---- > [!theorem] Theorem. ([[Weyl's theorem on complete reducibility]]) > **1.** Let $\mathfrak{g}$ be a [[semisimple Lie algebra|semisimple]] [[Lie algebra]] over a [[field]] of [[characteristic of a field|characteristic zero]]. Then every finite-dimensional [[Lie algebra representation|representation]] of $\mathfrak{g}$ is [[completely reducible]]. > > > **2.** If we are willing to invoke more machinery, namely, the [[the classification of complex irreducible semisimple Lie algebra representations]], then over $\mathbb{C}$ we can assert for the Weyl decomposition a specific form. Let $\Phi$ be the [[root system of a Lie algebra|root system]] [[classification of complex semisimple Lie algebras|of]] $\mathfrak{g}$, $\mathfrak{t} \subset \mathfrak{g}$ [[Cartan subalgebra|CSA]], $\Delta$ a [[root basis]], and $X \subset \mathfrak{t}^{*}$ the [[root and weight lattice of a root system|weight lattice]] of $\Phi$. For a highest weight $\lambda$, fix a basis $\{ v_{\lambda, i} \}$ for the $\mathfrak{t}$-action-[[simultaneously diagonalizable|simultaneous-eigenspace]] $V_{\lambda}=\{ v \in V : t \cdot v=\lambda(v) \text{ for all } t \in \mathfrak{t} \}$. > > Then $V$ decomposes into irreducible summands as follows: > > $V=\bigoplus_{\lambda \in X \text{ a highest weight}} \bigoplus_{i=1}^{\text{dim }V_{\lambda}} \mathcal{U}(\mathfrak{g}) \cdot v_{\lambda,i} \cong \bigoplus_{\lambda \in X \text{ a highest weight}} V(\lambda)^{\oplus \text{dim } V_{\lambda}},$ > > where $V(\lambda)$ [[the classification of complex irreducible semisimple Lie algebra representations|is the]] [[Verma module|unique irreducible]] [[highest weight module]] of highest weight $\lambda$. Note that if $v_{\lambda} \in V_{\lambda}$ is a [[on the weights of a representation|highest weight vector]] with highest weight $\lambda$, [[universal enveloping algebra|then]] $v_{\lambda}$ [[Lie algebra subrepresentation generated by a vector|generates]] a copy $\mathcal{U}(\mathfrak{g})\cdot v \cong V(\lambda)$ of $V(\lambda)$ in $\mathfrak{g}$. Also note that we could also have summed over all $\lambda \in X$ dominant (in light of the classification) if we're willing to permit some summands to be trivial ($\text{dim }V_{\lambda}=0$). > > The examples below should clarify how one can perform some decompositions 'in practice'. > --- > [!basicexample] > First we draw the weights of $V=V(\omega_{1}=\omega_{\alpha})$ in tan; per [[reading off the weights (with some information on multiplicity) of an irreducible semisimple Lie algebra representation|this result]] this can be done just by reflecting around $\omega_{\alpha}$. Plugging into the [[Weyl character formula|Weyl dimension formula]] confirms that $V$ is $4$-dimensional. > > The weights of $V \otimes V$ are just the various sums of weights of $\omega_{\alpha}$ (16-dimensional). They're depicted on the left in dark blue, with concentric circles indicating multiplicity. > > To decompose $V \otimes V$ into irreducibles, we start by focusing on the 'highest weight even among highest weights', i.e., the maximal element wrt the [[highest weight module|the order]] $\leq$. (this will always exist by finite-dimensionality), i.e., the top left corner $2\alpha+\beta$ (turns out to also be the highest root). This gives rise to an irreducible subrepresentation of $\mathfrak{g}$, whose weights may be [[reading off the weights (with some information on multiplicity) of