06-1350/Class Notes for Thursday November 16

Today's Agenda

Formulas are a Chore (Bore?)

• Sweeping clean a tree and ${\displaystyle {\mathcal {A}}(\Gamma )\equiv {\mathcal {A}}(\uparrow _{b_{1}(\Gamma )})}$.
• ${\displaystyle {\mathcal {A}}(\uparrow _{n})}$ is a VS-algebra (see more at VS, TS and TG Algebras).
• In the coordinates above, write the ${\displaystyle TR\Phi B}$ relations in various algebraic notations.
• R4: ${\displaystyle (1230)^{\star }B\cdot (1213)^{\star }B\cdot (1023)^{\star }\Phi =(1123)^{\star }\Phi \cdot (1233)^{\star }B}$ or ${\displaystyle (B_{1a}B_{2a}\Phi _{1a};B_{1b}B_{2b};B_{1c}B_{2a}\Phi _{1b};B_{2c}\Phi _{1c})=(\Phi _{2a}B_{3a};\Phi _{2a}B_{3b};\Phi _{2b}B_{3c};\Phi _{2c}B_{3c})}$.
• R3: ${\displaystyle (1230)^{\star }B\cdot (1213)^{\star }B\cdot (1023)^{\star }B=(1123)^{\star }B\cdot (1203)^{\star }B\cdot (1231)^{\star }B}$ or ${\displaystyle (B_{1a}B_{2a}B_{3a};B_{1b}B_{2b};B_{1c}B_{2a}B_{3b};B_{2c}B_{3c})=(B_{4a}B_{5a}B_{6a};B_{4a}B_{5b}B_{6b};B_{4b}B_{6c};B_{4c}B_{5c}B_{6a})}$.
• R2: ${\displaystyle (123)^{\star }B^{\pm }\cdot (132)^{\star }B^{\mp }=1_{3}}$ or ${\displaystyle (B_{1a}^{\pm }B_{2a}^{\mp };B_{2b}^{\pm }B_{3c}^{\mp };B_{1c}^{\pm }B_{2b}^{\mp })=(1;1;1)}$.
• R1: ${\displaystyle (B_{a}^{\pm };B_{b}^{\pm }B_{c}^{\pm })=(1;T^{\pm 2})}$.
• But for now, skip the writing of the following relations:
• Symmetry of ${\displaystyle \Phi }$ and of ${\displaystyle B^{\pm }}$.
• ${\displaystyle u}$, ${\displaystyle d}$ and ${\displaystyle \#}$
• Idempotence for ${\displaystyle T}$, ${\displaystyle R}$, ${\displaystyle \Phi }$ and ${\displaystyle B^{\pm }}$.
• ${\displaystyle B^{\pm }}$ in terms of ${\displaystyle \Phi }$ and ${\displaystyle R}$ and ${\displaystyle R}$ in terms of ${\displaystyle T}$.

Exponentiation is a Miracle

• Description of the problem.