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Mar
30
comment Counter-example to exponential law for locally compact [non-Hausdorff] spaces
@johndoe ...And Sierpinski set is, perhaps, a good choice for $Z$ — AFAIR, there is some theorem along the lines 'if the exponential law hold for $Z=\text{Sierpinski}$ it holds for all $Z$'.
Mar
30
comment Counter-example to exponential law for locally compact [non-Hausdorff] spaces
...the idea of taking finite $X$ and $Y$ is quite tempting — because finite topological spaces are locally-compact (and I don't know many example of non-Hausdorff locally-compact spaces). I've tried to play with some examples — but haven't succeeded.
Mar
30
comment Counter-example to exponential law for locally compact [non-Hausdorff] spaces
@johndoe Well, if either $X$, $Y$ or $Z$ is discrete the statement seems to be more or less obviously true. But slightly more generally...
Mar
30
comment Counter-example to exponential law for locally compact [non-Hausdorff] spaces
Related (but not answering the question): @ n-Category Cafe
Mar
21
comment Combinatorial definition of Hall–Littlewood polynomials (sum over SSYT?)
@IgorMakhlin (И про t-версию Бриона и т.п. мы бы с М.Б. с интересом послушали в какой-то момент.)
Mar
21
comment Combinatorial definition of Hall–Littlewood polynomials (sum over SSYT?)
@IgorMakhlin О, привет. Well, yes and no: there is an explicit description of the $t$-weight $\psi$ in Macdonald's book — but it's complicated and not terribly satisfying. So if you have a better answer, please explain it (here or iRL).
Mar
12
comment Geometric interpretation for sum of fourth powers
Well, yes, $1^k+2^k+...+n^k$ is the value of an Ehrhart polynomial for the 'hybercubic pyramid'. But does this help to compute the sum?
Jan
30
comment Is reduced homology a full functor on connected spaces?
The question is about the category of connected spaces.
Jan
30
comment Is reduced homology a full functor on connected spaces?
Uh? $\tilde H(pt)=0$.
Jan
22
comment Curious Binomial Coefficient Identity
it's a (yet another) form of Vandermonde's identity
Jan
19
comment Binomial Identity $\sum\binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$
see also math.stackexchange.com/q/1107465 for a bijective proof of an equivalent identity
Jan
19
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
come to think of it, substituting $a\to 2n+1$, $n\to 2m-2n$ (and $k\to m-2n+k$) one can see that these two identities are equivalent
Jan
19
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
Both identities are of the form $\sum_k\binom a{b-2k}\binom{c+k}d=\binom ef$ — and I strongly suspect there is a unified proof.
Jan
18
comment Does there exist $\mathbf{Q} \subset R \subset \mathbf{C}$, $R$ ring & not field
There is an injective map from $\mathbb Q[x]\to\mathbb R$ taking $x$ to $\pi$. Call the image of this map whatever you like — but that's an example.
Jan
17
comment Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$
cf. math.stackexchange.com/q/601940 etc
Jan
17
comment Elementary proof of the fact that any orientable 3-manifold is parallelizable
«The second surprise in this story» — that's, IMHO, the most interesting part...
Jan
17
comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$
@1234 $\frac{P(x)+P(-x)}2$ extracts even terms from any polynomial $P$, you know...
Jan
17
comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$
Well, do you know the answer for just $\sum \binom nlx^l$?
Jan
17
comment Elementary proof of the fact that any orientable 3-manifold is parallelizable
Do you know the standard proof («$w_1=w_2=0$ implies parallelizable by elementary obstruction theory» + «$w_1=0$ implies $w_2=0$ by Wu's formulas»)? It's not that hard — and at least the first part is, in a sense, the most straightforward approach possible (but I indeed don't know any intuitive explanation of the second part).
Jan
16
comment If $a,b,c$ are real numbers all less than or equal to $1$ such that $a+b+c=0$ , then is it true that $(1-a)(1-b)(1-c) \le 1$?
«which is true» — why, actually? ($ab$ can be negative)