What are all the generators of SU(5)?

[Anchovy]

I'm trying to find out what all the generators of the SU(5) group explicitly look like but I can't find them anywhere.
I know what the first 12 look like:

$$T^{1,2,3} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \sigma^{1,2,3} & \\ 0 & 0 & 0 & & \end{pmatrix},<br /> <br /> T^{4} = \begin{pmatrix} \frac{-1}{3} & 0 & 0 & 0 & 0 \\ 0 & \frac{-1}{3} & 0 & 0 & 0 \\ 0 & 0 & \frac{-1}{3} & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} \end{pmatrix},<br /> <br /> T^{\alpha = 5,...,12} = \begin{pmatrix} \lambda^{\alpha - 4} & & & 0 & 0 \\ & & & 0 & 0 \\ & & & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

where $\sigma^{1,2,3}$ are the Pauli matrices of SU(2) and $\lambda^{1,2,3,4,5,6,7,8}$ are the Gell-Mann matrices of SU(3). However, there are another 12 that I can't seem to find anywhere:
$$T^{\alpha=13,...,24} = \begin{pmatrix} 0 & 0 & 0 &m ^{13+n} & \\<br /> <br /> 0 & 0 & 0 & & \\<br /> <br /> 0 & 0 & 0 & & \\<br /> <br /> (m^{13+n})^{\dagger} & & & 0 & 0 \\<br /> <br /> & & & 0 & 0 \end{pmatrix} (n = 0, 1, ..., 11)$$

where $m^{13 + n}$ are $3\times2$ matrices. Can anyone show me?

 

[Orodruin]

I'm trying to find out what all the generators of the SU(5) group explicitly look like but I can't find them anywhere.

It is unclear what you mean by this. Any Hermitian traceless matrix is in the Lie algebra of SU(5). In order to have a complete set of generators you only need to write down a basis for this Lie algebra. The 3x2 matrices you are looking for are general complex matrices and are therefore spanned by 6x2 = 12 different generators.

 

[Anchovy]

It is unclear what you mean by this. Any Hermitian traceless matrix is in the Lie algebra of SU(5). In order to have a complete set of generators you only need to write down a basis for this Lie algebra. The 3x2 matrices you are looking for are general complex matrices and are therefore spanned by 6x2 = 12 different generators.

What I meant was, I'm wondering what those 3x2 matrices actually look like?

 

[Anchovy]

Ah I finally found what I was looking for (attached in case anyone searching for this might want it).

 

[Orodruin]

What I meant was, I'm wondering what those 3x2 matrices actually look like?

The 3x2 matrices are arbitrary complex matrices. They therefore form a 12-dimensional vector space and you can chose any set of basis vectors in this space which are linearly independent. The choice of basis in the document you found is just one possibility (just as the Pauli matrices are just one possibility for a basis in the Lie algebra of SU(2)), although perhaps the one most people would use.

 

[fzero]

I think what Oroduin was getting at, was, if you're already clever enough to use Gell-Mann matrics to cover the 3x3 block and Pauli matrices to cover the 2x2 block, then you only have the 2x3 blocks on the off-diagonal left. So you need a basis for 12 Hermitian matrices. Easiest to use are ones that are analogous to $\sigma_1$ and $\sigma_2$. Namely, there are 6 real matrices with a single element and its transpose image equal to 1, as well as 6 matrices with an $i$ in one element and $-i$ in the transpose image. I'm going to go out on a limb and guess that is precisely what your reference does.

 

[Anchovy]

That is what the reference does yes thanks.

 

[ChrisVer]

Think what you did with the Gellmann matrices:
$\lambda^{3} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 &0 \end{pmatrix}~~,~~\lambda^{8} =\frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &-2 \end{pmatrix}$
Diagonal elements.

And then you took the $\sigma^i$ and starting moving their entries along the off-diagonal positions. Eg $\sigma^1$:
$\lambda^1 = \begin{pmatrix} 0& 1 & 0 \\ 1 & 0 &0 \\ 0& 0 &0 \end{pmatrix} ~~,~~ \lambda^4 = \begin{pmatrix} 0& 0 & 1 \\ 0 & 0 &0 \\ 1& 0 &0 \end{pmatrix} ~~,~~\lambda^6 = \begin{pmatrix} 0& 0 & 0 \\ 0 & 0 &1 \\ 0& 1 &0 \end{pmatrix}$

You can do the same for the SU(5)... In fact you have $\sigma^{1,2}$ (2) off diagonal entries, which can be moved in $\frac{5^2-5}{2}=10$ positions. That is 20 , your T^a with a=5,...,24.
Of course that's in one basis, what you can also do is change your basis and complexify them... (in particular combine with $\pm i$ the ones that you made with $\sigma^1$ and those you made with $\sigma^2$).That way, as in SU(3), you can create the independent SU(2) subgroups.