In the many textbook of standard model, i encounter the relation \begin{align} SU(2)_L \times U(1)_L = U(2)_L \end{align} Here $L$ means the left-handness, (It is a physical meaning(representation), which states that fermion have left or right handness(chirality).) I wonder that above relation is true in general case, $i.e$, \begin{align} SU(2) \times U(1) = U(2) \end{align}


Not quite. There is a natural short exact sequence

$$1 \to SU(2) \to U(2) \xrightarrow{\det} U(1) \to 1.$$

This sequence doesn't have a natural splitting, but it does have a splitting given by

$$U(1) \ni z \mapsto \left[ \begin{array}{cc} z & 0 \\ 0 & 1 \end{array} \right] \in U(2).$$

Such a splitting exhibits $U(2)$ as a semidirect product $SU(2) \rtimes U(1)$, where the action of $U(1)$ on $SU(2)$ is given by conjugation with respect to the above splitting. The image of the splitting does not, and cannot, commute with the image of $SU(2)$ (we'd like to send $z$ to the diagonal matrix with entries $\sqrt{z}, \sqrt{z}$, but $\sqrt{z}$ isn't well-defined on $U(1)$), so this semidirect product cannot be refined to a direct product.

  • $\begingroup$ The center of $U(2)$ is $U(1)$, so we also have $U(2)/U(1)\cong SU(2)$, can we write $U(2)\cong U(1)\rtimes SU(2)$? $\endgroup$ – Yilong Zhang Aug 16 '15 at 16:13
  • 1
    $\begingroup$ Oh, no, I see a formula $U(n)/U(1)\cong SU(n)/Z_n$, so when $n=2$, we have $U(2)/U(1)\cong SU(2)/Z_2\cong SO(3)$. Compare with the fact $U(2)\cong SU(2)\rtimes U(1)$. The quotient of $U(n)$ by $U(1)$ is smaller than $SU(2)$. How to understand these facts? $\endgroup$ – Yilong Zhang Aug 16 '15 at 16:27

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