8
votes
Accepted
The sum of two roots when their pairing is zero
In a root system $R$ of type $B_2$, there are various pairs of short roots $\alpha, \beta$ with $(\alpha,\beta)=0$ but both $\alpha \pm \beta \in R$.
- 25.3k
7
votes
Accepted
Does every extension of a finite group by $\mathbb{R}^n$ split?
This is a standard cocycle averaging situation.
Choose an arbitrary (set-theoretic) section $s : G/N \to G$. The extent to which $s(G/N)$ fails to be a subgroup is measured by the cocycle $f : (G/N)^2 ...
- 10.7k
6
votes
Accepted
In what sense are Pauli matrices "intertwiners"?
I'm pretty sure that your objection is exactly right. That is, the Pauli matrices are literally a/the standard basis for the (complexified?!) Lie algebra $\mathfrak sl(2)$. Yes, $\mathbb R^3$ with ...
- 51.6k
6
votes
Can adding a single element to a Lie group make it infinite-dimensional?
There is no bound on the dimensionality of $G$.
In fact, a single element can generate a Lie group of arbitrarily large dimension.
Thus the $n$-dimensional torus $\mathbb T^n$ is the closure of the ...
- 440k
6
votes
Accepted
Why is the image of a Lie group representation an automorphism but a Lie algebra representation an endomorphism?
The short answer: Because $GL(V)$ is a group and $\mathrm{End}(V)$ is a Lie algebra.
In particular, $GL(V)$ has the following:
Every element has an inverse
There is no addition of matrices, only ...
- 2,810
6
votes
What is a simple way of understanding associated bundles?
I think that the right way to address this is to look at the example of the frame bundle of a vector bundle $E\to M$ in a nice description. If $V$ is the standard fiber of $E$, then the frame bundle $...
- 20k
5
votes
Finding the lie algebra of the symplectic lie group
A much simpler proof exists using standard properties of the matrix exponential. We wish to show that if $X^TJ + JX = 0$ holds, then $\exp(tX)^TJ\exp(tX) = J$ for all $t$. To that end, we note that ...
- 222k
5
votes
Geometry of Semi-Simple Lie Groups
The sentence "so it can be considered merely as a vector field in dimension one" makes no sense as a manifold is not the same thing as a vector field. There is no single vector field ...
- 76.3k
5
votes
Accepted
A compact connected Lie group with trivial center is a direct product of simple groups
This is true: a compact connected Lie group with trivial center is a product of simple compact connected Lie groups with trivial center.
Suppose $G$ is a compact connected Lie group. Then $G$ has a ...
- 49.8k
5
votes
Accepted
Prove all Lie group homomorphisms of the circle have certain form
I think you should just impose more. If $w\in \mathrm{ker}(\phi)$ such that $w^m=1$ then, passing to a power of $w$ if necessary, we can assume that $gcd(m,N)=1$, with $w^m=1$. The goal is to now show ...
- 2,662
5
votes
Accepted
Explicit Lie group embedding of $O(n)$ into $SO(n+1)$
I’ll just write up an answer with a few ways of attacking the problem, and let you fill in some of the details (which can easily be found if you search the site or any decent book). First, let me make ...
- 51k
5
votes
Accepted
Differential of conjugation map is smooth
This is basically by definition of smoothness: smooth means infinitely differentiable, so the derivative is still smooth. More precisely, suppose $M$ and $N$ are smooth manifolds and $f:M\to N$ is ...
- 326k
5
votes
Accepted
Geometric Intuition for Hamiltonian Actions
Per my comment, this answer is largely cribbed from an excellent text by Peter Woit called Quantum Theory, Groups and Representations: An Introduction. In my edition this discussion occurs in Chapter ...
- 5,565
5
votes
Trivial principal bundles and curvature.
In the non-abelian case, the curvature forms $F_{\mathcal{M}}^{A}$ and $F_{s}$ are generally not equal. This is because the connection form $A$ and the section $s$ transform differently under gauge ...
- 2,511
5
votes
Does the Mobius Strip have an homogenous embedding?
I think you're confounding "homogeneous" and something about embeddings.
The cylinder with the product metric is (metrically) homogeneous because for any point $(\alpha, C)$, the map
$$
(\...
- 92.4k
5
votes
Does the Mobius Strip have an homogenous embedding?
The accepted answer is simply wrong, there are no such things as "translations in the parameters (𝑠,𝑡), where 𝑠 is considered modulo 2𝜋." I will prove below that the open Moebius band ...
