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https://mathoverflow.net/questions/321210/matrix-operations-preserving-the-middle-coefficients-of-characteristic-polynomia
<p>Let <span class="math-container">$n$</span> be a positive integer. For a ring <span class="math-container">$A$</span> and matrix <span class="math-container">$M \in \mathrm{Mat}_{n \times n}(A)$</span>, let <span class="math-container">$\chi_{M}(t) = \det(M-t \operatorname{id}_{n}) = (-1)^{n}(t^{n} - \sigma_{M,1}t^{n-1} + \dotsb + (-1)^{n}\sigma_{M,n}) \in A[t]$</span> be the characteristic polynomial. Here <span class="math-container">$\sigma_{M,1}$</span> is the trace of <span class="math-container">$M$</span> and <span class="math-container">$\sigma_{M,n}$</span> is the determinant of <span class="math-container">$M$</span>. Recall that <span class="math-container">$\sigma_{M,1}$</span> is additive and <span class="math-container">$\sigma_{M,n}$</span> is multiplicative, i.e. <span class="math-container">$\sigma_{M_{1}+M_{2},1} = \sigma_{M_{1},1} + \sigma_{M_{2},1}$</span> and <span class="math-container">$\sigma_{M_{1}M_{2},1} = \sigma_{M_{1},1}\sigma_{M_{2},1}$</span>.</p>
<blockquote>
<p>Are there any matrix operations that preserve the middle coefficients <span class="math-container">$\sigma_{M,2},\dotsc,\sigma_{M,n-1}$</span> in any reasonable sense?</p>
</blockquote>
<p>This is related to the question <a href="https://mathoverflow.net/questions/33478/geometric-interpretation-of-characteristic-polynomial">Geometric interpretation of characteristic polynomial</a> but I don't yet understand whether it answers my question.</p>Fri, 18 Jan 2019 13:22:43 -0700Math Overflow Recent Questions: Dimensions of Lie algebra powers of irreducible representations
https://mathoverflow.net/questions/321209/dimensions-of-lie-algebra-powers-of-irreducible-representations
<p>Consider the following plethysms. For semisimple Lie algebras <span class="math-container">$L$</span>, if <span class="math-container">$A$</span> is the adjoint irrep, as far as I know, <span class="math-container">$d_1=|A^{\bigotimes 1}|=0$</span> (no lollipops), <span class="math-container">$d_2=|A^{\bigotimes 2}|=1$</span> (Schur's Lemma), <span class="math-container">$d_3=|A^{\bigotimes 3}|=1$</span> (structure constant tensor, 6j symbols).<br>
And if <span class="math-container">$A$</span> is the adjoint of a non-semisimple <span class="math-container">$L$</span>? I also read something like that the Killing form <span class="math-container">$K$</span> can be brought to <span class="math-container">$\mathrm{diag}(\{0,1\})$</span> in the right basis, implying <span class="math-container">$K^2=K$</span> and thus <span class="math-container">$d_2=2$</span> (unless <span class="math-container">$L$</span> is nilpotent, <span class="math-container">$K=0$</span> and thus <span class="math-container">$d_2=1$</span>). But I'm confused - rescale the basis and the Killing form might have more than two different eigenvalues for more than 4 generators? (Does it?)<br>
Assuming it miraculously hasn't, this means <span class="math-container">$d_3\leq 8$</span> (one or no Killing form pasted on the structure constant tensor). I already once jumped to conclusions :-) and thought <span class="math-container">$d_3=1$</span>, but I found a <span class="math-container">$L$</span> with <span class="math-container">$d_3=4$</span>. So, can you name a <span class="math-container">$L$</span> with <span class="math-container">$d_3=8$</span>?.<br>
(In graphic form the answers to my question seem obvious :-)</p>Fri, 18 Jan 2019 13:21:42 -0700Math Overflow Recent Questions: Generating random 6 digit numbers
https://mathoverflow.net/questions/321205/generating-random-6-digit-numbers
<p>We can generate a random 6 digit number <span class="math-container">$M$</span> by first generating a random distribution over the 10 digits and then sampling from that distribution. </p>
