Essay on Differential Geometry

Document Type:Research Paper

Subject Area:Geometry

Document 1

A parameterized curve Consider a parameterized curve: x = f(v) z = g(v) v ∈ (a,b) in the xz-plane, with f(v) > 0 everywhere. Let S be the surface of the revolution obtained by the revolving curve around z-axis. If u ∈ [0,2π] is the angular coordinate, calculate the induced metric on S ⊂ R3 in local coordinates (u,v). The parameterization of the surface is given by (x,y,z) = (f(v)cos(u),f(v)sin(u),g(v)). Therefore and Hence. The rationally symmetric metric Consider a rotationally symmetric metric: dr2 + (φ(r))2dθ2 where φ : [0,b) → [0,∞), with φ(0) = 0 and φ(r) > 0 for r > 0. To check for smoothness of the metric at the origin we need to change from polar to Cartesian coordinates: x = rcosθ y = rsinθ Write the metric in (x,y) coordinates and show that the components gxx, gxy = gyx, and gyy are smooth if and only if φ0(0) = 1 and φ(even)(0) = 0. Hint: Write φ(r) = rψ(r). Calculate gxx, etc. in terms of ψ(r). Show that smoothness at r = 0 implies ψ(0) = 1 and ψ(odd)(0) = 0. Assume that φ(r) is analytic so that it has a power series expansion φ(r) = a1r + a2r2 + a3r3 +. since φ(0) = 0. The function ψ(r) has a power series expansion ψ(r) = a1 + a2r + a3r2 +. Now r = px2 + y2 and θ = arctan(y/x). Therefore and Therefore. The metric is smooth if and only if is smooth. We immediately see that a1 = ψ(0) = : Then so a2 = 0. In fact, a power series in r is a smooth function of x and y if and only if all the odd degree terms vanish.

Sign up to view the full document!

By induction, this means that a4 = 0, a6 = 0, etc. e. To get f(0) = 0, the constant C must equal 0. One can then check that f(r) = tanh(r/2) also satisfies the algebraic equation (2). A proper affine function is a map f : R → R given by f(t) = bt+a for a ∈ R and b ∈ R>0. The set of proper affine functions forms a Lie group: as a smooth manifold it is just the upper half plane {(a,b) ∈ R2|b > 0}, and the group operation is given by composition. Therefore h(u1,u2),(v1,v2)i(a,b) = h(dL(a,b)−1)(a,b)(u1,u2),(dL(a,b)−1)(a,b)(v1,v2)i(0,1) = hb−1(u1,u2),b−1(v1,v2)i(0,1) = b−2(u1v1 + u2v2). We see that the left-invariant metric on U is the hyperbolic metric.

Sign up to view the full document!

Let ∇ be the connection on R2 that has vanishing Christoffel symbols Γkij with respect to Cartesian coordinates (x,y). Calculate Christoffel symbols with respect to polar coordinates (r,θ). In other words, calculate the coefficients of , , , and , in the basis. DXY N) ∈ TRN. Then we defineto be the orthogonal projection of ∇XY to TRN|M ⊂ TRN. Prove that ∇0 is the same as the Levi-Civita connection ∇ on M. We first show that ∇0 is a connection on M. The equations follow automatically from the corresponding equations for ∇: ∇0fX+gY Z = f∇0XZ + g∇0Y Z ∇0X(Y + Z) = ∇0XY + ∇0XZ = = = fproj(∇XY ) + X(f)proj(Y ) As required, proving that ∇0 is indeed a connection. xn,xn+1,. xN} on RN. Then where Xn+1,. XN all vanish on M since X|M = X. There is a similar expression for Y.

Sign up to view the full document!

e. find a formula for the parallel vector field that satisfies V (0) = v0. Let. Then where the last equality is due to the fact that y always equals 1 on the curve c. The unique solution with initial condition is. Prove that r(t)cosβ(t) is independent of t. This is called Clairaut’s relation. Let us first find a formula for cosβ(t). Suppose the parallel and the geodesic meet at the point (x,y,z) = (fcosu,fsinu,g). The unit tangent to the parallel is given by. Therefore for some constant C. Without loss of generality, let’s assume that r(t) = v achieves its minimum value v0 at t = 0. At this point the geodesic is tangent to the parallel and β(0) = 0. Since r(t)cosβ(t) must be constant, we see that for t > 0, β(t) > 0 and r(t) = v is increasing (the geodesic is “climbing”).

Sign up to view the full document!

On the other hand, for t < 0, β(t) < 0, and geodesic is “climbing” by reverse direction. Suppose X is given by an n × n matrix A, so that the vector field at a point (x1,. xn) is given by. The equation for a flow line (x1(t),. xn(t)) is then with solution. Thus the flow ψt is given by exp(tA) (which is defined using the power series for exp). We will prove the equation holds at p, and therefore by continuity it will hold everywhere. Since X(p) 6= 0 we can find a hypersurface (codimension one submanifold) S ⊂ M which is transversal to X at p, and also in a small neighbourhood of p. Let then (x1,. xn−1) be the local coordinates on S. There is a map S × (−δ,δ) → M which takes a point (q,t) to ψt(q) where ψt is the flow generated by X.

Sign up to view the full document!

 Kinematic Differential Geometry and Saddle Synthesis of Linkages, 2015, pp. Bernard, Lucas, and François Roux.  Emerging Topics on Differential Geometry and Graph Theory. Nova Science Publishers, 2010. Helgason, Sigurdur. Rutherford, D E.  Vector Methods Applied to Differential Geometry. Dover Publications, 2012. Schoen, Richard, and Shing-Tung Yau.  Lectures on differential geometry. Courier Corporation, 2013.

Sign up to view the full document!

From $10 to earn access

Only on Studyloop

Original template