      \delta^{(\sigma)}_{kl} &~~~k, l \in \{1, 2, \ldots, n\} - \{i, j\}\\
+      1 - \delta^{(\sigma)}_{kl} &~~~\text{either}~k \in \{i, j\} ~\text{or}~ l \in \{i, j\},
+    \end{cases}
+  \end{align*}
+
+  hence $I^-(\tau\sigma) = I^-(\sigma)$ and
+  \begin{align*}
+    I^+(\tau\sigma) &= 2(j - i -1) + 1 -  \sum_{\substack{l > k\\k \in \{i, j\} ~\text{or}~ l \in \{i, j\}}}\delta^{(\sigma)}_{kl}.
+  \end{align*}
+
+
+
+
+
+  First note that an inversion
+
+  First note that every inversion of $\tau$ involves either $i$ or $j$. Therefore if
+  neither $\sigma(k)$ nor $\sigma(l)$ is $i$ or $j$ then $\{k, l\}$ is an inversion of
+  $\tau_{i,j}\sigma$ if and only if it is an inversion of $\sigma$.
+
+  So what we need to do is count
+  \begin{align*}
+    \Big\{\{k, l\} ~\big |~ \{k, l\} \text{~is an inversion of~} \tau_{i,j}\sigma \text{~and~} k \in \{i, j\} \text{~or~} l \in \{i, j\}\Big\}.
+  \end{align*}
+
+
+  If $\sigma = 1$ then the claim is true since $|1| = 0$ is even and $|\tau\sigma| = |\tau|$ which we've already
+  proved is odd.
+
+  [incomplete and informal hereafter]\\
+
+  Recall the argument for $\tau_{i,j}$ being odd above: it involved the $\{i, j\}$ inversion
+  and $2|i - j|$ inversions involving $i$ or $j$ and one of the positions between $i$ and $j$.
+
+  Assume WLOG that $i < j$.
+
+  First note that if $\sigma$ does not touch any position in $i, i+1, \ldots, j$ then $\sigma$ and
+  $\tau$ do not interact and the claim is true: $\sigma$ does whatever it does, and $\tau$ adds its odd number
+  of inversions.
+
+  What remains is to prove the claim for any $\sigma$ that touches at least one position in the region
+  labeled $C$ here:
+
+    \includegraphics[width=400pt]{img/abstract-algebra--nf--2-b9b0.png}
+
+
+
+
+
+
+
+
+  Next, let's consider the simplest possible case that we haven't covered: $\sigma$ is a $2$-cycle that
 
-  First, suppose that the cycle decomposition of $\sigma$ does not contain $\tau$. Then
-  $\tau\sigma$ contains an inversion that $\sigma$ does not, so that $|I_{\tau\sigma}| = |I_\sigma| + 1$.
 
-  Alternatively, suppose that the cycle decomposition of $\sigma$ contains $\tau$, and write
-  $\sigma = \tau\sigma^-$, where $\sigma^-$ is the product of all cycles other than $\tau$ in the cycle decomposition
-  of $\sigma$. Note that exactly one of the inversions of $\sigma$ is due to $\tau$, so
-  that $|I_\sigma| = |I_\sigma^-| + 1$. We have $\tau\sigma = \tau^2\sigma^- = \sigma^-$, so
-  that $|I_{\tau\sigma}| = |I_{\sigma^-}| = |I_\sigma| - 1$ (this is non-negative, since the cycle decomposition
-  of $\sigma$ contains $\tau$, therefore $|I_\sigma| \ge 1$.)
 
-  In both cases, $|I_{\tau\sigma}| = |I_\sigma| \pm 1$, proving the claim.
 \end{proof}