Talk:Orthogonal group

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[edit] Over a finite field

I think it would be useful to discuss the orthogonal groups over a finite field here. If someone is capable of writing such a section it would be greatly appreciated. TooMuchMath 17:31, 15 April 2006 (UTC)

This is absolutely essential. It's a shame that there is no section here on it. My knowledge of these groups isn't good enough to do a section, however. We also need some words on O+ and O- etc. Triangle e 14:26, 9 December 2006 (UTC)

[edit] Incorrect

The definition of the orthogonal group given here is FALSE. The orthogonal group is defined only over the real numbers. When defined over the complex numbers it is known as the unitary group. Over quaternions it is known as the spin group.

Let $E$ the the $n \times n$ identity matrix and

$\mathbb{H} = \{a + ib + jc + kd : (a,b,c,d) \in \mathbb{R}^4, \ i^2 = j^2 = k^2 = -1, \ ij = k, \ jk = i, \ ki = j \} \ .$

Then the CORRECT definitions are given as

$O(n) = \{ X \in \mbox{Mat}(n,\mathbb{R}) : X^{\top}X = E \} \ ,$

$U(n) = \{ X \in \mbox{Mat}(n,\mathbb{C}) : \overline{X}^{\top}X = E \} \ ,$

$Sp(n) = \{ X \in \mbox{Mat}(n,\mathbb{H}) : \overline{X}^{\top}X = E \} \ .$

The definition of the orthogonal group comes from finding all matrices which preserve the standard euclidean dot product, so the preserve distances and angles. The definition of the unitary group comes from finding matrices which preserve the standard Hermitian product. We see that

$O(n) \subset U(n) \subset Sp(n) \ .$

In particicular, the statement that $\det(X) = \pm 1$ is also INCORRECT. It is true for the othogonal group since the definition $X^{\top}X = E$ means that $\det(X^{\top}X) = \det(E)$ which in turn shows that $det(X) 2 = 1.$

For the unitary group and the spin group we see that $\det(\overline{X}^{\top}X) = \det(E)$ this means that

$\overline{\det(X)}\det(X) = 1 \ ,$ which means that

| det(X) | = 1.

So for the unitary group, we see that

$\det(X) \in \{ z \in \mathbb{C} : |z| = 1 \} \neq \{\pm 1\} \ .$

Since $\mathbb{H}$ is not commutative, the idea of a determinant is not well defined!

—The preceding unsigned comment was added by 128.101.10.93 (talk • contribs) .

You are confusing the complex orthogonal group (preserving a symmetric form) with the unitary group (preserving a hermitean form). R.e.b. 18:52, 5 May 2006 (UTC)

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Talk:Orthogonal group

From Wikipedia, the free encyclopedia

[edit] Over a finite field

[edit] Incorrect

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