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SU(2) is a compact Hausdorff matrix group

Statement

Let $M_{2} (C)$ have its Euclidean topology as $R^{8}$ , write $Q^{†} : = {\overline{Q}}^{T}$ for conjugate transpose, and define

$S U (2) = {Q \in M_{2} (C) : Q^{†} Q = I_{2}, \det Q = 1} .$

Under matrix multiplication, $S U (2)$ is a compact Hausdorff topological group.

Statement references

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes

For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix

For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line

Topological group: multiplication and inversion are continuous

Facts & Assumptions

Given: The Euclidean topology on $M_{2} (C) ≅ R^{8}$ and the displayed subset defined by the conjugate-transpose and determinant equations.

[F1]

For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ : $\det (A B) = \det (A) \det (B)$ for square matrices over a commutative field.

[F2]

[F3]

Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: matrix multiplication and the identity matrix have their usual finite-sum entry formulas.

[F4]

Topological group: multiplication and inversion are continuous: a topological group is a group whose multiplication and inversion maps are continuous for the product and given topologies.

Proof

technique · direct

1.1givenF1F3F4algebra

If $Q, R \in S U (2)$ , then $(Q R)^{†} Q R = R^{†} Q^{†} Q R = I_{2}$ and $\det (Q R) = \det Q \det R = 1$ by [F1]; $I_{2}$ is in the set. From $Q^{†} Q = I_{2}$ , the columns of $Q$ are linearly independent and hence $Q$ is invertible; therefore $Q^{†} = Q^{- 1}$ and $Q Q^{†} = I_{2}$ . By [F1], $1 = \det (I_{2}) = \det (Q Q^{†}) = \det Q \det (Q^{†})$ , so $\det (Q^{†}) = 1$ and the inverse belongs to $S U (2)$ . Associativity is inherited from matrix multiplication. The product entries are polynomials in real and imaginary entries and the inverse $Q \mapsto Q^{†}$ is continuous, so these maps meet the topological-group definition [F4].

1.2F3algebra

The equations $Q^{†} Q = I_{2}$ and $\det Q = 1$ are inverse images of closed points under continuous functions, so $S U (2)$ is closed in $R^{8}$ . Each unitary column has squared norm one, hence the sum of the squared absolute values of all four entries is $2$ ; this bounds the subset.

2.1givenF2step 1.2construct∎

By [F2], the closed bounded subset is compact. Euclidean space is Hausdorff and every subspace of a Hausdorff space is Hausdorff, so $S U (2)$ is compact Hausdorff.

SU(2) is a compact Hausdorff matrix group

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources