You are apparently impassioned by submersions, but you don't have the tools to answer these sort of problems; earlier you posted a question in which you didn't realize you had access to Sard's theorem. I don't think this is a problem that you're going to solve by having people tell you the names of theorems or ideas. If you care about this, you should find a differential topology book to learn from - I like Guillemin and Pollack's book. Some of your questions...
...can be answered by material you'll learn from reading that book. Others from your list are less straightforward. In either case, if this is something you care about, you should try to learn it in context, as opposed to just solving a list of problems by asking people for solutions.
Yah it is an amazing and beautiful game, just started playing again. It is a great game. I SS most of the not "digital" levels, before DX (although I definitely did not SS all of them)
DX has all the worlds connected, in the old version you entered the books to get to the worlds, also I don't think they had map creators and stuff like that. Pretty sure they added levels and stuff. And yes to the digital levels (not sure what they are called)
though you do get some fun things when you do large $j$---namely, you get a certain notion of semiclassical asymptotics with $1/j$ as the small parameter
@TedShifrin If you like cool questions: Here's one from Hirsch. If $U \subset \Bbb R^3$ and $V \subset \Bbb R^2$ are open, and $f: U \to V$ is $C^1$ and surjective, does it have to have rank 2 somewhere?
@TedShifrin: Are there not problems with the order of smoothness? I forget.
I guess not, but one has to be a little bit of work; the constant rank theorem says that the images of small balls are nowhere dense, and you take a countable union and invoke Baire, I guess.
He has a question where he implicitly wants you to prove Morton Brown's theorem that an increasing union of $\Bbb R^n$s is diffeomorphic to $\Bbb R^n$. I have no idea how to do that.
Even proving a convex set in $\Bbb R^n$ is diffeo to $\Bbb R^n$ isn't that easy.
Right, @Semiclassic.
Of course, you have to decide what that really means.
user139655
Hello. I think that the following is not true, but I can't find a counterexample: "If $m_1$ and $m_2$ are two measures on some measurable space $(X, \Sigma)$, and if $E \in \Sigma$ such that $m_1(E) \le m_2(E)$, then for any subset $A$ of $E$, such that $A \in \Sigma$, we have $m_1(A) \le m_2(A)$". Do you have some suggestion?
@anon The part where I am confused is that X is a random variable from a distribution, why doesn't the answer to $E(X|Y)$ depend on that(distribution).
take for instance averaging two values on the number line. the average will be the midpoint, halfway in between them. if you shift the two values by c units, you shift the midpoint by c units too. that is, avg(a-c,b-c)=avg(a,b)-c.
the algebra works the same if you use more than two values, and also works the same if you use a probability distribution other than the uniform one
it might also help to think of center of mass in space. think of the density (in the physics sense) as a probability density function of a vector-valued random variable. the center of mass will be the expected value of this random variable. shifting the mass in space by a vector should shift the center of mass by the same vector.
I gave you three good metaphors to get intuition from (average height, midpoint, center of mass) and an indication of how to go about proving it algebraically. Have you been reading what I've written?
what does $E(X)$ mean? it means $\int p(x)x dx$. what does $E(X-a)$ mean? it means $\int p(x)(x-a)dx$. see anything you can do with that second integral?
simple: $=\int p(x)xdx-\int p(x)adx=E(X)-a$
it's the same algebra as the one I gave for averaging two values
but really if you understand the three examples I gave you it should be an intuitive fact, algebra or no algebra
Let $\{a_{n}\} $is decreasing sequence ,and $\sum_{n=1}^{+\infty}a_{n}=+1$Prove that $$\sum_{n=1}^{+\infty}\dfrac{a_{n}}{3^{\frac{a_{n}}{a_{n+1}}}}=+\infty$$
@TedShifrin @MikeMiller I saw that star-convex subset of R^n being diffeomorphic to R^n question. In R^2, that follows from Riemann mapping theorem, yeah?
I have no idea how to actually prove this, of course.
I've seen that algorithms like the QR algorithm only require that you have your matrix in upper Hessenberg form before you begin, but I've also seen this shifted variant that says $A^{0}$ should be a tri-diagonalization of $A$.
@Owatch It depends. Note that for an unsymmetric matrix, the methods for getting the similar tridiagonal matrix require the use of nonorthogonal (nonunitary in the complex case) transformations, which are potentially ill-conditioned.
But, Lanczos manages to work a number of times anyway.
But as soon as you have a tridiagonal matrix, you now have vastly reduced storage, and potentially an $O(n^2)$ method for the eigenvalues (temporarily ignoring the $O(n^3)$ work you did for the similarity transformations).
the majority of that article seems to be for the Brauer group of a field, with a passing remark on generalizing it to commutative rings before immediately going to the Brauer group of a scheme
this is useful because it explains why the opposite algebra should be the inverse in this weirdo group, since $A \otimes A^{\text{op}}$ for a free module $A \cong R^n$ is just $M_n(R)$.
@Ted It's not enough to throw out the rank 0 points. You see from this that the open set of rank 1 points has nowhere dense/measure 0 image. But we need to deal with the absurd possibility that the rank 0 points cover everything!
Eg, think of Cantor's staircase function, though this is of course not differentiable.
is the difference between a kernel and a stabilizer that: a stabilizer is elements in a group that commute with the subset of $G$ that is chosen. A kernel is the elements in a group that are the identity in the target set/domain?
