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Q&A for professional mathematicians
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Hottest Questions Today - MathOverflow
Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, ...
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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 $$\lVert(a_{ij})\rVert \le C \Bigl\lVert\s...
Mathoverflow.net news digest
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0 days
question about the widely used method for proving coarse Baum-Connes conjecture ...
The coarse Baum-Connes conjecture can be expressed in many ways. I choose the following, it states that$$e_{*}:\lim_{d\rightarrow\infty}K_{*}(C^{*}_{L}(P_{d}(X)))\rightarrow\lim_{d\rightarrow\infty}K_{*}(C^{*}(P_{d}(X)))$$induced by the evaluation map...
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0 days
Hasse derivative and Laurent Denis Lemma
In the article "Indépendance algébriques de logarithmes en caractéristique $p$" (BULL. AUSTRAL. MATH. SOC. VOL. 74 (2006) [461-470]), Laurent Denis stated the following Lemma 2.4 (I give the translation of the lemma and the proof)
I think that his argument of derivative is true too when one replaces the usual derivative by Hasse derivative of any order $\ge1$ with respect to the $X_i$ ($1\le i\le j$). If I am not wrong, this would allow to assert immediately that $Q(X_0,X_1,\cdots... -
0 days
Criterion of monogenic semigroup?
Let $p$ be a prime number.
Let $D$ be a finite semigroup such that $p$ divides the cardinality of $D$. Suppose that the intersection of all its non-empty subsemigroups is non-empty, and that every proper monogenic subsemigroup has a cardinality $c$ such that $c \equiv 1 \pmod... -
9 months
Sufficient condition for a density matrix to be positive definite
Note: since I have completely changed the content of my post compared to my deleted post with the same title, I figured it would make more sense to create another post (i.e. this post).
Let's say you have a multipartite quantum system consisting of $n$ qubits. Partition the qubits into 2 sets, say $A$ and $B$, so that $A$ consists of the first $n - 1$ qubits and $B$ is the $n$-th qubit....
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