# Tag Info

10

Short answer: If two Lie algebras are isomorphic, they have "the same" complex representations. A real semisimple Lie algebra and its complexification also have "the same" complex representations, but only to a certain extent: During complexification, the correspondence "forgets" which representations of the real Lie algebra were conjugate to which. Hence, ...

7

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

7

Yes, this holds for all complex simple Lie algebras. A reference is Theorem $A$ in the article On commutators in a simple Lie algebra. The result can be extended to simple Lie algebras over more general fields.

6

To deduces irreducibility of $V(\lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $W\subset V(\lambda)^*$ consider its annihilator $U:=\{v\in V(\lambda):\forall\phi\in W:\phi(v)=0\}$. This is clearly a linear subspace in $V(\lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $... 5 1) In addition to Brown Theorem above, it is worth noting the following "1.5" generators of such Lie algebras. Theorem: Let L be a simple Lie algebra over an infinite field k of char 0 (or char not 2,or 3). Then L=[L, a] +[L, b] for some a, b in L. In particular, every element of L is a sum of at most 2 commutators. Reference: G. Bergman, N. Nahlus, ... 5 A good reference are the course notes Lie algebras by Alberto Elduque. Pages$89-104$gives the classification of simple real Lie algebras in detail. 5 The Killing form can be explicitly calculated. Let$A,B\in \mathfrak{sl}_n(\mathbb{C})$, Killing form is $$\kappa(A,B) = 2n\text{tr}(AB)$$ it is straightforward to show this is non-degenerate. Briefly, if$A$is such that$\text{tr}(AB) = 0$for all$B\in \mathfrak{sl}_n(\mathbb{C})$, then testing with$B=e_{ij}, i\neq j$shows$a_{ji}=0$, testing with$B=e_{...

4

In order for this question not to stay on the list of unanswered, here is the full answer: Assume $L$ is a Lie algebra with a semisimple ideal $I$ such that $L/I$ is semisimple. Assume $S$ is a solvable ideal of $L$. The ideal $(S + I)/I$ in $L/I$ is isomorphic to $S/(S\cap I)$ and thus solvable. But since $L/I$ is semisimple, this means that it must be $0$...

4

The key is to look at just the vector space structure of $\mathfrak g$ and then use some nice topological properties. So for now, think of $\mathfrak g$ as just a real vector space, say $\mathbb R^n$, so that the Killing form is essentially an inner product. Since the Killing form is negative definite, there is a an element $A \in GL_n(\mathbb R)$ with all ...

4

Well, if you come to the definition of a Cartan subalgebra (in an arbitrary finite-dimensional Lie algebra over an arbitrary infinite field — denote by $d$ the dimension), you see that it is defined as $K_x=\mathrm{Ker}(\mathrm{ad}(x)^d)$, where $x$ is regular, and regular precisely means that $K_x$ has minimal dimension. So, the Cartan rank (I don't like ...

4

The following is an explicit argument building on the knowledge of the finite-dimensional irreducible representation of ${\mathfrak g}$. At its heart is the non-degeneracy of the Shapovalov-form and the description of its determinant, but I tried to keep the exposition elementary. Setup: Let ${\mathfrak g}={\mathfrak n}^-\oplus{\mathfrak h}\oplus{\mathfrak ... 4 Here's a diploma thesis on the topic: https://www.mat.univie.ac.at/~cap/files/wisser.pdf You should not expect much literature which focuses exclusively on Lie algebras, as the classification of semisimple algebraic / Lie groups is naturally connected to this. In my thesis, I did almost exclusively deal with Lie algebras. Although in later chapters I focus ... 4 It is the good idea, let$p:g\rightarrow g/I$the quotient map$p(rad(g))$is a solvable ideal, since$g/I$is semi-simple,$p(rad(g))=0$and$rad(g)\subset I$. 4 The proof is in the paper on Lie algebra cohomology by Chevalley and Eilenberg, in$1948$(and in Koszul's paper of$1950)$. Chevalley and Eilenberg proved that$H^3(\mathfrak{g},K)$is nonzero for every nonzero semisimple Lie algebra$\mathfrak{g}$and every field$K$of characteristic zero. This works as follows: Let$B(x,y)$denote the Killing form of$\...

4

Your confusion is understandable. It is true that the roots are originally defined as elements of $\mathfrak h^*$, which is a $\mathbb C$-vector space (and two-dimensional, hence abstractly isomorph to $\mathbb C^2$). However, note that there are only finitely many roots; and further, if you choose two of them, all the other roots are actually $\mathbb Z$-...

3

Theorem (Harish-Chandra 1949) for an arbitrary finite-dimensional Lie algebra over a field of characteristic, finite-dimensional representations separate points of the universal enveloping algebra. This is a deep result, proved in Chap 2 of Dixmier's book "enveloping algebras". It has Ado's theorem as corollary. In the semisimple case in characteristic ...

