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Prove that SU(n) is closed and bounded
Oh, so since SU(n) consists of n \times n matrices with complex entries, I first mapped each with points in C^{n^2}. And consequently, each of the points in C^{n^2} can be mapped to another point in R^{{2n}^2} since each entry of SU(n) can be written as a + bi. So we have coordinates of...- RickilusSmith
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- Forum: Calculus and Beyond Homework Help
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Prove that SU(n) is closed and bounded
Homework Statement Prove that SU(n) is closed and bounded Homework Equations The Attempt at a Solution So in order to prove this, I first mapped SU(n) to be a subset of R^{{2n}^2}. To prove the closed portion, I tried mapping a sequence in SU(n) to a sequence in R^{{2n}^2}. However, I...- RickilusSmith
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- Bounded Closed
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- Forum: Calculus and Beyond Homework Help
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Show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))?
Homework Statement Show that U(n) / Z(U(n)) = SU(n) / Z(SU(n))? Homework Equations Perhaps the first isomorphism theorem? That is, a group homomorphism ϕ of G onto G gives a 1-to-1 correspondence between G and G/(ker ϕ) that preserves products, that is, G is isomorphic to G/(ker ϕ)...- RickilusSmith
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- Forum: Calculus and Beyond Homework Help