# $SU(5)$ Gauge Field Theory, symmetry breaking

If I start with an $$SU(5)$$ gauge group and discover that the vacuum is preserved only by matrices of the form $$G$$ $$\begin{bmatrix} A & 0 \\ 0 & B \\ \end{bmatrix}$$ where the conditions on $$A$$ and $$B$$ ($$A$$ is a $$3\times 3$$ matrix and $$B$$ a $$2\times 2$$) are: that they must be Hermitian, and the matrix $$G$$ is traceless.

I don't understand why, if $$tr(A)=−tr(B)$$, then I have $$U(1)$$ and if I take $$trA = trB = 0$$ it is $$SU(3)\times SU(2)$$, so it means $$SU(3)\times SU(2)\times U(1)$$.

• Feb 2 at 17:24

There is no "if $$\operatorname {tr} G= 0= \operatorname {tr} A +\operatorname {tr} B$$": you already assumed it.
The traceless matrix diag$$(2I_3, -3I_2)$$ manifestly commutes with all 12 linearly independent Gs, so it is a sole generator commuting with the full algebra; hence it generates a U(1) by exponentiation, relating to all remaining generators in G by a Cartesian $$\times$$.
The remaining 11 Gs may now be made traceless by subtracting a suitable multiple of this U(1) generator from G, $$G'=G-(\operatorname {tr}A)\operatorname {diag}(I_3/3,-I_2/2)$$. Now both the A's and the B's are traceless.
3×3 hermitean traceless matrices comprise the algebra of SU(3), and 2×2 such the algebra of SU(2), commuting between these blocks by construction/posit, above. You are then left with the direct sum of 1+8+3 generators of $$SU(3)\times SU(2)\times U(1)$$.