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10

If $[x,e]=[e,x]$, then what does the antisymmetry of the Lie bracket tell you?


7

It would also contradict Jacobi's identity. If your Lie algebra is commutative, then clearly you can't have that $[x,e]=x$ unless $x=0$ since $[x,e]$ would have to be $0$. If it is non-commutative (meaning that for some $x,y$ in the Lie algebra we have that $[x,y]\neq 0$), then Jacobi's identity would say that $$0 = [x,[y,e]] + [y,[e,x]] + [e,[x,y]] = [x,y] ...


5

$$[x+y,x+y]=0=[x,x]+[x,y]+[y,x]+[y,y]=[x,y]+[y,x]\Rightarrow [x,y]=-[y,x]$$


4

For every field $K$ there are just two different $2$-dimensional Lie algebras. The first one is $K^2$, which is abelian (which means the bracket is zero). The second one is solvable, non-nilpotent, hence non-abelian, and the bracket of the basis $(x,y)$ is an arbitrary nontrivial linear combination of $x$ and $y$, i.e., $[x,y]=\alpha x+\beta y$ for some ...


4

We want to show that $$ \tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0} $$ We can start doing this by computing: \begin{eqnarray*} \left.\frac{d}{dt}\left(R_1(t)\Omega_2 R_1(t)^{-1}\right)\right|_{t=0} &=& \left.\left[\frac{d}{dt}R_1(t)(\Omega_2 R_1(t)^{-1})\right]\right|_{t=0} + \left.\left[(R_1(t)\Omega_2) ...


3

By definition, the Lie bracket of a Lie algebra is bilinear (see https://en.wikipedia.org/wiki/Lie_algebra), which means when you fix $v \in \mathfrak{g}$ the map $[-,v]: \mathfrak{g} \rightarrow \mathfrak{g}$ is a linear map. That is for any $w_1 , w_2 \in \mathfrak{g}$ and any $a \in \mathbb{C}$ we have the identity $[w_1 + a w_2,v] = [w_1,v] + ...


3

Since $\beta$ is a positive root, it is a positive linear combination of some of the $\alpha$'s. Since $\beta \ne \alpha_i$, then $\beta$ has some component $\alpha_j$ with positive coefficient where $j \ne i$ (since higher multiples of $\alpha_i$ cannot be roots). Therefore each element of $\beta + \mathbb{Z}\alpha_i$ has a component $\alpha_j$ with ...


3

If $[e,x]=x$ for all $x$, what can you conclude from $0=[e,e]$?


3

A compact Lie algebra $\mathfrak{g}$ for a compact Lie group $G$ always admits an $Ad(G)$-invariant inner scalar product, see Propositioon $4.24$ here. A compact Lie algebra need not be semisimple in general, but it is always reductive, i.e., a direct sum of the semisimple commutator ideal and the center. Hence the Killing form is negative semidefinite - see ...


3

The term representation comes from representation theory. In particular, Ad gives us a group representation, and ad gives us a Lie-algebra representation. In particular: given a Lie group $G$ and $A,B \in G$, $\operatorname{Ad}$ is a map from $G$ to $GL(\mathfrak{g})$ such that $$ \operatorname{Ad}_A \operatorname{Ad}_B = \operatorname{Ad}_{A B} $$ ...


3

no just consider the 1 by 1 case. i.e. real numbers. Let X=10 and the series doesn't converge.


2

The problem is explained here as follows: the $8$-dimensional real Lie algebra $\mathfrak{su}(3)$ has a basis $\lambda_1,\ldots ,\lambda_8$ as above, consisiting of Hermitian matrices. However, the Lie bracket of this subspace of $3\times 3$ matrices is not given by the commutator $AB-BA$, because the commutator of two Hermitian matrices is not Hermitian ...


2

Try looking at Vilenkin's introductory 1968 book on Representation Theory and Special Functions. Then look at his 3 volumes on the same subject with a more detailed treatment.


2

Since the Lie algebra $\mathfrak{g}=\mathfrak{gl}(2,K)$ is reductive, it is the direct sum of the commutator ideal and the center, i.e., we have $$ \mathfrak{gl}(2,K)=\mathfrak{sl}(2,K)\oplus Z(\mathfrak{gl}(2,K)). $$ Since $\dim \mathfrak{sl}(2,K)=3$, we obtain $\dim Z(\mathfrak{gl}(2,K))=1$. Because $I$ is in the center, it is a generator of the ...


2

A consideration of $\mathfrak{gl}(n,\Bbb F)$: Let $E_{ij} = e_i e_j^T$ denote the matrix with a $1$ in the $i,j$ entry. Let $A$ be a matrix with entries $a_{ij}$. We have $$ AE_{ij} = (Ae_i)e_j^T\\ E_{ij}A = e_i(e_j^TA) $$ Now, if $A$ is in the center, we must have for every $p,q$: $$ e_{p}^T[A,E_{ij}]e_q = 0 \implies\\ e_p^T\left((Ae_i)e_j^T - ...


2

Suppose $n>2$ and $[(a_1,\dots,a_n)]$ can be expressed (for general $a_1,\dots,a_n$) as $\mathbb Q$-linear combinations of brackets. This is equivalent to say $[(a_1,\dots,a_n)]$ is in the Lie algebra generated by $a_i$'s in $R:=\mathbb{Q}\langle a_1,\dots,a_n\rangle$ (the free $\mathbb Q$-algebra generated by $a_1,\dots,a_n$). But there is a standard ...


