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Lie algebras are algebraic structures which were introduced to study the concept of infinitesimal transformations. The term "Lie algebra" (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name "infinitesimal group" is used. Related mathematical concepts include Lie groups and differentiable manifolds.

2 votes

Recovering a Lie algebra from its affine Lie algebra

The extraction of a finite-type Lie subalgebra from an abstract affine Lie algebra is not functorial, because you have lots of automorphisms. Even if you are given a presentation with a Dynkin diagra …
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4 votes

faithful adjoint representation

Suppose $A \in PGL_n(\mathbb{R})$ lies in the kernel of the adjoint representation. Then for any lift $\tilde{A}$ of $A$ in $GL_n(\mathbb{R})$, and any traceless matrix $B$, we have $\tilde{A}B = B\t …
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What's the name of an algebra? is it isomorphic to $w_\infty \times w_\infty$?

If I'm not mistaken, this is the associative algebra of algebraic differential operators on the torus $\mathbb{G}_{m,\mathbb{C}}^2$. It is the tensor product of two copies of $w_\infty$, not the dire …
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2 votes

What is significant about the half-sum of positive roots?

If you have any free abelian group with an integral bilinear form embedded in the Lorentz space $\mathbb{R}^{n,1}$, you may consider the group of automorphisms generated by roots, i.e., reflections in …
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Kac Moody algebra defintion

I'll just elaborate on my comment from last year. A Kac-Moody algebra is defined by generators-and-relations, with starting data given by an $n \times n$ matrix $A$. It can be written as $L \rtimes …
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Several question on Affine Lie algebra

First question: Peter's (and Emerton's) argument works not only for affine algebras, but for arbitrary Kac-Moody algebras. Decompose your integrable representations into weights, and take the tensor …
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About localization theorem for affine Lie algebra?

The main problem seems to be that you think the global section functor for (twisted) D-modules on a singular variety depends on a choice of embedding into a smooth variety. This is not true - D-modul …
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2 votes

Invariant symmetric bilinear forms and H^4 of BG

One way to look at the invariant symmetric forms is by noting that they describe one dimensional central extensions of the loop algebra $L\mathfrak{g} = \operatorname{Maps}(S^1, \mathfrak{g})$. As a …
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Lie algabra of symmetric group

For any $n>1$, the lower central series for the symmetric group is $S_n > A_n\geq A_n \geq A_n \geq \cdots$, so the Lie ring formed by the sum of successive quotients is the group $\mathbb{Z}/2\mathbb …
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Lie group operation and tangent vectors

Here's another way to look at the problem. The derivative of a differentiable map at any point is a linear map of tangent spaces. We have five differentiable maps in play: The "pair of paths" map …
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twisted affine algebras

I don't have Kac's book next to me right now, but I was under the impression that representations of the affine algebra do not give rise to representations of the twisted algebra in any straightforwar …
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Is the centralizer of a torus in a Kac-Moody algebra always a Borcherds algebra?

The answer to your first question is "yes", and it follows from Theorem 1 in Borcherds's paper Central extensions of generalized Kac-Moody algebras, which is available online as number 11 on his paper …
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18 votes

About the definition of E8, and Rosenfeld's "Geometry of Lie groups"

The algebraic group $E_8$ is the group of automorphisms of the $E_8$ lattice vertex algebra, by Frenkel-Kac and Segal. This vertex algebra has a self-dual integral form, so the construction works ove …
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coset of affine Lie algebra

This is typically given by the commutant, or coset construction. You take the vector subspace of $\mathcal{R}_\text{vac}[\mathfrak{g}_k]$ spanned by vectors $v$ satisfying $Y(u,z)v \in \mathcal{R}_\t …
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2 votes
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Do rational points in a split reductive group act transitively on the orbits of the Cartan s...

Let $(G,T,M)$ be a split reductive group (over say, the integers), with Lie algebra $(\mathfrak{g}, \mathfrak{t})$, and let $R$ be a commutative ring. When $R$ is an algebraically closed field, it is …
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