@BalarkaSen There's a way of avoiding all that extension nonsense by defining something called a "mixed partial derivative" of a map $f:\Omega\subset\Bbb R^m\to M$. The map is uniquely defined by the equality of mixed second partials and the product rule.
Then $\partial_t^2c$ makes sense for a curve $c:I\to M$.
And the geodesic equation is just $\partial_t^2c=0$.
The only problem is that the uniqueness proof is not nice at all.
To define third partials one does need the extension theorem, it seems.
I've spent the last hour trying to understand the purpose of generating functions and am getting a bit caught up in "generating functions help us do cool things" (like find Binet's formula)
and what the 1/(1-z-z^2) in the question linked above is supposed to represent
hmm, I think I need to go back a bit further than I anticipated
so a taylor series allows you to calculate the value of a function at a particular point
thus, the generating function allows you to calculate the value of a series at a particular point...? @TobiasKildetoft I feel like I'm completely wrong here
I think I'm stumbling because I read: 1 + z + 2z^2 + 3z^3 + 5z^4 ... = 1/(1-(z+z^2)
and think: well, if z=2 then the two sides should equal each other
but they don't...
as that is clearly wrong - I then ask - what am I missing...
Is there exist any algorithm to find the kernal of the linear operator L if the vector space L is from is something that is continous and/or infinite like in the space of functions?
I noticed if L is a differential operator, the kernel finding becomes solving a homogenous differential equation. However due to the lack of background in functional analysis, I am wondering if this is what we expect, or there is a better way
Ok, here's a more specific example from a year ago, except phrased in terms of solving for eigenvalues
This question is inspired from accidentally misreading this question in my tutorial exercise in my linear algebra course because I forgot the 3 in $P_3(\mathbb{R})$
(The actual question can be easily solved by considering the matrix of $T$ and then the eigenvector is found to be the constant p...
Therefore, suppose $T$ is the shift operator in the space of polynomials $\mathcal{P}(\mathbb{R})$, do functional analysis provide an elegant way to solve $T(p)=0$?
It is easy to check 0 is a solution, but how do we know that it is the only one, since when I plug everything in, I got $p(t+1)=0$ which does not really say much on how the countably infinite many coefficients of p should behave?
ok that makes sense, only the zero polynomail has a degree of -1
So in that case for an eigenvalue equation $T(p)=\lambda p$, rearranging gives $(T-\lambda)p=0$. Therefore the LHS must be a zero polynomial regardless of p, and that is possible only when $(T-\lambda)$ is the zero map...ok I must have made a conceptual error, I am not getting anywhere...
no, mistake, $(T-\lambda)$ is not necessary the zero map
...
Anyway, I am more interested in the approach in solving this equation $T(p)=0$ where $p \in \mathcal{C}(\mathbb(R))$
Ok, nvm, my approach might be too weak because I am kinda make use of induction by starting from n=1, and then look at the pattern formed as I increase the size of the matrix n and note how the main diagonal is all zeros
However for $C(\mathbb{R})$ not even such weak approach is possible, therefore I am kinda stuck
>Hey folks, anyone care to check that an answer on a question I posted is sound? Thanks. (I need the result in order to answer another question on this site.)
@amWhy Thanks, I thought about it, but since it isn't my answer that needed feedback, I thought it wasn't appropriate - but I see that it is, thanks again! :)
@BalarkaSen In your studies of harmonic functions, did you find the precise conditions for the derivative $$\left.\frac{d}{dt}\right|_{t=0}E(f+t\eta,\Omega)$$ of the energy functional $E:W^{1,2}(\Omega)\to\Bbb R$ to exist?
(I think that's the right Sobolev space)
Jost claims it's in Appendix A.1, but I'm unable to find it.
shrug. I think the key theorem is that every de Rham cohomology class has a harmonic representative, the consequences of which I knew are of topological of nature.
But again, I don't know anything about it except being in a Hodge theory lecture and overhearing conversations.
although it's true that $\lambda = 0$ is the only way it can be an affine subspace, if you are looking at vector subspaces a quick way to write down a proof is to note it has to pass through the origin.
@BalarkaSen I picked up the "official" Ricci flow book when I went by the library earlier to see what kind of background I need. They make these claims on page 1 and give no references
The way I like to think about it is that in dimension 3 you get homotopical anomalies, rather than topological. E.g., there exists a noncompact contractible 3-manifold which is not homeomorphic to R^3.
hi @BalarkaSen just a quick question want to make sure I am sane. I want to show that H and K are normal subgroups of G such that $H \cap K = \{e\}$
I want to show that $HK \simeq H \times K$
finally
I considered the following map $\phi : HK \rightarrow H \times K$ given by $hk \mapsto (h,k)$
I proved such map is a homomorphism and it is bijective. I don't think I need to show that it is well-defined as well right ? because of $h_1k_1 = h_2k_2$ then $h_1 = h_2$ and $k_1 = k_2$ right ?
This question is inspired from accidentally misreading this question in my tutorial exercise in my linear algebra course because I forgot the 3 in $P_3(\mathbb{R})$
(The actual question can be easily solved by considering the matrix of $T$ and then the eigenvector is found to be the constant p...
