bio | website | |
---|---|---|
location | Buffalo, NY | |
age | ||
visits | member for | 3 years, 7 months |
seen | Apr 18 at 2:59 | |
stats | profile views | 620 |
Jan 15 |
asked | find a special group $G$ |
Jan 3 |
accepted | Restriction on the coefficients for an operator in the free group factor $ L(\mathbb{F}_2) $ |
Dec 24 |
revised |
Find a lower bound for a pre-invariant $Fol(L(F_m), X_m)$
Fix typo |
Dec 24 |
revised |
Find a lower bound for a pre-invariant $Fol(L(F_m), X_m)$
Fix typo |
Dec 20 |
awarded | Disciplined |
Dec 3 |
accepted | finite index, self-normalizing subgroup of $F_2$ |
Dec 3 |
comment |
finite index, self-normalizing subgroup of $F_2$
Thanks, I did not realize this simple observation. |
Dec 3 |
revised |
finite index, self-normalizing subgroup of $F_2$
added 25 characters in body |
Dec 3 |
asked | finite index, self-normalizing subgroup of $F_2$ |
Nov 27 |
accepted | noncommutative polynomials equality |
Nov 27 |
accepted | $Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$ |
Nov 16 |
answered | noncommutative polynomials equality |
Nov 13 |
comment |
$Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$
But then, it should be clear the only units of this quotient ring consist of $ \pm 1, \pm x^{\pm 1}, \pm y^{\pm 1}$ and their products. |
Nov 13 |
comment |
$Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$
You are right， the quotient ring is the two variable Laurent polynomial ring, which I misunderstood to be the polynomial ring. |
Nov 13 |
comment |
$Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$
there seems some tricky points concerning lifting automorphism from the quotient algebra to the original one, since $x\to x+y, y\to y$ gives us an automorphism of the quotient algebra $\mathbb{Z}G_2/I$, but we can check that $\alpha(x)=x+y, \alpha(y)=y, \alpha(z)=z$ is not an automorphism of $\mathbb{Z}G_2$ since it does not preserve the skew relation $yx=xyz^2$. So, in general, we have to determine two noncommutative polynomial $p, q$, s.t., $\alpha(x)=x+y+(z-1)p, \alpha(y)=y+(z-1)q, \alpha(z)=z$ and it preserves the above skew relation. |
Nov 13 |
asked | $Aut(\mathbb{Z}G)=?$ for $G=\mathbb{Z}^2\rtimes_n\mathbb{Z}$ |
Oct 31 |
comment |
Is it known that “hyperfinite length” cannot distinguish free group factors?
@Jon, roughly speaking, a result in equivalence relation setting is often easier to prove than its counterpart in Von Neumann algebras setting, but the result on the topological generators of the full group generated by the hyperfinite equivalence relation is proved rather later than its counterpart. This seems strange. |
Oct 30 |
comment |
Is it known that “hyperfinite length” cannot distinguish free group factors?
@Jon, it is interesting that Ge and Shen has already defined the "generating length" for a type II_1 Von Neumann algebra as an analogy of the definition of cost in the equivalence relation setting, you can find it in page 368 of the book" Third International Congress of Chinese Mathematicians, 1st part". But the relation between hyperfinite length and generating length has not been mentioned in that paper. |
Oct 29 |
comment |
Is it known that “hyperfinite length” cannot distinguish free group factors?
@Jon, roughly speaking, a similar result in the equivalence relation setting is used in the proof of theorem 4.10 of the paper "Topological Properties of Full Groups" by J.Kittrell and T.Tsankov, and F. Maitre refined this process in his paper on generator problems. I guess when Ge and Popa introduced this "hyperfinite length", they also expect it to be related to the generator problem for Von Neumann algebras? |
Oct 29 |
revised |
noncommutative polynomials equality
emphasize what I want |