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].
Cohomology with values in what? A sheaf of abelian groups? What is [math]A_{S^1}[/math]?
>What is ...?
The constant sheaf associated to A.
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)?
The lie algebra of a lie group is isomorphic to the tangent space of the group at the identity.
Isn't it rather trivial then?
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.
your mom is trivial xD
>linebreaks between greentext lines
reddit get out
I don't know.
>"""Advanced""" Math
stop posting my pictures everywhere jerk
gtfo
>Veeky Forums
>serious discussions by intelligent and educated people
Pick exactly one.
Cohomology is an absolute mystery for me.
Don't feed the summer fags.
We had this one already. Something something Baire category theorem.
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.
Pretty straightforward. H^i(X,F) is just the right derived functor of the global section functor of X on F.
Use chords to divide a circle into equal area pieces with no two pieces congruent.
>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.
you're on fire this week
that's the best description of cohomology i think i've ever seen
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.
trivial
>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.
>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,
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.
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.
Homological algebra. It stops being interesting, once you open a book on the topic.
>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.
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
Which definition of "non-curved" are you using? Are you familiar with metric tensors and curvature tensors?
the aforementioned triangle definition
geez, i just wanted a fun back-of-the-envelope calculation, not following rules
aka Lie's first fundamental theorem.
Veeky Forums needs more threads like these, this was really interesting to read.
>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.
the autism in that pic is off the charts
trivial
>want to study spectral problems with algebraic geometry but transcendental functions don't form a ring over any field
Not entirely sure, but you may want to look into non-commutative geometry.
Why do we define differential forms like we do it?
To capture the geometry of infinitesimal volume. You are basically applying determinants to tangent vectors.
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.
Thanks user but that's not it.
Think about tensors.
Properly, like a physicist, with all the indices and shit.
>Properly
>like a physicist
pick one
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.
>"applied" """"""""""mathematicians"""""""""""
>mfw this is what physicists actually believe
how to generalize gross-zagier theorem to higher derivatives in a nice enough way to solve BSD?
en.wikipedia.org
Using Injective resoltion do not allow you to understand cohomology. I'd say it's the opposite, cohomology can get you to understand resolutions.
Take your pedophile cartoons back to .
Thanks, that's very concise and it really answers my question. There is some doubt about an external derivative though.