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On the Wikipedia page on Gell-Mann matrices, it is said, that the $su(3)$ algebra has 3 independent $su(2)$ subalgebras ($\lambda_n$ denotes $n$th Gell-Mann matrix) $$\left\{\lambda_1,\lambda_2,\lambda_3\right\}\\ \left\{\lambda_4,\lambda_5,x\right\}\\ \left\{\lambda_6,\lambda_7,y\right\},$$ where $x$ and $y$ are some linear combinations of $\lambda_3$ and $\lambda_8$.

My question is: what are the independent $su(3)$ subalgebras in the $su(4)$ algebra? $$ \lambda_1 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_2 = \begin{pmatrix} 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_3 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ \lambda_4 = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_5 = \begin{pmatrix} 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_6 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ \lambda_7 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_8 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_9 = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}\\ \lambda_{10} = \begin{pmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \end{pmatrix} ,\qquad \lambda_{11} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} ,\qquad \lambda_{12} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \end{pmatrix} \\ \lambda_{13} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} ,\qquad \lambda_{14} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \end{pmatrix} , \qquad \lambda_{15} = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix} .$$

My attempt

Let us redenote the $su(4)$ generators as $\lambda_n$.

A first subalgebra can be seen pretty easily, $$\left\{\lambda_1,\lambda_2,..,\lambda_8\right\}.$$ Then, by analogy, one can see, that the 3-rd and 8-th Gell-Mann matrices are considered "colorless" in QCD, so we need to use $\lambda_{15}$ in the linear combination.

At this point I'm stuck, by analogy one can continue that $\left\{\lambda_9,\lambda_{10},..,\lambda_{14},x,y\right\}$.

With $x$ and $y$ being independent linear combinations of $\lambda_3$, $\lambda_8$ and $\lambda_{15}$, but I don't know how to prove it.

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  • $\begingroup$ @CosmasZachos would that help if I say, that I use the basis shown here: es.wikipedia.org/wiki/Grupo_unitario_especial $\endgroup$ Jul 17, 2023 at 22:10
  • $\begingroup$ Use Chapter 5 of this tutorial. $\endgroup$ Jul 18, 2023 at 0:12
  • $\begingroup$ Note "independent" is a mirage/exaggeration in your introduction. The 9 generators you wrote down are linearly dependent! The Cartan subalgebra is two dimensional: the third ones $x,y,\lambda_3$, are dependent, necessarily. $\endgroup$ Jul 18, 2023 at 0:40

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You can do this by looking at the root diagrams of $su(3)$ and $su(4)$.

First, start with $su(2)$ inside $su(3)$. The root diagram of $su(3)$ is enter image description here

with the fundamental roots $\alpha,\beta$ at $120^\circ$. (Image credit: Mark L MacDonald via Wikipedia)

The $su(2)$ subalgebras are spanned by opposite roots so three different subalgebra as spanned by $\{E_\alpha,E_{-\alpha},[E_\alpha,E_{-\alpha}]\}$, $\{E_\beta,E_{-\beta},[E_\beta,E_{-\beta}]\}$ and $\{E_{\alpha+\beta},E_{-\alpha-\beta},[E_{\alpha+\beta},E_{-\alpha-\beta}]\}$.

If I understand your question well, you now need to do a similar thing with $su(3)$ inside $su(4)$. In $su(4)$ there is a third simple root $\gamma$ that is also at $120^\circ$ from $\alpha$ and $\beta$. Thus you can form an $su(3)$ with simple roots $(\alpha,\beta)$ (this is the one of the figure and isomorphic to the one you have spanned by $\{\lambda_1,\ldots,\lambda_8\}$), $(\alpha,\gamma)$ and $(\beta,\gamma)$.

Note that root vectors are not hermitian but raising operators and so correspond to combinations such as $\lambda_1\pm i\lambda_2$. You can work on translating the root vectors into $\lambda$ matrices.

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Background: "independent" in your question is a slippery loose term, as it cannot mean "linearly independent" in the 3-vector sense with the dot product defined by the trace! This is obvious since there are only 8 independent such vectors in su(3), and so the sets of 9 vectors you wrote must be all, together, linearly dependent. Due diligence: compute x and y, $$ x=\frac{\sqrt{3}\lambda_8+\lambda_3}{2}, \qquad y=\frac{\sqrt{3}\lambda_8-\lambda_3}{2},\implies \\ \operatorname{Tr}(\lambda_3 x)=1= \operatorname{Tr}( x y)=-\operatorname{Tr}(\lambda_3 y). $$ So, not only do these three algebras fail to mutually commute, but they are not linearly independent as 3-vectors; not even two of them are independent, since their generators, $\lambda_3,x,y$, in the su(3) two-dim Cartan subalgebra must overlap!

In any case, if you wished to pursue this "monkey-see-monkey-do" game for su(4), for whatever undisclosed purpose, you might consider the following for its 15 generators.

Observe you may choose the sets lacking the first row and column, just as you did the ones lacking the last row and column in your first choice, $$\left\{\lambda_1,\lambda_2,..,\lambda_8\right\}.$$ So you have the obvious su(3) subalgebra, $$\left\{\lambda_6,\lambda_7,\lambda_{11},\lambda_{12},\lambda_{13}, \lambda_{14}, (\sqrt{3}\lambda_8-\lambda_3)/2, \left (2\sqrt{6}\lambda_{15}-\tfrac{3}{2}\lambda_3 -\tfrac{1}{2}\sqrt{3}\lambda_8\right )/3 \right\},$$ but you have still not explained what you'd use it for. Naturally, now four generators overlap with your original one, above.

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