""Advanced"" Math Questions Thread

[Anonymous No.8222605]

I will start.


Use the Mayer-Vietoris sequence, [math]\ldots \to {H^i}\left( {X,\mathcal{F}} \right) \to {H^i}\left( {U,\mathcal{F}} \right) \oplus {H^i}\left( {V,\mathcal{F}} \right) \to {H^i}\left( {U \cap V,\mathcal{F}} \right) \to {H^{i + 1}}\left( {X,\mathcal{F}} \right) \to \ldots [/math], to prove that [math]{H^1}\left( {{S^1},{A_{{S^1}}}} \right) \cong A[/math] for any abelian group [math]A[/math].

[Anonymous No.8222622]

>>8222605
Cohomology with values in what? A sheaf of abelian groups? What is [math]A_{S^1}[/math]?

[Anonymous No.8222635]

>>8222622
>What is ...?
The constant sheaf associated to A.

[Anonymous No.8222927]

>>8222605
What is the geometric picture that you have in your head for the lie algebra associated to a lie group (assuming you can picture the manifold structure of the lie group, like the simple case of S^1)?

[Anonymous No.8222933]

>>8222927
The lie algebra of a lie group is isomorphic to the tangent space of the group at the identity.

[Anonymous No.8223026]

>>8222635
Isn't it rather trivial then?

[Anonymous No.8223028]

>>8222927
Infinitesimal generators. Take a point on your Lie group. Imagine all directions where a flow on the Lie group (as a differential manifold) could take you. Each such flow is generated by a one-dimensional subspace of the Lie algebra.

[Anonymous No.8223053]

>>8223026
your mom is trivial xD

[Anonymous No.8223070]

>linebreaks between greentext lines
reddit get out

[Anonymous No.8223075]

>>8223026
I don't know.

[Anonymous No.8223078]

>>8222605
>"""Advanced""" Math

[Anonymous No.8223094]

>>8223078
stop posting my pictures everywhere jerk

[Anonymous No.8223202]

>>8223053
>>8223070
>>8223078
>>8223094
gtfo

[Anonymous No.8223231]

>>8223202
>4chan
>serious discussions by intelligent and educated people
Pick exactly one.

[Anonymous No.8223262]

>>8222605

Cohomology is an absolute mystery for me.

[Anonymous No.8223264]

>>8223202
Don't feed the summer fags.

[Anonymous No.8223274]

>>8223262
We had this one already. Something something Baire category theorem.

[OHP No.8223276]

>>8223262
Just think of cohomology as the association of spaces to their space of maps into some coefficient object. Homotopy is just the association to the space of maps from some coefficient object.

[Anonymous No.8223290]

>>8223262
Pretty straightforward. H^i(X,F) is just the right derived functor of the global section functor of X on F.

[Anonymous No.8223304]

>>8222605

Use chords to divide a circle into equal area pieces with no two pieces congruent.

[Anonymous No.8224180]

>>8223276
>>8223290
>algebraists think this is a viable intuition
Cohomology is about geometric obstructions. If something is a cocycle but not a coboundary, then it fails to behave nicely. The existence of such possible obstructions is a property of the underlying space.

[Anonymous No.8224243]

>>8224180
you're on fire this week
that's the best description of cohomology i think i've ever seen

[Anonymous No.8224524]

>>8224180
That doesn't work for sheaf cohomology. It is unclear exactly what are cocycles and coboundaries.

Sheaf Cohomology is defined using derived functors, [math]{H^i}\left( {X,\mathcal{F}} \right) = {R^i}{\Gamma _X}\left( \mathcal{F} \right)[/math].

Which means given any injective resolution [math]0 \to \mathcal{F} \to {\mathcal{I}^ \bullet }[/math] of the sheaf, then it looks like cohomology in the usual sense [math]{H^i}\left( {X,\mathcal{F}} \right) = {\mathcal{H}^i}\left( {{\Gamma _X}\left( {{\mathcal{I}^ \bullet }} \right)} \right) = \frac{{\ker \left[ {{\Gamma _X}\left( {{\mathcal{I}^i}} \right) \to {\Gamma _X}\left( {{\mathcal{I}^{i + 1}}} \right)} \right]}}{{\operatorname{im} \left[ {{\Gamma _X}\left( {{\mathcal{I}^{i - 1}}} \right) \to {\Gamma _X}\left( {{\mathcal{I}^i}} \right)} \right]}}[/math].

But if we want to think analogous the topological sense, then [math]{{\Gamma _X}\left( {{\mathcal{I}^ \bullet }} \right)}[/math] is our cochain complex.

But there practically no geometric intuition involved in this. Considering [math]{\mathcal{I}^ \bullet }[/math] is just some arbitrary complex of sheaves defined such that [math]0 \to \mathcal{F} \to {\mathcal{I}^ \bullet }[/math] is an exact sequence and [math]\operatorname{Hom} \left( { - ,{\mathcal{I}^i}} \right)[/math] is an exact functor for all i.

[Anonymous No.8226189]

>>8224524
trivial

[Anonymous No.8226326]

>>8224524
>It is unclear exactly what are cocycles and coboundaries.
For sufficiently nice spaces you can construct sheaf cohomology via Cech cohomology. Then it's immediately clear that the cohomology groups tell you something about the possible existence of certain bundles, e.g. orientability or the existence of spin structures.

