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

In summary: But not all J_{i} are linearly independent. In fact, they can be written in terms of the so-called (n-1)-dimensional generators J_{i}^{n} (which are independent) asJ_{ab} = \frac{1}{2}\sum_{i=1}^{(1/2)n(n-1)}J_{i}^{n}\epsilon_{iab} ,where \epsilon_{iab} are the structure constants of SO(n). Now, if you consider the commutator of a generic pair of elements of so(n) (i.e. of the generators J_{i}^{n
  • #1
spaghetti3451
1,344
33
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?
 
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  • #2
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.
 
  • #3
##\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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #8
failexam said:
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, [itex]\{|a\rangle \}, \ a = 1, 2, \cdots , n[/itex], which span the index space of tensors in [itex]SO(n)[/itex]. [tex]\langle b | a \rangle = \delta_{ab} .[/tex] Notice that normalization imposes [itex]n[/itex] real matrix constraints, and orthogonality leads to [itex](1/2)n(n-1)[/itex] real constraints. So, the number of independent real parameters, needed to specify the elements of [itex]SO(n)[/itex], is [tex]n^{2} - n - \frac{1}{2}n(n-1) = \frac{1}{2} n(n-1) .[/tex]
Let’s define the following hermitian antisymmetric operator [itex]J_{ab}[/itex] by
[tex]iJ_{ab} = |a\rangle \langle b| - |b\rangle \langle a | .[/tex] Notice that your generator [itex](A_{ab})_{mn}[/itex] is simply the matrix element of the operator [itex]J_{ab}[/itex] in the vector representation
[tex](A_{ab})_{mn} \equiv \langle m |J_{ab}|n \rangle = (-i) (\delta_{am} \delta_{nb} - \delta_{bm} \delta_{na}) .[/tex]
In fact, it is rather easy to show that the algebra of operators [itex]J_{ab}[/itex] is isomorphic to the Lie algebra [itex]so(n)[/itex].
[tex]
\begin{align*}
\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] \\
&+ \left[ |b\rangle \langle a| \ , \ |d\rangle \langle c| \right]-\left[ |a\rangle \langle b| \ , \ |d\rangle \langle c| \right] .
\end{align*}
[/tex]
Since
[tex]
\left[ |a\rangle \langle b| \ , \ |c\rangle \langle d| \right] = \delta_{cb} \ |a\rangle \langle d | - \delta_{ad} \ |c\rangle \langle b |,
[/tex]
you find
[tex]
(-i)\left[J_{ab},J_{cd}\right] = \delta_{ad}J_{cb} - \delta_{cb}J_{ad} + \delta_{ca}J_{bd} - \delta_{bd}J_{ca} .
[/tex]
Further more, if you write [tex]J_{(.)(.)} = \delta^{m}_{(.)} \ \delta^{n}_{(.)} \ J_{mn} ,[/tex] you get
[tex]
[J_{ab},J_{cd}] = i \ C^{mn}{}_{ab,cd} \ J_{mn} ,
[/tex]
where [tex]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) .
[/tex]
So, any element [itex]M \in SO(n)[/itex] can be written as exponent of a real linear combination of the generators [itex]J_{ab}[/itex] as follows
[tex]
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) .
[/tex]
 

What is a Lie algebra?

A Lie algebra is a mathematical structure that studies the algebraic properties of continuous symmetries. It is used extensively in physics and mathematics to describe the behavior of physical systems and to solve differential equations.

What is the significance of the so(n) Lie algebra?

The so(n) Lie algebra is one of the most important and widely used Lie algebras. It is the algebra of infinitesimal rotations in n-dimensional space and is used to study the dynamics of physical systems with rotational symmetry.

How do you derive the commutation relations of the so(n) Lie algebra?

The commutation relations of the so(n) Lie algebra can be derived using the structure constants of the algebra. These constants are defined by the structure equations, which relate the basis elements of the algebra to each other. By solving these equations, the commutation relations can be obtained.

What is the physical interpretation of the commutation relations of the so(n) Lie algebra?

The commutation relations of the so(n) Lie algebra have a physical interpretation in terms of angular momentum operators. These operators represent the rotational symmetries of a physical system and their commutation relations determine the properties of the system.

How are the commutation relations of the so(n) Lie algebra used in physics?

The commutation relations of the so(n) Lie algebra are used in physics to study the dynamics of physical systems with rotational symmetry. They are also used in quantum mechanics to describe the behavior of particles with spin. Additionally, they are used in the study of gauge theories and other areas of theoretical physics.

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