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visits | member for | 5 years |
seen | Jan 3 '14 at 9:16 | |
stats | profile views | 3,526 |
I'm interested in many things, but algebraic geometry I love.
Oct 4 |
comment |
The localization of a regular local ring is regular
I don't think you should. The stream-of-thought was quite nice. There certainly is a notion of a regular (not nec. local) ring with exactly the meaning you wrote. I'm actually quite pleased with the interpretation of the second paragraph. The intuition of switching from k[[x,y]] to k[[x]]((y)) being analogous to "something like going from a point to an irreducible subvariety going through that point" sits much less well with me. |
Oct 4 |
awarded | Critic |
Oct 4 |
comment |
The localization of a regular local ring is regular
Interesting. So the theorem is about localization by any set, not nec. by a prime ideal? |
Oct 4 |
awarded | Editor |
Oct 3 |
revised |
The localization of a regular local ring is regular
Tex is acting up. Had to remove parentheses from dollar signs |
Oct 3 |
revised |
The localization of a regular local ring is regular
Changed title to what I meant it to be; added 1 characters in body |
Oct 3 |
asked | The localization of a regular local ring is regular |
Oct 2 |
comment |
Is there a group-scheme equivalent of the theorem that any Lie group is diff. to a compact one cross R^n?
Yes, thanks Kevin. My question is: in what way does Chevalley's theorem answer the question? It's portrayed as a more general theorem than the one I stated in the question. I'm trying to understand why. In order to relate the two, I assumed that cfranc had it in his mind that $H$ is diff. to $\mathbb{R}^n$. Indeed you're right that there are problems with the definition of "diff." here, and even with "connected". Still -- for his answer to make sense he must have had an idea of how these two things are related in the classical case. |
Oct 2 |
comment |
Is there a group-scheme equivalent of the theorem that any Lie group is diff. to a compact one cross R^n?
You mean to say that every (connected) linear group over $\mathbb{R}$ is diffeomorphic to $\mathbb{R}^n$? This seems like a stronger version of the theorem that every linear group is rational, but I've never heard of it before. Also, in the theorem you state, is it implied that $H$ is connected? |
Oct 2 |
asked | Is there a group-scheme equivalent of the theorem that any Lie group is diff. to a compact one cross R^n? |
Sep 30 |
asked | Innovations in deformation theory |
Sep 12 |
comment |
Integrals from a non-analytic point of view
@Ricky: I forget. My last encounter with such ideas was roughly 13 years ago. But I'm looking for abstraction in a more categorical, rather than analytic, approach. |
Sep 12 |
comment |
Integrals from a non-analytic point of view
@Will: That shows that integrals, as we already defined, work to define cohomology. But it doesn't get to the essence of what we want integrals to do (and then maybe we can define integrals to be any definition that does that?). In any case, having done number theoretic algebraic geometry for so long, I wonder what the essence of integrals (a tool I don't see very often in my work) is. What function do they provide? Is being an alternate way of defining cohomology really their only function? |
Sep 12 |
asked | Integrals from a non-analytic point of view |
Aug 28 |
comment |
Is the strict transform a finite morphism?
Really? This perplexes me. This curve has only one tangent at the origin, and it is x=0. Indeed, the other affine would give (say v=1/s) y(v^2)=x-1 - which has no points above x=y=0. |
Aug 28 |
asked | Is the strict transform a finite morphism? |
Aug 2 |
awarded | Popular Question |
Jul 12 |
awarded | Nice Question |
Jun 12 |
awarded | Nice Question |
Jun 12 |
asked | In what ways is physical intuition about mathematical objects non-rigorous? |