176 reputation
9
bio website
location
age
visits member for 2 years, 2 months
seen May 31 at 1:17

Jul
14
awarded  Revival
Jul
2
awarded  Curious
May
31
asked Numbers of $m$-simplices in the barycentric subdivision of an $n$-simplex ($m \leq n$).
Mar
8
comment $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
@Dreamer: Do you mean that they are not square matrices? Then vous avez raison. Sorry for my confusion.
Mar
8
comment $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
@Dreamer: Not at all. Thomas uses the fact that $A^{\prime}$ is invertible (note that $A^{\prime} B = I$.)
Mar
7
comment Is a simply connected set connected?
I don't like to start a meta-discussion here, but why did someone give me a down vote? The first statement is a reminder of the definition, and the second statement is an easily exercise.
Mar
7
comment Is a simply connected set connected?
> the union of two open disjoint discs are simply connected< It is a disjoint union of simply connected sets, but since it is not path-connected, it cannot be simply connected by definition.
Mar
6
comment Simply connected and homotopic
>I see lots of proving left to do.< I quite agree. Besides, as I rewrote, we need many postulates to bypass pathology. So, this problem is not so easy as it appears.
Mar
6
revised Simply connected and homotopic
added 7 characters in body
Mar
6
revised Simply connected and homotopic
Reflection incorporated.
Mar
6
answered Is a simply connected set connected?
Mar
6
comment Simply connected and homotopic
>the sausage shape (assuming it includes the boundary) is not contractible< Do you mean the sausage to be $E$? Well $E$ is supposed to be a domain so must be open.
Mar
6
comment Simply connected and homotopic
On second thought if you meant, by `$C$ is homotopic to a point', that $C$ is shrinkable to a point of $E$ inside $E$, then that homotopy can be used to define a contraction of $E$ to the point inside $E$, thus $E$ is simply connected.
Mar
6
comment Simply connected and homotopic
You wrote "C is a closed curve that is homotopic to a point", didn't you?
Mar
6
answered Simply connected and homotopic
Mar
5
awarded  Supporter
Mar
5
revised R is a commutative ring with unity and prime characteristic p, show that $\phi: R \to R\,\,/\,\, \phi(a) = a^p$ is a homomorphism
added 1 characters in body
Mar
5
comment R is a commutative ring with unity and prime characteristic p, show that $\phi: R \to R\,\,/\,\, \phi(a) = a^p$ is a homomorphism
You are welcome. By the way a typo: $a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb \binom{p}{p - 1} + b^{p}$ should read $a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb + \binom{p}{p - 1}a b^{p -1} + b^{p}$
Mar
5
comment R is a commutative ring with unity and prime characteristic p, show that $\phi: R \to R\,\,/\,\, \phi(a) = a^p$ is a homomorphism
Alright. $(a + b)^{p} = a^{p} + \binom{p}{1} a^{p - 1}b^{1} + \dotsb \binom{p}{p - 1} + b^{p}$. Suppose $1 \leq k \leq p - 1$. Then we have $\binom{p}{k} = \frac{p (p -1)!}{k!(p - k)!} = p\times \frac{(p - 1)!}{k!(p - k)!}$. Note that there are no factors of $p$ appearing in $(p - 1)!, k!$ or $(p - k)!$. As this number must be an integer, it is divisible by $p$. Thus $\binom{p}{k} \equiv 0 \pmod p$.
Mar
5
revised R is a commutative ring with unity and prime characteristic p, show that $\phi: R \to R\,\,/\,\, \phi(a) = a^p$ is a homomorphism
added 9 characters in body