Recent content by OrderOfThings

The set of oriented lines in Rn is isomorphic to the tangent bundle of the n-1-sphere TS^{n-1}. At each point on a tangent plane of the unit sphere there is exactly one line intersecting orthogonally at that point. Choose one orientation of the line, say outwards, and this gives the isomorphism.

That's right. Just glue the two manifolds together. (But to be able to visualize this you might have turn one of the manifolds inside out.) When you cut along the nonbounding cycle, the resulting boundary is a two fold cover of the cycle. If the cycle is orientable then the cover is disconnected...

That would be possible if (and only if?) the base space has a two fold covering. But was the gluing in constructing the two fold cover clear enough? It should be done cross-wise: Cut the two manifolds along the nonbounding cycle. Label the two ends A1 and B1 on the first manifold and A2, B2...

The two manifolds should be glued together by a two fold covering of the nonbounding cycle.

Yes, it seems the gluing most be done cross-wise. Maybe it is possible to reason like this: The nonbounding cycle will also have nontrivial homotopy and will admit an antipodal fibration. This fibration gives a cross-wise gluing?

A Z_2-valued 1-form \Omega can be visualized as an n-1-dimensional submanifold N. The value of \Omega on a curve \gamma is given by \Omega(\gamma) = number of crossings with N mod 2. If the form is closed then the corresponding submanifold is a cycle. If the form is exact, the submanifold...

Well yes, since ds=1, the above fraction is identical to the fraction d^2y/dx^2. This is not a shorthand notation. But this fraction is only equal to the second derivative of y as a function of x when d^2x=0. The second derivative is computed by y''(x) =...

Take a curve in the x,y-plane and pick a tangent vector field along the curve. Then at a point (x,y), the tangent vector has coordinates (dx,dy) and the derivative of the tangent vector has coordinates (d^2x,d^2y). The derivative of y as a function of x can be calculated as \frac{dy}{dx}...

This is a correct identity. The fractions d^2 f/dx^2 and d^2 f/dy^2 equal the second derivative of f only under the assumption that d^2x and d^2y vanish. The general formula for the second derivative of f with respect to x is \frac{d^2\!f\,dx-df\,d^2x}{dx^3}

The action of SO(C,2) is transitive on each of the components (z,iz) and (z,-iz) separately. To move between the components you take the matrix \left(\begin{matrix}1 & 0 \\ 0 & -1\end{matrix}\right) which belongs to O(2,C). A quick generalization of this would be that the action of...

But the action of the whole orthogonal group, O(n,C), is transitive, yes?

A curve (x, y(x)) parameterized by x has tangent vectors (1, y'(x)) with lengths \sqrt{1+(y')^2}.

Well, yes, but the notation a+tv is a bit problematic. You are adding a vector to a point, so strictly speaking you are going off on a tangent. :smile: Instead you can restate the definition by first choosing any curve \gamma(t) such that \gamma(0)=a,\, \gamma'(0)=v and then defining the...

Indeed :wink:

Assume you have a two-dimensional smooth manifold M, a smooth function f and a vector (field) \mathbf{v}. And you would like to calculate the directional derivative of f at some point p. Since it is a manifold you have a coordinate system (x,y) around the point and since the function is...