I am trying to understand how to compute the Lie Algebra of a Lie Group and am having a bit of difficulty in answering a question.
The question asks to show that a suitable basis of $\mathfrak{so}(n)$ can be given by: $$(T_{ab})_{c}^{d} = i(\delta_{ac}\delta_{b}^{d} – \delta^{d}_{a}\delta_{bc})$$ Where $a,b,c,d \in 1,\dots,n$. And $\mathfrak{so}(n)$ is the Lie algebra of SO(n).
I'm unsure if it suffices to show that the generators of the algebra are traceless and anti-symmetric and show that all possible such matrices are linear combinations of $T_{ab}$?
Best Answer
There are many ways of dealing with lie algebras. For instance, we can consider two possible descriptions:
$\bullet $ From lie group: every lie algebra $\mathfrak{g}$ is the tangent space of the corresponding lie group $G$ at the identity: $\mathfrak{g}\cong T_e(G)$. Its relation is given by the exponential map: If $g\in \mathfrak{g} \ $,$\ \exists \ T \in G$ such that $g = \exp[T]$.
In fact, the third Lie's theorem ensures that every Lie algebra comes from a lie group.
The other possible description is via the abstract definition:
$\bullet$ From the very definition of an abstract lie algebra: A vector space with a commutation relation satisfying Jacobi identity. From this definition it sufficies to define generatos $\{T_a\}_{a=1,\cdots, \dim(\mathfrak{g})}$, i.e., the basis of the vector space, and the commutation relations $[T_a,T_b]=f_{\ ab}^cT_c$, which gives rise to the so-called structure constants and, in fact, they caries the information about the commutation relation.
As you may know, the lie algebra $\mathfrak{so}(n)$ carries this name because it is the lie algebra that comes from the $SO(n)$ lie group. Therefore, if $T\in\mathfrak{so}(n)$, $(e^T)^{tr}=(e^T)^{-1}$ and $\det(e^T)=1$, which implies respectively $T^{tr}=-T$ and from $\det(e^T)=e^{trace(T)}$, that $trace(T)=0$. This is the reason why $\mathfrak{so}(n)$ is the algebra of traceless anti-symmetric matrices.
This is fine. Now we can ask why $(T_{ab})^d_c = \delta_{ac}\delta^d_b - \delta^d_a\delta_{bc}$ is really a base for this algebra. To answer this, we first notice the notation. The $ab$ indices just organize the generators as entries of a $n\times n$ anti-symmetric matrix. In fact, as you can see, $T_{ab}=-T_{ba}$. It will be easily proven that $\{T_{ab}\}_{1\le a<b \le n}$ is a Linearly Independent set. As you can compute, there are $n(n-1)/2$ independent entries on an anti-symmetric matrix, and then, $T_{ab}$ will correspond to the $n(n-1)/2$ independent generators of the $\mathfrak{so}(n)$ algebra, that in fact has dimension $n(n-1)/2$.
The $cd$ are the matrix indices of the corresponding $T_{ab}$ matrix generator. All the $n(n-1)/2$ matrices $T_{ab}$ are anti-symmetric and traceless:
$$ (T_{ab})^c_{\ d} = \delta_{ad}\delta^c_b - \delta^c_a\delta_{bd} = -(\delta_{ac}\delta^d_b - \delta^d_a\delta_{bc} ) = -(T_{ab})^d_{\ \ c}$$
$$ trace(T_{ab}) = (T_{ab})^c_{\ c} = \delta_{ac}\delta^c_b - \delta^c_a\delta_{bc}=\delta_{ab}-\delta_{ab}=0$$
It remains to show that all the $n(n-1)/2$ generators $T_{ab}$ are linearly independent. Consider fixed $a,b$ such that $a<b$. From the formula, follows that $(T_{ab})^{\ \ b}_{a}=1$, $(T_{ab})^{\ \ a}_{b}=-1$ and the rest is zero. It is obvius that all the $T_{ab}$ for all $1\le a < b \le n$ are linearlly independent - in fact, given $a<b$, $(T_{a'b'})^a_b = 0$ for all $(a',b') \ne (a,b)$ and $a'<b'$.
Follows that $\{T_{ab}\}_{1\le a<b \le n}$ is a basis to $\mathfrak{so}(n)$. It is worthwhile to remember that $\{T_{ab}\}_{1\le a=b \le n}=0$ and $\{T_{ab}\}_{1\ge a>b \ge n}=-\{T_{ab}\}_{1\le a<b \le n}$, that is, the set $T_{ab}$ has only $n(n-1)/2$ L.I. vectors.
It was not asked, but it is important to consider why the generators are writen like this product of deltas. In fact, to answer this will give us the spirit of why the most general element of the lie algebra is like this.
Consider an orthogonal transformation on $\mathbb{R}^n$: $M \in GL(n,\mathbb{R})\ \ s.t. \ \ MM^T=1$. Then $M = e^{m}$ where $M\in SO(n)$ and $m \in \mathfrak{so}(n)$. It follows that $M = 1 + m + \cdots \approx 1 + m$. We know that $m$ is an anti-symmetric matrix. Now consider a small transformation on $x \in \mathbb{R}^n$
$$ x^\mu \longmapsto (\delta^\mu_{\ \ \nu} + m^\mu_{\ \ \nu})x^\nu $$ $$ \delta x^\mu = m^\mu_{\ \ \nu} x^\nu = \delta^{\mu\alpha}\delta^{\beta\nu}m_{\alpha \nu}x_\beta =\frac{1}{2}(\delta^{\mu\alpha}\delta^{\beta\nu}- \delta^{\mu\beta}\delta^{\alpha\nu})m_{\alpha \nu}x_\beta $$
In the last equality, we use the fact that $m_{\mu\nu}$ is anti-symmetric and now it becomes just a parameter. The real generators are indexed by the new indices $\alpha, \beta$:
$$ (T^{\alpha\beta})^\mu_{\ \ \nu} = \delta^{\mu\alpha}\delta^{\beta}_{\ \ \nu}- \delta^{\mu\beta}\delta^{\alpha}_{\ \ \nu} $$