Marek
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 Feb11 comment Product of polynomials with negative coefficients Product with what? Feb8 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). Feb8 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... Feb6 revised Question about a functional equation Added further discussion upon request Feb5 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. Feb5 comment Quotient map is closed @goobie: Interior. I thought this was standard notation, but now I am not sure. Feb5 answered Quotient map is closed Feb5 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). Feb5 answered Question about a functional equation Feb5 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? Feb5 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. Feb5 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.. Feb4 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). Feb4 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). Jan31 answered Is the inverse of a linear transformation linear as well? Jan29 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 $\sigma(\bar x)$. Jan29 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? Jan28 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. Jan28 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 $\bar V$. Jan28 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.