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Let U be set of all n×n matrices A with complex enteries s.t. A is unitary. then U as a topological subspace of $\mathbb{C^{n^{2}}} $ is

  1. compact but not connected.
  2. connected but not compact.
  3. connected and compact.
  4. Neither connected nor compact

I am stuck on this problem . Can anyone help me please..... I don't know where yo begin........

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    please guide me sir......2012-12-18

2 Answers 2

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First consider the continuous function $f: \mathbb{C}^{n^2} = M_{n\times n}(\mathbb{C}) \to \mathbb{C}$ that sends $A \mapsto |A|$. Since $f(U)$ is the unit circle $S^1$, which is closed, we know that $U = f^{-1}(S^1)$ is closed in $\mathbb{C}^{n^2}$. $U$ is also bounded (can you provide the details?), so by Heine-Borel, $U$ is compact.

Showing connectedness involves an explicit path from each $A \in U$ to the identity matrix, which can be done using the diagonalizability of $A$.

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    $|A|$ is the determinant of A? Maybe you should rather denote that by $det(A)$ instead?2012-12-18
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For connectedness, examine the set of possible determinants, and whether or not you can find a path of unitary matrices between two unitary matrices with different determinants.

For compactness, look at sequences of unitary matrices and examine whether or not one can be constructed to not have a convergent subsequence.

Once you have an affirmative or negatinve answer to the above paragraphs, you pick the corresponding alternative, and you're done.

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    @pankaj Then there isn't much I can do for you. I can give you a fish, but it will take me considerable effort and time which I do not have at the moment, and at the end of the day you're still expected to be able to fish.2012-12-18