- 93.8k
5
votes
Accepted
All compact, connected, complex Lie groups are Abelian.
Let $G$ be a compact, connected, complex Lie group. If $h\in G$, consider the map $\sigma_h\colon G\longrightarrow G$ defined by $\sigma_h(g)=h^{-1}gh$ (that is, $\sigma_h$ is the conjugation by $h$). ...
- 423k
4
votes
Accepted
Does SO$(V, Q)$ have a unique connected double cover?
This is false in indefinite signature. The Wikipedia article uses nonstandard notation here and I will use standard notation: for me $SO$ refers to the subgroup of the orthogonal group of elements ...
- 415k
4
votes
How can I prove that every complex lie group contains a two-dimensional subgroup.
After looking through Olver's book, here is the story.
He is sloppy and occasionally forgets to give definitions, forgets to assume some conditions, etc. Be aware of this when reading. For instance, ...
- 93.8k
4
votes
Accepted
Not all nonnegative integer combinations of simple roots are dominant weights
Consider the irreducible root system of type $G_{2}$ and fix a choice of its simple roots by $\alpha_{1},\alpha_{2}$ such that the Cartan matrix is
$$\left(\begin{matrix}
2&-3\\
-1&2
\end{...
- 1,076
4
votes
Accepted
Do the nilpotent linear operators form a Lie algebra
A counterexample exists in dimension $2$:
Let $X=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $Y=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)$, then $XY-...
- 2,810
4
votes
Accepted
Is $SU(2)\otimes SU(3)$ a subgroup of $SU(4)$?
I think that usually $\mathrm{SU}(2)\times\mathrm{SU}(3)$ is the direct product, where we can interpret its elements concretely as block-diagonal matrices with blocks from $\mathrm{SU}(2)$ and $\...
- 151k
4
votes
Accepted
Natural rep of $ SL(2,\mathbb{Z})$ and decomposing tensor powers into irreps
It's just the same as for $SL_2(\mathbb{C})$. Restricting an algebraic representation of $SL_2(\mathbb{C})$ to $SL_2(\mathbb{Z})$ preserves irreducibility. The key fact here is that $SL_2(\mathbb{Z})$ ...
- 11.1k
4
votes
Can every Lie group be realized as conformal group of smooth manifolds
First of all, one cannot talk about conformal transformations of a smooth manifold: For the notion of conformality to be defined you need an extra structure besides a smooth atlas. I will work with ...
- 93.8k
4
votes
What does $ 5 \bigotimes 5 = 10 \bigoplus 5$ mean for SU(5) Lie group?
Let $V=\Bbb C^5$ be the "natural" representation of $SU(5)$, with $\bar V$ the conjugate representation. Then
$V\otimes V$ splits into $V\odot V$ and $V\wedge V$, symmetric and skew-...
- 124k
4
votes
Accepted
Calculating the differential of the map $A \mapsto A A^T$, $A \in M(n\times n, \mathbb{R})$
Regarding your 2nd question: the tangent space of the space of matrices $M(n\times n, \mathbb R)$ at a point $A$ is, as you said, identified with $M(n\times n, \mathbb R)$ itself.
Regarding your 1st ...
- 418
4
votes
The Representation of $\mathbf{Sp}(V)$ on $V$ is irreducible?
You should try to prove something a little bit stronger: for any non-zero $x\in V$, there is a symplectic basis with $e_1=x$.
Then if $x,y\in V$ are non-zero, you can choose symplectic bases $\mathcal{...
- 24.9k
4
votes
Prove that if $M$ is simply connected and $H^2(g) = 0$, then $M$ is symplectomorphic to an adjoint orbit.
The point is that under your hypothesis, there exist a moment map $J=M\to g^*$, and this map is equivariant. This is called strongly hamiltonian in the literature.
Let $\bf X$ the space of hamiltonian ...
- 7,373
4
votes
A criteron for vanishing of Lie algebra cohomology
Aaaah, this was one of those things which I never had the pleasure to really write down in the detail it deserves. Thought this would be shorter, but to really lay it all out, you do require a bunch ...
- 1,643
4
votes
Accepted
Is there explicit formula for the smallest enclosing ball in $SO_3$
The key point is that there is a 2-to-1 mapping $S^3\to SO(3)$ where $S^3$ is viewed as the set of unit quaternions (see this wiki page for more on this map). This map preserves the standard geodesic ...
- 14.7k
Only top scored, non community-wiki answers of a minimum length are eligible