Mariano Suárez-Alvarez
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14h
comment Projective bundle of a sum of ample line bundles
Have you tried seeing if this works in the simplest example? :-|
23h
comment Scheme over a not algebraically closed field
If $n=1$, then $x$ is in $O(V)$ according to the OP 's definition. Is that inverttible there?
1d
answered Does every non-trivial $\mathbb{C}$-algebra contain an element which is not a square?
1d
comment Using Serre's Theorem, and showing lie algebra is 1-D
Again: these two problems are entirely unrelated.
1d
comment Field of char(F)=p, if p divides n then sl(n,F) is not simple
In fact, you have asked 4 questions almost at once! Do not do that. None of your questions shoe that you have even thought about the problems you posed, not even what you know about the subject.
1d
comment Field of char(F)=p, if p divides n then sl(n,F) is not simple
The two questions are completely unrelated. Please ask one question per question. Also, you have asked at least one other question: it is immensely better to ask one question at a time except in rare situations...
1d
comment Dimensions of classical lie algebras and showing $sp(4,C) \cong so(5,C)$
Surely you have some description of those Lie algebras. In particular, you have the vector spaces, so you should be able to compute their dimensions! That is pure linear algebra, not involving any Lie theory at all.
2d
comment Proofing that the Lucas numbers come closer to the Phi rounded numbers then the Fibonacci numbers.
Once one has a proof, it is not a bad idea to proof it just to be sure, of course!
2d
comment Proofing that the Lucas numbers come closer to the Phi rounded numbers then the Fibonacci numbers.
To proof something in English means to make something (like a boat) water-proof. To prove something is what you want.
2d
comment Real world uses or interesting facts about/for Associahedron or Permutohedron
Real life uses? I thought they had closed that place down...
2d
comment Why isn't the identity/unit matrix upright?
I don't think I know even one mathematician Who is aware the standard even exists, let alone if it prescribes a slanted I for the identity mateix, and I am sure no one I know cares.
2d
comment Why isn't the identity/unit matrix upright?
You should ask the ISO committee that decided on that.
2d
comment Computing relative de Rham cohomology?
The first rule of the cohomology club is that one simply does not compute cohomology directly.
Apr
28
comment Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$
It has been shown that nothing is gained from using abbeviations such as «plz» in this site. Please don't do it.
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
In fact, I think this definition is not that unusual, althought it is usually only considered in contexts where products do exist. It certainly makes sense in general. And it is a good one: how would you define a "topological space object in a category"? A sensible way is to say that it is an object together T with a factorization of hom(-,T) through the forgetful functor from spaces to sets. In this case, it is not clear what the "internal to the category" version of the definition is, if there is any at all!
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
Oh well. I guess you should report it to overloading police or something! ;-)
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
Zyx, I am sorry, I have no idea what you are talking about.
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
@Tsemo, indeed, that is precisely why it is sensible to suppose he is using the standard one. If he had made explicit a non-standard definition of group, using the standard one in answering him would be not a great idea.
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
Why would the set of bijections of a 1-element set not constitute a group?! It is certainly not empty. Not even the set of bijections of a 0-element set is empty!
Apr
27
comment What are the group objects in the category of finite sets and bijections, and its functor category?
You surely can, but the OP didn't in his question. If you ask a question and make explicit such a definition, I would probably have to think more.