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Q&A for professional mathematicians
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I hope everyone is doing well. Let $K \subset \mathbb{R}^n$ be a centrally symmetric convex body $(K = -K)$. Denote by $K \mid H$ the orthogonal projection of $K$ onto $H$, where $H$ is an $n - 1$
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Hottest Questions Today - MathOverflow
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Mathoverflow.net news digest
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0 days
How interesting are infinite dimensional manifolds?
I know that infinite-dimensional (smooth) manifolds are studied, but it seems to me that they are not treated widely. It looks unnatural for the following reason: infinite dimensional vector spaces are studied extensively as they form a very effective...
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0 days
Easier proof of Chapter 2, Theorem 9.1 in Neukirch's "Algebraic Number Theory"?
$\DeclareMathOperator{\Gal}{Gal}\DeclareMathOperator{\Aut}{Aut}\DeclareMathOperator{\Hom}{Hom}$Let $L|K$ be a Galois extension and $v$ a valuation on $K$. Then:
The Galois group $G=\Gal(L|K)$ acts transitively on the right on the set of extensions $E_v=\{w \ | \ w \text{ valuation on } L \text{ extending } v\}$.... -
0 days
Is weak homotopy equivalence of finite suspension posets decidable?
Say I hand you two finite posets $P,Q$. Is there an algorithm which allows you to decide whether or not the suspensions are weak homotopy equivalent?
For definiteness, you can take the suspension to be the join with $S^0$.
The nice thing about this question is that finite posets model all finite homotopy types, and the model is concrete enough that it shouldn’t matter how you encode them. Moreover, taking the suspension completely sidesteps the undecidability of the word... -
2 years
Defining the number of rightmost frozen digits of Graham's number
It is well-known that (in radix-$10$) Graham's number, $G$, can be expressed as a tetration with base $3$ and a very large hyperexponent $\tilde{b}$. Thus, we can write that $\exists! \hspace{1mm} \tilde{b} \in \mathbb{Z}^+ : G=g_{64}={^{\tilde{b}}3...
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