[Anonymous No.8226393]

>>8226326
>For sufficiently nice spaces you can construct sheaf cohomology via Cech cohomology.

I believe the are only always equal when the space is paracompact&hausdorff. This is obviously great for manifolds, but not so useful if you are working with varieties or schemes,

[Anonymous No.8227359]

Is there a visible analogy between the spectral theorem of operator theory and some homology or cohomology theory? The words are all the same and it bothers me. Every time I read the words "resolution of the identity" I have to remind myself I'm reading an analysis book.

[Anonymous No.8227389]

>>8224524
What is this? Manifold theory or something? It looks like spooky algebra and some geometry/topology words I saw in my algebra textbook. I can't wait to learn shit like this.

[Anonymous No.8227395]

>>8227389
Homological algebra. It stops being interesting, once you open a book on the topic.

[H No.8227412]

>>8227395
>It stops being interesting
So true. I was really interested in homological algebra until I started to read Weibel. There are more interesting books, but it is still boring in a rather fundamental way.

[Anonymous No.8227461]

>>8227389
https://www3.nd.edu/~eburkard/Talks/Category%20Theory%20Notes.pdf
http://www.math.toronto.edu/~jacobt/Lecture13.pdf

[Anonymous No.8227615]

I tried to prove that the euclidian space is non-curved
took 3 arbitrary points with general coordinates (is that the correct name for points such that no subset is linear dependent?) in space
w.l.o.g use (0,0,0), (1,0,0) and (a,b,0)
sum of angles between them must be 180°

is it a sufficient condition? or is it even included in the definition and im calculating in circles?
not done yet because i get terms with huge roots and i need to look up the relevant trig identities

[Anonymous No.8227624]

>>8227615
Which definition of "non-curved" are you using? Are you familiar with metric tensors and curvature tensors?

[Anonymous No.8227718]

>>8227624
the aforementioned triangle definition
geez, i just wanted a fun back-of-the-envelope calculation, not following rules

[Anonymous No.8227760]

>>8222933
aka Lie's first fundamental theorem.

[Anonymous No.8227778]

/sci/ needs more threads like these, this was really interesting to read.

[Anonymous No.8227789]

>>8226393
>if you are working with varieties or schemes,
there are results saying you can use cech cohomology for like separated noetherian schemes which is basically the same condition as the paracompactness/hausdorff but algebraized.

[Anonymous No.8227845]

>>8222605
the autism in that pic is off the charts

[Anonymous No.8227848]

>>8224524
trivial

[Anonymous No.8228394]

>want to study spectral problems with algebraic geometry but transcendental functions don't form a ring over any field

[Anonymous No.8228461]

>>8228394
Not entirely sure, but you may want to look into non-commutative geometry.

[Anonymous No.8228808]

>>8222605
Why do we define differential forms like we do it?

[Anonymous No.8228826]

>>8228808
To capture the geometry of infinitesimal volume. You are basically applying determinants to tangent vectors.

[Anonymous No.8229183]

>>8228808
The are defined in a way such that 1-forms are essentially dual to vector fields.

As for the motivation of higher k-forms, I don't know what the original motivation was but can give you some of my own input.

The way in which [math]{{\Omega ^k}M}[/math] results in the existence of a UNIQUE collection of linear maps, [math]{\left\{ {\operatorname{d} :{\Omega ^k}M \to {\Omega ^{k + 1}}M} \right\}_{k = 0,...,\dim M}}[/math], such that:

(i) If f is a function, then df is the usual differential.

(ii) d^2 = 0 always

(ii) If omega is a k-form and eta is a p-form, then [math]\operatorname{d} \left( {\omega \wedge \eta } \right) = \operatorname{d} \omega \wedge \eta + {\left( { - 1} \right)^k}\omega \wedge \operatorname{d} \eta [/math]


Where [math]\wedge :{\Omega ^k}M \times {\Omega ^p}M \to {\Omega ^{k + p}}M[/math] is the most natural choice of product of differential forms.

Along with many other things, the wedge product induces a cup product of the de rham cohomology.

[Anonymous No.8229380]

>>8228461
Thanks anon but that's not it.

[H No.8229387]

>>8228808
Think about tensors.
Properly, like a physicist, with all the indices and shit.

[Anonymous No.8229835]

>>8229387
>Properly
>like a physicist

pick one

[Anonymous No.8229855]

>>8228808
I just had a discussion about this with a differential geometer. The geometric idea is that they sort of define something orthogonal to the tangent space and give information about orientability.

[Anonymous No.8229932]

>>8229835
>"applied" """"""""""mathematicians"""""""""""

[Anonymous No.8229971]

>>8229835
>>8229932
>mfw this is what physicists actually believe

[Anonymous No.8230438]

how to generalize gross-zagier theorem to higher derivatives in a nice enough way to solve BSD?
https://en.wikipedia.org/w/index.php?title=Gross–Zagier_theorem

[Anonymous No.8230476]

>>8224524
Using Injective resoltion do not allow you to understand cohomology. I'd say it's the opposite, cohomology can get you to understand resolutions.

[Anonymous No.8230478]

>>8223262
>>8228394
Take your pedophile cartoons back to >>>/a/.

[Anonymous No.8230486]

>>8228826
Thanks, that's very concise and it really answers my question. There is some doubt about an external derivative though.