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Yes. The basic representation of $E_8$ has character $j(\tau)^{1/3} = q^{-1/3}(1+248q+4124q^2 + \cdots)$, and the 4124 decomposes as $1+248+3875$. By Frenkel-Kac-Segal, the basic representation has …

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 …

No. Let $G$ be $O_2(\mathbb{R})$, so the Lie algebra is one dimensional, and the center of the universal enveloping algebra is the symmetric algebra of the Lie algebra. Then the usual 2-dimensional …

Here is a counterexample: Let $a,b$ be distinct units in $k$ such that $ab \neq 1$, and let $G = SL_2 \times SL_2$ be given the usual block diagonal embedding into $GL_4$. Then the matrices $$A = \ …

This is not a complete answer, but it's a start. In the $2 \times 2$ case, the act of choosing the first column and clearing denominators describes a two-to-one map from orthogonal matrices to primit …

The answer to your question is "no". I've rewritten my previous answer to include details. In dimension $d$ at least 7, there are continuous positive-dimensional (non-isotrivial) families of nilpote …

You can extract an answer from the Wikipedia article (and the Diaconis-Shahshahani paper that is referenced). First, find/write a function that yields a uniform distribution of points on a unit spher …

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 …

We can say something stronger. Theorem: (Helling 1976) Consider the family of subgroups of $SL_2(\mathbb{C})$ that are commensurable with a conjugate of $SL_2(\mathbb{Z})$. The maximal elements o …

The answer is "yes". This is (the compact version of) Proposition 2.20 b on page 139 of the Deligne-Milne article on Tannakian categories in Hodge cycles, Shimura Varieties, and Motives (Springer LNM …

People are interested in the affine Grassmannian because it has nice properties that let you prove theorems. Sometimes, the full flag variety does not have the same nice properties. Perhaps most imp …

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 …

If we fix universal covering maps $H \to H_1$ and $H \to H_2$, then $G$ is uniquely defined as the image of the diagonally embedded $H \subset H \times H$ under the covering map to $H_1 \times H_2$. …

As Peter mentioned, reductive groups are determined by their root data, and the Langlands dual is given by switching weights with coweights, and roots with coroots. There is a "construction" of a gro …