|
- lie groups - Lie Algebra of SO (n) - Mathematics Stack Exchange
Welcome to the language barrier between physicists and mathematicians Physicists prefer to use hermitian operators, while mathematicians are not biased towards hermitian operators So for instance, while for mathematicians, the Lie algebra $\mathfrak {so} (n)$ consists of skew-adjoint matrices (with respect to the Euclidean inner product on $\mathbb {R}^n$), physicists prefer to multiply them
- Fundamental group of the special orthogonal group SO(n)
Also, if I'm not mistaken, Steenrod gives a more direct argument in "Topology of Fibre Bundles," but he might be using the long exact sequence of a fibration (which you mentioned)
- Why $\\operatorname{Spin}(n)$ is the double cover of $SO(n)$?
You can let $\text {Spin} (n)$ act on $\mathbb {S}^ {n-1}$ through $\text {SO} (n)$ Since $\text {Spin} (n-1)\subset\text {Spin} (n)$ maps to $\text {SO} (n-1)\subset\text {SO} (n)$, you could then use the argument directly for $\text {Spin} (n)$, using that $\text {Spin} (3)$ is simply connected because $\text {Spin} (3)\cong\mathbb {S}^3$ I'm not aware of another natural geometric object
- Homotopy groups O(N) and SO(N): $\\pi_m(O(N))$ v. s. $\\pi_m(SO(N))$
I have known the data of $\\pi_m(SO(N))$ from this Table: $$\\overset{\\displaystyle\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\quad\\textbf{Homotopy groups of
- Prove that the manifold $SO (n)$ is connected
The question really is that simple: Prove that the manifold $SO (n) \subset GL (n, \mathbb {R})$ is connected it is very easy to see that the elements of $SO (n
- How connectedness of $O(n)$ or $SO(n)$ implies the connectedness of . . .
So, the quotient map from one Lie group to another with a discrete kernel is a covering map hence $\operatorname {Pin}_n (\mathbb R)\rightarrow\operatorname {Pin}_n (\mathbb R) \ {\pm1\}$ is a covering map as @MoisheKohan mentioned in the comment I hope this resolves the first question If we restrict $\operatorname {Pin}_n (\mathbb R)$ group to $\operatorname {Spin}_n (\mathbb R
- Dimension of SO (n) and its generators - Mathematics Stack Exchange
The generators of $SO(n)$ are pure imaginary antisymmetric $n \\times n$ matrices How can this fact be used to show that the dimension of $SO(n)$ is $\\frac{n(n-1
- What is the relationship between SL (n) and SO (n)?
I'm in Linear Algebra right now and we're mostly just working with vector spaces, but they're introducing us to the basic concepts of fields and groups in preparation taking for Abstract Algebra la
|
|
|