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May
18
comment differential in AHSS for spin cobordism
...BTW could you please point out where exactly att. notes show that $d_2$ is non-trivial?
May
18
comment differential in AHSS for spin cobordism
...Higher differentials, on the other hand, are not coh. operations but higher coh. operations (secondary etc) — which are much less... tractable. (That's just my ignorance, but) I don't know if there is an explicit description even for $d_5$ in AHSS for complex K-theory...
May
18
comment differential in AHSS for spin cobordism
Well, for any cohomology theory the first non-trivial differential in AHSS ($d_2$ for MSpin, $d_3$ for BU...) is a stable cohomological operation. All cohomological operations are known and in low degree there are very few choices. In particular, there is only one non-trivial operation of deg 2 with Z/2-coefficients, $Sq^2$.
May
18
comment Proof that the subalgebra generated by any two elements of $\mathbb{O}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$
...that was the plan I had in mind — but actually, proving 'diassociativity' seems to be not that easy, sorry.
May
18
comment Proof that the subalgebra generated by any two elements of $\mathbb{O}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$
Indeed. Using a similar argument one can show that any associative division algebra is $\mathbb R$, $\mathbb C$ or $\mathbb H$. So if we can show the well-known fact that a subalgebra of $\mathbb O$ is associative, we're done...
May
18
comment Proof that the subalgebra generated by any two elements of $\mathbb{O}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$
Can you prove that a subalgebra of $\mathbb H$ generated by any non-real element is isomorphic to $\mathbb C$?
May
18
comment differential in AHSS for spin cobordism
(Re: Is this also the case here?) Sure — but one still needs to show that this operation is nonzero...
May
17
comment Determining if these surjections have sections
I wonder if $\pi^{-1}(A_5)$ is always isomorphic to the binary icosahedral group (and not to $S_5$)...
May
17
comment Proving $\sum_{k=1}^{n}\binom{n-1}{k-1}{\binom{n+k}{k}}^{-1}=\frac 12$ combinatorially
Since both sums have non-integral terms, it's not entirely clear what kind of combinatorial argument are you looking for...
May
14
comment Is Frobenius the only magical automorphism?
In the last proof $R\to R$ should be $R\to R_{\sigma^n}$, right?
May
13
comment How can I find $\sum\limits_{n=i+1}^\infty \binom{n-1}{i}\left (\frac{1}{3}\right)^{n}$?
en.wikipedia.org/wiki/Binomial_series#Special_cases
May
13
comment Restrictions on the coefficients that allow a polynomial in a field of characteristic 0 to be solvable by radicals and the associated formula.
An irreducible polynomial equation is solvable iff its Galois group is solvable — if this doesn't answer you question, what is you question exactly?
May
13
comment Finding a Monoid
Well, this is just colimit of sets of possible matrix ranks — i.e. of sets $\{0,...,n\}$ (wrt obvious inclusions) — and such colimit is just a union of all these sets
May
13
comment Finding a Monoid
Well, yes, it's path-connected, so $\pi_0$ is a 1-element set. Now you have to prove that all sets $\{A^2=A,\operatorname{rk}(A)=k\}\subset IM_n$ are path-connected, and basically you're done.
May
13
comment Finding a Monoid
Can you find, say, $\pi_0 GL_n(\mathbb C)$?
May
13
comment Trigonometric identity involving sum of “Dirichlet kernel like” fractions
Oh, seeing the word 'induction' I haven't tried to read your answer before — but now I see that my solution is essentially the same...
May
13
comment Finding a Monoid
well, that's a very simple colimit actually; but before taking colimits — can you find $\pi_0\mathop{\mathcal I}(M_n(\mathbb C))$?
May
13
comment Finding a Monoid
idempotent = projection on a subspace...
May
13
comment Verifying Touchard's Identity
What do you mean by «progress from here on»? — the linked page contains (a sketch of) a proof. If you don't understand the plan, perhaps you should read some general introduction to generating functions; otherwise — try to follow it.
May
13
comment Trigonometric identity involving sum of “Dirichlet kernel like” fractions
(Re: «Dirichlet kernel like») maybe «Fejér kernel like»...