Reputation
2,502
Next privilege 3,000 Rep.
Cast close and reopen votes
Badges
1 8 23
Newest
 Yearling
Impact
~17k people reached

Apr
24
reviewed Approve Do all singular $n\times n$ matrices form a vector subspace when $n\ge2$?
Apr
22
reviewed Approve How to calculate the arc length of y = C/x
Apr
22
reviewed Approve Fourier Series of Complex Valued Functions
Apr
21
reviewed Approve Multiple stage working out
Apr
20
reviewed Approve How to solve this integral by a simple way?
Apr
20
reviewed Approve Find the function satisfying the given condition
Apr
20
reviewed Reject generating system of the kernel of a module-transformation
Apr
20
reviewed Approve An example of a continuous function such that M
Apr
2
reviewed Approve Find the arc length.
Apr
1
reviewed Approve Constructing a one-to-one correspondence between closed interval and half open interval
Mar
31
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).
Mar
31
reviewed Reject Tangential space to the rational normal curve
Mar
30
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?
Mar
23
reviewed Approve List all possible subgroups of $A_4.$ Determine which subgroups of $A_4$ are normal.
Mar
23
reviewed Approve Prove that $|A|=n\implies\left|\mathcal P(A)\right|=2^n$
Mar
22
awarded  Yearling
Mar
19
reviewed Approve What's the difference between $|z|^2$ and $z^2$, where $z$ is a complex number?
Mar
14
awarded  Nice Answer
Mar
13
reviewed Approve Is there a more intelligent way to compute the determinant of the Killing form of $\mathfrak{sl}(3,F)$?
Mar
13
reviewed Reject Could you explain the failure of the Hodge decomposition to exist for non-compact manifolds?