7
$\begingroup$

I was working on the following problem from Stillwell's Naive Lie Theory.

Prove that $U(n)/Z(U(n))=SU(n)/Z(SU(n))$.

It was shown earlier in the text that $Z(U(n))=\{ e^{i\theta} \textbf{1}: \theta \in \mathbb{R} \} \cong S^{1}$ and $Z(SU(n))=\{ \omega \textbf{1}: \omega^{n}=1 \text{ and } \omega \in \mathbb{C} \}$. We have that $\textbf{1}$ denotes the identity matrix.

When thinking about this problem, I first considered the case where $n=2$. We have that unitary matrices in $U(2)$ are of the form \begin{equation*} e^{i\theta} \begin{bmatrix} \alpha &-\beta \\ \bar{\beta} & \bar{\alpha} \end{bmatrix} \end{equation*} We have that the relation which sends $e^{i\theta} \begin{bmatrix} \alpha &-\beta \\ \bar{\beta} & \bar{\alpha} \end{bmatrix}$ to $ \begin{bmatrix} \alpha &-\beta \\ \bar{\beta} & \bar{\alpha} \end{bmatrix}$ seems to be a well defined function from $U(2) \rightarrow SU(2)/Z(SU(2)$ since the center $Z(SU(2))$ consists of the matrices $\pm \textbf{1}$. This function even is a homorphism that has $Z(U(2))$ as it's kernel. From here we can conclude that $U(2)/Z(U(2))=SU(2)/Z(SU(2))$

$\textbf{However, I am having trouble generalizing from here}$

$\endgroup$
5
$\begingroup$

By the short exact sequence $$1\to SU(n)\to U(n)\xrightarrow{\det}S^1\to 1$$ $SU(n)$ is a normal subgroup of $U(n)$, and now identify $U(1)$ to $e^{i\theta}I$, where $I$ is the $n\times n$ identity matrix and $\theta\in[0,2\pi)$, and therefore $U(1)$ is a subgroup (and even normal) of $U(n)$.

Now, we have

(1) $SU(n)\cap U(1)=\{e^{ik2\pi/n}:0\le k\le n-1\}\cong \mathbb Z_n$;

(2) $SU(n)\cdot U(1)=U(n)$, since for each $A\in U(n)$, let $z=\det (A)$, so $A=(A/\sqrt[n]{z})\cdot (\sqrt[n]{z}\cdot I)\in SU(n)\cdot U(1)$.

By the 2nd Isomorphism Theorem for groups, we have $$(SU(n)\cdot U(1))/U(1)\cong SU(n)/(SU(n)\cap U(1)).$$ Combine the facts in (1) and (2), we have therefore proved $$U(n)/U(1)\cong SU(n)/\mathbb Z_n$$

@user135520 tried the special case $n=2$, it is right. But should be careful that the map should be $$U(2) \rightarrow SU(2)/\{\pm 1\}$$ $$e^{i\theta} \begin{bmatrix} \alpha &-\beta \\ \bar{\beta} & \bar{\alpha} \end{bmatrix} \mapsto \{ \begin{bmatrix} \alpha &-\beta \\ \bar{\beta} & \bar{\alpha} \end{bmatrix}, \begin{bmatrix} \alpha &\beta \\ \bar{\beta} & -\bar{\alpha} \end{bmatrix}\}$$

For this problem, there is still a different solution in here, who shows that $$U(n) \cong (SU(n) \times U(1))/ \mathbb{Z}_{n}.$$ But to use this isomorphism to solve our problem, it should be careful to take $U(1)$ from right hand side as a product to the left hand side as a quotient.

$\endgroup$
  • $\begingroup$ @user26857 Thanks for your edition, so for example, if I want to tap "kernel" of a map $\phi$, it should be $\ker \phi$ instead of $ker\phi$? $\endgroup$ – Yilong Zhang Aug 15 '15 at 21:13
  • $\begingroup$ Why was the kernel bigger than $U(1)$? How do you obtain that $U(n)/U(1) \cong SU(n)$? shoudn't it be $U(n)/SU(n) \cong U(1)$ from the SES $1 \rightarrow SO(n) \rightarrow U(n) \rightarrow U(1) \rightarrow 1$? $\endgroup$ – user135520 Aug 16 '15 at 14:34
  • 1
    $\begingroup$ @user135520 I see it, thanks for your remind. I'll soon get back to you and fix my solution. $\endgroup$ – Yilong Zhang Aug 16 '15 at 15:27
  • $\begingroup$ No problem, thanks. $\endgroup$ – user135520 Aug 16 '15 at 16:17
  • $\begingroup$ For a reader unfamiliar with short exact sequences, what is meant by $U(n)\xrightarrow{\det}S^1$, in particular, the "det"? $\endgroup$ – Doubt Feb 5 '18 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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