an irreducible semisimple Lie algebra representation|read off]] by first determining the other dominant weight ($\omega_{\beta}$) by noting $\omega_{\beta} \leq 2\alpha+\beta$ (mult=1 because only one positive root path to $2\alpha+\beta$), then reflecting all around. This gives the [[adjoint representation]]; equivalently $\text{sym}^{2}V$. The next 'highest-highest weight' is $\omega_{1}$. Repeat the above process; then we are just left with $0$, the trivial representation. So $V \otimes V=V(2\omega_{1})\oplus V(\omega_{2}) \oplus V(0)$. > > ![[Pasted image 20250524145245.png]] > ![[CleanShot 2025-05-31 at [email protected]]] **1.** One uses the picture of $B_{2}$ to calculate the first two rows of the following table. The second two rows are then deduced from the first two rows, and the [[Weyl character formula|Weyl dimension formula]] is employed thereon. ![[Pasted image 20250531090435.png|500]] We obtain: $\text{dim }V(\lambda)=\frac{(n_{1}+1)(n_{2}+1)(n_{1}+2n_{2}+3)(n_{1}+n_{2}+2)}{6}.$ **2.** The weights of $V(\omega_{1})$ are depicted in tan. The weights of $V(\omega_{2})$ are depicted in dark purple. The weights of $V(\omega_{1}+\omega_{2})$ are in dark red. The weights were ascertained using the [[reading off the weights (with some information on multiplicity) of an irreducible semisimple Lie algebra representation|recipe here]]: start with the assumed highest weight $\lambda \in \{ \omega_{1}, \omega_{2},\omega_{1}+\omega_{2} \}$, look at the dominant weights $\mu$ with $\mu \leq \lambda$, then add in all the Weyl conjugates. Each weight has multiplicity $1$, except the $W$-[[orbit]] of $\omega_{1}$ in $V(\omega_{1}+\omega_{2})$ which has multiplicity 2. For $V(\omega_{1})$ and $V(\omega_{2})$ this is because there are no $\mu$ other than $\omega_{i}$ in $X$ with $\mu \leq \omega_{i}$, so the only weights are the Weyl group conjugates of $\omega _i$. Since the [[Weyl group of a root system|Weyl group]] preserves multiplicity, [[highest weight module|and]] $\text{mult}(\text{highest weight of irrep})=1$, the weights must all of multiplicity 1. For $V(\omega_{1}+\omega_{2})$ one similarly has all the Weyl conjugates of $\omega_{1}+\omega_{2}$ have mult 1. There is also $\omega_{1} \leq \omega_{1}+\omega_{2}$ a dominant weight, and there are two positive root paths from it to $\omega_{1}+\omega_{2}$ ($\omega_{1}+\omega_{2}, \omega_{1}+\alpha_{2}+\alpha_{1}$). This says the multiplicity is at most 2. To get the exact number, we turn to the Weyl dimension formula: $\text{dim }V(\omega_{1}+\omega_{2})=\frac{2 \cdot 2 \cdot6 \cdot4}{6}=16.$ We know we have $12<16$ distinct weights total. To make up the difference, need multiplicity 2. ![[Pasted image 20250531104911.png|500]] **3.** The weights of a tensor product $V_{1} \otimes V_{2}$ of representations are given by the sum of multisets $\Pi(V_{1})+\Pi(W_{2})=(\mu_{1}+\mu_{2}: \mu_{1} \in \Pi(V_{1}), \mu_{2} \in \Pi(V_{2}))$. Looks like this (count 20 blues, as expected since $\text{dim }(V_{1}\otimes V_{2})=(\text{dim }V_{1})(\text{dim } V_{2}))$) ![[Pasted image 20250531105236.png|500]] Now we follow the strategy above to decompose into irreps. Start with the 'maximal highest weight' (here it turns out to be $\omega_{1}+\omega_{2}$); so we know $V(\omega_{1}+\omega_{2})$ will be a factor and can delete the weight spaces it contributes (note we have to know multiplicities of $V(\omega_{1}+\omega_{2})$ to do this) ![[Pasted image 20250531105457.png]] We see that $V(\omega_{1})\otimes V(\omega_{1})=V(\omega_{1}+\omega_{2}) \oplus V(\omega_{1})$ is the desired decomposition into irreducibles. Another technique would be to first recognize that we'll need a copy of $V(\omega_{1}+\omega_{2})$, and tentatively write $\underbrace{ V(\omega_{1}) \otimes V(\omega_{2}) }_{ \text{dim=20} }=\underbrace{ V(\omega_{1}+\omega_{2}) }_{ \text{dim}=16 } \oplus U$ for some $U$. So $\text{dim }U=4$. The Weyl dimension formula implies there's only one nontrivial irreducible representation with dim <= 4... and if $4=\frac{(n_{1}+1)(n_{2}+1)(n_{1}+2n_{2}+3)(n_{1}+n_{2}+2)}{6}$, then what can $n_{1}$ and $n_{2}$ be? There is only one choice: $n_{1}=1$, $n_{2}=0$. So $U=V(\omega_{1})$; done because we're out of dimensions **4.** As just said, the only nontrivial irrep of $\mathfrak{sp}_{4}$ (which has root system $C_{2} \cong B_{2}$) of dimension $\leq 4$ is $V(\omega_{1})$. We know the defining representation of $\mathfrak{sp}_{4}$ is irreducible of dimension $4$. So it's $V(\omega_{1})$. As for $\mathfrak{so}_{5}$ (which has root system $B_{2}$), we can again look at the Weyl dimension formula $5=\frac{(n_{1}+1)(n_{2}+1)(n_{1}+2n_{2}+3)(n_{1}+n_{2}+2)}{6}$ to see that if $V(n_{1}\omega_{1}+n_{2}\omega_{2})$ is a $5$-dimensional irrep of $\mathfrak{so}_{5}$, then necessarily $n_{1}=0$, $n_{2}=1$. Hence $\text{defining rep of }\mathfrak{so}_{5}=V(\omega_{2})$. The weight space decomposition of the adjoint representation is precisely the root space decomposition of $\mathfrak{so}_{5}$, so highest weight=highest root=$2 \alpha+\beta=2\omega_{1}$. That is, $\text{ad }\mathfrak{so}_{5}=V(2 \omega _1)$. > [!proof]- Proof. ([[Weyl's theorem on complete reducibility]]) > **1.** This argument goes very similar to the proof that [[any representation of sl2(C) is completely reducible]], and so we omit the details in this course. > > **2.** We know each summand of the decomposition of $\mathfrak{g}$ into irreducibles will be isomorphic to $V(\lambda)^{\oplus m_{\lambda}}$ for $\lambda \in X$ dominant, by [[the classification of complex irreducible semisimple Lie algebra representations|the classification]]. On the other hand, we also know that each highest weight vector $v_{\lambda}$ [[Lie algebra subrepresentation generated by a vector|generates]] a [[Lie algebra subrepresentation|subrepresentation]] $\mathcal{U}(\mathfrak{g}) \cdot v_{\lambda}$ of $\mathfrak{g}$. It is easy to see $\mathcal{U}(\mathfrak{g}) \cdot v_{\lambda} \cong V(\lambda)$;[^1] in particular it is irreducible and so must feature in the decomposition. Summarizing: any summand has the form $V(\lambda)$ for $\lambda$ a highest weight, and any highest weight $\lambda$ gives rise to a summand of the form $V(\lambda)$. The $m_{\lambda}$ come in because there may be highest weight vectors from > ---- #### [^1]: For example: the [[universal property]] of [[Verma module|Verma modules]] gives a [[surjection]] $M(\lambda) \twoheadrightarrow \mathcal{U}(\mathfrak{g}) \cdot v_{\lambda}$ sending $m_{\lambda} \mapsto v_{\lambda}$. [[characterization of quotienting a group|This factorizes]] into surjections $M(\lambda) \twoheadrightarrow V(\lambda) \twoheadrightarrow \mathcal{U}(\mathfrak{g}) \cdot v_{\lambda}$; the latter must have trivial kernel since $V(\lambda)$ is irreducible. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```