2

Your approach works without problems, if you write the condition $[Ax,Ay]=A[x,y]$ for all $x,y$ in terms of the $9$ coefficients of the matrix $A$. The polynomial equations in these $9$ unknowns over $\mathbb{R}$ quickly yield $\det(A)=0$, a contradiction. Another elementary argument is the following. $\mathfrak{sl}(2,\mathbb{R})$ has a $2$-dimensional ...


2

The answer to your question is, that for matrices $A$ and $B$ in $M_n(K)$ the standard Lie bracket is the commutator, i.e., $[A,B]=AB-BA$. The skew-hermitian matrices form a Lie subalgebra under this bracket, the unitary Lie algebra. For the commutator of Hermitian matrices see here. Then the usual commutator does not give a Lie subalgebra (the commutator of ...


2

Formula (1) is a bit strange indeed, $R_1(t)\Omega_2 R_1(t)^{-1}$ is not necessarily in $SO(3)$ ($\Omega_2$ could have a null determinant for example). One way to prove the result, is to consider that (for $t \to 0$): $$ R_1(t) = I + t \Omega_1 + t^2 W_1 + o(t^2)$$ $$ R_2(t) = I + t \Omega_2 + t^2 W_2 + o(t^2)$$ Then: $$ R_1(t)R_2(t) = I + t (\Omega_1 + ...


1

Here is a possibility as to how the professor was arriving at the result that $[\Omega_1, \Omega_2]$ is an element of the Lie algebra. Suppose that we take the above formula as the definition of the Lie Bracket: $$ \tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0} $$ There is a proposition that for any Lie group element ...


1

Found a solution, but it has nothing to do with $(1)$: $$ \begin{align} [\Omega_1,\Omega_2] & = \Omega_1\Omega_2-\Omega_2\Omega_1 = (-\Omega_1^T)(-\Omega_2^T) - (-\Omega_2^T)(-\Omega_1^T) \\ & = \Omega_1^T\Omega_2^T-\Omega_2^T\Omega_1^T=(\Omega_2\Omega_1)^T-(\Omega_1\Omega_2)^T = (\Omega_2\Omega_1-\Omega_1\Omega_2)^T \\ & = -[\Omega_2,\Omega_1] ...


1

$$\def\g{\mathfrak{g}} \def\h{\mathfrak{h}}$$ Let $\{e_{i}, f_{i}\}_{i=1}^{n}$ be the standard generators of the Kac Moody algebra $\g$ and let $\alpha_{1}^{\vee}, \ldots, \alpha_{n}^{\vee}$ be the simple coroots in its Cartan subalgebra $\h$. Then, by the defining relations of the Kac Moody algebra, the derived subalgebra $\g'(A)$ is the Lie subalgebra ...


1

This is a Q&A style answer not meant to be the final answer to the question. It completes the original technique for future readers. Thanks to Dietrich Burde for the motivation to continue with it. As above, suppose $A$ is an isomorphism from $\mathfrak{su}(2) \to \mathfrak{sl}(2, \mathbb{R})$. Then $$[AX, AY] = A[X, Y]$$ Let $A_i$ denote the column ...


1

It's not very hard to just do this explicitly, along the lines you started but keeping my comment in mind. Rather than $h, e, f$ I'll write $H, X, Y$. Suppose $I$ is a nonempty ideal. Then it contains some nonzero element $aH + bX + cY$. Applying $[H, -]$ to this element gives $$b [H, X] + c [H, Y] = 2b X - 2c Y$$ from which we conclude that either $a = 0$ ...


1

Yes, it can be confusing talking about modules one moment and then talking about representations the next. In your case there isn't much difference. In fact, the confusion shouldn't confuse you. A Lie algebra module is a vector space such that ... A Lie algebra module is irreducible is there are no invariant (non trivial) proper submodules. Since the ...


1

By definition, the dimension of $sl_n(K)$ is the dimension of the vector space of traceless $n\times n$ matrices. Just consider an obvious basis of this vector space. Take all matrices $E_{ij}$ with $i\neq j$, having entry $1$ at position $(i,j)$ and $0$ elsewhere, and $H_{i}=E_{ii}-E_{jj}$ for $i\neq j$. So we have $n^2-n$ matrices of the first kind, and ...


1

As I explained in the answer to this question, the Lie bracket of left-invariant vector fields is $[X,Y]=XY-YX$, whereas the Lie bracket of right-invariant vector fields has the opposite sign. (It's not a matter of convention. It's a computation of the Lie derivative in either case.) I don't know where your lecture got the second expression for the Lie ...


1

In general, given a subalgebra $\mathfrak{h}$ of $\mathbb{gl}_n(K)$ consisiting of matrices, one needs to compute all commutators of a basis $f_1,\ldots ,f_m$ of the vector space $\mathfrak{h}$. Using the canonical basis $E_{ij}$ of $\mathbb{gl}_n(K)$, there are formulas available for the commutators $[E_{ij},E_{kl}]$, i.e., $$ ...


1

Simply that with such definitions you have $[x,y]=-x$ (so that the equality $-x=-[x,y]$ is false). You can easily find at this point an isomorphism in terms of matrices to the standard form you prefer.


1

The space $N$ is called the normalizer of $A$ in $\mathfrak{g}$, and it is the largest subalgebra of $\mathfrak{g}$ containing $A$ as an ideal. If $A$ is already an ideal in $\mathfrak{g}$, then $N_{\mathfrak{g}}(A)=\mathfrak{g}$ of course. If $N_{\mathfrak{g}}(A)=A$, then we only know that $A$ is self-normalising.



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