Proof: This will be perfectly analogous to the proof for the first definition of . It is clear that each component is linear, so we only check that the Lie brackets for and are preserved: e, f, h First we check that : Next we show that : Finally we show that : Now that we have arrived at the same conclusion that we did earlier with regards to maintaining the commutator relations, we can update what our matrices for the actions of and should e, f, h look like: At this we present one of the small, yet important, results in the paper: Under the new definition of and the complex polynomial space will be Theorem 4: e, f, h, C[x], reducible, iff is a nonnegative integer. m Proof: (←) Here we assume is some nonnegative integer. contains some vector of the m C[x] v form , where is in by definition. This polynomial will have + 1 terms. We will show in a C bi n similar manner to the proof of theorem 3 that we must eventually be left with a monomial after at most n applications of e. Let = 0: If we let = 0 then of course we have just the constant term, If we apply f, then we n n b0. get 0 + If we let = 0, then of course we get zero. This gives us the module with only the mx. m constant basis element contributing to the span so we have the simplest nontrivial submodule, which makes is reducible. C[x]
Proof: This will be perfectly analogous to the proof for the first definition of . It is clear that each component is linear, so we only check that the Lie brackets for e, f, and h are preserved: First we check that : Next we show that : Finally we show that : Now that we have arrived at the same conclusion that we did earlier with regards to maintaining the commutator relations, we can update what our matrices for the actions of e, f, and h should look like: At this we present one of the small, yet important, results in the paper: Theorem 4: Under the new definition of e, f, and h, the complex polynomial space C[x], will be reducible, iff m is a nonnegative integer. Proof: (←) Here we assume m is some nonnegative integer. C[x] contains some vector v of the form , where bi is in C by definition. This polynomial will have n + 1 terms. We will show in a similar manner to the proof of theorem 3 that we must eventually be left with a monomial after at most n applications of e. Let n = 0: If we let n = 0 then of course we have just the constant term, b0. If we apply f, then we get 0 + mx. If we let m = 0, then of course we get zero. This gives us the module with only the constant basis element contributing to the span so we have the simplest nontrivial submodule, which makes C[x] is reducible.
sidoli_thesis.docx