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54 votes

Is SO(4) a subgroup of SU(3)?

No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument. If $H \subset G$ are Lie groups with $H$ ...
mme's user avatar
  • 9,463
41 votes
Accepted

Is $O_n({\bf Q})$ dense in $O_n({\bf R})$?

There's an easy argument based on the Cayley transform: If $a$ is a skew-symmetric $n$-by-$n$ real matrix, then $I_n+a$ is invertible (since $(I_n-a)(I_n+a)=I_n-a^2$ is a positive definite symmetric ...
Robert Bryant's user avatar
38 votes

Is SO(4) a subgroup of SU(3)?

Maybe the simplest argument, if you know something about compact Lie groups, is that SO(4) and SU(3) both have rank 2, i.e., they each contain a maximal torus, which is $S^1\times S^1$. Since all ...
Robert Bryant's user avatar
31 votes

Is SO(4) a subgroup of SU(3)?

No. Indeed $\mathrm{SO}(4)$ satisfies the following condition (which is a first-order existential formula) but not $\mathrm{SU}(3)$: $$\exists w,x,y,z: [x,w]\neq 1\neq [y,z],\; [w,y]=[w,z]=[x,y]=[x,z]$...
YCor's user avatar
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22 votes

Is SO(4) a subgroup of SU(3)?

It isn't too bad to describe the irreducible representations of $SO(4)$. We can realize $SO(4)$ as $SU(2) \times SU(2) / \langle (- \text{Id}, - \text{Id})\rangle$. To see this, identify $\mathbb{R}^4$...
David E Speyer's user avatar
21 votes

Is SO(4) a subgroup of SU(3)?

Here is a very minimalistic argument. $SO(4)$ contains a rank $3$ elementary abelian $2$-group (product of three groups of order two), namely the group of diagonal matrices in $SO(4)$. $SU(3)$ does ...
Tom Goodwillie's user avatar
18 votes

Is $O_n({\bf Q})$ dense in $O_n({\bf R})$?

By Cartan-Dieudonné's theorem, every element of $O_n({\bf R})$, resp. $O_n({\bf Q})$ is a product of at most n hyperplane reflections $\sigma_u$ for u in ${\bf R}^n$, resp. u in ${\bf Q}^n$. Now it ...
Name's user avatar
  • 2,006
16 votes

Is $O_n({\bf Q})$ dense in $O_n({\bf R})$?

Yes, here'a a proof by induction, granted the $n=2$ case (which is the only one where [basic] arithmetic occurs). Let $G$ be the closure. I first claim that $G$ acts transitively on the sphere. Indeed,...
YCor's user avatar
  • 61.8k
13 votes
Accepted

Continuous version of the fundamental theorem of invariant theory for the orthogonal group

Yes. It suffices to show that if one has a sequence $\vec v^{(n)} = (v^{(n)}_1,\dots,v^{(n)}_m) \in E^m$ whose Gram matrix $(\langle v^{(n)}_i, v^{(n)}_j \rangle)_{i,j=1,\dots,m}$ converges to a Gram ...
Terry Tao's user avatar
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12 votes
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Invariant polynomials under diagonal action of the orthogonal group

The invariants are generated by the quadratic polynomials $(u,u)$, $(u,v)$, and $(v,v)$ where $(.,.)$ is the scalar product defining $O(n)$. This pattern generalizes to arbitrary many copies of $\...
Friedrich Knop's user avatar
12 votes

Is SO(4) a subgroup of SU(3)?

In addition to all the completely correct answers to this question, it's probably worth spelling out some of the misconceptions that are common to physicists who work in this area and giving the ...
Aaron Bergman's user avatar
11 votes
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Separating closed $SO(p,q)$ orbits by invariant polynomials

The group $SO(p,q)$ is not per se an algebraic group. Rather there is an algebraic group $G$ such that $SO(p,q)=G(\mathbb R)$ is its group of real points. The main point is that also $G(\mathbb C)$ is ...
Friedrich Knop's user avatar
10 votes

Invariant polynomials under diagonal action of the orthogonal group

This is covered in Chapter 2, Section 9 (starting on page 52) in Weyl's The Classical Groups, Their Invariants and Representations. As already answered, there you will find: Theorem (Theorem 2.9.A) ...
Sean Lawton's user avatar
  • 8,414
10 votes

Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$

If it's the same $n$ then yes this can happen. For example, the lattice $D_4$ (consisting of all integer vectors in ${\bf Z}^4$ with even sum) has more isometries than ${\bf Z}^4$. If we allow ...
Noam D. Elkies's user avatar
9 votes
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A pair of non-conjugate subgroups: a simple proof

I think that the elements $g = \dfrac1{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}^{\oplus3}$ and $h = \dfrac1 2\begin{pmatrix} 1 & 1 & 1 & 0 & 1 & 0 \\ -1 & 1 ...
LSpice's user avatar
  • 11.7k
8 votes

