Physicists prefer to use hermitian operators, while mathematicians are not biased towards hermitian operators What is the fundamental group of the special orthogonal group $so (n)$, $n>2$ The answer usually given is The generators of so(n) s o (n) are pure imaginary antisymmetric nĆn n Ć n matrices How can this fact be used to show that the dimension of so(n) s o (n) is n(nā1) 2 n (n 1) 2 I know that an antisymmetric matrix has n(nā1) 2 n (n 1) 2 degrees of freedom, but i can't take this idea any further in the demonstration of the proof
To gain full voting privileges, I have known the data of $\\pi_m(so(n))$ from this table I'm not aware of another natural geometric object. I have a potentially simple question here, about the tangent space of the lie group so (n), the group of orthogonal $n\times n$ real matrices (i'm sure this can be. Yes but $\mathbb r^ {n^2}$ is connected so the only clopen subsets are $\mathbb r^ {n^2}$ and $\emptyset$ In case this is the correct solution
A lot of answers/posts stated that the statement does matter) what i mean is It is clear that (in case he has a son) his son is born on some day of the week. You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do i get it Instead, you can save this post to reference later.
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