# Sanchez

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I really appreciate it when you take time to answer my questions. Thanks!

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 Dec29 comment Understanding induced representations It's nice that the two answers below convinced you that induced rep is a good notion, although I'm personally curious what you mean by "there are many induced rep in contrast to restriction" - in fact I can't see how you may cook up a $G$-rep naturally from a $H$-rep other than the extension of scalars above. Dec28 comment Understanding induced representations Why is $K[G] \otimes_{K[H]} V$ unintuitive to you? Dec27 comment contractible and simply connected Consider loop 1 formed by the starting point of your free homotopy, loop 2 formed by the ending point of the free homotopy. Then at time $t$, traverse along loop 1 up to time $t$, go through the homotopy at time $t$, and come back along loop 2. This is then a based homotopy to the trivial loop at based point. Dec25 revised Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ added 22 characters in body Dec25 answered Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Dec24 comment 1 dimensional representations of $S_n$ A hint for the complex number case: It suffices to show that every homomorphism $\chi: S_n$ to $\mathbb{C}^*$ is real, which then forces any such homomorphism to be either identity or sign, depending on the value at the cycle (12). To show that it's real, it suffices to show that $\chi(g) = \overline{\chi(g)}$ for any $g$, i.e. $\chi(g) = \chi(g^{-1})$. Try to show that $g$ and $g^{-1}$ are actually conjugates which lead to the result. Dec23 comment Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ This is wrong. Use AM-GM or other tools, you can show that $xy+yz+zx \ge 3$. Dec22 comment Is noetherianity a local property? @PeteL.Clark, Ah, thanks! That was what I was thinking of indeed. Dec22 comment Is noetherianity a local property? @BenjaLim, I believe that they are equivalent. Dec18 comment On convergence of nets in a topological space @BrianM.Scott, Excellent! Thanks. Dec18 comment On convergence of nets in a topological space Thanks! Is there any (succint) reference that compares nets and filters? I would like to know more about their similarities and differences. Dec18 comment On convergence of nets in a topological space I just read online that filter was another proposal to generalize sequences. Do filters capture the topology of a space in the same sense? Dec18 comment On convergence of nets in a topological space @Elias, it's net, not network if I understand your question correctly. A quick google search returns this: planetmath.org/encyclopedia/… Dec18 comment On convergence of nets in a topological space Properties defined by sequences (eg sequentially compact) are often different from those not defined by sequences (eg compact) especially in spaces that are not first countable. Nets give a more general framework that corresponds to the latter concept, e.g. compactness is equivalent to every net having a convergent subnet. Dec18 comment Is there a fundamental reason that $\int_b^a = -\int_a^b$ @asmeurer, Signs come up to reflect orientation issues. This doesn't have anything to do with measure, so there's no such phenomenon for Lebesgue integrals, but there would be something similar for integration over manifolds. Dec18 comment Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$ I meant to ask what a necessary and sufficient condition for a possible extension of one vector to an orthogonal basis would be. I doubt that entries being coprime would be enough. Dec18 comment Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$ What about an orthogonal basis, as mentioned by OP? Dec18 comment Primes in arithmetic progression @ThomasAndrews, Ah I see your point. Yet, as OP is thinking about Dirichlet's theorem, I would guess that he meant for all $j$, blah blah blah instead. Dec18 comment Primes in arithmetic progression @Ethan, you definitely need some restrictions on $j$ you consider, like $(j,a) = 1$. Second, infinitude of primes for one such $j$ is enough, since you can divide $j$ to the other side, as in the other thread you posted. Dec18 comment Primes in arithmetic progression @ThomasAndrews, it's not a simultaneous equation, so subtracting doesn't make sense.