Computing Haar measure of matrices sampled from SO(n)

Indeed, the distribution function of the eigenphases of a random matrix in $\operatorname{SO}(n)$ has a peak at 0 and at $\pm\pi$. It only becomes uniform for large $n$. The joint distribution ...
Carlo Beenakker's user avatar
6 votes

Nearest matrix orthogonally similar to a given matrix

Not a full answer, but some pointers on how to get a numerical method: If $\|\cdot\|_2$ denotes the spectral norm, then, by unitary invariance, your problem is equal to $$ \min_{T\in O(n)}\|AT-TB\|_2. ...
Dirk's user avatar
  • 12.4k
6 votes
Accepted

Upper bound on the sectional curvature of the orthogonal group

The $n = 3$ case is a straightforward computation using the identification of the cross product on $\mathbb{R}^{3}$ with the Lie bracket on $\mathfrak{so}(3)$. Namely, defined $\hat{x}:\mathbb{R}^{3} \...
Dan Fox's user avatar
  • 2,140
6 votes
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Grand tour of the special orthogonal group

[EDIT: Dan Asimov notified me that this construction is similar to a construction in his 1985 paper entitled "The Grand Tour: a Tool for Viewing Multidimensional Data". The construction in ...
Adam P. Goucher's user avatar
6 votes

Does $O(4,\mathbb{Q})$ have an exceptional outer automorphism?

$\def\QQ{\mathbb{Q}}\def\Out{\mathrm{Out}}\def\HH{\mathbb{H}}$I think tim penttila's answer may be misleading: $\Out(O(4, \QQ))$ is nontrivial and quite large, and the exceptional automorphism of $W(...
David E Speyer's user avatar
5 votes
Accepted

Bounding the non-multiplicativity of isometric projection

If the constant is allowed to depend upon the dimension, the estimate is simple. Let $A=O_AP_A,B=O_BP_B$, then $AB=O_AO_B(O_B^*P_AO_B)P_B$ and we are left with showing that if the product of two ...
fedja's user avatar
  • 60.6k
5 votes
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On the number of involutions in some groups

Let $q$ be an odd prime power and say that $q\equiv\epsilon\pmod4$, where $\epsilon=\pm1$. Using Geoff's suggestion I calculated the number of involutions in $\Omega_7(q)$ (the simple group; called $...
Richard Lyons's user avatar
5 votes

Does $O(4,\mathbb{Q})$ have an exceptional outer automorphism?

No. See [Edward A. Connors, The Automorphisms of $O^+_4(V)$ in the Anisotropic Case Proceedings of the American Mathematical Society, Vol. 54, No. 1 (Jan., 1976), pp. 16-18].
tim penttila's user avatar
5 votes

Probability distribution of vectors obtained from Gram-Schmidt process on i.i.d. Gaussian vectors

The orthogonalization requires $N\leq K$. For $N=K$ the resulting $N\times N$ unitary matrix is distributed according to the Haar measure, see On asymptotics of large Haar distributed unitary matrices ...
Carlo Beenakker's user avatar
4 votes

Nearest matrix orthogonally similar to a given matrix

Depends what you mean by "technique". Your problem is a quadratically constrained quadratic program. To be precise, the objective function is $$\mbox{tr} (A-TBT^t)(A^t - TB^t T^t) = \mbox{tr} ( AA^t +...
Igor Rivin's user avatar
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4 votes
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Motivation and reference for Brauer algebras

For motivation I would advise starting with Brauer's original paper. You'll need a JSTOR login though: https://www.jstor.org/stable/1968843?origin=crossref&seq=1#metadata_info_tab_contents
Simon Wadsley's user avatar
4 votes
Accepted

Explicit computation of spinor norm

Posted from the comments (1 2 3), by request. Of course, the spinor formula itself arguably is an explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries,...
LSpice's user avatar
  • 11.7k
4 votes

Grand tour of the special orthogonal group

Take a parameterization of $SO(n)$ that has domain a hypercube of dimension $\binom{n}{2}$, e.g. https://math.stackexchange.com/questions/965451/. Now take the image of a Brownian motion in the ...
Steve Huntsman's user avatar
4 votes
Accepted

Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$

Let $n=m^2$ be a square. Then the vector $(1,\dots,1)\in\mathbf{Q}^n$ has norm $m$, as does $(0,\dots 0,m)$. Hence by Witt's theorem there exists an element $u$ of $\mathrm{O}_n(\mathbf{Q})$ mapping $(...
YCor's user avatar
  • 61.8k
4 votes

Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$

A theorem which you might be looking for is that, if $G$ is any finite subgroup of $O_n(\mathbb{Q})$, then there is a lattice $\Lambda$ in $\mathbb{Q}^n$ which is preserved by $G$. However, that ...
David E Speyer's user avatar

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