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I have known the data of $\\pi_m(so(n))$ from this table The generators of $so(n)$ are pure imaginary antisymmetric $n \\times n$ matrices I'm not aware of another natural geometric object.

Artofit

Welcome to the language barrier between physicists and mathematicians Assuming that they look for the treasure in pairs that are randomly chosen from the 80 Physicists prefer to use hermitian operators, while mathematicians are not biased towards hermitian operators

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. 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). 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. 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. 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

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. Each of 20 families selected to take part in a treasure hunt consist of a mother, father, son, and daughter