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Everyone: I am trying to make sense of these statements (doubts are ones contained within asterisks **); both $M$,$N$ are surfaces.:

Setup: $N$ is a surface trivially-embedded in $X^4$, a 4-manifold. $M$ is a compact surface embedded in $X^4$ , with $\partial M$ contained in $N$, and ** $M$ is normal to $N$ along its boundary *, (1)and, * Interior of $M$ is transverse to $N$ ** (2)

This is what I (think) I understand:

(1) *$M$ is normal to $N$ alongs its bdry *: I guess this means that the curve $\partial M$ is in a position such that at each point $p$ where $\partial M\cap N$ is not empty, $T_p(\partial M)$ is perpendicular to $T_pN$. Is it? I know the normal line is (a) perpendicular to the tangent so I am extrapolating that if two manifolds intersect each other normally, that their respective tangent spaces at points (as local linearizations) of intersection are perpendicular to each other, i.e., the tangent space to one of the manifolds is normal to the tangent space of the other manifold.

(2)*"The interior of $M$ is transverse to $N$ *" Here, by interior of $M$, we mean $M-\partial M$. Since both $M$,$N$ are embedded in $X^4$, and are transverse to each other, they intersect at points (since codim(M)+codim(N)=4), and do not intersect in some other way.

Is this correct? Is there something else I should consider?

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By the way, you can use standard TeX stuff if you put stuff inside dollar-signs, and it makes it much easier to read. I edited the question just to fix this, but apparently I'm at some middle level of rep points where I can edit but it still needs to be peer-reviewed or something like that. –  Aaron Mazel-Gee Apr 4 '11 at 7:43
    
Sorry, Aaron, I have been way too lazy and inconsiderate about this. I will change next time , if more than one-or-two tags are necessary. Sorry to all who have read my posts. –  gary Apr 4 '11 at 7:56
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Two things: For (1), the phrase "$M$ is normal to $N$ along its boundary" is slightly ambiguous about whose boundary we're talking about, except that $M$ is explicitly defined as having a boundary so I think you're right about that. But for (2), note that it's stronger to say that two submanifolds of complementary codimension intersect transversely than to say that they intersect in points. Remember that a transverse intersection is one where the local model is what you'd expect; a standard counterexample is the pair of curves $y=0$ and $y=x\cdot \sin(1/x)$ in $\mathbb{R}^2$.

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Aaron: by "what you'd expect", are you referring to the tangent space issue, i.e., that the tangent spaces TpM and TpN at the point of intersection should span TpX, the tangent space of the ambient manifold? –  gary Apr 4 '11 at 8:00
    
No, this is where geometry gets fancier than linear algebra. In this case, an intersection is regular if every point in it has a coordinate neighborhood $U$ such that $M\cap U=\{x_1=x_2=0\}$ and $N\cap U=\{x_3=x_4=0\}$. If you think about it, you'll see that this precludes the existence of cluster points in $M\cap N$, which your suggested interpretation still allowed. –  Aaron Mazel-Gee Apr 4 '11 at 8:30
    
Sorry for the delay, Aaron, I had some confusion registering, and could not vote. –  gary Jul 7 '11 at 7:05
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