# Trying to get my head around tangent bundles

1. Nov 11, 2009

### mikeph

Hello,

Say you have a function f on the domain R^n, and an integral transform P which integrates f over all possible straight lines in R^n. I am lead to believe that the range of this is R^(2n), or a tangent bundle, which I am having MASSIVE problems visualising!

Am I right in saying the tangent bundle can be described by the multiplication of a vector on the unit sphere in R^n with a vector in R^n, ie. all points, then from each point, subtending all angles?

But surely this creates duplication? ie. for n=3, the line passing point (0,0,0) parallel to (1,0,0) must be the same as the line passing through (1,0,0) parallel to (1,0,0). So I am trying to picture a more "efficient" way to specify the range of this transform...

Is it completely described by all vectors in R^n perpendicular to each plane described by the vector on the unit sphere in R^n? How many are there per plane?

SO confused! But intrigued....

Thanks,
Mike

2. Nov 11, 2009

### 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.

Last edited: Nov 11, 2009
3. Nov 11, 2009

### mikeph

Thankyou,

Can I confirm, does isomorphism mean all the lines parallel to that line?

I would think the space which this transform maps to is R^(n-1)^2...

The reason being, each hyperplane in R^n can be described as perpendicular to a point on the unit sphere in R^n, requiring n-1 scalars. And then for each plane you can completely parametrise the perpendicular lines crossing it using R^(n-1), since the plane itself is parametrised this way.

So pick a direction, then pick a point on the plane perpendicular to this, and you get a unique straight line through R^n.

Is that correct?

Thanks,