20

Bott's result is best formulated by saying that the sequence of spaces
$$
\mathbb Z\times BO,\quad O,\quad O/U,\quad U/Sp,\quad \mathbb Z\times BSp,\quad Sp,\quad Sp/U,\quad U/O
$$
are related by the property that each one is the loop space of the previous one (mod 8).
From that, you get for example that $\pi_4(O)=\pi_3(O/U)=\pi_2(U/Sp)=\pi_1(Z\times BSp)=\...

answered Jan 6 '15 at 22:25

André Henriques

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You might like to look at my thesis, where a lot of this kind of thing is recalled in a consolidated way:
http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf
Some key points:
The homotopy groups of the space $O$ are the same as the homotopy groups of the spectrum $KO$, shifted by one. It is easier to think about $KO$, because it has a ring ...

answered Jan 6 '15 at 22:50

Neil Strickland

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13

This is a long comment on André's answer. Here is a way to organize the spaces that appear in it using the categories of $\mathbb{Z}_2$-graded modules over Clifford algebras.
Let $\text{Cliff}(p, q)$ denote the Clifford algebra obtained by adjoining $p$ anticommuting square roots of $1$ and $q$ anticommuting square roots of $-1$ to $\mathbb{R}$, and let $\...

answered Jan 6 '15 at 23:24

Qiaochu Yuan

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No. E.g. $\psi=\begin{pmatrix}0&0&-1&1\\0&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}$ is nilpotent: $\psi^3=0$. If your $\phi=\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&b&0\end{pmatrix}$ was $Q\psi Q^{-1}$, we would have $\phi^3=\begin{pmatrix}0&-a^3&0&0\...

answered Aug 11 '15 at 15:36

Francois Ziegler

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There is a complete characterization of matrices that belong to at least one
orthogonal group. It reads as follows over any arbitrary field $\mathbb{F}$
with characteristic different from $2$ (with algebraic closure denoted by $\overline{\mathbb{F}}$:
Given a matrix $M \in \mathrm{GL}_n(\mathbb{F})$, there exists an
invertible symmetrix matrix $\beta$ ...

12

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 matrix $(\langle v_i, v_j \rangle)_{i,j=1,\dots,m}$ of a tuple $\vec v = (v_1,\dots,v_m) \in E^m$, then after applying linear isometries to each of the $\vec v^...

10

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 $\mathbb R^n$. This is called the first fundamental theorem for the orthogonal group.

9

A nice way to do this calculation can be extracted from the papers
[ABS] Atiyah, Bott, Shapiro: ''Clifford modules''
[AtKR] Atiyah: ''K-Theory and Reality''.
Let $A_n$ be the group of $Cl^n$-modules, modulo $Cl^{n+1}$-modules, as introduced by Atiyah, Bott, Shapiro. The sum of these is a graded ring, and the main result of [ABS] is that the homomorphism
$...

answered Jan 7 '15 at 10:23

Johannes Ebert

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Consider the entries of $X^t \beta X - \beta$ as linear equations in the entries of the symmetric matrix $\beta$, and solve. For $n \times n$ matrices these are
$(n+1)n/2$ equations in $(n+1)n/2$ unknowns. If there is a nontrivial solution, compute the rank of $\beta$ (using generic values for the free variables).
EDIT: It probably should be mentioned ...

8

Such maps are conformal. A theorem of Liouville says that if $n\geq 3$,
the only conformal maps (defined in some region in $R^n$) are Mobius. A Mobius map is a composition of inversions
in spheres. For example $x\mapsto x/|x^2|$ is the inversion in the unit sphere.
Inversions in all spheres generate the Mobius group.
Derivative of a conformal map is a ...

answered Jan 21 '16 at 20:01

Alexandre Eremenko

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8

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) Every orthogonal invariant in vectors $x^1,...,x^m$ in $\mathbb{R}^n$ is expressible in terms of the $m^2$ scalar products $(x^i,x^j)$.

8

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 defined which leads to two different concepts of orbits. First, there are the orbits $G(\mathbb R)v$ which you are probably after.
But then there are also the ...

7

Interesting question !
Let us denote $L_X:\beta\mapsto X^T\beta X$. This is an endomorphism of ${\bf Sym}_n({\mathbb C})$.
Lemma. The characteristic polynomial of $L_X$ is
$$\prod_{i\le j}(t-\mu_i\mu_j)$$
where $\mu_1\ldots,\mu_n$ are the eigenvalues of $X$.
Sketch of the proof : the characteristic polynomial has degree $n(n+1)/2$. Suppose that $X$ ...

6

Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).
Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\...

6

The question is brief, but there are a great many unknown features of the modular representation theory of these families of groups (for the prime $p=2$). The approach via algebraic groups pioneered by Steinberg is conceptually attractive but far from able to deal effectively with small primes at this point. There might also be approaches via finite ...

answered Nov 19 '15 at 17:31

Jim Humphreys

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6

$\DeclareMathOperator{\SO}{SO}$$\DeclareMathOperator{\RP}{RP}$$\DeclareMathOperator{\Stab}{Stab}$There is a difference between the sets $C = [-1,1]^3$ and $E = [0,1]^3$. Note that $\Stab(C)$, the stabilizer of $C$ inside of $\SO(3)$, is called the cube group and it has 24 elements. Note that $\Stab(E)$ has three elements.
Are you sure you want $E$ and not ...

