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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?

Please help.

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3 Answers 3

up vote 7 down vote accepted
  • $M_n(\mathbb C)$ 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 $1$).
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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 $(\mathfrak{gl}(\mathbb{C},n),||\cdot||).$

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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.

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1  
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. –  Jason DeVito May 15 '12 at 0:39

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