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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.

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    $\begingroup$ What does it mean to describe a subgroup of $\operatorname{SU}(3)$ starting with a basis of $\mathfrak{su}(3)$? Also, I can never remember real-group stuff like when the exponential is surjective, but it is true that $\operatorname N_{\operatorname{SU}(3)}(\mathfrak h)$ is an extension of $\operatorname C_{\operatorname{SU}(3)}(\mathfrak h)$ by a discrete group (which is the same as saying that $\operatorname W(\operatorname{SU}(3), \mathfrak h)$ is discrete). $\endgroup$
    – LSpice
    Jun 16, 2020 at 21:02
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    $\begingroup$ For a more explicit description, you'll have to say which Cartan subalgebra $\mathfrak h$ you have in mind—they're all conjugate, but the explicit description will depend on which one you pick. $\endgroup$
    – LSpice
    Jun 16, 2020 at 21:04
  • $\begingroup$ @LSpice I edited the question to address your comments. "What does it mean to describe a subgroup of SU(3) starting with a basis of su(3)?" - I just meant that every subgroup should be representable in terms of a subalgebra via exponentiation. The Cartan subalgebra I have in mind is spanned by $\lambda_3$ and $\lambda_8$ Gell-Mann matrices. This $\mathfrak{h}$ is easy to exponentiate explicitly. $\endgroup$
    – Rudyard
    Jun 16, 2020 at 21:44
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    $\begingroup$ Since $\operatorname N_{\operatorname{SU}(3)}(\mathfrak h)$ is an extension of $\operatorname C_{\operatorname{SU}(3)}(\mathfrak h)$ by a discrete group, it has the same dimension as the centraliser; and the centraliser has Lie algebra $\mathfrak h$, so is 2-dimensional. (Now that I think about it, of course $\operatorname C(\mathfrak h)$ equals $C = e^{i\mathfrak h}$; it's generated by $C$, but $C$ is a group because $\mathfrak h$ is Abelian.) Because of this, you won't capture any more of $\operatorname N(\mathfrak h)$ by exponentiating a subalgebra. $\endgroup$
    – LSpice
    Jun 16, 2020 at 21:51
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    $\begingroup$ On looking at en.wikipedia.org/wiki/Gell-Mann_matrices , $C = \operatorname C_{\operatorname{SU}(3)}(\mathfrak h)$ for your $\mathfrak h$ is the diagonal torus in $\operatorname{SU}(3)$, and $\operatorname N_{\operatorname{SU}(3)}(\mathfrak h)$ is generated by $C$ and the permutation matrices (which are unitary). $\endgroup$
    – LSpice
    Jun 16, 2020 at 21:56

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Reproduced from my comment; 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, 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? and Tits's article Normalisateurs de tores I linked there—sadly there was never a II—for more details.)

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    $\begingroup$ Thanks. I'm confused about one aspect. Are the permutation matrices special unitary? If I take the 3 dimensional representation as shown explicitly at the top of page 26 of math.lsa.umich.edu/~kesmith/rep.pdf, it seems to me that 2 out of 3 have determinant -1, so they are not an element of $SU(3)$ and I thought $N(\mathfrak{h})\subset SU(3)$ $\endgroup$
    – Rudyard
    Jun 17, 2020 at 16:11
  • $\begingroup$ I'm sorry; you are right. Just change one of the +1s to a -1 to get determinant 1. (There's no well defined choice of lift, but that's usual for Weyl groups, where usually the best you can do is lift modulo 2-torsion. In any case, all the choices are congruent modulo $\operatorname C(\mathfrak h)$.) $\endgroup$
    – LSpice
    Jun 17, 2020 at 16:32
  • $\begingroup$ I see. That works. "There's no well defined choice of lift" - by lift do you mean writing these 3 'permutation' matrices in exponentiated form in terms of $\mathfrak{su}(3)$ algebra elements? I thought exponentiating a compact simply-connected Lie group such as $SU(3)$ was a well defined mapping. $\endgroup$
    – Rudyard
    Jun 17, 2020 at 16:54
  • $\begingroup$ By lift, I mean choice of an element of $\operatorname N_{\operatorname{SU}(3)}(\mathfrak h)$ that projects to a given element of $\operatorname W(\operatorname{SU}(3), \mathfrak h)$. Exponentiation is well defined for any Lie group (not surjective in general, but I think so for compact Lie groups? As I say, I can never remember how real groups work). $\endgroup$
    – LSpice
    Jun 17, 2020 at 17:09

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