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Today in a differential geometry lecture, the lecturer put down a question for us to think about:

Given a regular curve (differentiable function $\alpha:I\to\Bbb R^3$ on an open interval with $\alpha'$ nonzero everywhere) and a point $t\in I$, we consider the two points $\alpha(t+h),\alpha(t-k)$ close to $\alpha(t)$ where $h,k\gt0$. Hopefully when $h,k$ are small enough, the two points are distinct. Then we can define a straight line $L$ through $\alpha(t+h),\alpha(t-k)$, call this line $L(h,k)$. He then asks whether $$\lim\limits_{(h,k)\to(0^+,0^+)}L(h,k)$$ exists, and if it exists, what it would be.

As far as I know, defining limits require the notion of topology. How do we define a good topology on the set of all straight lines (required to be infinite on both sides) in $\Bbb R^3$? I have heard of Grassmannian, which parametrises all vector subspaces of a fixed dimension, but this does not fully address my question because I do not require the lines to pass through a specific point.

I have some idea on defining a good topology. First given a fixed unit vector $v\in\Bbb R^3$, for each point $x\in\Bbb R^3$, define the line $L(x,v):=\{x+tv\in\Bbb R^3:t\in\Bbb R\}$. Then the family of all lines $L(x,v)$ with $v$ fixed can be parametrised by $\Bbb R^2$. Denote this family of lines $S(v)$. Note that $S(v)=S(-v)$. The set of all straight lines in $\Bbb R^3$ is the disjoint union of all the $S(v)$, with $v$ parametrised by $\Bbb RP^2$ because of $S(v)=S(-v)$.

Can the set of all straight lines be seen as a plane bundle over $\Bbb RP^2$? And explicitly how is the plane bundle defined? Is it simply the tangent bundle? Please assume that I know little about smooth manifolds.

Edit: I should state some desirable properties of the topology explicitly. For the set $S(v)$ as a subspace of space of all lines, I hope $S(v)$ would be homeomorphic to $\Bbb R^2$. For example, if $v=(0,0,1)=e_3$, I expect the map $P\to S(e_3),(x,y,0)\mapsto L((x,y,0),e_3)$ is a homeomorphism, where $P=\{(x,y,0)\in\Bbb R^3:x,y\in\Bbb R\}$. For each point $x\in\Bbb R^3$, Define $T(x)$ as the set of all lines through $x$. I expect that the subspace $T(x)$ is homeomorphic to $T(0)$ (this $0$ is zero vector), i.e. the Grassmannian/$\Bbb RP^2$. Since no line is special (not even the origin is special), I also expect the whole space of all lines is homogeneous in this sense, that for every two "points" $x,y$ in this space, there is a self-homeomorphism sending $x$ to $y$.

Edit 2: There is a related question for a similar problem on $\Bbb R^2$. See A topology on the set of lines?.

By the way, if there is any, I would like to see reference on similar problems with solutions, i.e. how to define natural topologies on the set of $k$-dimensional affine subspaces in $\Bbb R^n$, and on a family of curves in $\Bbb R^n$, and on a family of surfaces in $\Bbb R^n$, etc.

edm
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    If you want limits that are at all nice then you surely want to find a metric on the set of lines and let that induce a topology, rather than trying to explicitly construct a topology – Dan Robertson Sep 05 '17 at 11:15

2 Answers2

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You can consider every line in the plane $\mathbb{R}^2= \mathbb{R}^2 \times \{0\}$ as the intersection of $\mathbb{R}^2$ with a (unique) plane passing through $(0,0,1)$. This will make the set of lines in $\mathbb{R}^2$ as a subset of all the planes in $\mathbb{R}^3$ passing through a given point, so a subspace of a grassmanian.

But the general idea is the following: if you can parametrize your set with elements of a topological space ( perhaps now in a unique way), the natural topology would be the quotient topology. So, if you parametrize the lines with triples $(a,b,c)$, with $(a,b)\ne 0$, then you have a have from a subset of $\mathbb{R}^3$ to your set, so you should have the quotient topology on it.

orangeskid
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  • I already found a topology for the set of all lines in $\Bbb R^2$ as answered in another question. I want to find a topology for the set of all lines in $\Bbb R^3$. Perhaps I can use your idea to parametrise all lines in $\Bbb R^3$ by seeing it as a subspace of Grassmannian of higher dimensional spaces. – edm Sep 06 '17 at 00:28
  • @edm: oh, i thought you meant $\mathbb{R}^2$. Yes, they are to be seen as a subset of the Grassmanian. While at it, there is a(n equivalent) way to parametrize lines in $\mathbb{R}^3$ using Plucker coordinates, it ls used even in technical applications ( mechanisms, robotics, etc), check Plucker coordinates of a line. – orangeskid Sep 06 '17 at 00:38
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The constructions in the answer here can be generalised. Every line falls into at least one of the following categories:

$1$. those that are not parallel to $xy$-plane

$2$. those that are not parallel to $yz$-plane

$3$. those that are not parallel to $xz$-plane

Each of the above categories of lines can be parametrised as $U\times\Bbb R^2$, where U is the open hemisphere. The bijective map between the set of lines in category $1$ and the set $U\times\Bbb R^2$ is given by using $U$ to parametrise the direction of the lines, and using $\Bbb R^2$ to parametrise the horizontal position of the line when a direction of line is fixed. The rest is to write down the bijective maps concretely and show the transitions are smooth.

Declaring the three bijective maps with the three categories of lines shall be coordinate charts, the set of lines is a smooth manifold (this requires some detail-checking). Then we can show that this manifold is a vector bundle on $\Bbb RP^2$, by using the three charts above.

edm
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