@MatheinBoulomenos Ehresmann's lemma is more general. Proper surjective submersions are bundles (in particular proper surjective local homeomorphisms are covering maps). This fails badly non-properly.
I'm thinking of the 2D version of $(-\epsilon, 1+\epsilon) \to S^1$ sending $x$ to $e^{2\pi i x}$.
Oh sorry not $z^2$
I need something better lmao
Ah OK can do it with torus I think
Take $\Bbb C \to \Bbb T^2$, the universal covering map. Thicken the unit square a bit and restrict the map there to get a surjective submersion $U_{\epsilon}(I^2) \to \Bbb T^2$ which is holomorphic
With the weaker definition of a covering map, if the target is connected then the map will be surjective, and this has epsilon more content than the analogous statement for the definition which requires surjectivity
@Daminark Idea (from which you can reconstruct the proof): Let $f : M \to N$ be a proper surjective submersion. Let's just take $M$ to be compact for now (how to remove this assertion?). Let $\gamma : [0, 1] \to N$ be a smooth path downstairs. $f$ is transverse to $\gamma$ (Warning! What does being transverse to a manifold with boundary mean?), so $P = f^{-1}(\gamma)$ is a smooth submanifold of $M$ which is a cobordism between $F_0 = f^{-1}(0)$ and $F_1 = f^{-1}(1)$. Then $f : P \to [0, 1]$ is a smooth function between manifolds-with-boundary sending boundary to boundary and with no critica…
This is a sketch and hardly a proof but basically the proof is the same idea (but by flowing along lifts of the coordinate vector fields in $N$ which would be $n$ commutative flows upstairs, and rigging local triviality by hand)
This is just the first thing I came up with when I encountered the theorem, which I still like as a way to explain why it should be true
If $M$ is compact one can do the following; for $x \in N$, $f^{-1}(x)$ is a submanifold of $M$. Then $f_*(T_x N)$ is the normal bundle to $f^{-1}(x)$ in $M$. Now $f_*(T_x N) \cong f^{-1}(x) \times \Bbb R^n$ is actually the trivial bundle
Hm, maybe harder to set up the commutative diagram of exponential maps than I thought
Ah no I can do this, let $U_F$ be a normal bundle of $F= f^{-1}(x)$ in $M$. Then I have the map $(\pi, f) : U_F \to F \times N$ where $\pi : U_F \to F$ is tubular projection.
For any point $x \in F$, $d(\pi, f)(x) = (d\pi(x), df(x))$ which is nonzero ($\pi$ is smooth retraction to $F$, and $f$ is submersion). So by inverse function theorem I can choose a subneighborhood $V_F \subset U_F$ of $F$ such that $(\pi, f) : V_F \to F \times \Bbb R^n$ is an diffeomorphism.
@RyanUnger finite-dimensional Morse theory is basically ODE + linear algebra
@RyanUnger @TedShifrin If I have a submanifold $N \subset M$ of a Riemannian manifold and I look at $\rho(x) = \text{dist}(x, N)^2$ where dist is Riemannian distance, then gradient flowlines of $\rho$ are geodesics in $M$, right?
$$\sum_{i=h}^{2h}i=\frac{3}{2}h(h+1)\implies\sum_{i=h+1}^{2(h+1)}i=\frac{3}{2}(h+1)(h+2)$$ so $$\sum_{i=h+1}^{2(h+1)}i=\sum_{i=h+1}^{2h+2}i=\sum_{i=h}^{2h}i+\sum_{i=1}^{2}i$$
Let $k$ be an even integer and $\Gamma \subset \mathrm{SL}_2(\Bbb Z)$ be a congruence subgroup and let $X(\Gamma)$ be the compactification of $\Bbb H/\Gamma$, then the Eichler-Shimura correspondence states an isomorphism of $H^1(X(\Gamma),\Bbb{C}[X,Y]_{k-2}) \cong S_k(\Gamma)\oplus \overline{S_k(\Gamma)}$ where $S_k(\Gamma)$ is the space of weight $k$ cuspidal modular forms of level $\Gamma$ and $\Bbb{C}[X,Y]_{k-2}$ is the space of homogenous polynomials of degree $k-2$
The Hodge decomposition is pretty explicit, right? you just decompose a harmonic form into its $(p,q)$-components. I need to work with the isomorphism explicitly, because I need it to commute with some actions of the Hecke algebra
Let $f(x,y)$ define a surface $S$
in $\mathbb{R}^3$ with a unique local minimum at $b \in S$.
Suppose gradient descent from any start point $a \in S$
follows a geodesic on $S$ from $a$ to $b$.
(Q1.)
What is the class of functions/surfaces
whose gradient-descent paths are geodesics?
Certainly if ...
This establishes the Eichler-Shimura isomorphism for weakly holomorphic modular forms of level 1 and then you can get the usual Eichler-Shimura isomorphism by looking at a part of the weight filtration
I think that's still doable using Brown and Hain's strategy The main problem is that Brown and Hain use a very explicit model of M_{1,\vec{1}} over M_{1,1} and it's a little tricky to get all this to work explicitly at higher level
@RyanUnger That is usually not going to happen with my functions. I mean that's not even true for $f(x) = d(x, N)^2$. $df$ vanishes on the directions which "swirl inwards" towards $N$
@Balarka ok, but its really simplistic and im very tired right now, but the basic idea was to first look at $\Bbb R^2$ mod a finite group action, then there are some bad points where a bunch of $\Bbb R^2$s get glued together.
