--- title: "Problem Sets: Lie Algebras" subtitle: "Problem Set 4" author: - name: D. Zack Garza affiliation: University of Georgia email: dzackgarza@gmail.com date: Fall 2022 order: 4 --- # Problem Set 3 ## Section 7 :::{.problem title="Humphreys 7.2"} $M=\mathfrak{sl}(3, \FF)$ contains a copy of $L\da \liesl(2, \FF)$ in its upper left-hand $2 \times 2$ position. Write $M$ as direct sum of irreducible $L$-submodules ($M$ viewed as $L$ module via the adjoint representation): \[ V(0) \oplus V(1) \oplus V(1) \oplus V(2) .\] ::: :::{.solution .foldopen} Noting that - $\dim V(m) = m +1$ - $\dim \liesl_3(\FF) = 8$ - $\dim(V(0) \oplus V(1) \oplus V(1) \oplus V(2)) = 1+2+2+3 = 8$, it suffices to find distinct highest weight elements of weights $0,1,1,2$ and take the irreducible submodules they generate. As long as the spanning vectors coming from the various $V(n)$ are all distinct, they will span $M$ as a vector space by the above dimension count and individually span the desired submodules. Taking the standard basis $\ts{v_1,\cdots, v_8} \da \ts{x_1, x_2, x_3, h_1, h_2, y_1, y_2, y_3}$ for $\liesl_3(\FF)$ with $y_i = x_i^t$, note that the image of the inclusion $\liesl_2(\FF) \injects \liesl_3(\FF)$ can be identified with the span of $\ts{w_1,w_2,w_3} \da \ts{x_1, h_1, y_1}$ and it suffices to consider how these $3\times 3$ matrices act. Since any highest weight vector must be annihilated by the $x_1\dash$action, to find potential highest weight vectors one can compute the matrix of $\ad_{x_1}$ in the above basis and look for zero columns: \[ \ad_{x_1} = \left(\begin{array}{rrrrrrrr} 0 & 0 & 0 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{array}\right) .\] Thus $\ts{v_1 = x_1, v_2 = x_2, v_8 = y_3}$ are the only options for highest weight vectors of nonzero weight, since $\ad_{x_1}$ acts nontrivially on the remaining basis elements. Computing the matrix of $\ad{h_1}$, one can read off the weights of each: \[ \ad_{h_1} = \left(\begin{array}{rrrrrrrr} 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right) .\] Thus the candidates for highest-weight vectors are: - $x_1$ for $V(2)$, - $x_2$ for one copy of $V(1)$, - $y_3$ for the other copy of $V(1)$, - $h_1$ or $h_2$ for $V(0)$. We can now repeatedly apply the $y_1\dash$action to obtain the other vectors in each irreducible module. For $V(2)$: - $v_0 = x_1$ which has weight 2, - $v_1 = y_1.v_0 = [y_1, x_1] = -h_1$ which has weight 0, - $v_2 = {1\over 2} y_1^2.v_0 = {1\over 2} [y_1, [y_1, x_1]] = -y_1$ which has weight -2. Since we see $h_1$ appears in this submodule, we see that we should later take $h_2$ as the maximal vector for $V(0)$. Continuing with $V(1)$: - $v_0 = x_2$ which has weight 1, - $v_1 = y_1.v_0 = [y_1, x_2] = x_3$ which has weight -1. For the other $V(1)$: - $v_0 = y_3$ with weight 1, - $v_1 = -y_2$ with weight -1. For $V(0)$: - $v_0 = h_2$. We see that we get the entire basis of $\liesl_3(\FF)$ this way with no redundancy, yielding the desired direct product decomposition. ::: :::{.problem title="Humphreys 7.5"} Suppose char $\FF=p>0, L=\mathfrak{s l}(2, \FF)$. Prove that the representation $V(m)$ of $L$ constructed as in Exercise 3 or 4 is irreducible so long as the highest weight $m$ is strictly less than $p$, but reducible when $m=p$. > Note: this corresponds to the formulas in lemma 7.2 parts (a) through (c), or by letting $L\actson \FF^2$ in the usual way and extending $L\actson \FF[x, y]$ by derivations, so $l.(fg) = (l.f)g + f(l.g)$ and taking the subspace of homogeneous degree $m$ polynomials $\gens{x^m, x^{m-1}y, \cdots, y^m}$ to get an irreducible module of highest weight $m$. ::: :::{.solution} The representation $V(m)$ in Lemma 7.2 is defined by the following three equations, where $v_0 \in V_{ m}$ is a highest weight vector and $v_k \da y^k v_0/k!