<p>Formally, let <span class="math-container">$x = (x_0, x_1, x_2, \ldots, x_9)$</span> be a point sampled from the uniform distribution on the unit sphere in 10 dimensions.</p>
<p>Then <span class="math-container">$x$</span> can be used to define a new RV <span class="math-container">$D_x$</span> on the set <span class="math-container">$D = [0, 1, 2, \ldots, 9]$</span> such that <span class="math-container">$P[D_x=i] = {x_i}^2$</span>.</p>
<p>Now consider the RV <span class="math-container">$M_{D_x}$</span> which is defined by first sampling <span class="math-container">$D_x$</span> 6 times (with replacement) to get a set <span class="math-container">$\{s_0, s_1, s_2, \ldots, s_5\}$</span> and then taking the sum <span class="math-container">$\sum_{j = 0}^5 10^j s_j$</span> (that is, use the 6 numbers generated from <span class="math-container">$D_x$</span> as digits of a number between 0 and 999999).</p>
<p>My question is: what is <span class="math-container">$P[M = n]$</span> for <span class="math-container">$n$</span> between 0 and 999999? Is this a uniform distribution over 6 digit integers?</p>Fri, 18 Jan 2019 12:48:41 -0700Math Overflow Recent Questions: How do we introduce a signed finite measure on the space of curves confined into the box $[0,1]^{n}$?
https://mathoverflow.net/questions/321204/how-do-we-introduce-a-signed-finite-measure-on-the-space-of-curves-confined-into
<p>Given <span class="math-container">$\Omega_{n} = \{\alpha:[0,1]\rightarrow[0,1]^{n}\,|\,\alpha\,\,\text{is smooth}\}$</span>, consider the equivalence relation:
<span class="math-container">\begin{align*}
& \alpha_{1} \sim \alpha_{2} \Leftrightarrow \int_{0}^{1}\langle G(\alpha_{1}(t)),\alpha^{\prime}_{1}(t)\rangle\,\mathrm{d}t = \int_{0}^{1}\langle G(\alpha_{2}(t)),\alpha^{\prime}_{2}(t)\rangle\,\mathrm{d}t,\\
\end{align*}</span></p>
<p>Where <span class="math-container">$\alpha_{1}(0) = \alpha_{2}(0)$</span> and <span class="math-container">$\alpha_{1}(1) = \alpha_{2}(1)$</span>. It is assumed that <span class="math-container">$G:[0,1]^{n}\rightarrow[0,1]^{n}$</span> is known. Such set can naturally be considered as a metric space (thus a topological space) in accordance to the norm:</p>
<p><span class="math-container">\begin{align*}
\lVert\alpha\rVert = \max_{0 \leq t \leq 1}\lVert\alpha(t)\rVert_{2}
\end{align*}</span></p>
<p>Let us define <span class="math-container">$\Omega := \Omega_{n}/G$</span> as the quotient space according to the above-mentioned equivalence relation. Thus we can introduce a topology on <span class="math-container">$\Omega$</span>. Precisely speaking, the quotient topology:
<span class="math-container">\begin{align*}
\tau_{\Omega} := \{O\subset\Omega\,|\,\pi^{-1}(O)\in\tau_{\Omega_{n}}\}
\end{align*}</span></p>
<p>Where <span class="math-container">$\pi$</span> is the map which associates each <span class="math-container">$\alpha\in\Omega_{n}$</span> to <span class="math-container">$[\alpha]\in\Omega$</span>: <span class="math-container">$\pi(\alpha) = [\alpha]$</span>. Finally, given the topological space <span class="math-container">$(\Omega,\tau_{\Omega})$</span>, we can construct its associated Borel <span class="math-container">$\sigma$</span>-algebra <span class="math-container">$\Sigma$</span>. </p>
<p>Here is my question: how do we introduce a signed finite measure on <span class="math-container">$(\Omega,\Sigma)$</span>? Precisely, I would like to define a triple <span class="math-container">$(\Omega,\Sigma,\mathbb{P})$</span> such that <span class="math-container">$\mathbb{P}([\alpha]) = -\mathbb{P}([\alpha^{-}]) \geq 0$</span>, where <span class="math-container">$\alpha^{-}(t) = \alpha(1-t)$</span>.</p>
<p>Such problem makes part of my research project on negative probabilities. I apologize if the question does not fit into Math Overflow context. Any help is appreciated. Thanks in advance.</p>Fri, 18 Jan 2019 12:48:04 -0700Math Overflow Recent Questions: Is the metric completion of a Riemannian manifold always a geodesic space?
