1
$\begingroup$

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 $\mathbb{R}^2$?

enter image description here

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.