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Q&A for professional mathematicians
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Let $A$ be a $C^*$-algebra and $(a_{ij}) \in M_n(A)$ be a positive matrix. Does there exist a constant $C \ge 0$ (not depending on the $a_{ij}$) such that $$\lVert(a_{ij})\rVert \le C \Bigl\lVert\s...
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Hottest Questions Today - MathOverflow
Stack Exchange Network Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledg...
Mathoverflow.net news digest
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0 days
Some simplicial invariants associated to a finite group
I participated in a talk about abstract combinatorial simplicial complex. During the talk the following question came to my mind.
Let $G$ be a finite group. We construct a simplicial complex $\Delta$ whose set of vertices is $G$. A face $F \in \Delta$ is a subset $F \subset G$ whose elements mutually commute. I am curious about the topological realization $X_G=|\Delta|$ of $\Delta... -
6 months
Eigenvector quadratic form inequality
Given matrix $B\in\mathbb{R}_+^{n\times n}$ and scalar $\alpha \in \mathbb{R}_{+}$, let $A:=\alpha B+B^T/\alpha$. Note that $B$ and $A$ have nonnegative entries and that $\alpha$ controls degree of asymmetry of $A$. Let $v \in \mathbb{R}^n_+, \Vert v...
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0 days
Matrix formula for point of intersection of three circles through six points
Suppose we have the six points in the cartesian plane $(x_i, y_i)$ for $1 \leq i \leq 6$. Further suppose that we draw three circles through them, so that each circle passes through three of the points - say circles 123, 345, and 561.
Is there a matrix determinant formula to determine whether these three circles are concurrent?The long way of solving this problem (explicit formulas for the circles and their intersection points) is not what I need - it's too untenable.... -
0 days
Is the solvable closure of $\mathbb{F}_p(t)$ PAC?
It is a famous open problem in field arithmetic whether $\mathbb{Q}^{\mathrm{solv}}$, the solvable closure of $\mathbb{Q}$, is pseudo algebraically closed (PAC). That is, whether every absolutely irreducible $\mathbb{Q}$-variety admits a rational point...
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