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6

Solution 1. Take any generator of $\text{SO}(N)$. It corresponds, in the fundamental representation, to rotating a plane defined by two of the coordinates. By rotating or relabeling these coordinates appropriately, we can redefine the generator to rotate $x^1$ and $x^2$. The generator can then be taken as $\sigma_2$, in the vector representation, with zeros ...

3

It is well-known that the Killing form of $\mathfrak{so}(n)$ satisfies $$Tr (ad(x)ad(y))=(n-2)Tr(xy),$$ see here. The first formula comes form setting $T=ad(x)=ad(y)$ in the Killing form above. For elements $x$ in the Cartan subalgebra we know that $ad(x)$ is a diagonal matrix. This makes a computation easy. We also obtain $$Tr(ad(x)^2ad(y)^2)=(n-6)Tr(x^... 2 Let V be the defining 2-dimensional representation of SL_2(\mathbb{R}). It has symmetric powers S^n (V) for all positive integers n, and you can verify that when n is even, the kernel of S^n(V) is precisely \pm 1, so all of these are faithful linear representations of PSL_2(\mathbb{R}). More generally, if G is any connected Lie group, ... 2 The dimensions of the irreducible representations of \mathfrak{e}_8 are$$ 1, 248, 3875, 27000, 30380, 147250, 779247, 1763125, 2450240, 4096000, 4881384, 6696000, 26411008, 70680000, 76271625, 79143000, 146325270, 203205000, 281545875, 301694976, 344452500, 820260000, 1094951000, 2172667860, 2275896000, 2642777280, 2903770000, 3929713760, 4076399250, ...

2

A 1-dimensional represenation $V$ is always invertible, in the sense that there is another representation $V^{-1}$ such that $V\otimes V^{-1}$ and $V^{-1}\otimes V$ are both isomorphic to the trivial representation. Indeed, you can tale $V^{-1}$ equal to the dual (that is, contragredient) repreentation of $V$. It follows from that that if $V\otimes W$ is ...

2

Let $H \subset G$ be a closed subgroup. First: $C_G(H)$ is positive-dimensional if and only if the adjoint representation has nontrivial fixed points when restricted to $H$. To see this, consider the tangent space $\mathfrak{c_h} \subset \mathfrak g$; I claim that the action of $H$ preserves this. This is nothing particularly clever; if $h \in H$, then by ...

2

This is exercise (2.17) of Isaacs' Character Theory of Finite Groups. With Lemma (2.29) of that book you can see that $\chi$ vanishes outside $H$. Now look at the subgroup $N = \langle g \in G : \chi(g) \neq 0 \rangle$. Obviously $N \subseteq H$ hence $N$ is abelian. This subgroup is normal (conjugation does not change the character value) and non-trivial (...

1

I think that if you define things properly, you will see that the statement is obvious. First of all, let $P_n=F[x_1,\ldots,x_n]$ be the polynomial ring in $n$-variables (here $F$ is an arbitrary field, but could also be taken to be any commutative ring). There is a surjective homomorphism $P_{n+1}\twoheadrightarrow P_n$ determined by the mapping \$f(x_1,\...

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