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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
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)
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
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
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
How to find the difference between the sons and mothers age if it . . . A son had recently visited his mom and found out that the two digits that form his age (eg :24) when reversed form his mother's age (eg: 42) Later he goes back to his place and finds out that this whole 'age' reversed process occurs 6 times And if they (mom + son) were lucky it would happen again in future for two more times
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