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Jan
11
comment Group Duality with respect to Generators and Relations
(2/2) On the other hand, there certainly are some notions of duality that arose from mathematical questions, which you may be interested in looking up. A simple one would be the concept of a dual vector space (vector spaces are, of course, continuous groups too), which arises by considering functions on that space (or matrix transposition). A harder one: Pontryagin duality comes from trying to do Fourier analysis on certain groups, and crops up very naturally in a bunch of places, including e.g. the character theory of finite abelian groups.
Jan
11
comment Group Duality with respect to Generators and Relations
(1/2) I think my real point is: in order for such a definition to be mathematically meaningful, its conception needs to be mathematically well-motivated. You can't just say "I want duals to exist; where are they?". Better questions: what do you want $G^D$ to do? Or how do you want it to look relative to $G$? I see no mathematical reason to swap the generators and relations of a group (and no guarantee that it could be made to work). My geometric proposal was just something slightly more well-motivated (but, if you work through the details, you'll see it's also not very interesting).
Jan
10
comment Group Duality with respect to Generators and Relations
Suppose $G = \langle x | x^2 \rangle$. What does your proposed object $G^D$ mean? (But, if you want your 'dual' group to be the group of symmetries of the respective 'dual' polyhedron - which seems like quite a reasonable request to me - then the right definition is probably simpler than you think.)
Jan
10
comment How to prove that a set spans a plane
@Pow: aha, okay - take an element (x,y,z) on your plane. Supposing $c \neq 0$, we can more helpfully write this as (x,y,$\frac{-ax-by}{c}$). Now it suffices for you to find r and s (in terms of a, b, c, x and y) such that r(b,-a,0) + s(0,c,-b) = (x,y,$\frac{-ax-by}{c}$) - shouldn't be too hard. (Then of course you have to deal with the case $c = 0$...) :)
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
(You might also find it helpful to note that the "natural" multiplication on $\mathbb{R}^2$, namely $(u,v)\cdot (x,y) = (ux, vy)$, does not agree with the natural multiplication on $\mathbb{C}$ for any choice of $g\in \mathrm{GL}_2(\mathbb{R})$.)
Jan
25
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
@laovultai: Let $f$ be the map $\mathbb{R}^2\to \mathbb{C}$, $(a,b) \mapsto a+ib$, and let $g$ be any element of $\mathrm{GL}_2(\mathbb{R})$. Then $f\circ g$ is an isomorphism $\mathbb{R}^2\to \mathbb{C}$ as $\mathbb{R}$-vector spaces. Now simply "pull back" the multiplication from $\mathbb{C}$ to $\mathbb{R}^2$ along the map $f\circ g$ (e.g. when $g$ is the identity map, the multiplication inherited is $(u,v)\cdot (x,y)=(ux−vy,uy+vx)$), and you get an isomorphism of rings (or, equivalently, $\mathbb{C}$-vector spaces). Does that answer your question?
Jan
17
comment How should I understand $R[x]/(f)$ for a ring $R$?
f should be irreducible, otherwise it is not true that R[alpha] = R[x]/f.
Jan
17
comment What's the difference between $\mathbb{R}^2$ and the complex plane?
@laovultai: Because they're not "equal". They're isomorphic, but in order to prove that, I have to choose an isomorphism. I chose the "obvious" one, $(a,b) \mapsto a+ib$. (There are lots more, e.g. $(a,b) \mapsto 2a + b - 5ia$, or $(a,b) \mapsto ia - b$.)
Aug
2
comment Surjection/Injection in Product of Linear Transformation
Also, why not just pick an explicit example? The first example I think of is $$S = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\\0&0&0\end{pmatrix}, T = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}.$$ What properties does this have?
Aug
2
comment Surjection/Injection in Product of Linear Transformation
"Say, S(abc)=abcd, also S(abc)=pqrs." But this can never happen! Functions don't work like this. If you plug in an input, you get one output.
Jul
29
comment Should we simplify or not for function domain?
Do you mean "what is the domain of $f$"? Personally, if I was being very strict, I would say that the definition of $f$ was "take $x$, square it, and then divide the result by $x$", which always gives you $x$ back except when $x = 0$, when the calculation doesn't make sense. So the domain is the set of non-zero real numbers. (In practice, I am never this strict, except when I am teaching students how to be this strict.)
Jul
28
comment linear map vs operator: raised to power
Your example illustrates the point nicely, but of course, it's slightly worse than that: even if $V\cong W$, there's no obvious way of defining $T^2$. (Matrices are evil notation in that they hide the implicitly chosen bases. The OP might want to consider a map from $V$, the space of real polynomials of degree at most 4, and $W$, the space of real symmetric 3*3 matrices of trace 0. Both are $\cong \mathbb{R}^5$, and there are clearly lots of nice linear maps between them, but there's no obvious way of applying one twice.)
Jul
28
comment linear map vs operator: raised to power
What does $T^2$ mean, when $T: V\to W$? If I try to plug in a vector $v\in V$, I get a little confused: $T^2 v = T(T(v))$, but $T(v)\in W$, so I can't apply $T$ again.
Jul
28
comment what is the difference between sections and germs in a sheaf?
Sections are "functions on a large open set". Germs are "functions around a point". If this isn't obvious, why don't you tell us what your definitions are?
Jul
28
comment Fixed point of a continuous map
For the first problem, consider the map $g:x\mapsto f(x) - x$ on $[-1, 1]$. What is $g(0)$? What is $g(1)$? Where does $f$ have a fixed point?
Jul
28
comment Help to prove bijection between subset of $S^2$ and $\mathbb{R}^2$
@user1620696 No problem!
Jul
28
comment Help to prove bijection between subset of $S^2$ and $\mathbb{R}^2$
I've edited my post and added the surjectivity argument.
Jul
28
comment Find equation that represents $m$ as the subject
@Nick The answer is, essentially, that you need to plug numbers into formulas, or perhaps write a computer program to do so if you can't face doing it yourself. If you try to solve it algebraically, you'll end up with something disgusting like George's formulas above no matter how you spin it (Wikipedia's methods are all different ways of saying the same thing). I like George's recursive algorithm, but of course it (almost certainly) won't give you exact answers. Still, let me know if I can help with anything.
Jul
27
comment Find equation that represents $m$ as the subject
Perhaps the most appropriate tag is "algebra-precalculus". (Though I wouldn't expect anyone to learn how to solve a cubic equation until well after calculus, admittedly, but it has the same flavour as a quadratic equation, just much uglier...)
Jul
27
comment Why do I disagree with my calculator?
I agree with GEdgar, but if remembering that numbers have to be packaged with the sign to the left of them is hard, it's enough to remember that + and - are to be given the same priority, and so you should work from left to right, which always gives the right answer.