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I looked up [Hi, 11], which was constructing a cofibrantly generated model structure on $M^C$, where $M$ has a cofibrantly generated model structure and $C$ is a category. … In order to apply this to our setting, where the domain category is also a dg-category, we need some enriched version of the [Hi, 11] I suppose, but I am confused how this can be done. …

Thus, $K$ is duo. [] For the next results, recall that a subset $I$ of a semigroup $H$ is a left (resp., right) ideal if $HI \subseteq I$ (resp., $IH\subseteq I$); some people would also require that $ …

Some HI spaces fail the approximation property, which implies that no finite codimensional subspace has a Schauder basis. …

Hi I would like to grasp the theory behind Maude [1], [2] Are there any recommended video lecture notes, talks or introductory notes? …

Excellent Algorithm Hi @fedja, after streamlining the JavaScript program a little bit and some performance testing I'm really happy with the solution provided by you. …

Hi I just happened to have a small question. If we have $$\frac{\int_\Omega x}{|\Omega|}=\frac{\int_{\partial\Omega} x}{|\partial\Omega|}$$ for a simply connected set $\Omega$ with analytic boundary. …

Hi! …

=1,\ldots,6$, $\text{SO}(2N+1)_{2}$ for $N=1,\ldots,3$, etc Fusion rings related to subfactors, e.g., a ring which we called Pseudo $\text{PSU}(2)_6$ (for now) Extensions of the above, such as $\text{HI …

Bernstein--Chernoff inequality, \begin{equation*} q_k\le\exp\Big\{-h(1+\de)\frac{k(k+1)}{2}\Big\} E\exp h\sum_{i=1}^k iY_i \\ =\exp\Big\{-h(1+\de)\frac{k(k+1)}{2}\Big\}\prod_{i=1}^k\frac1{1-hi … So, letting now $h=u_0/k$ and using \eqref{35} with $u=hi$, we get \begin{equation*} q_k\le q_{k,\ep,u_0} :=\exp\Big\{-u_0(1+\de-g(u_0))\frac{k+1}{2}\Big\}, \tag{40}\label{40} \end{equation*} and …

(Hi David, thanks for the mention.) I think the result I described in the previous paragraph is the one David is thinking of. …

Then the slit $S$ will be mapped to $[0,hi]$ for $h$ comparable to $3a$ and the mapping will be bi-Lipshitz on the slit and $\omega$ will be mapped to the harmonic measure $\omega'$ in the half-plane with … When $z$ is in $I$, the corresponding point $w$ is in the "middle part" of $[0,hi]$, so the factor in front of $|dw|$ is comparable to $1$. …

The length of $HI$ measures $\sqrt6/2$, congruent with $AG$. The remaining cut from $I$ bisects angle $GIH$ and the remaining two cuts (from $D$ and $E$) meet at the center of the hexagon. …

$\def\R{\mathbb R}$$\def\aha{{1/2}}$$\def\maha{{1/2}}$ Hi everyone, I am interested in the following problem: Let consider the heat equation problem: $$\forall (t,x) \in \mathbb{R}_+\times\mathbb{R}, ~ …

(Tuesday, Sept 5:) For a number field $Fˣ$ and a number ring $Oˣ$ it is common to define: $Z(f,χ) = ʃ_{Fˣ} f(x) χ(x) dˣ x$ $g(ω,ψ) = ʃ_{Oˣ} ω(x) ψ(x) dˣ x$ where $dˣx$ is the multiplicative Haar measu …

mathfrak{f}$ denotes the conductor $(\mathfrak{O}_E \mathfrak{O}_F : \mathfrak{O}_{EF})$, one has $\mathfrak{f} \cdot \mathfrak{O}_L$ $=$ $\prod \mathfrak{e}_{\sigma,L}$ where $\sigma$ runs over the set $HI …