https://mathoverflow.net/questions/321201/is-the-metric-completion-of-a-riemannian-manifold-always-a-geodesic-space
<p>A length space is a metric space <span class="math-container">$X$</span>, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve <span class="math-container">$c: [0,1] \rightarrow X$</span> is the sup of
<span class="math-container">$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + \cdots + d(c(t_{N-1}), c(1)) $$</span>
over all <span class="math-container">$0 < t_1 < t_2\cdots < t_{N-1} < 1$</span> and <span class="math-container">$N > 0$</span>.</p>
<p>A geodesic space is a length space, where for each <span class="math-container">$x,y \in X$</span>, there is a curve <span class="math-container">$c$</span> connecting <span class="math-container">$x$</span> to <span class="math-container">$y$</span> whose length is equal to <span class="math-container">$d(x,y)$</span>.</p>
<p>A Riemannian manifold <span class="math-container">$M$</span> and its metric completion <span class="math-container">$\overline{M}$</span> are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.</p>
<p>But is <span class="math-container">$\overline{M}$</span> necessarily a geodesic space? If not, what is a counterexample?</p>
<p>This was motivated by my flawed answer to <a href="https://mathoverflow.net/q/317523/613">Minimizing geodesics in incomplete Riemannian manifolds</a></p>
<p>Also, note that if <span class="math-container">$\overline{M}$</span> is locally compact, then it is a geodesic space by the usual proof. One example of <span class="math-container">$M$</span>, where <span class="math-container">$\overline{M}$</span> is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.</p>Fri, 18 Jan 2019 11:55:31 -0700Math Overflow Recent Questions: Under which conditions do ellipsoids have a focal property?
https://mathoverflow.net/questions/321199/under-which-conditions-do-ellipsoids-have-a-focal-property
<p>Given an ellipsoid <span class="math-container">$E$</span>, we consider the trajectory that light would do inside if the surface of <span class="math-container">$E$</span> would be a mirror. In other words, we consider trajectories <span class="math-container">$\gamma:[0,\infty)\rightarrow E$</span> such that that <span class="math-container">$\gamma'$</span> is locally constant for any <span class="math-container">$t$</span> such that <span class="math-container">$\gamma(t)\in \mathrm{int}\,E$</span> and such that <span class="math-container">$\gamma'(t)$</span> becomes its reflection with respect the hyperplane <span class="math-container">$T_{\gamma(t)}\partial E$</span> for <span class="math-container">$t$</span> such that <span class="math-container">$\gamma(t)\in\partial E$</span>.</p>
<p>When <span class="math-container">$E$</span> is an ellipse, any such trajectory that start at a foci will return to a foci after a reflection. In one generates a 3-ellipsoid by rotating the ellipse around the axis containing the foci, the same property continue to be true.</p>