3

Suppose that $K$ is the base field. Fix $h\in\mathfrak{h}$. Let $h_i\in\mathfrak{g}_i$ be such that $h=\sum\limits_i\,h_i$ (the $h_i$'s are unique). We claim that $h_i\in\mathfrak{h}$ for all $i$. Let $t\in\mathfrak{h}$. Write $t=\sum\limits_{i}\,t_i$ with $t_i\in\mathfrak{g}_i$. For $i\neq j$, $$\left[t_i,h_j\right]\in\mathfrak{g}_i\cap\mathfrak{g}_j=\... 3 Here's a proof from Humphrey's excellent book: First of all, show that I=ad(\mathfrak g)\subset\partial\mathfrak g is an ideal of the Lie algebra \partial\mathfrak g: if x\in\mathfrak g and \delta\in\partial\mathfrak g, then$$[\delta,ad(x)]=\cdots\in\partial\mathfrak g$$This implies that the Killing form on \mathfrak g coincides with that ... 3 It is not vacuous. Consider \mathfrak{g} = \mathfrak{sl}_2(\Bbb C), \mathfrak{h} = \Bbb C x for any non-semisimple element x\in \mathfrak{g}: It satisfies 1, but not 2, and is not a Cartan subalgebra. Maybe it is noteworthy that a subalgebra which consist of ad-diagonalisable elements is automatically abelian (cf. Humphreys' Introduction to Lie ... 3 You can't prove it, since it is false. Take any finite-dimensional represention \pi of \mathfrak{sl}(2). And now consider the representaion \pi^\star that it induces on V\oplus V. Then, whatever the dimension d of the eigenspace of the highest weight \Lambda of \pi is, the dimension of the eigenspace of the highest weight \Lambda of \pi^\... 3 Not all. If x is semisimple and contained in a subalgebra isomorphic to \mathfrak{sl}_2, then since every semisimple element of \mathfrak{sl}_2 is regular, x is proportional to a standard Cartan element h. As \mathfrak{g} is a finite-dimensional \mathfrak{sl}_2-representation, the eigenvalues of ad(h) are integers, and hence the eigenvalues ... 2 These notions are all equivalent, as you surmise. The various equivalences are not all trivial, though. It is pretty clear that 3. \implies 2. \implies 1. (A connected Lie subgroup is detected by its Lie algebra, and a connected algebraic subgroup is necessarily also a connected Lie subgroup.) So one approach to getting the equivalence is to show that ... 2 Let \phi_i\colon L\rightarrow L\cap L_i be the canonical projection to the i-th factor. We can use now the following lemma: Lemma: Let \phi:L\rightarrow L' be a surjective Lie algebra homomorphism, and H be a Cartan subalgebra of L. Then \phi(H) is a Cartan subalgebra of L'. Proof: See Lemma 3.6.2 here. The proof works also for not ... 2 I suppose that f(x,y)=\mathbf{Trace}(XY) means in fact$$f(x,y)=\mathbf{Trace}(ad_xad_y) since I don't get otherwise what $X,Y$ mean since $f$ is function of $x,y$... in other words I assume that we are speaking about the Killing form. Then the thing is quite easy: suppose $\mathfrak{g_1}$ is an ideal, then consider $\mathfrak{g}=\mathfrak{g_1}\oplus\... 2 Take any semi-simple algebra$g$which is not compact, the killing form is not a scalar product, there exists$x\in g$such that$\langle x,x\rangle =0$,$Vect(x)$is a subalgebra of$g$and$x$is in the orthogonal of$Vect(x)$. 2 Hint: Recall that$X_\alpha$,$X_{-\alpha}$and$[X_\alpha,X_{-\alpha}]$span a Lie subalgebra of$\mathfrak g$that is isomorphic to$\mathfrak{sl}(2,\mathbb C)$. Now given another root$\beta$, you have the$\beta$-string through$\alpha$, which defines a representation of this subalgebra that you can analyze using the representation theory of$\mathfrak{...

2

This is true (assuming $\mathfrak{g}$ is finite dimensional). Indeed, assume every finite dimensional representation is completely reducible. Then, since the adjoint representation is completely reducible, it follows that $\mathfrak{g}$ is reductive. If $Z(\mathfrak{g})\neq 0$, then fix $z\in Z(\mathfrak{g})\backslash\{0\}$ and decompose $\mathfrak{g}=\... 2 If you are asking why the$u_i$are in$H$, here is an argument: Take another arbitrary element$y\in H$. Write$y = v_1+...+v_t$. Then because$H$is abelian we have$[x,y] =0$, and because the$L_i$are ideals and the decomposition is direct, this implies$[u_i,v_i] = 0$for all$i$. But then, again because the$L_i$are ideals and the decomposition is ... 2 No. For example, the Lie algebra of type$D_4$is the algebra$\mathfrak{so}_8$of$8\times 8$skew-symmetric matrices. It is contained in the Lie algebra of type$A_7$, that is, the algebra$\mathfrak{sl}_8$of$8\times 8$trace-zero matrices, even though$D_4$is not a subgraph of$A_7$. 2 Let$\Phi$be a root system with inner product$\langle, \rangle$which w.l.o.g. is chosen as invariant under the automorphism group$A(\Phi)$. As remarked in a comment, the question is equivalent to: for a given root$\alpha$, of what type is the root system$\alpha^\perp := \lbrace \gamma \in \Phi: \gamma \perp \alpha\rbrace\$? Remark first that we can ...

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