Hanno

Yes -- it's just $k$-valued functions on a finite set. I suggest you sit down and work it out for a bit yourself?

You can just write down bases for both sides and see that the canonical map is an isomorphism.

A $0$-dimensional manifold is just a discrete space.

Yes, the discrete structure. What the authors are ultimately after is a Hopf-algebra(-like) structure on the smooth functions on a Lie group, but as a motivation, they consider the much simpler special case of a finite group.

If $G$ is finite, then ${\mathcal F}_k(G)\otimes_k {\mathcal F}_k(G)\to {\mathcal F}_k(G\times G)$ is indeed an isomorphism, for the naive algebraic tensor product. Only if you want to carry over this idea to other contexts, e.g. $G$ being a Lie group, then you'd surely need a different version of tensor product. I don't know if the book elaborates on that later on.

Oh, you see, the authors assume $G$ to be finite here! So then there's no issue with completion of the tensor product.

I'm sorry, I cannot help you share the document, if only because I don't know if it is meant to be shared. I'm afraid I have to stop at my initial comment that I suspect that some technical details are missing here.

I can't open that, either :-(

Sorry for the confusion, too, but the doc you ref'd stops at p.37 for me.

this does not include a Chapter 8?

Can you provide a link to the lecture notes? On first sight I agree that this seems to be true only for some form of completed version of the tensor product.

Hi all

@s.harp alright :)

yes and if I is not prime you instead consider the (minimal) primes containing it

then its the statement that passing to the 'zero set' of an element does not drop dimension by more than one

@s.harp you can reduce to the case where A is local and I is the maximal ideal

but its also less frustrating ;)

computer science is fun but - at least in what i have been doing last year - lacking the depth and beauty of math

though i at times thinking about trying to return

@TobiasKildetoft not really, i think about math from time to time and enjoy it, but no research at the moment

thx, u2!

no worries

I'm sorry I don't think I'm the best person to ask for this stuff. Maybe Mike or Tim will answer to your comment-question on MO?

I think so

uh.. yea sounds technical at first. did you try just mimicking the set construction of the dependent product in the category of k spaces?

sure @ ask sth diff, what do you want to know?

so an element of dep(f) is a function assigning to each x of X an element of Z lying over x

as usual: a map f: Z -> X is viewed as a family of sets parametrized over X by sending x\in X to its preimage

now you just have to make the translation from families of sets to maps of sets

If you have a function f with domain some set X, you may think of f as a family of sets varying over x. the dependent product is then just prod_{x\in X} f(x)

hey

I guess I won't have time to look at it in detail. Got to go now, have fun, bye

just thinking out loud why intuitively an eff desc morph is necessarily effective epi

While for effecitve epimorphisms, I am concerned with describing morphisms B -> C as special morphisms A -> C. Maybe the connection is that to any morphism f: A -> C you can assign graph(f) together with the projection graph(f) -> A, thereby getting into the 'sheaf' context

Ok, the contexts for effective epimorphism and effective descent morphism seem to be different at first. For effective descent A -> B, I consider objects C -> A which I somehow might think of as sheaves over A, and I require that I have some kind of transport between fibers over points of A that sit over the same point of B. Then the assertion is that such a sheaf actually comes from / descents to a sheaf C -> B

I'll medidate over it ;)

no problem, didn't contribute much

I don't get the def of eff desc morph intuitively either - currently I can only grasp the equivalent formulation of a stable effective epimorphism

I don't know the details very well I have to admit, but I agree that intuitively it sounds good

yes that sounds reasonable

but (1) is far too strong, right? being isomorphic to the multiplicative action should mean being a trivial torsor

hm, and how do you want to judge whether your definition is suitable? do you have certain simple properties or equivalences in mind you'd like to have?

so what do you want to assume?

ah, so your point is that you are seeking for a suitable definition of torsor for categories more general than regular ones?

yes

which is much weaker than the existence of a global point 1 -> X (i.e. a section of X -> 1)

I think it should be 'X -> 1' is an epimorphism

hm, enough for what?

From my understanding there is no special situation of 'working over a point'. The general definition is concerned with objects acted upon by some group object, and should include that the object surjects onto the terminal object

hi