Mariano Suárez-Alvarez
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4h
comment Contractible chain complex
Where exactly in the verification of the claim you are asking about do you get stuck?
4h
comment A normal subgroup that is not a characteristic
@xndrme, what is the simplest abelian group you know which is not cyclic? Does that provide an example for what you want?
5h
comment Can $ {L^{1}}(G) $ be a $ C^{*} $-algebra?
Your definition of $f^*$ has $x$ free on the left hand side and no free $x$ on the right hand side!
5h
comment A given ring of matrices has an infinite number of invertible elements
Instead of «bringing outside» information you invoked the gods to tell you a solution of infinite order — and they probably used the solution of Pell's equation to find it...
1d
comment Why are quadrants defined the way they are?
They are numbered the order you encounter them when you move in the direction angles increase. There is nothing more.
2d
comment Aren't there obvious patterns in the primes that no one makes use of and what about this…
Questions of the form «can X be used to prove Y?» with Y a statement for which no one knows any proof are mostly useless, for the only way to answer them affirmatively is to exhibit a proof of Y using X, and no one knows a proof of Y whether it uses X or not, and there is pretty much no way to answer them in the negative.
2d
comment Intuition Behind the Riesz Transform
Do you have any intuition to explain the constant $1/\pi$ appearing in from of the Hilbert transform? Exactly the same intuition applies here (that $\pi$ is half the length of the unit circle, just as the constant here is $1/v$ with $v$ half the volume of the $n$-sphere). Now, if you do not have any intuition about that $1/\pi$, then you should be asking about that, really :-)
2d
comment Intuition Behind the Riesz Transform
Plainly, no, we are not integrating over the sphere! You yourself typed the domain of the integral...
2d
comment Intuition Behind the Riesz Transform
The constant is related to the volume of the sphere...
2d
comment Complex irreducible representation of solvable lie algebra
$v$ cannot be the zero vector (if we allowed $v$ to be zero in Lie's theorem, the theorem would be a triviality!)
2d
comment Ceil () and Floor()
That does not tell us what you are talking about.
2d
reviewed Approve Ceil () and Floor()
2d
comment Ceil () and Floor()
If you do not tell us what you are talking about, we can only guess. Can you see the many ways in which this is not great?
2d
comment Ceil () and Floor()
Where did you get this from? It seems you are talking about the Ceil and Floor functions of some programing language, and in that case which programming language you are talking about certainly seems to be a significant piece of information useful for anyone who is trying to help you... Then again, you may be not talking about a programming language at all. Who knows! :-)
2d
comment Complex irreducible representation of solvable lie algebra
Take an irreducible representation $V$. Using Lie's theorem, you have found a $1$-dimensional subspace which is in fact a subrepresentation of $V$. Since $V$ is irreducible, then it does not have proper non-zero subrepresentations so...
2d
comment Why is it called the cotangent bundle?
Co- is used throughout to denote relations of duality. Algebra and coalgebra,module and comodule, and so on.
2d
comment How can we memorize the formula for the determinant of a $4\times4$ matrix?
Memorizing the formula for the 4x4 determinant is only less absurd than memorizing the formula for the 5x5 determinant.
2d
comment $\mathbb Q [\sqrt{2} i]$ contains neither $\sqrt[4]{2}$ nor $\sqrt{2}$
Let $f=X^4-2$. The polynomial $f(X-2)$ is irreducible over $Q$ because of Eisenstein at the prime $2$.
2d
answered $\mathbb Q [\sqrt{2} i]$ contains neither $\sqrt[4]{2}$ nor $\sqrt{2}$
2d
comment Definition of adjoint functor and locally small categories
You need to make precise what exactly you mean by «works»...