In the Standard model electroweak theory, the Higgs field is a complex doublet field, which couples to the $SU(2)$ gauge field. Suppose we replace the complex doublet with a complex triplet $\Sigma$:
$$\Sigma=\left( \begin{array}{cc} \phi_1+i \phi_2 \\ \phi_1+i \phi_2 \\ \phi_1+i \phi_2 \end{array} \right) \tag{1}$$
which couples to the $SU(2)$ gauge field. The Lagrangian reads
$$\mathcal{L}=(D_\mu \Sigma)^\dagger (D^\mu \Sigma)-\mu^2 \Sigma^\dagger \Sigma +\lambda (\Sigma^\dagger \Sigma)^2 \tag{2}$$
where
$$D_\mu=\partial_\mu-igA_\mu^aT^a \tag{3}$$
with
$$T^1=\left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0 \\
\end{array}
\right),\ \ T^2=\left(
\begin{array}{ccc}
0 & -\frac{i}{\sqrt{2}} & 0 \\
\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\
0 & \frac{i}{\sqrt{2}} & 0 \\
\end{array}
\right), \ \ T^3=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1 \\
\end{array}
\right) \tag{4}$$
Before the spontaneous symmetry breaking (SSB), the system has the symmetry of $SU(2)$. After the SSB, we can set
$$(a) \langle \Sigma \rangle = \left( \begin{array}{c} 0 \\ \frac{v}{\sqrt{2}} \\ 0 \end{array} \right),\quad (b) \langle \Sigma \rangle = \left( \begin{array}{c} 0 \\ 0 \\ \frac{v}{\sqrt{2}} \end{array} \right) \tag{5}$$
I find that:
For case $(a)$, the gauge bosons' mass term are
$$ \frac{1}{2}g^2 v^2 [(A_\mu^1)^2+(A_\mu^2)^2] \tag{6}$$
so two gauge bosons obtain mass. The $T^3$ generator is unbroken.
For case $(b)$, the gauge bosons' mass term are
$$ \frac{1}{4}g^2 v^2 [(A_\mu^1)^2+(A_\mu^2)^2+2(A_\mu^3)^2] \tag{7}$$
where all three gauge bosons obtain mass.
We denote $G$ as the symmetry group before SSB; $H$ is the symmetry group after SSB; and $G/H$ is the coset space. $n_\text{G}$ as the number of the original generators; $n_\text{BG}$ as the number of broken generators.
For a relativistic system, the Goldstone theorem dictates
$$ n_\text{G}-n_\text{H}=n_\text{BG}=\text{dim}(G/H) \tag{8}$$
$\textbf{My question:}$
For either case (a) or (b) in Eq.(5), we obtain 5 massless bosons.
(1) In case (a), only two generators are broken, so we have 2 goldstone bosons. Then, what's the meaning of other three massless particles? (And would we know which of the two particles are goldstone bosons, and eaten by gauge field). And would the residual symmetry group $H=SO(2)$? How to understand this?
(2) In case (b), similar question with (a). But now, would the residual symmetry $H=1$?
Some of my personal ideas: I think goldstone bosons are eaten by gauge field are related with the field parameterization and gauge choice. In the complex doublet case, we can write
$$H=e^{i(\alpha^a T^a)}\left( \begin{array}{cc} 0 \\ v+h(x) \end{array} \right) $$
then we take the unitary gauge.
Best Answer
For reasons to become evident in a moment, I'll relabel your six scalar fields as follows, $$\Sigma=\left( \begin{array}{cc} \phi_1+i \xi_1 \\ \phi_2+i \xi_2 \\ \phi_3+i \xi_3 \end{array} \right) .\tag{1}$$ It is then evident that, for g=0, your Lagrangian is O(6) symmetric, and your Higgs potential breaks this symmetry spontaneously to O(5), so the obvious broken generators in O(6)/O(5) correspond to the five massless Goldstone fields detailed in the linked question, in this language, for $\langle \phi_3\rangle = v/\sqrt 2$ : $~~~~\phi_1, \phi_2,\xi_1,\xi_2,\xi_3$. Think of this as the global custodial symmetry structure of the potential.
However, this large symmetry is explicitly broken by your gauging terms to SU(2) ~ O(3) in your lagrangian, and there is a way to see the $\phi$s separate from the $\xi$s in the bilinears through the gauging terms as follows.
You utilized the spherical basis for the (real) adjoint representation, but you may switch to the real antisymmetric basis, instead, utilized in classical rotations, which does not mix the real and imaginary components of Σ. (The above vev corresponds to your "case" (a) which, as explained in the comment, is identical to your (b), and preserves ${\mathbb I} \otimes T^3$.)
In this basis, the covariant completion in (3) reads $$-igA^a_\mu T^a=g \left(\begin{array}{ccc}0&-A^3_\mu &A^2_\mu \\ A^3_\mu &0&-A^1_\mu \\-A^2_\mu &A^1_\mu &0\end{array}\right) , $$ antisymmetric and so antihermitean, just like the gradient.
Plugging it into the bilinear terms in A in the Lagrangian, it yields the mass terms (6), as you observed, leaving $A^3_\mu$ massless, as the v.e.v. $\langle \Sigma\rangle= (0,0,v)^T/\sqrt 2$ is a null vector of the above completion: $T^3$ remains unbroken, and SU(2)/U(1) is two-dimensional. (As indicated, your case (b) is a mirage, and (7) is flat wrong.)
So, two of the previous Goldstone bosons, now of SU(2)/U(1), $\phi_1$, $\phi_2$, have been "eaten" (gauge-absorbed) by these two gauge fields, $ A^\mu_1$, $A^\mu_2$, and the shifted $\phi_3$ is massive: the "Higgs". (This is a choice: we could have chosen $\xi_1$ & $\xi_2$, instead. You may have noticed the global U(1) interchanging φs and ξs.)
But what about the stranded massless $\xi_{1,2,3}$, you are asking. They are still massless at tree level, but they are not goldstons anymore. (I think, but am not sure, that they might develop masses due to radiative corrections, not being Goldstone-protected.)
Your exponential cosets' separation does not easily work in this setting.