/sci/ - Science & Math

>>12475346
One can approach the subject from two directions. For simplicity, let's say we have a finite group [math]G[/math] and some commutative ring (unital) [math]R[/math]. We can then think of all the groups of square matrices with non-zero determinants and with entries in [math]R[/math], denote those by [math]GL_n(R)[/math], where the subscript tells you the dimension. If the dimension is too small, then there will be non-trivial relations (or the homomorphism [math]\rho \colon G \to GL_n(R)[/math] will have a non-trivial kernel, and if the dimension is too big, it it will provide no information we did not already have. Let's simplify even more and assume our ring [math]R[/math] is a field [math]k[/math]. Then [math]GL_n(k) = \text{Aut}(k^n)[/math], so actually we obtain an action of the group on the n-dimensional vector space over the field of choice. We may then ask what is the minimal dimension such that the kernel is trivial, and that for example will give us an invariant that can be used to distinguish between groups.

The other direction is that we start by constructing the group algebra [math]RG[/math] and then the modules over this group algebra are obtained like above. Take the free [math]R[/math]-module [math]R^n[/math]. Its automorphisms are the nxn-matrices with non-zero determinants, so map your group into that, and then use the projections [math]R^\mathscr{J} \to M[/math] to obtain all the possible actions of [math]G[/math] on [math]R[/math]-modules. Again using a field, we have the actions of our group on the vector spaces. Actually, this time we don't think about the action itself, but we think of it as the scalar multiplication of [math]kG[/math] (as [math]R=k[/math] for simplicity again). Then one can use all sorts of ring theoretical tricks to compare different groups, like for example ask what are the simple or indecomposable modules like, and so on.

Two corresponding ways to answer the same questions with different tools. Hope this helps.

>>12462288
When I speak of homotopy theory, I mean [math]\pi_*(- ; G) \colon \textbf{Top}_* \to \textbf{Top}_*[/math], where the group [math]G[/math] is some [math]\mathbb{Z} /n\mathbb{Z}[/math] with n=0 included for the sake of ordinary stuff. Everything but those are "other" types for me. I could of course think of rational stuff as "non-other", but that would be highly irrational because the homotopy groups I would encounter are mostly finite and then they would be killed by the UCT [math] 0 \to \pi_n(X) \otimes_\mathbb{Z} \mathbb{Q} \to \pi_n(X; \mathbb{Q}) \to \text{Tor}_1^\mathbb{Z}( \pi_{n-1}(X), \mathbb{Q}) \to 0[/math].

>>12173851
Yes. I think that works. Sorry for the flow of thought post, I'm thinking as I'm writing. We need reflexivity, but that is guaranteed by [math]x=y \Rightarrow d(x, y)=0[/math], we need symmetry, but that is also given by the axioms of a pseudometric, [math]d(x, y) = d(y, x)[/math], and transitivity. Now this thing uses the triangle inequality. If [math]d(x, y) = 0 = d(y, z)[/math], then [math]0 \le d(x, z) \le d(x, y) + d(y, z) = 0 + 0 = 0[/math], so defining [math]x \sim y \Leftrightarrow d(x, y) = 0[/math] does indeed give an equivalence relation. Now, this gives quotient space [math]X/\sim[/math] with all the equivalence classes as its points. And yes, we do get an induced metric [math]\tilde{d} \colon X/\sim \times X/\sim \to [0, \infty)[/math] which will be given by [math]\tilde{d}([x], [y]) = d(x, y)[/math] for some representatives of the classes. Is it well-defined? If [math]x, x'\in [x], y, y'\in [y][/math], then [math]\tilde{d}([x], [y]) = d(x, y) \le d(x, x') + d(x', y') + d(y', y) = 0 + d(x', y') + 0 = d(x', y') = \tilde{d}([x], [y])[/math], so yes it is! Very good. The answer is yes.

>>12148906
Imagine being almost 30 and still a zoomer.

>>11819333
I just found it after posting. Thanks anyway.

>>11819329
Where's the time stamp?

>>11703875
I have no idea. My supervisor said that I would get the Fields medal if I proved the Moore conjecture, but I can't remember what that was about. Something about spheres having homotopy exponents, I think. Do that and become a superstar.

>>11703880
I would but I don't know how to. How about you show us how to start a new one?

>>10161007
>he downloads textbooks but doesn't download anime or drugs

Ok, one black hole with mass of mountain will produce light energy output 10 Gwt in first billion years and much-much more until the end (10 billions years). You can literaly create star for your purpose. But how realisticly(theoretical) we can create one(with mass of mountain +/-)?