Deriving the commutation relations of the so(n) Lie algebra

[spaghetti3451]

The generators $(A_{ab})_{st}$ of the $so(n)$ Lie algebra are given by:

$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$,

where $a,b$ label the number of the generator, and $s,t$ label the matrix element.

Now, I need to prove the following commutation relation using the definition above:

$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$.Here's my attempt.

$([A_{ij},A_{mn}])_{st}$
$= (A_{ij})_{sp}(A_{mn})_{pt}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j]p}\delta_{p[m}\delta_{n]t}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$

Could you please suggest the next couple of steps? Should I expand all the antisymmetrised Kronecker delta's, or is there some sneaky shortcut to get to the answer?

 

[fresh_42]

I don't know the difference between $δ_{as}$ and $δ_{a|s}$. But setting $(m,n) = (i,j)$ doesn't seem $[A_{ij},A_{mn}]$ to be $0$.
But I may be wrong. I'd use another notation.

 

[spaghetti3451]

$\delta_{a|s}$ is meaningless - I did not use a vertical bar $|$; rather, I used a square bracket.

$(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = \delta_{s[a}\delta_{b]t}$ gives the definition of the square brackets $[$ and $]$.

$\delta_{s[a}\delta_{b]t}$ means that you first take $\delta_{as}\delta_{bt}$ and then flip the order of the indices $a$ and $b$ to form $\delta_{bs}\delta_{at}$. Then , you subtract one from the other.I did not set $(m,n) = (i,j)$. I simply switched $(m,n)$ with $(i,j)$ and vice-versa in the second term. I did this because the definition of the commutators is such that the indices are bound to switch places in the second term.

 

[fresh_42]

I asked you to explain $δ_{a|b}$ and not $δ_{a|s}δ_{b|t}$ because you don't always use it paired.
However, if I interpret $A_{i|j}δ_{m|n}$ which you did not define neither as $A_{ji}δ_{mn} - A_{jn}δ_{mi}$ according to your pairing with the $δ$ I still don't get, e.g.$[A_{12},A_{12}] = 0$ which it should. But the index salad confuses me. I might have been wrong. Nevertheless you should check this first.

 

[spaghetti3451]

All the statements in my original post are taken straight out of Bincer's 'Lie Groups and Lie Algebras' Chapter 5 page 42.

I just need to fill in the last couple of steps which are not in the textbook.

If anyone in the forum can provide a hint, that will be helpful.

 

[JorisL]

When in doubt expand the brackets.
It'll make the calculation tedious but also easier to track any mistakes.

Once you've done such a thing several times it will get more natural.

I haven't looked at the details as I'm getting ready for bed, I'll revisit this tomorrow if I find the time.

@fresh_42 he uses the short-hand as defined here: https://en.wikipedia.org/wiki/Antisymmetric_tensor#Notation
Minus the prefactor that is. It is quite common and powerful.

 

[fresh_42]

Phh ... coordinates ... @JorisL thank you

 

[samalkhaiat]

The generators $(A_{ab})_{st}$ of the $so(n)$ Lie algebra are given by:

$(A_{ab})_{st} = -i(\delta_{as}\delta_{bt}-\delta_{at}\delta_{bs}) = -i\delta_{s[a}\delta_{b]t}$,

where $a,b$ label the number of the generator, and $s,t$ label the matrix element.

Now, I need to prove the following commutation relation using the definition above:

$([A_{ij},A_{mn}])_{st} = -i(A_{j[m}\delta_{n]i}-A_{i[m}\delta_{n]j})_{st}$.


Here's my attempt.

$([A_{ij},A_{mn}])_{st}$
$= (A_{ij})_{sp}(A_{mn})_{pt}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j]p}\delta_{p[m}\delta_{n]t}-(ij \iff mn)$
$= -\delta_{s[i}\delta_{j][m}\delta_{n]t}+\delta_{s[m}\delta_{n][i}\delta_{j]t}$

Could you please suggest the next couple of steps? Should I expand all the antisymmetrised Kronecker delta's, or is there some sneaky shortcut to get to the answer?

It is much easier for you to work with the abstract generators than with the matrix elements of the vector (fundamental) representation. The column vectors of a real orthogonal matrix are ortho-normal to each other. That is, there exists a complete orthomormal set of basis vectors, $\{|a\rangle \}, \ a = 1, 2, \cdots , n$, which span the index space of tensors in $SO(n)$. $$\langle b | a \rangle = \delta_{ab} .$$ Notice that normalization imposes $n$ real matrix constraints, and orthogonality leads to $(1/2)n(n-1)$ real constraints. So, the number of independent real parameters, needed to specify the elements of $SO(n)$, is $$n^{2} - n - \frac{1}{2}n(n-1) = \frac{1}{2} n(n-1) .$$
Let’s define the following hermitian antisymmetric operator $J_{ab}$ by
$$iJ_{ab} = |a\rangle \langle b| - |b\rangle \langle a | .$$ Notice that your generator $(A_{ab})_{mn}$ is simply the matrix element of the operator $J_{ab}$ in the vector representation
$$(A_{ab})_{mn} \equiv \langle m |J_{ab}|n \rangle = (-i) (\delta_{am} \delta_{nb} - \delta_{bm} \delta_{na}) .$$
In fact, it is rather easy to show that the algebra of operators $J_{ab}$ is isomorphic to the Lie algebra $so(n)$.
$$\begin{align*}<br /> \left[iJ_{ab},iJ_{cd}\right] &= \left[ |a\rangle \langle b| \ , \ |c\rangle \langle d| \right]-\left[ |b\rangle \langle a| \ , \ |c\rangle \langle d| \right] \\<br /> &+ \left[ |b\rangle \langle a| \ , \ |d\rangle \langle c| \right]-\left[ |a\rangle \langle b| \ , \ |d\rangle \langle c| \right] .<br /> \end{align*}$$
Since
$$\left[ |a\rangle \langle b| \ , \ |c\rangle \langle d| \right] = \delta_{cb} \ |a\rangle \langle d | - \delta_{ad} \ |c\rangle \langle b |,$$
you find
$$(-i)\left[J_{ab},J_{cd}\right] = \delta_{ad}J_{cb} - \delta_{cb}J_{ad} + \delta_{ca}J_{bd} - \delta_{bd}J_{ca} .$$
Further more, if you write $$J_{(.)(.)} = \delta^{m}_{(.)} \ \delta^{n}_{(.)} \ J_{mn} ,$$ you get
$$[J_{ab},J_{cd}] = i \ C^{mn}{}_{ab,cd} \ J_{mn} ,$$
where $$C^{mn}_{ab,cd} = \left(\delta_{ad}\delta^{m}_{c}\delta^{n}_{b} - (a,d) \leftrightarrow (c,b) \right) + \left(\delta_{ca}\delta^{m}_{b}\delta^{n}_{d} - (c,a) \leftrightarrow (b,d) \right) .$$
So, any element $M \in SO(n)$ can be written as exponent of a real linear combination of the generators $J_{ab}$ as follows
$$M = \exp \left(\frac{i}{2}\sum_{a,b}^{n}\omega^{ab}J_{ab}\right) =\exp \left(\sum_{i}^{(1/2)n(n-1)}i \alpha_{i}J_{i}\right) .$$