# Can we parameterise $SU(3)$ in such a way that there are clearly 2 parameters corresponding to the cartan torus?

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.

• Why not $\theta_i\mapsto e^{\theta_iT_i}$? – doetoe Sep 5 at 15:28
• 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? – h_m Sep 5 at 15:30
• Plug into the generic expression for H an arbitrary combination of your two Cartan generators, of course. – Cosmas Zachos Sep 5 at 15:39
• Ah I wasn't aware of this formula, thanks! – h_m Sep 5 at 15:52
• 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. – Cosmas Zachos Sep 5 at 16:12

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:

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