<p>Now, imagine you rotate the ellipse with respect the other axis of symmetry. Then the image of pair of foci becomes a circle <span class="math-container">$C$</span>. The question is the following: Do any light trajectory starting at <span class="math-container">$C$</span> come back to <span class="math-container">$C$</span> after a reflection? If this is not true, does this hold for the convex hull of <span class="math-container">$C$</span>?</p>
<p>More generally, one can we said about ellipsoids, light trajectories and subsets with the property that light trajectories become back to them after one reflection?</p>
<p>[This was a question made to me by an experimental physicist, but after thinking about it I have been unable to find any reference about it despite its elementary looking form].</p>Fri, 18 Jan 2019 10:56:43 -0700Math Overflow Recent Questions: Quadrics over the univariate function field with discriminant of minimal degree
https://mathoverflow.net/questions/321198/quadrics-over-the-univariate-function-field-with-discriminant-of-minimal-degree
<p>Consider a non-degenerate quadric <span class="math-container">$Q(x,y,z) \subset \mathrm{P}^2$</span> over the univariate function field <span class="math-container">$\mathbb{F}_p(t)$</span>, where <span class="math-container">$\mathbb{F}_p$</span> is a prime finite field, <span class="math-container">$p > 2$</span>. For simplicity assume that coefficients of <span class="math-container">$Q$</span> lie in <span class="math-container">$\mathbb{F}_p[t]$</span>.</p>
<p>Let <span class="math-container">$\Delta_Q(t) \in \mathbb{F}_p[t]$</span> be the discriminant of <span class="math-container">$Q$</span>, i.e., the determinant of the corresponding symmetric matrix.</p>
<p>How to understand is there or not (at least in some cases) a quadric <span class="math-container">$Q^\prime$</span> isomorphic over <span class="math-container">$\mathbb{F}_p(t)$</span> to <span class="math-container">$Q$</span> such that the polynomial <span class="math-container">$\Delta_{Q^\prime}(t)$</span> has strictly smaller degree than <span class="math-container">$\Delta_Q(t)$</span>. And how can I explicitly construct a linear transformation (over <span class="math-container">$\mathbb{F}_p(t)$</span>) between <span class="math-container">$Q$</span> and <span class="math-container">$Q^\prime$</span>?</p>Fri, 18 Jan 2019 10:54:47 -0700Math Overflow Recent Questions: On permanents involving trigonometric functions
https://mathoverflow.net/questions/321195/on-permanents-involving-trigonometric-functions
<p>Here I pose my conjectures on permanents involving trigonometric functions.</p>
<p><strong>Conjecture 1.</strong> Let <span class="math-container">$p$</span> be an odd prime. Then both
<span class="math-container">$$c_p:=\sqrt p\,\text{per}\left[\cot\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2}$$</span>
and
<span class="math-container">$$t_p:=\frac1{\sqrt p}\text{per}\left[\tan\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$</span> are integers. Moreover,
<span class="math-container">$$c_p\equiv1\pmod p\ \ \text{and}\ \ t_p\equiv (-1)^{(p+1)/2}\pmod p.$$</span></p>
<p><em>Remark 1</em>. Via Mathematica I find that
<span class="math-container">\begin{align}&c_3=1,\ c_5=-4,\ c_7=22,\ c_{11}=1816,\ c_{13}=-5056,
\\&c_{17}=-2676224,\ c_{19}=58473280.