6

This is at least a partial answer. There are two distinct viewpoints here: the concrete one involving forms and automorphism groups (which came first historically and usually requires characteristic $\ne 2$ to avoid tricky points) and the much more general one involving simple algebraic groups over a field $k$ (where Borel and Tits covered a great deal of ...

answered Jul 30 '14 at 16:25

Jim Humphreys

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6

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
$$
\min_{T\in O(n)}\|AT-TB\|_2.
$$
As such, the objective is the concatenation of the linear map $T\mapsto AT-TB$ and the norm and hence, convex. Subgradients of the objective can be ...

6

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 $\hat{x}:\mathbb{R}^{3} \to \mathbb{R}^{3}$ by $\hat{x}y = x \times y$. Then $|\hat{x}|^{2} = 2|x|^{2}$, and the Jacobi identity implies $[\hat{x}, \hat{y}] = \widehat{x \times y}$, so ...

5

If $q$ is a quadratic form over a field $k$, the orthogonal group $\mathrm{O}(q)$ doesn't change if you replace $q$ by $\lambda q$ for $\lambda\in k^*$. In rank 3, since $\lambda ^3\equiv \lambda $ (mod. $k^{*2}$), this implies that you can consider only forms of discriminant 1. Over $\mathbb{Q}_p$ $(p>2)$ this leaves only the invariant $\varepsilon $ (...

5

This has to do with SL(2) representation theory. The answer for $O(2l)$ is $d=2l-1$. First of all, let $W$ be the irreducible representation of $SL(2)$ of dimension $2l-1$. This preserves a non-degenerate quadratic form and hence the image of $SL(2)$ lies in $SO(2l-1)\subset SO(2l)$. The image of the nontrivial unipotent of $SL(2)$ is "regular" in $SO(2l-1$ ...

5

The smallest degree faithful representation of $S_n$ is $n-1$ in characteristic does not divide $n$. It is Theorem 22 of Chapter 19, Section 8 of Y. G. Berkovich, E. M. Zhmud; Characters of finite groups. Part 2.
Translated from the Russian manuscript by P. Shumyatsky, V. Zobina and Berkovich. Translations of Mathematical Monographs, 181. American ...

answered Oct 9 '15 at 10:23

Benjamin Steinberg

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5

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 positive definite self-adjoint operators $X=O_B^*P_AO_B$ and $Y=P_B$ is $\delta$-close to a unitary operator $V$, then it is $C\delta$-close to $I$. Now, we do not ...

5

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 $O_7(q)$ in the Atlas) to be
$$
\frac12 q^5(q^4+q^2+1)(q+\epsilon) + \frac12 q^6(q^4+q^2+1)(q^2+1) + \frac12 q^3(q^3+\epsilon).$$
The summands give the ...

4

$\def\RR{\mathbb{R}}$A brute force approach shows that there are no other $C^2$ solutions. Let $F: \RR^n \to \RR^n$ have orthogonal Jacobian everywhere. We will show that the Hessian of $F$ vanishes everywhere, so $F$ is linear.
It is enough to show that Hessian vanishes at $0$, since there is nothing special about $0$.
Translating and rotating our ...

answered Jan 21 '16 at 19:59

David E Speyer

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4

This is a confusing concept, as acknowledged
in the paper you cite,
"On the definition and computation of rectilinear convex hulls"
(Elsevier link).
(Image from Wikipedia.)
It may help to look at this more modern paper:
Alegría-Galicia, Carlos, Tzolkin Garduño, Areli Rosas-...

answered Sep 24 '15 at 23:00

Joseph O'Rourke

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4

Depends what you mean by "technique". Your problem is a quadratically constrained quadratic program. To be precise, the objective function is
$$\mbox{tr} (A-TBT^t)(A^t - TB^t T^t) = \mbox{tr} ( AA^t + TBB^tT^t - A TB^t T^t - TBT^t A^t) =\\ = \|A\|_F^2 + \|B\|_F^2 - 2 \mbox{tr} (ATB^t T^t),$$
so you are trying to maximize the trace of $ATB^t T^t$ under the ...

answered Mar 7 '18 at 15:56

Igor Rivin

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3

More useful than embedding it in some higher-dimensional space will be to give it as a manifold with identifications, I think. Topologically, $SO(3)/\!\sim$ is the same space as what you get by taking an octahedron and gluing the opposite faces with a $1/3$ turn. This is related to the 24-cell, one of the 6 different regular 4-dimensional polytopes (...

3

Your ultimate question was answered by Gauss: $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ is nonempty if and only if the class of $f$ is ambiguous (i.e. its square is the trivial class).
Indeed, $f(x,y)$ is improperly equivalent to itself if and only if $f(x,y)$ is properly equivalent to $f(y,x)$. As the classes of $f(x,y)$ and $f(y,x)$ are inverses to ...

3

This is really just a very long comment.
First, if you haven't run across the Orthogonal Procrustes Problem before, you may find it interesting. (It's very similar, and has an efficient algorithm.)
Second, a little observation. As Igor Rivin points out, this problem is equivalent to maximizing
$$\max_{ T\in O(n,\Bbb R)} \mathrm{Tr}[TB^{t}T^{t}A]$$
...

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