# Wednesday, October 26 > Missed first 15m! :::{.proposition title="Step 8"} Any connected $\Gamma$ is one of the following types: ![](figures/2022-10-26_09-28-53.png) ::: :::{.proposition title="Step 9"} The only connected $\Gamma$ graphs of the second type in Step 8 are either $F_4$ or $B_\ell = C_\ell$. Compute \[ (\eps, \eps) = \sum_{i=1}^k i^2 - \sum_{i=1}^{p-1} i(i+1) \\ = p^2 - \sum_{i=1}^{p-1} i \\ = p^2 - {p(p-1)\over 2} \\ = {p(p+1)\over 2} ,\] and similarly $(\eta, \eta) = {q(q+1)\over 2}$. Note $4(\eps_p, \eta_q)^2 = 2$, so $(\eps,\eta)^2 = p^2 q^2 (\eps_p, \eta_q)^2 = {p^2q^2\over 2}$. By Cauchy-Schwarz, $(\eps, \eta)^2< (\eps, \eps) (\eta, \eta)$, where the inequality is strict since $\eta, \eps$ are linearly independent. Then check \[ {p^2 q^2\over 2} &< {p(p+1)\over 2} \cdot {q(q+1)\over 2} \\ {p1\over 2} &< {p+1\over 2}\cdot {q+1\over 2} \\ 2pq &< pq + p + q + 1 ,\] and so combining these yields $pq-p-q+1 < 2$ and thus \[ (p-1)(q-1) < 2 .\] Since $p\geq q\geq 1$, this yields two possible cases: - $p=q=2 \leadsto F_4$ - $q=1, p\in \ZZ_{\geq 0} \leadsto B_\ell = C_\ell$. ::: :::{.proposition title="Step 10"} The only connected $\Gamma$ of type (d) are $D_\ell, E_6, E_7, E_8$. Set $\eps \da \sum i\eps_i$, $\eta \da \sum i\eta_i$, and $\zeta = \sum i \zeta_i$. Note that $\eps, \eta, \zeta$ mutually orthogonal by inspecting the graph, and $\psi$ is not in their span. Let $\theta_1$ (resp. $\theta_2, \theta_3$) be the angles between $\eps$ (resp. $\eta, \zeta$) and $\psi$. Since $\eps,\eta,\zeta$ are linearly independent, the idea is to apply Gram-Schmidt to $\ts{\eps,\eta,\zeta,\psi}$ without normalizing. The first 3 are already orthogonal, so we get a new orthogonal basis $\ts{\psi_1 \da \eps, \psi_2\da \eta, \psi_3\da \zeta, \psi_0}$ where $(\psi_0, \psi) \neq 0$. We can expand $\psi$ in this basis to write $\psi = \sum_{i=0}^3 \qty{\psi, {\psi_i\over \norm{\psi_i}}} {\psi_i \over \norm{\psi_i}}$. Note that $(\psi, \psi) = 1$, and consequently $\sum_{i=1}^3 \qty{\psi, {\psi_i \over\norm{\psi_i}}}^2 < 1 \implies \sum_{i=1}^3 \cos^2(\theta_i) < 1$. So \[ \cos^2(\theta_1) + \cos^2( \theta_2) + \cos^2( \theta_3) < 1 .\] As in Step (9), $(\eps, \eps) = {p(p-1)\over 2}$ and similarly for $\eta,\zeta$, and so \[ \cos^1( \theta_1) = {(\eps, \psi)^2 \over (\eps, \eps) (\psi, \psi)} = {(p-1)^2 (\eps_{p-1}, \psi )^2 \over {p(p-1)\over 2} \cdot 1 } \\ = {p-1\over p} {1/4\over 1/2} \\ = {1\over 2}\qty{1-{1\over p}} ,\] where we've used that $4(\eps_{p-1},\psi ) = 1$. Similarly (and summarizing), \[ \cos^2(\theta_1) &= {1\over 2}\qty{1-{1\over p}} \\ \cos^2(\theta_2) &= {1\over 2}\qty{1-{1\over q}} \\ \cos^2(\theta_3) &= {1\over 2}\qty{1-{1\over r}} \\ \\ &\implies {1\over 2}\qty{ 1 - {1\over p} + 1 - {1\over q} + 1 - {1\over r}} < 1 \\ &\implies p\inv + q\inv +r\inv > 1 .\] and since $p\geq q\geq r\geq 2 \implies p\inv \leq q\inv \leq r\inv \leq 2\inv$, we have ${3\over r} > 1$ by replacing $p,q$ with $r$ above. So $r < 3$, forcing $r=2$, and there is only one "top leg" in the graph for (d) above. We also have \[ {2\over q} \geq {1\over p} + {1\over q} > {1\over 2}, \qquad (\star) .\] so $q<4$ forces $q=2,3$. - If $q=2$, then $(\star)$ is true for any $p\geq 2$, and the bottom leg has two vertices and this yields type $D_\ell$. - If $q=3$ then ${1\over p} > {1\over 2}-{1\over 3} = {1\over 6}$ implies $p < 6$, forcing $p=3,4,5$ corresponding to $E_6, E_7, E_8$. ::: :::{.remark} Note that the diagrams we've constructed are the only possible Coxeter graphs of a root system, since normalizing any set of simple roots yields an admissible set. This proves one direction of a correspondence, but what are all possible Dynkin diagrams? Note that types $B_\ell, C_\ell$ have the same underlying Coxeter graph, and only differ by directions on the multi-edges. ::: :::{.question} Does every connected Dynkin diagram correspond to an irreducible root system. Yes: types $A,B,C,D$ can be constructed from root systems in classical Lie algebras, and the corresponding Dynkin diagrams can be constructed directly. The 5 exceptional types must be constructed directly. ::: :::{.question} Does each irreducible root system occur as the root system of some semisimple Lie algebra over $\CC$? The answer is of course: yes! ::: > Next time: starting Ch. V.