$: 1. $h\cdot v_i=(m-2 i) v_i$ 2. $y \cdot v_i=(i+1) v_{i+1}$ 3. $x \cdot v_i=(m-i+1) v_{i-1}$. Supposing $m< p$, the vectors $\ts{v_0, v_1,\cdots, v_m}$ still span an irreducible $L\dash$module since it contains no nontrivial $L\dash$submodules, just as in the characteristic zero case. However, if $m=p$, then note that $y.v_{m-1} = (m-1+1) v_m = 0 v_m = 0$ and consider the set $\ts{v_0, \cdots, v_{m-1}}$. This spans an $m\dash$dimensional subspace of $V$, and the equations above show it is invariant under the $L\dash$action, so it yields an $m\dash$dimensional submodule of $V(m)$. Since $\dim_\FF V(m) = m+1$, this is a nontrivial proper submodule, so $V(m)$ is reducible. ::: :::{.problem title="Humphreys 7.6"} Decompose the tensor product of the two $L$-modules $V(3), V(7)$ into the sum of irreducible submodules: $V(4) \oplus V(6) \oplus V(8) \oplus V(10)$. Try to develop a general formula for the decomposition of $V(m) \otimes V(n)$. ::: :::{.solution} By a theorem from class, we know the weight space decomposition of any $\liesl_2(\CC)\dash$module $V$ takes the following form: \[ V = V_{-m} \oplus V_{-m+2} \oplus \cdots \oplus V_{m-2} \oplus V_m ,\] where $m$ is a highest weight vector, and each weight space $V_{\mu}$ is 1-dimensional and occurs with multiplicity one. In particular, since $V(m)$ is a highest-weight module of highest weight $m$, we can write \[ V(3) &= \hspace{5.6em} V_{-3} \oplus V_{-1} \oplus V_1 \oplus V_3 \\ V(7) &= V_{-7} \oplus V_{-5} \oplus V_{-3} \oplus V_{-1} \oplus V_1 \oplus V_3 \oplus V_{5} \oplus V_{7} ,\] and tensoring these together yields modules with weights between $-3 -7 = -10$ and $3+7 = 10$: \[ V(3) \tensor V(7) &= V_{-10} \oplus V_{-8}\sumpower{2} \oplus V_{-6}\sumpower{3} \oplus V_{-4}\sumpower{4} \oplus V_{-2}\sumpower{4} \\ \qquad& \oplus V_0\sumpower{4} \\ \qquad& \oplus V_2\sumpower{4} \oplus V_4\sumpower{4} \oplus V_6\sumpower{3} \oplus V_8\sumpower{2} \oplus V_{10} .\] This can be more easily parsed by considering formal characters: \[ \ch(V(3)) &= e^{-3} + e^{-1} + e^{1} + e^3 = \\ \ch(V(7)) &= e^{-7} + e^{-5} + e^{-3} + e^{-1} + e^1 + e^3 + e^5 + e^7 \\ \ch(V(3) \tensor V(7)) &= \ch(V(3))\cdot \ch(V(7)) \\ \\ &= (e^{-10} + e^{10}) + 2(e^{-8} + e^{8}) + 3( e^{-6} + e^6) \\ &\qquad + 4( e^{-4} + e^4) + 4(e^{-2} + e^2) +4 \\ \\ &= (e^{-10} + e^{10}) + 2(e^{-8} + e^{8}) + 3( e^{-6} + e^6) \\ &\qquad + 4\ch(V(4)) ,\] noting that $\ch(V(4)) = e^{-4} + e^{-2} + e^{2} + e^{4}$ and collecting terms. To see that $V(3) \tensor V(7)$ decomposes as $V(4) \oplus V(6) \oplus V(8) \oplus V(10)$ one can check for equality of characters to see that the various weight spaces and multiplicities match up: \[ \ch(V(4) \oplus V(6) \oplus V(8) \oplus V(10)) &= \ch(V(4)) + \ch(V(6)) + \ch(V(8)) + \ch(V(10) \\ \\ &= \qty{e^{-4} + \cdots + e^4} + \qty{e^{-6} + \cdots + e^6} \\ &\quad +\qty{e^{-8} + \cdots + e^8} + \qty{e^{-10} + \cdots + e^{10}} \\ \\ &= 2\ch(V(4)) + (e^{-6} + e^6) \\ &\,\, + \ch(V(4)) + (e^{-6} + e^6) + (e^{-8} + e^8) \\ &\,\,+ \ch(V(4)) + (e^{-6} + e^6) + (e^{-8} + e^8) + (e^{-10} + e^{10})\\ \\ &= 4\ch(V(4)) + 3(e^{-6} + e^6) \\ &\,\, + 2(e^{-8} + e^8) + (e^{-10} + e^{10}) ,\] which is equal to $\ch(V(3) \tensor V(7))$ from above. More generally, for two such modules $V, W$ we can write \[ V\tensor_\FF W = \bigoplus _{\lambda \in\lieh\dual} \bigoplus _{\mu_1 + \mu_2 = \lambda} V_{\mu_1} \tensor_\FF W_{\mu_2} ,\] where we've used the following observation about the weight of $\lieh$ acting on a tensor product of weight spaces: supposing $v\in V_{\mu_1}$ and $w\in W_{\mu_2}$, \[ h.