# Compactness of the set of all unitary matrices in $M_2(\mathbb{C})$

Is the set of all unitary matrices in $M_2(\mathbb{C})$ is compact? I can show that as determinant map is continuous so unitary matrices are closed but how to show they are bounded?

• $M_n(\mathbb C)$ is a finite-dimensional space, so it's enough to show that $U_n(\mathbb C)$ is closed and bounded.
• The maps $f_1\colon U\mapsto U^*U-I$ and $f_2\colon U\mapsto UU^*-I$ are continuous since so is the map $U\mapsto U^*$, so $U_n(\mathbb C)=f_1^{-1}(\{0_n\})\cap f_2^{-1}(\{0_n\})$ is closed as an intersection of such two sets.
• $U_n(\mathbb C)$ is bounded for the euclidian operator norm, since for each $x$ and $U$ unitary $$\langle x,x\rangle=\langle U^*Ux,x\rangle=\langle Ux,Ux\rangle$$ (hence $U$ is an isometry, in particular its norm is $1$).
• What is the norm of a unitary matrix as an element of the set of all $n\times n$ complex matrices rather than as an operator? is it $\sqrt{n}$?
– DpS
Dec 7, 2016 at 6:50
• Just a little supplement: Taking $x=(1,0,\cdots,0)^T$. Then $1=\langle x,x\rangle=\langle Ux,Ux\rangle=\langle U (1,0,\cdots,0)^T,U (1,0,\cdots,0)^T\rangle$. The last term on the R.H.S is the square root of the sum of the square of the entries in the first column of $U$. So the sum of the square of the entries in the first column of $U$ equals $1$. Similarly we can find the same bound for all other columns. So the Euclidean norm of dimension $n^2$ of $U$ is bounded above by $\sqrt n$ (in fact it exactly equals $\sqrt n$). Apr 28, 2021 at 14:55
• @DpS Yes, it's $\sqrt n$. See my last comment. Apr 28, 2021 at 14:59

They are isometries for the hermitian form (i.e. $u^* M^*Mv=u^*v,\ \forall u,v\in\mathbb{C}^n$), so their operatorial norm is $1,$ (i.e. $||M||:=\sup_{|u|=1}|Mu|=1$.)
Hence $U(n)$ is included in the unit sphere of the normed vector space $(\mathfrak{gl}(\mathbb{C},n),||\cdot||).$

One of the definitions of a unitary matrix is that its rows (or columns) form an orthonormal basis with respect to the standard inner product on $\mathbb{C}^n$; the set of orthonormal frames in $\mathbb{C}^n$ is obviously bounded. Topologically it's a torus, I think. [?]

This makes sense to me, since the eigenvalues of unitary matrices lie on the unit circle.

• Sorry to be late to the party. Topologically, $U(2)$ is $S^3\times S^1$, so it's not a torus in the usual sense. May 15, 2012 at 0:39
• so then it's also connected?
– DpS
Dec 7, 2016 at 6:58