When $X$ also happens to be a submanifold of $Y$, then, as in the mod $2$ case, we define its intersection number with $Z, I(X, Z)$, to be the intersection num­ber of the inclusion map of $X$ with $Z$. If $X \pitchfork Z$, then $I(X, Z)$ is calculated by counting the points of $X \cap Z$, where a point $y$ is included with a plus sign if the orientation of $X$ and $Z$ (in that order!) ''add up'' at $y$ to the orientation of $Y$; otherwise $y$ is counted with a minus sign (Figure 3-9).

  • My question: So how the sign $+1$ or $-1$ is assigned? I am just guessing it is counterclockwisely? I don't feel this is right, not to mention this idea can not expand to higher dimensions.

  • My progress (hopefully): From Does "Add up" just means oriented counterclockwisely? I know to take a postively oriented basis in $X$, add to it a postively oriented basis of $Z$ and check whether or not they give a positively oriented basis of the tangent space. So in this case, I realize the two basis element contributed by $X$ and $Z$, but what is the default orientation of $\mathbb{R}^2$?

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An orientation is a map that, given a basis of the tangential space, returns either $+1$ or $-1$. By the transversality assumption, the tangential space $T_Y(p)$ of $Y$ in the point $p$ of intersection equals the direct sum of the tangential spaces of $X$ and $Z$, i.e. $T_Y(p)=T_X(p)\oplus T_Z(p)$. Thus a positively oriented basis of $T_X(p)$ and a positively oriented basis of $T_Z(p)$ give us a basis of $T_Y(p)$. The latter is either positively or negatively oriented and this does not depend on the specific choice of basis in $T_X(p)$ or $T_Z(p)$.

The default orientation in $\mathbb R^n$ is that for which the standard basis $(e_1,\ldots , e_n)$ is postively oriented. Especially, for $\mathbb R^2$, the basis $(e_1,e_2)$ is positively oriented. With the usual conventions on how to visualize $\mathbb R^2$, $e_1$, and $e_2$, this means that counterclockwise order of basis vectors is positive orientation.

  • $\begingroup$ Hmmm~ love it~ So clear yet so rigorous! Thanks a lot Hagen!@->- $\endgroup$ – 1LiterTears Jul 21 '13 at 22:27
  • $\begingroup$ How about in $\mathbb{R}^3$, when a $1$-dimensional submanifold $X$ with basis element $\{v\}$ transverse a $2$-dimensional submanifold $Z$ with basis element $\{e_1, e_2\}$? So I should compare $\{v, e_1, e_2\}$ with $\{e_1, e_2, e_3\}$? So if $v$ points up, it is equivalent to $-e_3$ when computing the orientation; otherwise negative? $\endgroup$ – 1LiterTears Jul 21 '13 at 23:00
  • $\begingroup$ And we also shall assume $Z$ is positively oriented right? Otherwise, if $Z$ is with basis element $\{e_2, e_1\}$, the sign would change, again. $\endgroup$ – 1LiterTears Jul 21 '13 at 23:03

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