\end{align}</span></p>
<p><strong>Conjecture 2.</strong> Let <span class="math-container">$p$</span> be an odd prime. Then both
<span class="math-container">$$s_p:=\frac{2^{(p-1)/2}}{\sqrt p}\text{per}\left[\sin2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$</span>
and
<span class="math-container">$$r_p:=\frac{\sqrt p}{2^{(p-1)/2}}\text{per}\left[\csc2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}$$</span>
are integers. Moreover,
<span class="math-container">$$s_p\equiv(-1)^{(p+1)/2}\pmod p\ \ \text{and}\ \ r_p\equiv1\pmod p.$$</span></p>
<p><strong>Conjecture 3.</strong> Let <span class="math-container">$p$</span> be an odd prime. Then
<span class="math-container">$$a_p:=\frac1{2^{(p-1)/2}}\text{per}\left[\sec2\pi\frac{jk}p\right]_{1\le j,k\le(p-1)/2}\in\mathbb Z.$$</span> If <span class="math-container">$p>3$</span> and <span class="math-container">$p\equiv3\pmod 4$</span>, then</p>
<p><span class="math-container">$$\text{per}\left[\sec2\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2}\equiv \text{per}\left[\cos2\pi\frac{jk}p\right]_{1\le j,k\le (p-1)/2}
\equiv-1\pmod p.$$</span></p>
<p><em>Remark 2</em>. Using Galois theory, I have shown that <span class="math-container">$a_p,c_p,r_p,s_p,t_p$</span> are rational numbers for any odd prime <span class="math-container">$p$</span>.</p>
<p>Your comments are welcome!</p>Fri, 18 Jan 2019 10:19:59 -0700Math Overflow Recent Questions: Enumerating all partitions induced by Voronoi diagrams for clustering
https://mathoverflow.net/questions/321194/enumerating-all-partitions-induced-by-voronoi-diagrams-for-clustering
<p>a classical results by M. Inaba et al. in "Applications of Weighted Voronoi Diagrams and Randomization to Variance-Based k-CLustering" (Theorem 3) says</p>
<blockquote>
<p>The number of Voronoi partitions of <span class="math-container">$n$</span> points by the Euclidean Voronoi diagram generated by <span class="math-container">$k$</span> points in <span class="math-container">$d$</span>-dimensional space is <span class="math-container">$\mathcal{O}(n^{dk})$</span>, and all the Voronoi partitions can be enumerated in <span class="math-container">$\mathcal{O}(n^{dk+1})$</span>.</p>
</blockquote>
<p>They basically divide the <span class="math-container">$d$</span>-dimensional space into equivalence classes where two sets of center <span class="math-container">$\mu^1$</span> and <span class="math-container">$\mu^2$</span> are equivalent if they lead to the same Voronoi Diagram. Then they show that the arrangement of the <span class="math-container">$nk(k-1)/2$</span> surfaces</p>
<p><span class="math-container">$$ ||x_i-\mu_j||^2- ||x_i-\mu_{j'}||^2 = 0 $$</span></p>
<p>for each point <span class="math-container">$x_i$</span> and two cluster center <span class="math-container">$\mu_j$</span> and <span class="math-container">$\mu_{j'}$</span> coincides with the equivalence relation from Voronoi partitions.</p>
<p>Next they argue that the combinatorial complexity of arrangements of <span class="math-container">$nk(k-1)/2$</span> constant-degree algebraic surfaces is bounded and that this implies and algorithm with running time <span class="math-container">$\mathcal{O}(n^{dk+1})$</span>. Unfortunately, the cited source (Evaluation of combinatorial complexity for hypothesis spaces in learning theory with application, Master's Thesis, Department of Information Science, University of Toko, 1994) I cannot find anywhere. More precisely I cannot see the two following things.</p>
<ol>
<li>Where can I find a bound for the combinatorial complexity of the arrangement of <span class="math-container">$nk(k-1)/2$</span> constant degree algebraic surfaces and</li>
<li>How does this help me to compute the arrangement?</li>
</ol>
<p>For 2. I found the Bentleyâ€“Ottmann algorithm, however that only works for line segments and not degree 2 polynomials. How can this algorithm be generalized?</p>
<p>Thanks so much!</p>Fri, 18 Jan 2019 10:05:26 -0700Math Overflow Recent Questions: False constantes in first Hardy-Littlewood conjecture?
