Spontaneous symmetry breaking with 1 massive scalar but three unbroken generators? Consider some theory of four real scalar fields, $\phi_1$, $\phi_2$, $\phi_3$ and $\phi_4$, that is invariant under a global $$SO(4)\cong SU(2)_L\times SU(2)_R$$ symmetry. We can rewrite the real scalars in terms of the $2 \times 2$ matrix $$\Sigma = \frac{1}{\sqrt{2}}\left(\mathbb{1}\cdot\phi_4+i\sigma^i\phi_i\right)$$ which transforms as$$\Sigma\rightarrow U_L\Sigma U_R^\dagger\quad.$$
Let's say the theory exbibits spontaneous-symmetry breaking where the potential is $$V[\phi] = -\frac{1}{2}\mu^2|\phi|^2(\lambda|\phi|^2-4\mu^2).$$ We pick some vacuum: $$<\phi>=\begin{pmatrix} v \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
Now when expanding this in the potential one finds that the potential has one massive scalar and 3 Goldstone bosons. But when I look at the group $SO(4)$, it has 6 generators. And by requiring the condition of an unbroken vacuum, we get a set of three unbroken generators. Which leave us with 3 broken ones.
I am confused: on one way I found 1 massive scalar and 3 Goldstones. On the other way I found 3 massive scalar and 3 Goldstones. I know that the first one should be the right one as we started with 4 degrees of freedom, but how come the second method gives a different result.
 A: This is something like the inverse fallacy. :)
Goldstone's theorem tells you that for every broken symmetry generator, you must find one massless mode in the spectrum of the theory.
It does not tell you for every unbroken symmetry generator, you must find one massive mode in the spectrum of the theory.
If that were the case, then if you had chosen a symmetry-preserving potential and therefore had all 6 generators unbroken, you would have to have 6 massive modes, which you obviously don't by construction.
So the number of massive modes will be the difference between the original number of degrees of freedom and the number of broken symmetry generators. You can convince yourself that it won't be possible for your potential to break more generators than there are degrees of freedom. (As an easy-to-visualize example, even though your vacuum is now singling out some direction in the symmetry space, there are still azimuthal-like rotations around that direction which must be preserved.)
A: *

*For what it's worth, it is useful to rewrite OP's scalar field $$\phi~\in~\mathbb{H}~\cong~\mathbb{R}^4\tag{1}$$
as taking values in the quaternions,
$$ {\cal L}~=~|\partial\phi|^2+m^2|\phi|^2-\lambda(|\phi|^2)^2. \tag{2}$$


*The 6-dimensional global symmetry group is
$$G~=~U(1,\mathbb{H})_L\times U(1,\mathbb{H})_R.\tag{3}$$
Here
$$\begin{align}U(1,\mathbb{H})~:=~&\{q\in\mathbb{H}\mid |q|=1 \}\cr
~\cong~& SU(2)\end{align} \tag{4} $$
is the group of unitary $1\times 1$ matrices with quaternionic entries, cf. e.g. my Phys.SE answer here.


*The group $G$ acts on the field $\phi$ as
$$\phi~\to~ q_L\phi \bar{q}_R, \qquad (q_L, q_R)~\in~G. \tag{5}$$


*One may check that the VEV
$$\langle\phi\rangle~=~v~=~\frac{m}{\sqrt{2\lambda}}~\in~\mathbb{R}\backslash\{0\}\tag{6}$$
breaks spontaneously the group $G$ down to a 3-dimensional diagonal isotropy subgroup
$$\begin{align}H~=~&\{(q_L, q_R) \in G \mid q_Lv\bar{q}_R=v\}\cr
~=~&\{(q_L, q_R) \in G \mid q_L= q_R\}\cr
~\cong~&U(1,\mathbb{H}),\end{align}\tag{7}$$
i.e. we have $6-3=3$ broken generators.


*Furthermore, one may show that the fluctuations $$\eta~=~\phi-v\tag{8}$$
have 3 massless imaginary Goldstone modes, and 1 massive real mode,
$$\begin{align} {\cal L}~=~&|\partial\eta|^2+\frac{1}{2}m^2v^2\cr
&-2m^2({\rm Re}\eta)^2+{\cal O}(\eta^3),\end{align} \tag{9}$$
in accordance with Goldstone's theorem.
