I am reading some papers in physics. I don't know some notations in those papers. For example, $SU(1, 1)$, $U(1)$. I think these are Lie groups which consist of matrices. But I don't know what kind of matrices are in these groups.

What are elements in $SU(1, 1)$ and how to show that $U(1)$ is the maximal subgroup of $SU(1, 1)$?

Thank you very much.

  • 1
    $\begingroup$ The elements in $SU(1,1)$ consist of matrices of unit determinant which preserve the quadratic form $(x,y) = \bar{x}_{1} y_1 - \bar{x}_{2} y_2$ over $\mathbb{C}$, that is, if $A \in SU(1,1)$, then $(Ax,Ay) = (x,y)$ and $\det A = 1$. $\endgroup$
    – user02138
    Nov 5, 2012 at 2:44
  • $\begingroup$ $U(1)$ is usually the unitary group which can be written as the unit-circle in the complex plane; $e^{i\theta} \in U(1)$. This can also be written as $2 \times 2$ matrices of the form $\left[ \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right]$. Perhaps the latter is what you seek. $\endgroup$ Nov 5, 2012 at 2:50
  • $\begingroup$ @JamesS.Cook, thank you very much. Does "1" in $U(1)$ mean 2 by 2 matrices? What about two "1"'s in $SU(1, 1)$? $\endgroup$
    – LJR
    Nov 5, 2012 at 3:10
  • $\begingroup$ @user02138, thank you very much. I search on Wiki and find that the definition of $SU(1)$ is the same as you said. What are differences between $SU(1, 1)$ and $SU(1)$? $\endgroup$
    – LJR
    Nov 5, 2012 at 3:12
  • $\begingroup$ @user9791 complex numbers can be formulated by several different objects. $a+ib = (a,b)$ with $(a,b)(c,d) = (ac-bd,ad+bc)$ as Gauss proposed, or $\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right]$ the regular representation. Both of these allow the algebra of complex numbers which in a nutshell is $i^2=-1$ then proceed normally. That said, if you're looking for a copy of $U(1)$ embedded in $2 \times 2 $ matrices then I guess the choice is clear. Oh, the $1$ is because $U(1)$ multiplies 1-dim's complex vectors. $\endgroup$ Nov 5, 2012 at 4:21

2 Answers 2


If $n = p + q$, then $U(p,q)$ means the group of invertible complex $n \times n$ matrices which preserve the inner product $(z_1,\ldots,z_n) \cdot (w_1,\ldots,w_n) = z_1 \bar{w}_1 + \cdots + z_p \bar{w}_p - z_{p+1} \bar{w}_{p+1} - \cdots - z_n \bar{w}_n.$

In the special case where $p = n$ and $q = 0$, we just write $U(n)$ rather than $U(n,0)$.

$SU(p,q)$ or $SU(n)$ denotes the subgroup of matrices of det. $1$.

There is an additional complication that when people write that $U(1)$ is the maximal compact subgroup of $SU(1,1)$, they don't mean this literally (they couldn't, since $1\times 1$ matrices aren't a subgroup of $2 \times 2$ matrices), but rather that $U(1)$ is isomorphic to the maximal compact subgroup of $SU(1,1)$, in this particular case via the isomorphism that maps $a$ (a complex number of abs. value $1$, which is what $U(1)$ consists of) to the matrix $\begin{pmatrix} a & 0 \\ 0 & \bar{a} \end{pmatrix}$.

A further complication is that there are some non-obvious isomorphisms among these groups, and between some of them and some other matrix groups. E.g. there is an isomorphism between $SU(1,1)$ and the group $SL(2,\mathbb R)$ of $2\times 2$ matrices with real entries and det. equal to $1$. (This question has some discussion of this isomorphism, and googling (which is how I found this link) will produce many others, I'm sure.)

Finally, there is a rich theory of these sorts of groups (semisimple Lie groups) and their representations, and any discussion of them in the literature might be explicitly or implicitly drawing on that literature. So don't expect to learn everything about them just by knowing the definitions! There is (much!) more to be said about them then you would guess just from their definitions as matrix groups.

  • $\begingroup$ anyway, what book that contain all material about unitary group? $\endgroup$ Nov 23, 2017 at 13:20

The group $\rm{SU}(1,1)$ has a faithful representation as the group of complex matrices $$\left\{ \begin{bmatrix} a & b \\ \bar{b} & \bar{a}\end{bmatrix} \text{ st. } |a|^2 - |b|^2 = 1 \right\} .$$

The group $\rm U(1)$ is a natural subgroup of $\rm{SU}(1,1)$, in the representation above $$ \mathrm{U}(1) \simeq \left\{\begin{bmatrix} a & 0 \\ 0 & \bar{a}\end{bmatrix}, a \in \mathbb U\right\}.$$

Let's show it is a maximal compact subgroup.

Any subgroup of $\rm{SU}(1,1)$ containing strictly $\rm U(1)$ contains a matrix $\mathrm{M} = \begin{bmatrix} a & b \\ \bar{b} & \bar{a}\end{bmatrix}$ with $|a|^2 - |b|^2 = 1$ and $a$ and $b$ both not zero.

This implies that $|a| > 1$ and so $\rm M$ has an eigenvalue strictly greater than $1$. Hence $||\mathrm M||^n \to \infty$ and the group is not compact.


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