Reputation
2,539
Top tag
Next privilege 3,000 Rep.
Cast close and reopen votes
Badges
3 21
Impact
~24k people reached

12h
accepted Tangent bundle of manifold with no odd dimensional sub-bundles
12h
comment Tangent bundle of manifold with no odd dimensional sub-bundles
I realized after asking the first question that it was rather obviously false; thanks for the simple counterexample. I guess that the orientable double cover is an odd-dimensional orientable subbundle of the double cover of $M$, which is impossible since it's entirely contained in one of the copies of $M$.
2d
asked Tangent bundle of manifold with no odd dimensional sub-bundles
2d
accepted $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes
2d
comment $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes
Thanks! I see the mistake I was making before - I didn't check that $(1+ax+x^2)^{-1}(1+x)^{n+1}$ could have degree less than $n-1$ in larger cases. Quite silly of me
2d
revised $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes
added info
2d
asked $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes
Mar
23
awarded  Popular Question
Mar
4
comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
that being said, now it's unclear what the problem statement actually means, though we assume the more difficult one evidently
Mar
4
revised Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
deleted 15 characters in body
Mar
4
comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
@rah4927, I think you're right; my original line of thinking was that if they're all distinct, then the LHS is bounded by something smaller than the RHS, but I think I messed up. I'll revert it.
Mar
3
comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
yes, you're right. my mistake!
Mar
3
comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
For $n=3,$ there are two solutions: $(1,2,3)$ and $(2,2,2)$ so wolframalpha lied to you. @rah4927, yes, there are infinitely many solutions over the reals. Just pick $a_i \le i$ for $i < n$ and then by continuity you can always find $a_n$ that works.
Mar
3
comment Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
A remark: $a_i\mid 2i$ for each $i$ and note that solutions with $a_1 = 1$ are in bijective correspondence with solutions having $a_1 = 2$ just by taking the $n$-tuple with $2i/a_i$ in place of $a_i,$ so you can assume $a_1 = 1$ without loss of generality. Also, you can reduce it to an integer equation, but I don't know how helpful that is.
Mar
3
revised Positive integer solutions of $\frac{1}{a_1}+\frac{2}{a_2}+\frac{3}{a_3}+\cdots+\frac{n}{a_n}=\frac{a_1+a_2+a_3+\cdots+a_n}{2}$
clarifying problem statement, as $a_i$'s are not necessarily distinct
Mar
2
answered Minimizing the product $xy$ subject to a polynomial constraint on $x, y$
Mar
1
reviewed Approve Complex Number Maths
Feb
14
comment Bound on $L^1$ norm of pairwise sums
yes @RoryDaulton
Feb
14
revised Bound on $L^1$ norm of pairwise sums
expanded on question
Feb
14
comment Bound on $L^1$ norm of pairwise sums
haha yes, sorry