# Tag Info

53

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 $H$ closed in $G$, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a ...

35

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 maximal tori are conjugate in a compact Lie group, if SO(4) were a subgroup, then, after a conjugation in SU(3), you can assume that SO(4) contains the diagonal ...

29

No. Indeed $\mathrm{SO}(4)$ satisfies the following condition (which is a first-order existential formula) but not $\mathrm{SU}(3)$: $$\exists w,x,y,z: [x,w]\neq 1\neq [y,z],\; [w,y]=[w,z]=[x,y]=[x,z]$$ (it just says there are two commuting pairs of non-abelian subgroups). Indeed $\mathrm{SO}(4)$ even contains a copy of the direct product of two non-abelian ...

22

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 $\mathbb{R}^4$ with the quaternions, and identify $SU(2)$ with the norm $1$ quaternions. For $(u,v) \in SU(2) \times SU(2)$, the map $q \mapsto uqv^{-1}$ from the ...

20

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie subgroup, and $\mathrm{SL}(2,\mathbb{R})$ cannot be a Lie subgroup of any compact Lie group. To see this, note that any compact Lie group has a bïinvariant ...

18

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 not. Any abelian subgroup of $SU(3)$ is conjugate to a group of diagonal matrices, and among the diagonal matrices there are only four whose squares are the ...

12

Here is a cool fact about $\mathrm{SL}(2, \mathbb{R}) / \mathrm{SL}(2, \mathbb{Z})$; it is homeomorphic to the complement of the trefoil knot in the three-sphere. Apparently this was first proved by Quillen - see page 84 of Introduction to algebraic K-theory by Milnor. For a discussion with further links, see here: Why S^3-K and SL(2,R)/SL(2,Z) are ...

11

These days, many popular quotients of ("real") hyperbolic three-space are obtained as $\Gamma\backslash SL_2(\mathbb C)/SU(2)$, where $\Gamma$ is the group of units in a quaternion division algebra over a complex-quadratic field. More generally, with a field $k$ with exactly one complex archimedean completion, and the rest real, a quaternion ...

10

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 physics answer. The main thing that arises all the time is that physicists often conflate representations of a group and of its Lie algebra. The reason one can get ...

8

This is an answer to questions 7 and 8 (I have to say, having 8 questions in one post is way too much for my taste): Suppose that $M$ is a finite-volume quotient of $H^3$ or a compact quotient of $H^2\times {\mathbb R}$ by a discrete group of isometries. Then $M$ cannot be homogeneous (in the non-Riemannian sense!), meaning that there is no (finite-...

8

$\def\Spin{\text{Spin}}$The cover $\Spin(n) \to SO(n)$ is $2$ to $1$, the nontrivial element kernel is a central element of $\Spin(n)$ which I'll call $z$. Since $z$ is central, it acts on any irrep of $\Spin(n)$ by $\pm 1$; if $z$ acts by $1$, then the irrep factors through $SO(n)$, if $z$ acts by $-1$, then it does not. Irreps of Lie groups are usually ...

6

I think, the statement "the set of Borel subalgebras of a semi-simple Lie algebra $\mathfrak g$ forms a Zariski-closed subset of a Grassmannian" can be understood in more than one way and then the question becomes less trivial than one thinks. Everything is over $\mathbb C$, by the way, otherwise things become even more complicated. First ...

6

It is a standard fact that the automorphism group of $G=\mathrm{SL}_2(\mathbf{R})$ as topological group equals $\mathrm{PGL}_2(\mathbf{R})$, which is also the automorphism group of the Lie algebra viewed as $\mathbf{R}$-algebra. In particular, the outer automorphism group as topological group is cyclic of order 2. I claim that this is the same as abstract ...

5

YCor's comment contains the essential idea needed for the proof, but maybe a few more details would be helpful. (If YCor does provide something similar later, feel free to award YCor's answer the bounty.) Consider the mapping $\sigma:\mathrm{GL}(2,\mathbb{H})\to M_2(\mathbb{H})$ given by $$\sigma(A) = A^* A$$ where $A^*$ is the conjugate transpose of $A$ ...

4

(Comment converted to answer per request:) The “surprising result” about simply connected homogeneous spaces in your (currently) last paragraph is Montgomery's Theorem (1950): generally (in your notation) if $G/H$ is compact and $H$ closed connected, then $K$ is transitive on $G/H$. The theorem is also discussed in Samelson (1952, p. 17).

3

If $G$ is a reductive Lie group, then it may be viewed as a group of symmetries for a geometry whose salient objects are the coset spaces $G/P$ with respect to parabolic subgroups $P$. This point of view is alluded to in TWF249, and worked out in detail for $G = \operatorname{PGL}(3)$ (not, as mentioned in TWF250, the isogenous group $\operatorname{SL}(3)$, ...

3

The second version given in the question is correct: \begin{align} \mathrm{Pin}^+(4k+1) &\cong \mathrm{Spin}(4k+1) \times \mathbb{Z}_2 \\ \mathrm{Pin}^-(4k+1) &\cong (\mathrm{Spin}(4k+1) \times \mathbb{Z}_4)/\mathbb{Z}_2 \\ \mathrm{Pin}^+(4k+3) &\cong (\mathrm{Spin}(4k+3) \times \mathbb{Z}_ 4)/\mathbb{Z}_2 \\ \mathrm{Pin}^-(4k+3) &\...

1

The quadratic form whose matrix is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_n \\ 0 & I_n & 0 \end{pmatrix}$ gives an embedding of $\operatorname{SO}(2n + 1, \mathbb C)$ in $\operatorname{GL}(2n + 1, \mathbb C)$ whose derivative is your specified embedding $\mathfrak{so}(2n + 1, \mathbb C) \to \mathfrak{gl}(2n + 1, \mathbb C)$. Under ...

1

$\newcommand{\mk}{\mathfrak}$Let $\mk{h}_{2n+1}$ be the $(2n+1)$-dimensional Lie algebra (basis $x_1,\dotsc,x_n,y_1,\dotsc,y_n,z$, nonzero brackets $[x_i,y_i]=z$). The $n$-dimensional abelian Lie algebra $\mk{a}_n$ (basis $a_1,\dotsc,a_n$) acts on it by $a_i\cdot x_i=x_i$, $a_i\cdot y_i=-y_i$, rest of the action being zero. Through the quotient map \$\mk{h}_{...

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