Cullen-regular functions have an open mapping theorem so you just mimic the complex analysis proof. Cullen-regularity is the same as saying complex analytic on the C-slices
@Thorgott you have R
you do R - 0
guess what thats a stupid disconnected space
so no
the loopy proof is actually a generalization of the intermediate value theorem lmao
@BalarkaSen anyway when I first learned those things I was coming from the model theory angle, so to me the amazing result was that being elementarily equivalent to the real numbers actually has more intrinsic algebraic characterizations
Say I have a principal bundle with a connection. How do I write down an explicit formula for the projection onto the vertical subbundle in terms of the connection form?
@BalarkaSen Cheeger and Colding prove that the eigenvalues of the Laplacian are continuous for limits under the Gromov-Hausdorff(-with-measure) topology
@BalarkaSen You start with a complete connected Riemannian manifold (probably satisfying some global lower bound on Ric) but GH limits usually leave that category
Apparently the right thing to think about isn't $(X,d_g)$ (with the distance metric induced by the Riemannian metric) but rather $(X, d_g, d\text{vol}_g)$
Then there's a notion of measured Gromov-Hausdorff convergence where you also need to converge in measure as well
On some spaces $(X,\mu)$ where $X$ is a metric space and $\mu$ is a measure (Radon?) there's some notion of rectifiability, where I think the limit of Riemannian manifolds is always a rectifiable measured metric space, and on these beasts one can define a Laplacian
the barbell with shrinking tube is an example of an unnormalized Ricci flow on S^2, so keeps bulging at the +ve Ricci parts and keeps pinching at -ve Ricci parts
idk
@MikeMiller yeah alright
I guess there are many ways to have singularities, but I am just not seeing one at this moment
"irregardless of whether this would garner me points on the following test, I am nonplussed about utilizing phantasmagorical words to galvanize my audience"
Suppose we have an oriented, orthonormal frame $e_1,...,e_n$, which gives a section $s\colon U\rightarrow SO(TM)$. Let $\pi\colon SO(TM)\rightarrow SM$. Can we say anything about $d(\pi\circ s)\vert_p(e_i(p))$?
@Thor: First, I assume $\pi(x,e_1,\dots,e_n) = (x,e_1)$. So, let's see. $\pi\circ s(x) = e_1$, so you're asking for $(de_1)_p(e_j) = \sum_{k=2}^n \omega_1^k(e_j)e_k$, all evaluated at $p$. Note that $e_2,\dots,e_n$ span the tangent space of $SM$ at $(x,e_1)$.
@MikeMiller I resemble that remark. My best friend in San Diego and I always say "irregardless" to one another ... as sarcastic humor. It's funny when other friends correct us, not realizing apparently that we of course know better.
@Thorgott: Sorry, I meant the tangent space of the fiber, yes.
But since you're looking at how $e_1$ twists as you move in a certain direction tangent to $M$, that's not going to move you tangent to the manifold, just to the fiber.
Hmm, but let $\tilde{\pi}\colon SM\rightarrow M$. Then $d\tilde{\pi}\vert_{\pi(s(p))}(d(\pi\circ s)\vert_p(e_i(p)))=d(\tilde{\pi}\circ\pi\circ s)\vert_p(e_i(p))=d\operatorname{id}\vert_p(e_i(p))=e_i(p)$ isn't zero, so $d(\pi\circ s)\vert_p(e_i(p))$ isn't vertical, no?
So the point is that I was inconsistent in my own notation. I should have written $\pi\circ s(x) = (x,e_1(x))$, and so that recovers what you just wrote. I wrote only the fiber component, you're right.
If $R$ is a commutative unitary ring and $M$ is an ideal in $R$ generated by elements $\alpha_1,..,\alpha_m$, is it true that $M$ is maximal in $R$ iff I can find a composition of surjective homomorphisms $\pi_1 \circ... \circ \pi_m$ where $\pi_1$ has domain $R$ and $\pi_m$ has codomain $S$ such that $R/M \cong S$ with $Ker(\pi_1 \circ... \circ \pi_m) = M$?
Actually, I only need to care about the vertical part, so that formula will suffice for me, but I don't quite see why it's true. It resembles the identity that's defining the connection forms, but I'm not really sure how to make the transition.
@Lozansky $R$ a field, $M=(0)$, $\alpha_1=0$, $\pi_1=\operatorname{id}$
@Thor: So, yes, I'm using $\nabla e_i = \sum\omega^j_i e_j$. The question is: did it mean anything when I wrote $(de_i)_p$ earlier? I meant $d$ in the setting of derivative of a map, not as exterior differential. So I presume what's bothering you is why $de_1$ in that sense should be $\nabla e_1$?
That's actually an interesting point to raise, as nowhere have we necessarily mentioned a connection in this story.
Although there's a natural connection on the frame bundle, the question you asked is just on the level of Riemannian manifolds with no connection. I'll have to ponder this.
@Ted I don't think there's an issue with this per se, though, since the $\omega_i^j$ exist canonically on $SO(TM)$ for any Riemannian manifold. It's not like your identity depends on a choice of connection, but it should only work with the connection forms coming from the natural connection, no?
Yes, there is the canonical connection (Levi-Civita), but what's bothering me is that, aside from using the metric to define $SM$ (or $SO(TM)$), your question lives in the world of smooth maps between manifolds. So should I be able to answer it without referring to the connection?
I mean, I need a connection to differentiate sections of a vector bundle, so something I said is nonsense.
It's not intrinsic — it depends on your trivialization. There is no canonical "horizontal" subspace unless you have a connection. Indeed, that's one of the definitions of a connection.
How exactly do you make sense of $\omega_1^k(e_j)$. $\omega_1^k$ is a form on $SO(TM)$, so acts on tangent vectors of $SO(TM)$, but $e_j$ is a tangent vector of $M$. What identification am I missing?
