📝 improve Weyl's theorem page

content/lie_algebra/schur.md

@@ -1,6 +1,6 @@
 +++
 title = "Schur's Lemma"
-date = "2024-12-8"
+date = 2024-12-08
 [taxonomies]
 tags = ["Lie algebra"]
 +++

content/lie_algebra/weyl.md

@@ -1,17 +1,78 @@
 +++
 title = "Weyl's Theorem"
-date = "2024-12-8"
+date = 2024-12-08
+updated = 2024-12-10
 [taxonomies]
 tags = ["Lie algebra"]
 +++
 
 ## Claim
 
-Let $L$ be a semisimple Lie algebra, $V$ a finite dimensional linear space, $\phi: L \to \mathfrak{gl}(V)$ a representation. Then, $\phi$ is completely reducible.
+Let $L$ be a semisimple Lie algebra over $\mathbb{F}$, $V$ a finite dimensional linear space, $\phi: L \to \mathfrak{gl}(V)$ a representation. Then, $\phi$ is completely reducible.
 
 ## Proof
 
-to be updated
+We show every submodule $W \subset V$ has a complement, i.e. $\exists X \subset V$ submod, s.t. $V = W \oplus X$ using induction over $\dim{W}$.
+
+### Case 1: $\dim V/W = 1$
+
+The action of $L$ on $V/W \cong \mathbb{F}$ is trivial, so there is a exact sequence
+$$ 0 \to W \to V \to \mathbb{F} \to 0$$
+
+#### Case 1-a: $W$ is reducible
+
+Let $0 \neq W' \subsetneq W$ be a submodule, then $\dim((V/W') / (W/W')) = \dim(V/W) = 1$, so
+$$ 0 \to W/W' \to V/W' \to \mathbb{F} \to 0$$
+is exact.
+
+Since $\dim (W/W') < \dim W$, we can use the induction hypothesis to conclude $W/W'$ has a complement, i.e.
+$\exists \tilde{W}/W' \subset V/W'$ submod, s.t. $V/W' = W/W' \oplus \tilde{W}/W'$.
+
+Further, $\dim (\tilde{W}/W) = 1$, and
+$$ 0 \to W \to \tilde{W} \to \mathbb{F} \to 0$$
+is exact.
+
+Applying induction hypothesis again to get $\exists X \subset \tilde{W}$ subsp, s.t. $\tilde{W} = W' \oplus X$.
+
+Then,
+
+- $\dim X = 1, \dim W = \dim V - 1.$ Therefore, $\dim V = \dim W + \dim X$
+- $W \cap X = W \cap (X \cap W') = (W \cap W') \cap X \subset W' \cap X = 0.$
+
+$\therefore V = W \oplus X$
+
+#### Case 1-b: $W$ is irreducible
+
+We can assume $\phi$ is faithful, since if not, $\phi' : L/\ker \phi \to \mathfrak{gl}(V)$ is so.
+
+Let $c := c_\phi$ be the Casimir element of $\phi$. Then $v \mapsto cv$ is an L-mod endmorphism.
+
+($\because \forall x \in L, \forall v \in V, c (\phi(x).v)= \phi(x) (cv)$)
+
+- $\ker c \subset V$ is an L-submod.
+- $c$ maps $V$ into $W$. In particular, $\mathrm{Tr}_{V/W} (c) = 0$
+  - The action of $L$ on $V/W \cong \mathbb{F}$ is trivial, so $L.V \subset W$, i.e. $\forall x \in L, \phi(x) V \subset W$. Therefore, $cV \subset W$
+
+- $c$ acts as a nonzero scalar on $W$
+  - $W$ is irred, $c$ is the Casimir element, Shur's lemma.
+  - $\mathrm{Tr}_V(c) \neq 0, \mathrm{Tr}_{V/W}(c) = 0.$ Therefore, $\mathrm{Tr}_W (c) \neq 0.$
+
+Hence,
+
+- $\dim \ker c= \dim V - \dim W = 1$
+  - $c: V \to W$ is a surjective L-mod morphism.
+- $W \cap \ker c = 0$
+
+i.e. $V = W \oplus \ker c$
+
+### Case 2: General case
+
+Let $Hom(V, W)$ be the set of linear maps from $V$ to $W$. Then, $Hom(V, W)$ is an L-mod with action
+$$(x.f)(v) := x.f(v) - f(x.v)$$
+
+Let $\mathcal{V} := \set{f \in Hom(V, W);  f|_W \text{is a scalar mult.}}$. Then, $\mathcal{V}$ is an L-mod.
+
+...To be updated...
 
 ## Reference