Tom Cooney
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 Apr24 reviewed Approve Do all singular $n\times n$ matrices form a vector subspace when $n\ge2$? Apr22 reviewed Approve How to calculate the arc length of y = C/x Apr22 reviewed Approve Fourier Series of Complex Valued Functions Apr21 reviewed Approve Multiple stage working out Apr20 reviewed Approve How to solve this integral by a simple way? Apr20 reviewed Approve Find the function satisfying the given condition Apr20 reviewed Reject generating system of the kernel of a module-transformation Apr20 reviewed Approve An example of a continuous function such that M Apr2 reviewed Approve Find the arc length. Apr1 reviewed Approve Constructing a one-to-one correspondence between closed interval and half open interval Mar31 comment Commutant of algebra of multiplication operators First of all, Proposition 5.3.2 in Pedersen is about unbounded operators and $\mathfrak A$ contains only bounded operators. But $\mathfrak A$ is just the set of self-adjoint operators $\mathfrak M_{sa}$ in the von Neumann algebra $\mathfrak M = \{M_f: f\in L_\infty(X)\}$ and, yes, there is a bijection between $\mathfrak M_{sa}$ and the real-valued functions in $L_\infty(X)$. This bijection is the restriction of the $*$-isomorphism between $\mathfrak M$ and $L_\infty(X)$ to their self-adjoint parts, where this bijection is the one guaranteed by the Borel functional calculus (Section 4.5). Mar31 reviewed Reject Tangential space to the rational normal curve Mar30 comment Commutant of algebra of multiplication operators Your $\mathfrak A$ is basically $L^\infty(X,R)$ (the real bounded, measurable functions on $X$) acting on $L^2(X) =L^2(X,\mathbb R) + i L^2(X,\mathbb R)$. It is not a von Neumann algebra because von Neumann algebras are closed under multiplication by complex scalars and your $\mathfrak A$ is not. This suggests that it is a real von Neumann algebra? Mar23 reviewed Approve List all possible subgroups of $A_4.$ Determine which subgroups of $A_4$ are normal. Mar23 reviewed Approve Prove that $|A|=n\implies\left|\mathcal P(A)\right|=2^n$ Mar22 awarded Yearling Mar19 reviewed Approve What's the difference between $|z|^2$ and $z^2$, where $z$ is a complex number? Mar14 awarded Nice Answer Mar13 reviewed Approve Is there a more intelligent way to compute the determinant of the Killing form of $\mathfrak{sl}(3,F)$? Mar13 reviewed Reject Could you explain the failure of the Hodge decomposition to exist for non-compact manifolds?