0
$\begingroup$

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 $SU(3)\times SU(2)\times U(1)$.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

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 $SU(3)\times SU(2)\times U(1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.