The normalizer $N(\mathfrak{h})$ of the Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{su}(3)$ is defined as
$$N=\left\{ x \in SU(3)\;|\; x^\dagger\mathfrak{h}x \in \mathfrak{h}\right\}$$
I would like to know explicitly what this subgroup looks like; for example, in terms of the exponentiation of generators of $\mathfrak{su}(3)$in Gell-Mann basis, where $\mathfrak{h}= \mathrm{Span}\{\lambda\_3,\lambda\_8\}$. Clearly $e^{i\mathfrak{h}}\subset N(\mathfrak{h})$, is there anything else?


I often see in the literature how $N(\mathfrak{h})/C(\mathfrak{h})$ is the discrete permutation group $S\_3$ or the Weyl group $W(SU(3))$, where
$$ C(\mathfrak{h}) =\left\{ x \in SU(3)\;|\; x^\dagger h x = h, \quad \forall h \in \mathfrak{h}\right\} $$
is the centraliser. As all Cartan subalgebra elements commute with each other by definition, again clearly $e^{i\mathfrak{h}}\subset C(\mathfrak{h})$. This makes me wonder if $N(\mathfrak{h})$ is really just $e^{i\mathfrak{h}}$ plus some discrete elements..


Finally I will include here the application I have in mind for my purposes. Consider 2 general (8 component) $\mathfrak{su}(3)$ 'vectors' $r=r^i \lambda^i$ and $s=s^i \lambda^i$, where the $\lambda^i$ are the Gell-Mann basis. I want to see how many independent directions I can eliminate by the action
$$r\rightarrow x^\dagger r x, \quad s\rightarrow x^\dagger s x, \quad \mathrm{for \; some} \; x\in SU(3) .$$
If I start by focusing on $r$, it is known that one can make it so that $x^\dagger r x \in \mathfrak{h}$, thus eliminating 6 components. In order not to spoil this, while still trying to eliminate components of $s$, the residual freedom in the transformation is reduced to precisely
$$r\rightarrow x'^\dagger r x', \quad s\rightarrow x'^\dagger s x', \quad\mathrm{for \; some} \; x'\in N(\mathfrak{h}). $$
Hence why I am intersted in this subgroup.



Explicit Normalizer of SU(3) Cartan subalgebra
Reproduced from my [comment](https://mathoverflow.net/questions/363272/explicit-normalizer-of-su3-cartan-subalgebra#comment916426_363272); please let me know if it does not answer the question.


For your choice of $\mathfrak h$ (as the diagonal Cartan subalgebra in $\mathfrak{su}(3)$), we have that $C = \operatorname C\_{\operatorname{SU}(3)}(\mathfrak h)$ is the diagonal torus in $\operatorname{SU}(3)$, which equals $e^{i\mathfrak h}$, and that $\operatorname N\_{\operatorname{SU}(3)}(\mathfrak h)$ is generated by $C$ and the permutation matrices (which are unitary). (EDIT: As @Rudyard points out in the [comments](https://mathoverflow.net/questions/363272/explicit-normalizer-of-su3-cartan-subalgebra/363287#comment916641_363287), half the permutation matrices are unitary but have determinant $-1$, so one must change the sign of an appropriate entry. This introduces a 2-torsion ambiguity, but in general one can't do any better; the fact that one can always do this well is called the "Tits lift". See [Can we realize Weyl group as a subgroup?](https://mathoverflow.net/questions/13139/can-we-realize-weyl-group-as-a-subgroup) and Tits's article [Normalisateurs de tores I](https://doi.org/10.1016/0021-8693(66)90053-6) linked there—sadly there was never a II—for more details.)


