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 ...


Only top voted, non community-wiki answers of a minimum length are eligible