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visits member for 4 years, 5 months
seen Jun 26 at 4:00
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May
28
awarded  Necromancer
May
27
awarded  Yearling
May
27
answered Can the unsolvability of quintics be seen in the geometry of the icosahedron?
May
6
awarded  Popular Question
Dec
16
awarded  Nice Answer
Jul
29
awarded  Nice Question
Jun
25
awarded  Excavator
Apr
19
comment Giving $Top(X,Y)$ an appropriate topology
For the non-Hausdorff case, see ncatlab.org/nlab/show/exponential+law+for+spaces.
Mar
8
comment Topological spaces determined by generalized metric spaces
Notice that this Arens' space is slightly different from the one given in Wikipedia (en.wikipedia.org/wiki/Arens%E2%80%93Fort_space), since in this space the set $\{c\}\cup\{a_{n,m}:n,m\in\omega\}$ is not open.
Feb
20
awarded  Yearling
Jan
25
asked Nontrivial copies of SO(r) in SO(n)
Jan
24
comment Suggestions for good notation
@Ben, the index in the coordinate expression $\frac{\partial f}{\partial x^j}$ for the 1-form $df$ is clearly in the low position! In fact, this is the main reason that I see for having to put the indexes of the coordinates in the high position as we do, instead of doing everything in the opposite way, which would be better in some way: we could write $f=x_1^2+x_3$ instead of $f=(x^1)^2+x^3$.
Jan
24
comment Suggestions for good notation
Regarding differential geometry: If $f:M\to\mathbb R$ is a smooth function on a manifold and $x:M\to\mathbb R^n$ is a chart, I prefer $\left(\frac{\partial f}{\partial x}\right)_j$ or $\left(\frac\partial{\partial x}\right)_j f$ (or even $\partial_j f$ if the choice of the particular chart is clear or irrelevant). Because the notation $\frac{\partial f}{\partial x^j}$ suggests that $\frac{\partial f}{\partial g}$ could be defined using only $g$, and in fact you need to know that you are restricting to the curve along which the other coordinates $x^i$ are constant.
Jan
21
revised Why is a topology made up of 'open' sets?
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Jan
21
revised Why is a topology made up of 'open' sets?
deleted 2 characters in body
Jan
14
comment real symmetric matrix has real eigenvalues - elementary proof
I think that the main difference is that Alexander extremises $x^tAx$ and I extremise $y^tAx$. That the two situations are not trivially equal is the subject of p.32 of Conway.
Jan
14
revised real symmetric matrix has real eigenvalues - elementary proof
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Jan
13
revised real symmetric matrix has real eigenvalues - elementary proof
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Jan
13
comment real symmetric matrix has real eigenvalues - elementary proof
Alexander, when you said that the minimum is an eigenvalue, did you mean to prove it by applying the Lagrange multiplier equation to the function $f(x)=x^tAx$ restricted to a level set of $g(x)=x^tx$, or did you have a different idea in mind?
Jan
13
comment real symmetric matrix has real eigenvalues - elementary proof
But Lagrange multipliers is, in my opinion, different from the argument above, which in fact was originally designed to deal with bounded operators, as explained in the comment. Can Lagrange multiplier be used to prove that $\pm\|T\|$ is an approximate eigenvalue of a bounded operator T? If not, is this enough to conclude that the proofs are different?