https://mathoverflow.net/questions/321193/false-constantes-in-first-hardy-littlewood-conjecture
<p>Let <span class="math-container">$q$</span> be a prime number, and <span class="math-container">$m$</span> an even number.</p>
<p>Let <span class="math-container">$\displaystyle\mathcal{B}_q=\{b \in \mathbb{N}^{*} \, | \, b \wedge {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}=1 \text{ and } b \leq {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} \}$</span></p>
<p>Let <span class="math-container">$I_{q,m}(n)$</span> denote the number of elements <span class="math-container">$(b,b+m)$</span> less than <span class="math-container">$n$</span> and coprime to <span class="math-container">$\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}$</span></p>
<p>Let <span class="math-container">$q(n)$</span> be the small prime verify <span class="math-container">$n+m < \displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a}}$</span></p>
<p>Let <span class="math-container">$b_1$</span> be the prime that the gap <span class="math-container">$(b,b+m)$</span> is appear the first time in <span class="math-container">$\mathcal{B}_{b_1}$</span> and <span class="math-container">$s = \#\{(b,b+m)\in\mathcal{B}_{b_1}^2\}$</span></p>
<p>I proove that <span class="math-container">$I_{q(n),m}(n) \sim \dfrac{\theta_m}{\displaystyle{\small \prod_{\substack{3 \leq a < b_1 \\ \text{a prime}}} {\normalsize (a-2)}}} \; 2 \, C_2 \; \dfrac{n}{\ln(\ln(n))^2} e^{-2 \gamma}$</span></p>
<p>With <span class="math-container">$\theta_m$</span> is a constante verify <span class="math-container">$\dfrac{s}{b_1 - 2} \leq \theta_m \leq \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}$</span> , and <span class="math-container">$C_2 = \displaystyle{\small \left( \prod_{\substack{3 \leq a \\ \text{a prime}}} {\normalsize \frac{a (a-2)}{(a-1)^2}} \right)}$</span></p>
<p>Show that <span class="math-container">$\theta_2 = \theta_4 = 1$</span></p>
<p>If we denote <span class="math-container">$I_n$</span> the number of elements less than <span class="math-container">$n$</span> and coprime to <span class="math-container">$\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$</span> (<span class="math-container">$q(n)$</span> the small prime verify <span class="math-container">$n < \displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$</span>)</p>
<p>I prove that <span class="math-container">$I_n \sim \dfrac{n}{\ln(\ln(n))} \, e^{-\gamma}$</span></p>
<p>Show that : <span class="math-container">$\dfrac{n}{\ln(\ln(n))} \, e^{-\gamma} = \dfrac{n}{\ln(n)} \dfrac{\ln(n)}{\ln(\ln(n))} \, e^{-\gamma} \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$</span></p>
<p>Then if we use the same analogy with the case <span class="math-container">$I_{q(n),m}(n)$</span> :</p>
<p><span class="math-container">$I_{q(n), m}(n) \sim \pi_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$</span></p>
<p>Then : <span class="math-container">$\pi_m(n) \sim \dfrac{\theta_m}{\displaystyle{\small \prod_{\substack{3 \leq a < b_1 \\ \text{a prime}}} {\normalsize (a-2)}}} \; 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$</span></p>
<p>Well that's just a conjecture (i don't proove <span class="math-container">$I_{q(n), m}(n) \sim \pi_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$</span>)</p>
<p>But the Hardy-Littlewood conjecture state that <span class="math-container">$\pi_m(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$</span></p>
<p><a href="http://lagrida.com/prime_numbers_construction.html" rel="nofollow noreferrer">http://lagrida.com/prime_numbers_construction.html</a></p>
<p><a href="http://lagrida.com/Calculate_Gaps_Cardinality.html" rel="nofollow noreferrer">http://lagrida.com/Calculate_Gaps_Cardinality.html</a></p>
<p><a href="http://lagrida.com/Fondamentale_Conjonctures_Prime_Numbers.html" rel="nofollow noreferrer">http://lagrida.com/Fondamentale_Conjonctures_Prime_Numbers.html</a></p>Fri, 18 Jan 2019 10:04:24 -0700