## geometry – Evenly spaced polygon dilation

I would like to calculate a scaled polygon, such that its sides would be evenly distanced from the original polygon ones.

I’ve tried to use scaling, but it looks like I don’t get a perfect result:

For rectangles, it works just fine, but with polygons, it seems that the sides are not spaced evenly. Furthermore, the angles don’t look equal.

• Is this a normal situation with scaling?
• Is there any better way to achieve my goal?
• Maybe there’s something wrong with my calculations?

Here are the listings of code which I used:

``````    public static float() getBypassPoints(Polygon polygon, float offsetX, float offsetY) {
Rectangle bbox = polygon.getBoundingRectangle();

float width = bbox.getWidth();
float height = bbox.getHeight();
float origScaleX = polygon.getScaleX();
float origScaleY = polygon.getScaleY();
float xScale = ((width + offsetX * 2) / width) * origScaleX;
float yScale = ((height + offsetY * 2) / height) * origScaleY;

setPolygonOrigin(polygon);
polygon.setScale(xScale, yScale);

float() transformedVertices = polygon.getTransformedVertices();
float() bypassPoints = transformedVertices.clone();

polygon.setScale(origScaleX, origScaleY);
return bypassPoints;
}

public static void setPolygonOrigin(Polygon polygon) {
float() vertices = polygon.getVertices();
int countOfCoords = vertices.length;

Vector2 centroid = polygonCentroid(
vertices, 0, countOfCoords, new Vector2(polygon.getX(), polygon.getY()));

polygon.setOrigin(centroid.x, centroid.y);
}
``````

``````    public float() getTransformedVertices () {
if (!dirty) return worldVertices;
dirty = false;

final float() localVertices = this.localVertices;
if (worldVertices == null || worldVertices.length != localVertices.length) worldVertices = new float(localVertices.length);

final float() worldVertices = this.worldVertices;
final float positionX = x;
final float positionY = y;
final float originX = this.originX;
final float originY = this.originY;
final float scaleX = this.scaleX;
final float scaleY = this.scaleY;
final boolean scale = scaleX != 1 || scaleY != 1;
final float rotation = this.rotation;
final float cos = MathUtils.cosDeg(rotation);
final float sin = MathUtils.sinDeg(rotation);

for (int i = 0, n = localVertices.length; i < n; i += 2) {
float x = localVertices(i) - originX;
float y = localVertices(i + 1) - originY;

// scale if needed
if (scale) {
x *= scaleX;
y *= scaleY;
}

// rotate if needed
if (rotation != 0) {
float oldX = x;
x = cos * x - sin * y;
y = sin * oldX + cos * y;
}

worldVertices(i) = positionX + x + originX;
worldVertices(i + 1) = positionY + y + originY;
}
return worldVertices;
}
``````

## dg.differential geometry – Nonexistence of harmonic \$S^2\$

Let $$M$$ be a complete Riemannian manifold with $$pi_2(M)=0$$. Can we prove that there is no nonconstant minimizing harmonic map from $$S^2$$ to $$M$$?

Notice that if $$M$$ is compact and $$pi_2(M) ne 0$$, it follows from the work of Sacks-Uhlenbeck that there exists a nonconstant minimizing harmonic $$S^2$$. I was wondering if the converse holds.

## riemannian geometry – Discs bundles along a curve and positive curvature

If $$(M,g)$$ is a smooth Riemannian manifold and $$c : (a,b) to M$$ is a smooth embedded simple curve on $$M$$, it is always possible to choose locally a Riemannian metric $$g_0$$ on $$M$$ for which $$c$$ is a geodesic for $$g_0$$. As I understand, this can be done by pulling back a tubular neighborhood of $$c$$ to a disc bundle along $$c$$. There is always a flat Riemannian metric for which the $$0$$-section ($$c$$ itself) is a smooth geodesic. My question is: is it possible to locally realize $$c$$ as a geodesic to a metric such that the tube is positively curved? For instance, one of constant sectional curvature? I mean, can $$g_0$$ be positively curved? If it helps, $$M$$ can be assumed to be a surface.

## ag.algebraic geometry – About union of two algebraically independent sets

My question might be very simple for those that have a deep understanding for algebraically independent sets.

Let $$P_1$$ and $$P_2$$ are two algebraically independent sets and $$P_1cap P_2=emptyset.$$ Assume there exist $$a,bin Qsetminus{0}$$ such that $$P^*=aP_1+bsubset J$$ for some open interval $$J.$$ Does the following hold ?

I. $$P^*$$ is algebraically independent ?

II. $$P^*cup P_2$$ is algebraically ?

I believe that I is correct since just rescaling and shifting by rational numbers
About, II I do not find an example that shows it is incorrect or proof it.

Any help will be appreciated greatly.

## differential geometry – Is it possible to reverse engineer the metric to find the basis vectors of the manifold

For a given surface that has metric components $$g_{munu}$$ is it possible to find it’s basis vectors and more importantly the exact surface that space is on. Since we know that
$$g_{munu} = e_mu cdot e_nu$$

However, for an intrinsic geometry, the transformation that gives it its geometry is given by a general transformation $$x’^mu b_mu = G^mu(x^mu) b_mu$$ if we take the partial derivative with respect to $$x^nu$$ we get that
$$e_nu=partial_nu G^mu(x^mu) b_mu$$
The new space that these intrinsically curved coordinates live in is measured with orthogonal basis vectors $$b_mu$$. This gives us
$$g_{munu}=partial_mu G^alpha(x^mu) b_alpha cdot partial_nu G^beta(x^mu) b_beta$$
$$g_{munu}=partial_mu G^alpha(x^mu) partial_nu G^beta(x^mu) b_alpha cdot b_beta$$
$$g_{munu}=partial_mu G^alpha(x^mu) partial_nu G^beta(x^mu) delta_{alphabeta}$$

With this information is it possible to find the transformation $$G^mu(x^mu)$$ in terms of the metric components?

