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## Hot answers tagged lie-algebras

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 ...

5

Yes, Lie algebras (over any commutative ring) are the algebras of a monad over $\text{Set}$. This is true more generally for any kind of algebraic object defined by some operations satisfying some equational identities. The monad is built out of the free object functor. In other words, the forgetful functor $$\text{LieAlg} \to \text{Set}$$ has a left ...

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 $$Tr(ad(x)^2ad(y)^2)=(n-6)Tr(x^... 3 Your statement about the determinant of the diagonal matrices is wrong! \operatorname{det} \operatorname{diag}(a_1,\ldots, a_n)=\operatorname{det}\operatorname{diag}(a_2,a_1,a_3,\ldots,a_n)=\prod_{i=1}^n a_i. Re the addendum: it's not correct that a matrix S conjugating those two matrices must be a permutation matrix. Indeed, in the case n=2 the ... 2 The dimensions of the irreducible representations of \mathfrak{e}_8 are$$ 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

If I understand the question correctly, if we have any linear operator $A$ for which the exponential $\exp(A)$ is meaningful, then $A$ commutes with $-A$, and so $$\exp(A) \exp(-A) = \exp(A + (-A)) = \exp(\mbox{the zero operator}) = I.$$ The inverse of $(I + B)^{-1}$ (again, if the latter is defined) is $(I+B)$.

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 ...

1

In coordinates, you can write the vector fields, $$X_1 = f_1^i \frac{\partial}{\partial x^i}, Y_1= g_1^i \frac{\partial}{\partial y^i}$$ And similarly for $X_2$ and $Y_2$, where the $x^i$ are coordinates on $M$ and the $y^i$ on $N$. In these coordinates it's easy to see that $$X_j \oplus Y_j = f_j^i \frac{\partial}{\partial x^i} +g_j^i \frac{\partial}{... 1 For future reference, as it is remarked in Gerstenhaber's On the Deformation of Rings and Algebras (but not proved, as it is analogous to the associative case; thanks to @DietrichBurde for the reference, really nice paper, by the way) the correct complex is the Chevalley-Eilenberg complex of the Lie algebra \mathfrak{g} (with coefficients in itself), which ... 1 I don't know about using the Zassenhaus formula, but I did obtain the following expression for the terms that are first-order in d:$$\sum_{n\geq1}\frac{c^{n-1}t^n}{n!}\sum\limits_{r,s\geq0,\ \ r+s=n-1}\hat{X}^r\hat{Y}\hat{X}^s. $$Basically, expand e^{t(c \hat{X} + d \hat{Y})} as a power series:$$e^{t(c \hat{X} + d \hat{Y})}=\sum_{n=0}^\infty \frac{t^...

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