@TedShifrin which of the following are incorrect: (i) $x_s\in \frak gl(g)$ and $x_n\in \frak gl(g)$, (ii) the domain of $\varrho$ is $\frak g$, (iii) $\varrho(x)$ only makes sense if $x$ is in the domain of $\varrho$
well ${\rm ad}\rho(x_s)$ and ${\rm ad}\rho(x_n)$ don't make sense either, because they're elements of $gl(gl(V))$, while the text says $\varrho(x_s)$, $\varrho(x_n)$ are in $gl(V)$. why don't you just set me straight and tell me what $\varrho(x_s)$ and $\varrho(x_n)$ are? :)
well ${\rm ad}(x_n)$ and ${\rm ad}(x_s)$ are in $gl(gl(g))$, which is even further from the domain of $\varrho$ (namely $g$) than $gl(V)\ni x_n,x_s$ is.
in order for ${\rm ad}(\rho(x))$ to make sense, ${\rm ad}$ needs to be able to take in $\rho(x)$ as an argument, which means the domain of ${\rm ad}$ must be ${\frak gl}(V)$. and $\rm ad$ takes a lie algebra to its own general lie algebra, so ${\rm ad}:{\frak gl}(V)\to{\frak gl}({\frak gl}(V))$
@blue: I haven't thought about representation theory in a while, but I think this is just notation, and you should be reading a more thorough exposition. So, I believe $\rho(\text{ad}(x))\in{\frak gl}(V)$ is the linear map given by $\rho(\text{ad}(x))v = \text{ad}(\rho(x))(v) = [\rho(x),v]$.
@anon: I suppose one way out of this is to think of $\frak gl(g) = \frak g^*\otimes\frak g$ and use the tensor product of the two representations. But then we don't end up in ${\frak gl}(V)$. GRR.
@Pedro: You didn't done answer me. You feeling better?
@JasperLoy It depends on where you want to go with it. I think you can get a good number of examples of the ideas to be used with Banach and Hilbert spaces.
so $\sum_{k=1}^n a_k$ looks like $\sum_{i=1}^m i(i+1)/2$ which is $\sim m^3/6$ (don't need to work out the lower order terms), whereas $n^\alpha\sim m^{2\alpha}/2^\alpha$. hence $\alpha=3/2$ and $\beta=2^{3/2}/6$. (we'd need to show the limit does indeed exist with some bounding arguments, but that's bookkeeping)
@BalarkaSen The problem is that "easy" may be interpreted depeding on the persons you are referring at. I also think that my experience plays an important part here. I probably say "easy" to many problems that are considered pretty hard by others.
@Tunk-Fey I suspect you're right, as $\displaystyle\int_0^\pi \phi^{n-2}d\phi=\frac{\pi^{n-1}}{\color{Red}{n-1}}$ (p.s. you need the LaTeX in chat bookmark linked on the starboard to see latex in chat)
wait a sec i am doing some typing in another forum. crank handling.
OK, here goes. are you with me @blue?
The problem is to prove that if $xy$ divides $x^2 + y^2 + 1$ then $\cfrac{x^2 + y^2 + 1}{xy} = 3$. This looks very similar to the other problem about $\cfrac{x^2+y^2}{1+xy}$ which is done by boring methods like Vieta jumping and inequalities so I suspected that it's from Olympiad. My approach was somewhat like this
Let $\cfrac{x^2 + y^2 + 1}{xy} = k$. Then $x^2 + y^2 + 1 = kxy$ which is $x^2 - (ky)x + (y^2 + 1) = 0$ after rearranging. The disc of this polynomial is $(k^2-4)y^2 - 4$ which we require to be a square.
Thus one arrives at a Pell-like equation $m^2 - (k^2 - 4)n^2 = - 4$. A little modular arithmetic confirms that $k$ is one of $\pm\{1 , 3\}$ modulo $8$ but that's all I have found right now.
Given a projective plane (P,L) we can form an equivalence relation on (P choose 2) as {a,b}~{c,d} if the line incident with a,b equals the line incident with c,d. We can identify lines in L with equivalence classes in (P choose 2). In this way, we can think of line sets as partitions of (P choose 2). I'm wondering what conditions we need on such a partition to guarantee it's a valid line set.
@Chris'ssis That's a strong statement! :) I love your responses on MSE and I really want to develop similar abilities. This book may be just the right thing
@blue @anon: OK, I've done a bit of research for you. So, given $x\in\frak g$, there are $x_s,x_n\in\frak g$ so that $\rho(x_s)$ is the semisimple part of $\rho(x)$, and correspondingly for $x_n$.
By the way, Fulton/Harris make a big deal of saying $\frak g$ has to be semisimple. So the key point is that when $\frak g$ is semisimple, $\text{ad}:\frak g\to\frak gl(g)$ gives an embedding. So we take the Jordan decomposition of $\text{ad}(x)$ and these elements come back to unique elements of $\frak g$.
P.S. @blue @anon: I highly recommend Fulton/Harris's Representation Theory: A First Course. These guys are among the best expositors of algebra/algebraic geometry.
He was quoting a poorly written source on representation theory which didn't explain things and, to make things worse, seems to have omitted a necessary hypothesis. But I couldn't explain off the top of my head and only made things worse. Of course, why blue would ask me a question about algebra is beyond me ...
Would I be right in stating that say $V$ is a normed space, with subset $W$. Let wcl$(W)$ be the weak closure of $W$ in $V$, then if we take $x$ in wcl$(W)$ it follows that there exists a net $(x_{i})_{i}$ which converges weakly to $x$ in $V$. So this result states that at least one such directed set exists so as to define a net which converges weakly?
@JohnDoe You mean a net in $W$, presumably. Yes, in every topological space, the closure of a set $A$ is precisely the set of points such that some net in $A$ converges to that point.
Yes a net in $W$. So this differs with sequences since in sequences you know that the directed set in the set of natural numbers here you can't say exactly what the directed set is?
