What is a transversal intersection? Can it be explained without tangent spaces and tangent bundles?

Question

What is a transversal intersection? Can it be explained without tangent spaces and tangent bundles?

Background: Transversal intersection was used to explain If the interior of two convex manifolds intersect, what is the dimension of their intersection?. I would like to know what transversal intersection is, but the resources use the concepts of tangent bundles and tangent spaces to explain it.

Is there a simple explanation of transversal intersection that doesn't use tangent bundles and tangent spaces? Perhaps explaining it in a limited context (e.g. convex sets in $\mathbb R^n$). Or, if not: Is there a simple explanation of enough background to understand transversal intersection?

(If so, I expect this Q&A may be very useful for others in the future as well.)

Answer 1

Roughly speaking, two spaces (manifolds) $X,Y \subseteq \mathbb R^n$ intersect transversally if $X \cap Y \neq \emptyset$ and, at any intersection point, the union of all possible lines tangent to either $X$ or $Y$, generates $\mathbb R^n$.

To make some examples let's consider first $\mathbb R^2$ as ambient space, then any pair of incident lines intersect transversally as two independent vectors are enough to generate $\mathbb{R}^2$.

A non-transverse intersection can be obtained by considering a quadric in $\mathbb R^2$ and the tangent line at some point of its: for example take $y - x^2 = 0$ and $y = 0$. These varieties (and manifolds) intersect in $(0,0)$, but here the tangent line to the parabola is precisely $y = 0$, which of course does not generate the plane.

Lastly, as open subsets of $\mathbb R^2$ are 2-dimensional manifolds, they contain all possible tangent directions at any point of theirs, so they intersect transversally any other manifold in $\mathbb R^2$.

Answer 2

You can describe transverse intersection by describing models where it holds and models where it does not hold.

Here's two examples:

• The $x$ and $y$ axes in $\mathbb R^2$ intersect transversely.
• The $x$ and $y$ axis in $\mathbb R^3$ do not intersect transversely.

What's the difference?

Well, in $\mathbb R^2$, if I jiggle one of the two lines, say if I move the $x$-axis up or down a bit, the picture does not change qualitatively, in particular the jiggled version of the $x$-axis still intersects the $y$-axis.

Whereas in $\mathbb R^3$, if I jiggle the $x$-axis a little bit in the $z$-direction, just the tiniest bit, then it has no intersection with the $y$-axis.

We can now use this model to describe transverse intersection of curves in two situations. First, given two curves in $\mathbb R^2$ which intersect one another at some isolated point, and which are modeled near that point by the $x$ and $y$ axes in $\mathbb R^2$, are said to intersect transversely at that point. And the "jiggling" statement still holds: if you jiggle those two curves, that crossing point does not go away.

On the other hand two curves in $\mathbb R^3$ which intersect at an isolated point and which are modelled by the $x$ and $y$ axes in $\mathbb R^3$ are not transverse. The jiggling analysis still applies in this case, but with opposite effect, namely it is possible to jiggle the two curves near their intersection point so that point goes away.

One can formalize this "modelling" feature exactly as the definition of a submanifold is formalized: by requiring special kinds of coordinate charts that are invertible and smooth in both directions. More formally, two curves $C,D$ in $\mathbb R^2$ intersect transversely at a point $x \in C \cap D$ if there exists an open set $U$ around $x$, an open ball $B$ around the origin, and a smooth invertible map $f : U \to B$ with smooth inverse, such that $f(U \cap C) = (\text{$x$-axis}) \cap B$ and $f(U \cap D) = (\text{$y$-axis}) \cap B$.

So now the question is how to do this in $\mathbb R^n$ with coordinate functions $x_1,...,x_n$. The model for a single $m$-dimensional manifold is a single $m$-dimensional coordinate subspace of $\mathbb R^n$, determined by singling out a list of $m$ of the coordinate functions, say $x_{i_1},...,x_{i_m}$, letting those vary freely, and setting the rest equal to $0$. Let's also take an $l$-dimensional coordinate subspace $L$, determined by singling a list of $l$ coordinate functions $x_{j_1},...,x_{j_l}$. Clearly $M$ and $L$ intersect at the origin; in general $M \cap L$ will itself be a coordinate subspace of some kind, defined by the common coordinate functions amongst the two lists. Now jiggle $M$ and $N$, examine how $M \cap L$ changes, and ask yourself: is the intersection still a subspace of the same dimension? If so, $M$ and $L$ are said to be transverse. And with a little bit of linear algebra, you can determine exactly when this happens, namely when the union of the two sets of coordinate functions $x_{i_1},...,x_{i_m},x_{j_1},...,x_{j_l}$ contains all $n$ of the coordinate functions. This happens if and only if $$\text{dimension}(M) + \text{dimension}(L) - \text{dimension}(M \cap L) = n $$ Finally, given an $m$-dimensional and $l$-dimensional submanifold intersecting at a point $P$, that intersection is transversal if and only if their intersection near $P$ is locally modelled by a transverse pair of coordinate subspaces as defined above.