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:
enter image description here

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;

    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
gamma(t) =
gamma_1(2t) &text{for } t in (0,1/2)\
gamma_2(2(1 – t)) & text{for }t in (1/2,1)

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!

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?