Let U denote the set of all nxn matrices A with complex entries such that A is unitary. Then U as a topological subspace of $C^{n^2}$ is

a) compact but not connected.

b) connected but not compact.

c) connected and compact.

d) neither connected nor compact.

we can say a set is compact if it is closed and bounded.how to find a matrix that will be a limit point of the set to show that it is closed, and also how to prove or disprove it is connected...provide a hint..please as i am preparing for my exam..


1 Answer 1


Consider the map $F: GL_n(\mathbb{C})\rightarrow \mathbb{C}$, $F(A)=AA^*$. Show this is a continuous map. $U=F^{-1}(1)$, and hence closed. To show it is bounded observe that from the definition, norm of the columns are bounded hence norm of the elements.

For connectedness, show that every element can be joined by a path to the identity matrix as follow. Take an unitery matrix $A$. Diagonalize it.Then the diagonal elements are $e^{i\theta_r}$. Join them to $1$ by a paths $\gamma_r$ in the circle and consider the path of matrices with diagonal element $\gamma_r$.

  • $\begingroup$ Just a little correction: The codomain of $F$ should also be $GL_n(\mathbb C)$. And in case ones don't know why $F$ is continuous, I want to point out that the continuity of $F$ follows from the continuity of the map $A\to A^*$. $\endgroup$
    – Sam Wong
    Commented Apr 28, 2021 at 14:34

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