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We can parameterise the lie algebra of $SU(3)$ using the Gell-Mann matrices, so that a general element of LA is $\theta_i T_i$, where $T_i=\lambda_i/2$ and $\lambda_i$ are the Gell-Mann matrices.

Then, since $SU(3)$ is compact and (simply) connected, we can express any element of it in the form $e^{\theta_i T_i}$. However, actually computing this exponential is rather difficult, since it is just a generic traceless hermitian matrix.

The nice thing about this parameterisation is that two of the matrices in the basis generate the Cartan Subalgebra of $\mathcal{L}(SU(3))$, and so it is clear which of the coordinates $(\theta_i)$ correspond to the toroidal subgroup of $SU(3)$.

Is there some way to explicitly parameterise $SU(3)$ (and not just the Lie algebra) so that it is clear which parameters/coordinates correspond to the cartan torus? Essentially, the problem with the above parameterisation is that I can't explicitly compute the exponential, and therefore I can't get an explicit form of the $SU(3)$ matrices.

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  • $\begingroup$ Why not $\theta_i\mapsto e^{\theta_iT_i}$? $\endgroup$ – doetoe Sep 5 at 15:28
  • $\begingroup$ sorry, maybe I wasn't very clear. What I'd like is an explicit general matrix in the lie group $SU(3)$. The map given is fine of course, but to actually get the group element explicitly you need to compute the exponential, which is hard. Is there a way to explicitly represent $SU(3)$ matrices which still makes the cartan torus clear? $\endgroup$ – h_m Sep 5 at 15:30
  • $\begingroup$ Plug into the generic expression for H an arbitrary combination of your two Cartan generators, of course. $\endgroup$ – Cosmas Zachos Sep 5 at 15:39
  • $\begingroup$ Ah I wasn't aware of this formula, thanks! $\endgroup$ – h_m Sep 5 at 15:52
  • $\begingroup$ Are you looking for a two-parameter expression, or an 8-parameter one? The 2-parameter one is a trivial diagonal matrix, but the 8-parameter one is recondite: I'm not sure you can get the Cartan torus simply stick out of the 8-d manifold. $\endgroup$ – Cosmas Zachos Sep 5 at 16:12
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The group manifold $SU(3)$ can be parametrized almost everywhere according to:

$$ SU(3) \ni g = v(u, \bar{u}) e^{i \theta_3 \lambda_3 + i \theta_8 \lambda_8}$$

where $v(u, \bar{u})$ is a unitary matrix depending on a three dimensional complex vector $u = (u_1, u_2, u_3)$. This matrix parametrizes the flag manifold $SU(3)/(U(1)\times U(1))$ (which is therefore 6-dimensional). It is given for example in: Daoud and Jellal's work (equation (23)), included in this answer for completeness:

$$v(u, \bar{u}) = \begin{pmatrix} \frac{1}{\sqrt{\Delta_1}} & -\frac{\bar{u_1}+u_2\bar{u_3}}{\sqrt{\Delta_1\Delta_2}} & -\frac{\bar{u_3}-\bar{u_1}\bar{u_2}}{\sqrt{\Delta_2}} \\ \frac{u_1}{\sqrt{\Delta_1}} & \frac{1+|u_3|^2 - u_1u_2\bar{u_3}}{\sqrt{\Delta_1\Delta_2}} & -\frac{\bar{u_2}}{\sqrt{\Delta_2}} \\ \frac{u_3}{\sqrt{\Delta_1}}& \frac{u_2+u_2|u_1|^2 - u_3\bar{u_1}}{\sqrt{\Delta_1\Delta_2}} & \frac{1}{\sqrt{\Delta_2}} \end{pmatrix}$$

with:

$$ \Delta_1(u, \bar{u}) = 1 + |u_1|^2+|u_3|^2$$ $$ \Delta_2(u, \bar{u}) = 1 + |u_2|^2+|u_3-u_1u_2|^2$$

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