Now I look at a path $\gamma$ in this space. Lift $\gamma$ at a good point and suppose the lift enters a bad point $x$. I will say $\gamma$ is differentiable if I can extend from another point $gx$ so that the derivative from the right of $\gamma$ is the same as the derivative from the left of $g\gamma$
this gives me an idea of what is differentiable, and at the bad points i have a tangent space that is like a disjoint union of tangent spaces in the cover (mod some gluing)
@AlessandroCodenotti because I have a more complicated statement: descompose $\sum_{i=h+1}^{2(h+1)}i$ using the fact that $\sum_{i=h}^{2h}i=\frac{3}{2}h(h+1)$. I have no idea what I need to write
hello, i gave a quick question is the open ball of center x and radius r, with the distance |exp(x)-exp(y)| when $r-exp(x)\leq0$ is it $(-\infty, ln(r+exp(x))) $???
@s.harp Take the standard orbifold picture, where $\Bbb Z_n$ acts on $\Bbb R^2$ by rotation. Then $\Bbb R^2/\Bbb Z_n$ is a cone where the angle at the vertex is $2\pi/n$, basically. What's the correct tangent space at the vertex? $T_0 \Bbb R^2$, the $2\pi/n$-sector of $T_0 \Bbb R^2$, or the $2\pi/n$-sector of $T_0\Bbb R^2$ with two rays separated by an angle of $2\pi/n$ glued togather (not linear anymore)?
Usually this isn't how one thinks of tangential structures on stratified spaces but it's an interesting idea.
@BalarkaSen here a (regular) path going through the cone point is differentiable if there is a lift in $\Bbb R^2$ that has a kink of angle $k\,2\pi/n$ at $0$ (I think that angle is correct). I cannot see the tangent structure since my brain is fried, but I think it should be $\Bbb R^2/\Bbb Z_n$ itself
I linked a paper which said that if you give $G$ the left-invariant metric coming from $\mathfrak g$, then the only isometries are of the form $R_x L_y$ for elements $x, y \in G$ (right and left translation). I am trying to choose an inner product on $\mathfrak g$ so that all elements of the isometry group are $L_y$.
It suffices to show that $R_x L_{x^{-1}}$ is not an isometry for any $x$
So we are now in an abstract situation. You have a compact group $G$ with an action on a vector space $V$, and you want to ask if there's an inner product on $V$ not preserved by any element of $G$.
@MikeMiller If you give $\Bbb R^n$ the usual additive structure then any translation invariant metric should have an isoemtry group isomorphic to $O(n)\ltimes\Bbb R^n$ though, right?
OK. $G$ is compact, so choose an invariant inner product to begin with, and enumerate an orthonormal basis $\langle e_1, \cdots, e_n\rangle$. Replace this with a metric so that $\langle e_i/i \rangle$ is an orthonormal basis. Now, "norm of an element on the new unit sphere with respect to the original norm" has essentially a unique minimum; actually it has two, and the corresponding function on $\Bbb{RP}^n$ has a unique minimum, $[e_1]$.
Therefore, any element $g \in G$ which gives an isometry with respect to the new metric necessarily sends $e_1$ to $\pm e_1$.
So you pass to the orthocomplement and iterate the argument. This should imply that the set of $g \in G$ which can act as isometries with respect to both metrics are a subgroup of $(\Bbb Z/2)^{\dim G}$.
I don't know what to say from here. Probably nothing.
Probably this doesn't work.
Ah yeah note in particular that this could never work for $T^n$ lol
hello, i have a quick question is the open ball of center x and radius r, with the distance |exp(x)-exp(y)| when $r-exp(x)\leq0$ is it $(-\infty, ln(r+exp(x))) $???
What's the intuition behind this theorem? I'm looking up surface integrals
Let $M$ be an $N-1$ dimensional oriented manifold and $F: M \to \mathbb R^N$ a smooth vector field, $F = (F_1, \ldots, F_n)$, and $M$ has a unit normal $\hat n(x)$, then: $\langle F(x), \hat n(x) \rangle = \sum_{i = 1}^N F_i(x) \star \mathrm dx_i$
Assuming I wrote it down correctly in my class notes
Question: I've got triples like $(a, b, s)$ repesenting ranges of $\lbrace x * s : a \le x * s \lt b, x \in \mathbb{Z} \rbrace$ where $s \in \mathbb{Z}, s \ne 0$, basically (start, end, step), and I'm trying to figure out how to convert that to sets of digit sequences, like $(3, 20, 4)$ to $\lbrace \lbrace 3 \rbrace, \lbrace 7 \rbrace, \lbrace 1, 1 \rbrace, \lbrace 1, 5 \rbrace, \lbrace 1, 9 \rbrace \rbrace$, without evaluating the sequence directly.
The degenerate cases of 1 and -1 for me are pretty obvious, as well as a few cases like 2. But I'm struggling to find any leads on a general solution. Admittedly, I'm not well-versed in number theory.
@IsiahMeadows: I assume $*$ means multiplication. I dunno why you write it in math. But I have no idea what you're talking about with converting sequences.
