Wikipedia:Reference desk/Archives/Mathematics/2016 August 21

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August 21

Topology - properties of closure

Let X be an (irreducible) affine variety, U a nonempty open subset of X, Y an affine subvariety of X and $f:U\to \mathbb {P} ^{r}$ a morphism. Denote the closure of the graph of f in $X\times \mathbb {P} ^{r}$ by X', and denote the graph of $f|_{U\cap Y}$ by G. Why is the closure of G in $Y\times \mathbb {P} ^{r}$ equal to the closure of G in X'? (There should be an easy reason...) — Preceding unsigned comment added by [[User:{{{1}}}|{{{1}}}]] ([[User talk:{{{1}}}|talk]] • [[Special:Contributions/{{{1}}}|contribs]])

Y is closed in X, so $Y\times \mathbb {P} ^{r}$ is closed in $X\times \mathbb {P} ^{r}$. Therefore the closure of any subset of $Y\times \mathbb {P} ^{r}$ is also closed in $X\times \mathbb {P} ^{r}$. Sławomir Biały (talk) 12:30, 21 August 2016 (UTC)

Thank you! I'll write the full details: the claim follows from the following general proposition. Let X be a topological space, Y a closed subset of X, V an arbitrary subset of X, $G=V\cap Y$, then $Cl_{Y}G=Cl_{Cl_{X}V}G$. Indeed, we can substitute $X\times \mathbb {P} ^{r}$ for X, $Y\times \mathbb {P} ^{r}$ for Y, and the graph of f for V. We prove the general claim by noting both sides are equal to $Cl_{X}G$, since in general if $A\subseteq Y\subseteq X$ where Y is closed in X, then $Cl_{Y}A=Cl_{X}A$. — Preceding unsigned comment added by 212.179.21.194 (talk) 15:48, 21 August 2016 (UTC)

Intersection of Ellipsoid and Hyperplane

I have an n-dimensional ellipsoid E and a hyperplan H. This hyperplane cuts E into two parts: E1 and E2 (whose disjoint union is E). I want to find another ellipsoid E' that has minimal hyper-volume and contains E1. To do this, I fist thought to formulate it as an optimization problem, but I am having difficulty with formulating it, as I don't know how to formulate the containment (of E1 in E') constraint. Could anyone help me formulating it or pointing out another way to do it (suggesting some program, or an algorithm, etc.). 213.8.204.9 (talk) 11:31, 21 August 2016 (UTC)

Here's an approach. A diameter of E1 is a line segment of maximal length that lies inside E1. We construct an ellipsoid E' inductively, as follows. An ellipsoid is determined by a set of orthogonal axes. The first axes of E' shall be a diameter of E1. Now, orthogonally project E1 onto the normal plane of this diameter. This gives a new truncated ellipsoid in one dimension lower. Repeat the construction there to get the remaining axes of the ellipsoid E'. (What is not immediately clear is if E' actually contains E1, much less that it has the least volume. But at least this gives a possible approach to the problem.) Sławomir Biały (talk) 12:25, 21 August 2016 (UTC)

This is a great idea! But now I am facing a new porblem: how to find an ellipsoid's diameter? 213.8.204.9 (talk) 12:47, 21 August 2016 (UTC)

That's a constrained optimization problem. There are two parts to it: one under the assumption that one endpoint of the diameter is on the intersection of the hyperplane H and (the solid ellipsoid) E and the other is on the boundary of E (lying on the E1 side), and the other where both endpoints lie on the boundary of E (in which case, the diameter is a diameter of the original ellipsoid, which can be found by the principal axis theorem). Sławomir Biały (talk) 13:44, 21 August 2016 (UTC)

There's also a possibility that both ends of the diameter belong to the hyperplane. --CiaPan (talk) 07:33, 24 August 2016 (UTC)

Here's a hint for another possible approach.

• If E' is the solution ellipsoid relative to the ellipsoid E and the hyperplane H, and if φ is an affine isomorphism of your n-dimensional Euclidean space, then the solution ellipsoid relative to the ellipsoid φ E and the hyperplan φ H is just the ellipsoid φ E' .
• Any ellipsoid is an affine image of an Euclidean ball.

Therefore you can assume E is a ball.

pma 18:39, 22 August 2016 (UTC)