## dg.differential geometry – Product rule for vector bundle (Leibniz rule)

Let $$pi:Eto Y$$ be a vector bundle. Write $$mathrm{T}(E/Y)subset mathrm TE$$ for its vertical bundle. Write $$Phi:mathrm{T}(E/Y)cong pi^ast E$$ for the vector bundle isomorphism over $$E$$ given fiberwise by the canonical isomorphism between a vector space and its tangent space at a point.

Trying to carry over the product rule for differentiable maps between vector spaces, I have arrived at the following “formula”, where $$fin C^infty_Y$$ is a real function, $$s$$ is a local section of $$pi$$, and $$+_mathrm{T},cdot_mathrm{T}$$ are the addition and scalar multiplication of the secondary vector bundle structure $$mathrm TEto mathrm TY$$. $$mathrm T_y(fcdot s)(dotdelta)overset{?}{=}(fcirc delta)^prime(0)cdotPhi^{-1}(sy,sy)+_mathrm{T}f(y)cdot_{mathrm T}mathrm T_ys(dotdelta)$$

Question 1. Is this formula correct? If so, the RHS lies in $$mathrm T_{((fcirc delta)^prime(0)+f(y))s}V$$, which looks a bit strange…

Question 2. For differentiable maps between vector spaces, the product rule is a consequence of the chain rule along with the additional structures of sums and powers. I doubt it, but just in case – is there a coordinate free way of arriving at this formula?

## differential geometry – Homotopic paths in a simply connected space

In my algebraic topology course, we define a simply connected topological space as a topological space $$X$$ whose fondamental group $$pi_1(X)$$ is reduces to the trivial group $${1}$$ and my teacher told me that in such a space, if we consider two paths $$gamma_1:(0,1) rightarrow X$$, $$gamma_2:(0,1) rightarrow X$$ such that $$gamma_1(0) = gamma_2(0) = x_0$$ and $$gamma_1(1) = gamma_2(1) = x$$, then $$gamma_1, gamma_2$$ are homotopic.

I do not really understand why this is true. Of course, my first intuition was to concatenate these paths such that
$$begin{equation*} gamma(t) = begin{cases} gamma_1(2t) &text{for } t in (0,1/2)\ gamma_2(2(1 – t)) & text{for }t in (1/2,1) end{cases} end{equation*}$$
and by simply connectedness of $$X$$, we conclude that $$gamma$$ is homotopic to the constant loop, but from here I do not see how to conclude that $$gamma_1$$ and $$gamma_2$$ are homotopic. Could someone give me a lead ? Thank you!

## ag.algebraic geometry – Local surjectivity of word maps

Let $$F$$ be a free group on 2 generators and $$G = SL(d,mathbb{C})$$.
A word $$w in F$$ induces the word map $$ev_w: G times G to G$$.

Can we find some (generic) conditions on $$A,B in G$$ such that, for all $$w in F$$,
the differential of $$ev_w$$ is surjective at $$(A,B)$$ ?

## algebraic geometry – Divisor which is principal on the generic fiber of (flat ?) morphism

We have $$pi : A’ longrightarrow S$$ a (flat if necessary ?) morphism between a variety $$A$$ which is projective, normal and irreducible, and $$S$$ which is smooth, projective, irreducible, both defined over a number field $$k$$. I take $$D in Div(A)$$ a divisor, which verify that, if we denote by $$eta$$ the generic point of $$S$$ and $$A_eta$$ the generic fiber of $$pi$$, then $$D_eta sim^{lin} 0$$, where $$D_eta$$ is any divisor on the restriction of the divisor class $$D$$ to $$A_eta$$. We have supposed that $$A_eta$$ is a smooth abelian variety (irreducible).

Then we want to have the following : it exists $$D” in Div(A)$$ such that $$D” sim 0$$ (linearly) and $$pi(Supp(D”)) subsetneq S$$.

As we have $$D_eta = div(f_eta)$$ for some $$f_eta$$ a rational function of $$A_eta$$, I wanted to lift $$f_eta$$ on the function field of $$A$$, which I denote by $$k(A)$$. But $$A_eta = A times_S Spec(k(S))$$ where $$k(S)$$ is the function field of $$S$$. I don’t see how I can lift $$f_eta$$ since the function field of $$A_eta$$ seems to be eventually much bigger than the function field of $$A$$

Thank you for the help !

## algebraic geometry – inverse image presheaf of the inclusion of a point

Let $$mathscr{F}$$ be a presheaf over a topological space $$Y$$. If $$f:Xto Y$$ is a continuous map, then we have a natural presheaf on $$X$$ given by
$$f^{-1}mathscr{F}(U):= operatorname{colim}_{f(U)subset V}mathscr{F}(V).$$
(Most people will say that the inverse image is the sheafification of that but it’ll be enough to consider this for now.)

If $$f:{p}to X$$ is the inclusion of a point and $$mathscr{G}$$ is a presheaf on $$X$$, I want to calculate the presheaf $$f^{-1}mathscr{G}$$. The topological space $${p}$$ has only two open sets. By definition, $$Gamma({p},f^{-1}mathscr{G})$$ is the stalk $$mathscr{G}_p$$. Now, I wonder what is $$Gamma(varnothing,f^{-1}mathscr{G})$$.

It surely is the colimit of $$mathscr{G}(V)$$ where $$V$$ runs over all the open sets of $$X$$ but I can’t say nothing more precise. Does this object has a better description?