(v\tensor w) &= (hv)\tensor w + v\tensor(hw) \\ &= (\mu_1 v)\tensor w + v\tensor (\mu_2 w) \\ &= (\mu_1 v)\tensor w + (\mu_2 v)\tensor w \\ &= (\mu_1 + \mu_2)(v\tensor w) ,\] and so $v\tensor w \in V_{\mu_1 + \mu_2}$. Taking $V(m_1), V(m_2)$ with $m_1 \geq m_2$ then yields a general formula: \[ V(m_1) \tensor_\FF V(m_2) = \bigoplus _{n=-m_1-m_2}^{m_1+m_2} \bigoplus_{a + b = n} V_a \tensor_\FF V_b = \bigoplus_{n = m_1-m_2}^{m_2 + m_1} V(n) .\] ::: ## Section 8 :::{.problem title="Humphreys 8.9"} Prove that every three dimensional semisimple Lie algebra has the same root system as $\mathfrak{s l}(2, \FF)$, hence is isomorphic to $\mathfrak{s l}(2, \FF)$. ::: :::{.solution} There is a formula for the dimension of $L$ in terms of the rank of $\Phi$ and its cardinality, which is more carefully explained in the solution below for problem 8.10: \[ \dim \lieg = \rank \Phi + \size \Phi .\] Thus if $\dim L = 3$ then the only possibility is that $\rank \Phi = 1$ and $\size \Phi = 2$, using that $\rank \Phi \leq \size \Phi$ and that $\size \Phi$ is always even since each $\alpha\in \Phi$ can be paired with $-\alpha\in \Phi$. In particular, the root system $\Phi$ of $L$ must have rank 1, and there is a unique root system of rank 1 (up to equivalence) which corresponds to $A_1$ and $\liesl_2(\FF)$. By the remark in Humphreys at the end of 8.5, there is a 1-to-1 correspondence between pairs $(L, H)$ with $L$ a semisimple Lie algebra and $H$ a maximal toral subalgebra and pairs $(\Phi, \EE)$ with $\Phi$ a root system and $\EE\contains \Phi$ its associated Euclidean space. Using this classification, we conclude that $L\cong \liesl_2(\FF)$. ::: :::{.problem title="Humphreys 8.10"} Prove that no four, five or seven dimensional semisimple Lie algebras exist. ::: :::{.solution} We can first write \[ \lieg = \lien^- \oplus \lieh \oplus \lien^+,\qquad \lien^+ \da \bigoplus _{\alpha\in \Phi^+} \lieg_{ \alpha},\quad \lien^- \da \bigoplus _{\alpha\in \Phi^+} \lieg_{ -\alpha} .\] Writing $N\da \lien^+ \oplus \lien^- = \bigoplus _{\alpha\in \Phi} \lieg_{\alpha}$, we note that $\dim_\FF \lieg_{ \alpha} = 1$ for all $\alpha\in \Phi$. Thus $\dim_\FF N = \size \Phi$ and \[ \dim_\FF \lieg = \dim_\FF \lieh + \size \Phi .\] We can also use the fact that $\dim_\FF \lieh = \rank \Phi \da \dim_\RR \RR \Phi$, the dimension of the Euclidean space spanned by $\Phi$, and so we have a general formula \[ \dim_\FF \lieg = \rank \Phi + \size \Phi ,\] which we'll write as $d=r+f$. We can observe that $f\geq 2r$ since if $\mcb\da \ts{ \alpha_1, \cdots, \alpha_r}$ is a basis for $\Phi$, no $-\alpha_i$ is in $\mcb$ but $\ts{\pm \alpha_1, \cdots, \pm \alpha_r} \subseteq \Phi$ by the axiomatics of a root system. Thus \[ \dim_\FF \lieg = r+f \geq r + 2r = 3r .\] We can now examine the cases for which $d = r+f = 4,5,7$: - $r=1$: as shown in class, there is a unique root system $A_1$ of rank 1 up to equivalence and satisfies $f=2$ and thus $d=3$, which is not a case we need to consider. - $r=2$: this yields $d \geq 3r = 6$, so this entirely rules out $d=4,5$ as possibilities for a semisimple Lie algebra. Using that every $\alpha\in \Phi$ is one of a pair $+\alpha, -\alpha\in \Phi$, we in fact have that $f$ is always even -- in other words, $\Phi = \Phi^+ \disjoint \Phi^-$ with $\size \Phi^+ = \size \Phi^-$, so $f \da \size \Phi = 2\cdot \size \Phi^+$. Thus $d = r+f = 2 +f$ is even in this case, ruling out $d=7$ when $r=2$. - $r\geq 3$: in this case we have $d \geq 3r = 9$, ruling out $d=7$ once and for all. :::