# Apparent Contradiction to Weyl's Theorem

Let $L$ be $sl(2)$, i.e., $L=span\{h,e,f\}$, where $[h,e]=2e$,$[h,f]=-2f$,$[e,f]=h$. This is semi-simple. Suppose I create a module $V=span\{v_1,v_2,v_3\}$ and define actions as follows: $$h(v_1)=v_1,h(v_2)=-v_2,h(v_3)=v_1$$ $$e(v_1)=0,e(v_2)=v_1,e(v_3)=0$$ $$f(v_1)=v_2,f(v_2)=0,f(v_3)=v_2$$ Then I can test the brackets on each basis vector of $V$. For $v_1$: $$2e(v_1)=0,[h,e](v_1)=h(e(v_1))-e(h(v_1))=0$$ $$-2f(v_1)=-2v_2,[h,f](v_1)=h(f(v_1))-f(h(v_1))=h(v_2)-f(v_1)=-2v_2$$ $$h(v_1)=v_1,[e,f](v_1)=e(f(v_1))-f(e(v_1))=e(v_2)=v_1$$ For $v_2$: $$2e(v_2)=2v_1,[h,e](v_2)=h(e(v_2))-e(h(v_2))=h(v_1)-e(-v_2)=2v_1$$ $$-2f(v_2)=0,[h,f](v_2)=h(f(v_2))-f(h(v_2))=0$$ $$h(v_2)=-v_2,[e,f](v_2)=e(f(v_2))-f(e(v_2))=-f(v_1)=-v_2$$ For $v_3$: $$2e(v_3)=0,[h,e](v_3)=h(e(v_3))-e(h(v_3))=0-e(v_1)=0$$ $$-2f(v_3)=-2v_2,[h,f](v_3)=h(f(v_3))-f(h(v_3))=h(v_2)-f(v_1)=-2v_2$$ $$h(v_3)=v_1,[e,f](v_3)=e(f(v_3))-f(e(v_3))=e(v_2)=v_1$$ All of these brackets agree, so $V$ is a module, but if $W=span\{v_1,v_2\}$, then $W$ is an irreducible submodule, but $C=span\{v_3\}$ is not a submodule (since $L(v_3)\subseteq W$). So $V$ is not completely reducible.

This seems to contradict Weyl's theorem, that every module of a semi-simple Lie algebra is completely reducible, but I don't see the error.

• Why would the span of v3 not being a submodule contradict complete reducibility? Semi simplicity guarantees the existence of a complement, not that that complement is some particular space – Matthew Towers Mar 26 '16 at 1:14
• Oh, that makes sense! So if I take $C=span\{v_1+v_3\}$, then since the action of $L$ on $v_1$ is the same as on $v_3$, $L(C)=\{0\}$, and $L=W\oplus C$. – Sam Jaques Mar 26 '16 at 1:17
• I think you need the span of $v_1-v_3$ as your complement to $W$. For $h\cdot (v_3-v_1)=v_1-v_1=0$ and $f\cdot (v_3-v_1)=v_2-v_2=0$ and $e\cdot (v_1-v_3)=0-0=0$. – Matthew Towers Mar 26 '16 at 9:51

$W$ does have a complement, it just isn't $\operatorname{span}\{v_3\}$ which is you say is not a submodule. You can check that $\operatorname{span}\{v_3-v_1\}$ is a submodule (isomorphic to the trivial module) which is a complement to $W$.