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Feb
11
comment Product of polynomials with negative coefficients
Product with what?
Feb
8
comment Formula for evaluation of character on a transposition
@Alexander: ah, right, thanks. I confused this with characters that are homomorphisms (as opposed to just functions, like here).
Feb
8
comment Formula for evaluation of character on a transposition
So, for $\lambda = \lambda^t$ we get zero on the RHS. That seems strange. Also, what is $\chi(1)$ supposed to mean? My characters use to have $\chi(1) = 1$. Unless I am misunderstanding something trivial, this formula is very weird...
Feb
6
revised Question about a functional equation
Added further discussion upon request
Feb
5
comment Quotient map is closed
@goobie: $W$ intersects $X$ in the boundary. The moving of $W$ is supposed to cover all the new pieces of the boundary that $D$ covers (I am not sure this helps, a picture would be worth thousand words here). Anyway, looking forward to your answer. You're welcome.
Feb
5
comment Quotient map is closed
@goobie: Interior. I thought this was standard notation, but now I am not sure.
Feb
5
answered Quotient map is closed
Feb
5
comment Quotient map is closed
@Arthur: That $q$ is closed is what one wants to prove here, so that's not helpful. Also, projections are quotient maps which are not closed (they are open though).
Feb
5
answered Question about a functional equation
Feb
5
comment Question about a functional equation
But regarding your two ODEs for $A$ and $B$, if you plug $B(t,T) = 0$ into them, you obtain contradictory $\partial_t A = 0$ and $\partial_t A = -1$. Aren't you missing a $\partial_t B$ term in the second one?
Feb
5
comment Question about a functional equation
Finally, the open segments thesis is clear. If $B(t) = 0$ for some $t$ and, as we have by $(2)$, $\partial_t B = -1$ then $t$ is an isolated zero of $B$ and therefore the set of all $t$ where $B$ and $B^2$ is dependent consists only of isolated points. Therefore its complement is open and dense.
Feb
5
comment Question about a functional equation
Well, $(2)$ is clear, if $B(t, \cdot) = 0$, then almost all the terms die and you are left with $(\partial_t B + 1) = 0$. But $(3)$ is puzzling. It is true that from it it follows that $B(t,T) = B(t)$ is independent of $T$. I think what is meant here is that there exists $t$ such that $B(t) = 0$. Existence of such a $t$ probably follows from the stuff above which I don't really understand..
Feb
4
comment Union of two self-intersecting planes is not a surface
Regarding 4 connected components, it's correct. But to prove this, you'd need something on the order of Jordan Curve theorem (to show that number of components outside a simple curve in $\mathbb R^2$ is at most two).
Feb
4
comment Union of two self-intersecting planes is not a surface
Regarding "but V is any arbitrary open set.", it's not arbitrary. Since $V$ is homeomorphic to $S$, it must be connected and simply connected, i.e. it is a ball. Minus a point, it is an annulus. So $\pi_1(V \setminus \{f^{-1}(p)\}) = \mathbb Z$. But $\pi_1(X)$ is something else (try to show this).
Jan
31
answered Is the inverse of a linear transformation linear as well?
Jan
29
comment Embedded surface in $\mathbb{R}^3$
Because you can work locally and pick a neighbourhood of $\sigma(\bar x$) small enough, so that it looks like (flat) circle, and the tubular neighbourhood like a cylinder. In this cylinder, everything should be regular, but by definition of $\bar x$, there are singular points arbitrarily close (in the normal direction) to $\sigma(\bar x)$.
Jan
29
comment Solving to get free falling coordinate as function of arbitrary coordinate
@Aftnix: I think all that's going on here is plugging into the equation the general Taylor expansion (up to second order terms) and comparing coefficients. Do you need an explanation how to do this?
Jan
28
comment Gauss-Bonnet theorem for spheres that almost look like a torus
I think it's also worth emphasizing that the first part is actually quite negligible, since it's enough to cut out arbitrarily small area of negative curvature. It is the second part (i.e. bending the torus inwards) that accounts for the difference.
Jan
28
comment Embedded surface in $\mathbb{R}^3$
sorry, I don't have time right now, that's why I'm only posting comments. Suppose for contradiction that no such $\epsilon$ exists. Produce a sequence of points $(u,v)$ that map under $\tau_{\delta}$ to a singular point. Now, you can pick a convergent subsequence in $\bar V$ and this will under $\sigma|_{\bar V}$ map to a singular point of $\sigma(\bar V)$, a contradiction. Now the general case follows by restricting to an open subset $V$ of $\bar V$.
Jan
28
comment Embedded surface in $\mathbb{R}^3$
Yes, that's correct. See the picture here: en.wikipedia.org/wiki/Tubular_neighborhood . Now, try to think of conditions that guarantee that self-intersections don't occur. The (relative) compactness, (i.e. boundedness) is essential here.