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Jul
6
comment How to embed $U(1)$ (or other groups) into a bigger group, using Dynkin diagrams
@Oscar Maybe the way to think about it is that when you are erasing a node, you are reducing the dimension of the root vector space. So you lose all the root vectors that had a component parallel to the root you erased. However, as you said, the node you erased had a corresponding generator in the maximal torus, which is sitting at the origin in the root vector space (i.e. its weight is the zero vector). So it can be included in the subspace, but will commute with everything else in the subalgebra that is left.
Jun
19
answered How to embed $U(1)$ (or other groups) into a bigger group, using Dynkin diagrams
Dec
8
awarded  Caucus
Dec
1
revised Function such that $f(x) f(\pi/2 - x) = 1$
edited tags
Dec
1
awarded  Scholar
Dec
1
accepted Function such that $f(x) f(\pi/2 - x) = 1$
Dec
1
revised Function such that $f(x) f(\pi/2 - x) = 1$
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Dec
1
comment Function such that $f(x) f(\pi/2 - x) = 1$
@StevenStadnicki okay, I see. However, I have to choose the $C^\infty$ function such that its power series works at the midpoint. Setting all the derivatives to zero is like making it locally the constant function, and I could also locally make it match a $\tan$ function. Maybe I should have asked for how many analytic functions satisfy the requirement, since each analytic function would characterize the type of power series the $C^\infty$ function can have at the midpoint.
Dec
1
revised Function such that $f(x) f(\pi/2 - x) = 1$
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Dec
1
comment Function such that $f(x) f(\pi/2 - x) = 1$
by smooth I mean $C^\infty$. Looking at the second derivative at $x=pi/4$ (or I guess $1$ in your example) I think gives a nontrivial restriction on the function at that point, $f''-(f')^2=0$. Higher derivatives give still more restrictions, and I'm wondering if $tan(x)$ and the constant function are the unique solutions satisfying all these smoothness conditions.
Nov
30
comment Function such that $f(x) f(\pi/2 - x) = 1$
I changed the interval to $0<x<\pi/2$, I think it makes more sense that way.
Nov
30
revised Function such that $f(x) f(\pi/2 - x) = 1$
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Nov
30
revised Function such that $f(x) f(\pi/2 - x) = 1$
added 62 characters in body
Nov
30
comment Function such that $f(x) f(\pi/2 - x) = 1$
@StevenStadnicki Are you sure this implies smoothness of the function at $\pi/4$? I'm not seeing it immediately...
Nov
30
comment Function such that $f(x) f(\pi/2 - x) = 1$
@KristofferRyhl oh whoops, I should probably say that it doesn't have to hold at $x=0$ since I want to allow $f$ to diverge as $x\rightarrow \pi/2$.
Nov
30
awarded  Student
Nov
30
asked Function such that $f(x) f(\pi/2 - x) = 1$
Oct
3
answered Möbius Transformation interchanging two preassigned points in the upper half plane
Feb
16
revised Vector Calculus Identities Using Differential Forms
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Feb
14
revised Vector Calculus Identities Using Differential Forms
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