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.