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visits member for 3 years, 11 months
seen Aug 24 at 13:58

Aug
21
asked Reference on calculation of 2nd cohomology group
Aug
11
accepted vanishing higher cohomology group for property T group?
Aug
9
comment vanishing higher cohomology group for property T group?
@IgorBelegradek, thanks a lot for clarification!
Aug
9
comment vanishing higher cohomology group for property T group?
@IgorBelegradek, I am not able to download the paper you mentioned right now, if I understand your remark correctly, you mean there exists a property $T$ group $G$ with $H^2(G, \mathbb{Z})\neq 0$?
Aug
8
comment vanishing higher cohomology group for property T group?
@YCor, you mean for n=2, $G=SL_3(\mathbb{Z})$, the cohomology group vanishes?
Aug
8
asked vanishing higher cohomology group for property T group?
Jul
3
asked seek another proof of a result in Fourier analysis
Jul
2
awarded  Curious
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.