I am reading through Guiellman and Pollack's Differential Topology. Ultimately I would like to understand the Euler characteristic. On page 116, they define the Euler characteristic of a compact oriented manifold $Y$ as the oriented self-intersection number of the diagonal $\Delta$ in $Y\times Y$. This definition is based on the definition of oriented intersection number of a pair of oriented submanifolds $X\subset Y$, $Z\subset Y$. When $X$ is transversal to $Z$ (and we have dimensional complementarity so that $Y=X\oplus Z$), I understand the concept of the oriented intersection number: At each intersection point of $X,Y$, just concatenate an oriented basis of $X$ with an oriented basis of $Z$ and see if you get an oriented basis of $Y$. If you do, then the local intersection number at that point is positive, otherwise it's negative. Then add up the local intersection numbers from each point of intersection. (Note: if you are familiar with the intersection number $I(f,Z)$ as defined for a function $f$ mapping into $Y$ instead of $I(X,Z)$ for a submanifold $X$, the latter notion can be derived from the former by setting $f$ to be the inclusion map $X\hookrightarrow Y$.)

Main Question

Where I am struggling to understand is when $X$ is not transversal to $Y$ (and I think this will basically always be the case when $Z=X$ as in the Euler characteristic definition). I gather that the thing to do is to slightly deform $X$ so it becomes transversal to $Z$ and then look at the intersection number of the deformed version of $X$ with $Z$ as defined above. But we would need to know that the resulting number doesn't depend on how we deform $X$ (within some neighborhood). This is where I am confused.

It may be useful to consider an example where $X=Z$ is the x-axis in $\mathbb{R^2}$ such that the positive direction (i.e. moving from left to right) is considered positively orientated. It would appear to me that the intersection number we get after deforming $X$ (leaving $Z$ alone) would depend on whether we rotate $X$ clockwise or counterclockwise. If clockwise, I think we would get a positively orientated basis for $\mathbb{R^2}$ (so an intersection number of 1) and if counterclockwise we would get a negatively oriented basis of $\mathbb{R^2}$ (so an intersection number of -1). I think there is some mistake in my understanding here? Note that I am thinking of the tangent space of $X$ as the first element of the basis and of the tangent space of $Z$ as the second element of the basis (of $\mathbb{R^2}$).

Possibly relevant: I have read through the proof that the intersection number (also apparently called degree) is homotopy invariant (from Guilleman and Pollack and from Milnor) and I think I understand this proof (which uses the classication of 1-manifolds) but I don't see how the non-transversal case is accounted for.

As a secondary question, I am also not sure how to define the oriented intersection number when $X$ is transveral to $Z$ but their tangent spaces overlap as in $\mathbb{R^3}$. A third question is how to define oriented intersection number when $X$ is not transveral to $Z$ but where their tangent spaces do not overlap as with two perpendicular lines intersecting at a point in $\mathbb{R^3}$.

Note: I have not had the time to properly work through Guilleman and Pollack line-by-line from the beginning. I have skipped over several parts so it is possible that I have missed some important concept that is typically presumed when presenting this material.


Yes, you need to read the earlier parts of the book. The crucial thing you’re missing is that $X$ must be compact (and $Z$ closed). Indeed, in your example, you can push the $x$-axis up and have no intersections at all.

  • $\begingroup$ Thanks, I think I see what's going on here now. A better example is where 𝑋=𝑍 is the equator on the sphere. Then the oriented intersection number is zero no matter how you deform 𝑋. This leads me to think if the oriented intersection number in the non-transversal case is always zero? To show this we would try to pull $X$ away from $Z$ along the dimension (backwards or forwards?) that is not being covered by the combined tangent spaces--such a dimension exists by non-travsersality. Will have to think about a rigorous proof. $\endgroup$ – Smithey Mar 24 '20 at 3:54
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    $\begingroup$ No. When $\dim X$ is odd, it will be $0$, but not when it's even. Go back to where your question started. Take the diagonal in $S^2\times S^2$ or take $\Bbb CP^1$ in $\Bbb CP^2$. $\endgroup$ – Ted Shifrin Mar 24 '20 at 6:42
  • $\begingroup$ Αfter reading the lefschetz section, I see that $\chi$ is 2 for $S^2\times S^2$ since the map that moves everything south keeping poles fixed is homotopic to identity. I don't understand whether we need to restrict to boundaryless 𝑋? E.g. in $X=[0,1]$ I'm confused about the corners of the square. Self-int # of diagonal seems to vary by 1 as we perturb to add new crosses keeping end pts fixed: Adding a new cross of Δ, we also change the direction from which we are approaching one of the corners. The latter change would be a $\pm 2$ which would overcompensate for the $\pm1$ of new cross? $\endgroup$ – Smithey Mar 25 '20 at 5:16
  • $\begingroup$ If $X$ has boundary, intersection number is no longer a homotopy invariant. If you require that no intersection occurs at the boundary and that homotopy fix the boundary, perhaps something works. I’ve never thought about it. $\endgroup$ – Ted Shifrin Mar 25 '20 at 22:34

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