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Some obvious things are less obvious than others.... Let $X$ be a rank two graph consisting of a disjoint pair of circles plus an arc with one endpoint on each circle. Then no matter what $v \in X$ you pick, $\pi_1(X,v)$ is not generated by simple closed curves through $v$.


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No, in any characteristic. Let $f:\mathbb{P}^1\times\mathbb{P}^1=X\to Y=\mathbb{P}^2$ be such a morphism. Then, the map can not be finite, since it is of degree one and $Y$ is smooth would imply that $f$ is an isomorphism. This is not true by Picard group considerations or many other considerations. So, one must have an irreducible curve $E\subset X$ such ...


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An example of a frame operator depends of course on the frame. For a frame $(f_n)_{n \in \mathbb{N}}$ in a Hilbert spaces $\mathcal{H}$, the frame operator $S : \mathcal{H} \to \mathcal{H}$ is, as you mentioned, just defined as $$ S f = \sum_{n \in \mathbb{N}} \langle f, f_n \rangle f_n. \quad \quad (*)$$ To my knowledge, this is the only form in which ...


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I've found the source of this interview. It's in Paul Erdos' biography by Paul Hoffman, " The Man Who Love Only Numbers"(1998). Starting from page 78, the book describes Erdos left Hungary for Cambridge in 1934 due to the raging Hungarian Fascism. It was at Cambridge, the second day of his arrival, that he met G. H. Hardy, and inquired him about Ramanujan. ...


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Some remarks on infinite-dimensional manifolds. There are two approaches which to me still feel very manifold-ish (there are others yet: Frolicher spaces and diffeological spaces, which feel a bit less so); the approach of Banach manifolds and the much more general approach of Frechet manifolds. The former is almost precisely the same theory as the theory ...


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By the universal UHF algebra people usually mean the UHF algebra with the associated supernatural number: $$\prod_{p\in \mathbb{P}}p^\infty.$$ In other words, UHF algebra is the the unique UHF algebra $A$ with $K_0(A)=\mathbb{Q}$. A good reference is Rørdam's Classification of Nuclear C${}^\ast$-Algebras.


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Two possible techniques. First technique: use other people's work. The distribution in question appears to be, from @BGM's comment, a multivariate hypergeometric distribution, which is a very enjoyable name to pronounce very quickly 5 times in a row. More precisely, $(X,Y,k)\sim \operatorname{MultivHypergeom}_3(\underbrace{(\gamma N,\gamma N,(1-2\gamma) ...


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I've got a strange result for expectation, so feel free to downvote it: At first, lets split $k$ balls taken into two parts - one with red and blue balls, and other - with green balls. So we get the random variable $U$: $$P(U=u) = Hypergeometric(u, k,2\gamma N,N) $$ which means that drawing $k$ balls from the box with $N$ balls that contains $2\gamma N$ ...


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This is functional in general. Let $X$ be a set an let $\mathcal V\subseteq\wp(X)$. Then the collection of finite intersections of sets in $\mathcal V$ automatically has the basic properties of a base of a topology. This in the understanding that the empty intersection equals $X$. Denoting the collection by $\mathcal{V}^{\stackrel{\cap}{f}}$ we have ...


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We have: $$\left(\bigcup_{i \in I} A_i\right) \cap \left(\bigcup_{j \in J} B_j\right) = \bigcup_{(i,j) \in I \times J} (A_i \cap B_j)$$ so that the intersection of two unioned families is again a unioned family, over a bigger index set, and if the $A_i,B_j$ come from a family which is closed under finite intersections, the latter is also a union from that. ...


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With the question by the OP asking for a reference I will try to do just that, providing links to the OEIS. Surprisingly enough even the OEIS does not offer the usual variety of references here, suggesting that this problem is open. We compute generating functions $T_{\le h}(z)$ for the height being at most $h$ and the desired count is then given ...


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Let $f$ be a twice differentiable function. Observe $$ \text{det} \left[ \begin{array}{ccc} y & y' & y'' \\ f(x) & f'(x) & f''(x) \\ e^x & e^x & e^x \end{array} \right] =0$$ is a 2nd order linear ODE with horrible coefficients which takes $y=f(x)$ as a solution (and $y=e^x$ as a second solution). But, $f(x)$ is nearly arbitrary, so ...


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By the Schreier conjecture, now generally accepted as a corollary of the classification of finite simple groups, the outer automorphism group of a finite simple group is a solvable group. Thus a finite almost simple group is an extension of a solvable group by a simple group. This is how far one can get, since it is very general.


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I found the assignments from the dead link previously posted. Here is the list in case the link dies again. HOMEWORK Assignment 1: Due in class, Thursday August 25, 2011 required: Chapter 1: 1, 2, 5, 6, 10, 11, 13, 14, 15, 20, 25 extra: Chapter 1: 16, 17, 22 Assignment 2: Due 4 pm, Friday September 2, 2011 required: Chapter 2: 1,2, 3 parts a ...


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From $$\alpha(t)=\int_0^t e^{is^2}\>ds$$ it follows that $\dot \alpha(t)=e^{it^2}$ has absolute value $1$, hence $\alpha$ is a locally isometric immersion. From ${\rm arg}\bigl(\dot\alpha(t)\bigr)=t^2$ it then follows that the curvature $$\kappa(t)={d\over dt}{\rm arg}\bigl(\dot\alpha(t)\bigr)=2t$$ is strictly monotonically increasing for ...



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