10
$\begingroup$

I'm trying to follow a proof about immersing/embedding $\mathbb{RP}^n$ into $\mathbb{R}^{n+1}$, which goes roughly as follows:

Write $\tau=T\mathbb{RP}^n$. The normal bundle $\nu$ has rank 1, so its Steifel-Whitney class is $w(\nu)=1$ or $w(\nu)=1+x$. In every case, we need $w(\nu)\cdot w(\tau) = w(\nu \oplus \tau) = w(\epsilon^{n+1})=1$. If $w(\nu)=1$, then $w(\tau)=(1+x)^{n+1}=1$, so $n+1=2^r$. If $w(\nu)=1+x$, then similarly $(1+x)^{n+2}=1$ so $n+2=2^r$. If the immersion is an embedding, the former case must hold.

Why is this true? I feel like there should be an easy reason, but none of the people I talked with were able to nail down anything precise. This could be wrong, but it seems like this is tacitly saying that a codimension-1 embedding of a closed manifold must be in fact of an orientable manifold, which is the same as saying that the the normal line bundle has trivial $w_1$ (since line bundles are totally classified by their orientability, i.e. by $w_1$). Is this true?

  • 0
    Is your question about $M^n$ or about $\mathbb {RP}^n$? If the former, the Möbius band embedded into $\mathbb R^3$ says the answer is no.2010-11-13
  • 0
    I don't know about Stiefel-Whitney classes, but I can answer the question in the title. Yes (for closed M). If $f\colon M\to\mathbb{R}^{n+1}$ is a smooth embedding and M is not orientable, then it would be possible to construct a map $g\colon S^1\to\mathbb{R}^{n+1}$ such that f,g have (mod 2) intersection number 1, which is a contradiction. Just take a closed curve in f(M) along which it is not orientable.2010-11-13
  • 0
    to me it looks like you're missing one or two tools. Look up the Jordan-Brouwer Separation Theorem and the Tubular Neighbourhood theorem.2010-11-13
  • 0
    @ Josh: My question was about closed manifolds in general, it just happened to be motivated by stuff about projective space. Thanks for the catch.2010-11-14
  • 0
    @ Ryan: The JBST I found on wikipedia only says that the image of an n-sphere in $\mathbb{R}^{n+1}$ separates. I'm having trouble seeing what this buys me in the general case...?2010-12-27
  • 0
    @Aaron: sorry, at one point I had edited in the appropriate generalization of JBST into the wikipedia page but it appears to have vanished. The conclusion of the generalized JBST holds for any compact connected boundaryless co-dimension 1 submanifold of a sphere. Specifically, the complement has two connected components. The proof in this generality is the same as the one outlined by Jim. In fact, you can replace the ambient sphere by any simply-connected manifold in Jim's argument.2011-12-02
  • 0
    Yes, [any closed smooth hypersurface of $\mathbb R^n$ is orientable.](http://math.stackexchange.com/a/879449/3217) See also [here](http://math.stackexchange.com/a/864058/3217).2015-07-25

3 Answers 3

15

The normal bundle of a codimension 1 embedding of a compact closed manifold $M$ in $\mathbb R^{n+1}$ is indeed trivial. Otherwise, you could find a simple closed curve in $\mathbb R^{n+1}$ that intersects $M$ in a single point. This implies that both the curve and $M$ represent nontrivial mod $2$ homology classes in $\mathbb R^{n+1}$; this is because the intersection product is dual to the cup product and cannot be nonzero on trivial homology classes. However there are no nontrivial homology classes in $\mathbb R^{n+1}$ since it is contractible. So we get a contradiction.

Once the normal bundle is seen to be trivial, one can use the ambient orientation of $\mathbb R^{n+1}$ to locally orient the manifold, since there is a well defined positive normal direction.

  • 0
    Would you mind expanding your line "Otherwise, you could find a simple closed curve..."? As George says in the comments, one can just take a closed curve in M which witnesses the nonorientability of M. But why should this curve (made generic) intersect M in exactly one point? More specifically, where does the hypothesis that M be closed come in to the picture?2010-11-14
  • 0
    This is beautifully clean. George mentioned something like it as a comment, but I didn't fully understand what he meant.2010-11-14
  • 0
    @Jason: The intersection number is 1 (mod 2) if M is not orientable along the given curve. This is true regardless of whether M is closed or not. However, if M is closed, then we can shrink the curve down to a point in $\mathbb{R}^{n+1}$ without passing through a boundary point of M. This shows that the intersection number must be zero - the required contradiction. You can try the same argument with a Mobius band embedded in $\mathbb{R}^{n+1}$ in which case it shows that you can't shrink the curve to a point without passing through the boundary of the band.2010-11-15
  • 0
    @Aaron: Yes, this method is what I was thinking of. The great thing is that it applies to all continuous embeddings of non-orientable closed topological manifolds (smoothness is not required). There is a fair amount of machinery required to define intersection numbers and determine their basic properties. That's why I liked Plop's answer. It is much more elementary, although it doesn't generalize as easily.2010-11-15
  • 0
    @ George: I <3 algebraic topology, so this is my favorite answer. I'm glad there's a geometric proof for the smooth case, though.2010-11-15
  • 0
    Sorry, went to bed, so I haven't kept up with the comments. The reason you can find such a simple closed curve is that if the normal bundle is not a product, you can send a normal vector along some path so that it becomes its opposite. Connect the trace of the tip of the normal vector with a curve connecting the two oppposite normal vectors, which pierces $M$ once. $M$ needs to be closed, or it wouldn't represent a (mod 2) homology class.2010-11-15
8

Any compact, connected manifold $M$ of dimension $n$ embedded in $\mathbb{R}^{n+1}$ is orientable. This follows from the fact that there exists a smooth function $f$ on $\mathbb{R}^{n+1}$ s.t. $M$ is the zero locus of $f$ and the derivative of $f$ does not vanish on $M$ (so the gradient of $f$ gives a smooth normal vector field). This also proves Jordan-Brouwer's theorem ($M$ cuts $\mathbb{R}^{n+1}$ in two parts: $f>0$ and $f<0$).

The existence of $f$ is a "technical lemma" ;) First, using compactness, you can show that there is an $\epsilon > 0$ and a covering $\left( V_x \right)_{x \in M}$ of $M$ such that $B=\left\{ y \in \mathbb{R}^{n+1}\ |\ d(y,M) \lt \epsilon \right\}$ is the disjoint union of the $\left\{ x + t n_x \ |\ -\epsilon \lt t \lt \epsilon \right\}$ ($n_x$ being a vector of norm one orthogonal to $T_x M$), and so $B$ is locally $V_x \times ]-\epsilon,\epsilon[$. From there you can deduce that $\mathbb{R^{n+1}}$ has an open covering $\left(A_i\right)_i$ with smooth functions $f_i$ on $A_i$ s.t. whenever $A_i \cap A_j \neq \emptyset$, $f_i = \pm f_j$ locally on $A_i \cap A_j$, and $\cup_i f_i^{-1}(0) = M$ has empty interior (for this you need to take a smooth non-decreasing function $\lambda : \mathbb{R} \rightarrow \mathbb{R}$ s.t. $f(x)=-1$ when $x \lt -\epsilon/2$, $f(x)=1$ when $x \gt \epsilon/2$, and $f$ is increasing inbetween. then for $(y,t) \in V_x \times ]-\epsilon,\epsilon[$, define $f_i(y,t)=\lambda(t)$, and take the $f_i$s to be $1$ outside the $\epsilon/2$-neighbourhood of $M$). This, together with the simple connectedness of $\mathbb{R}^{n+1}$, gives a smooth function $f$ on $\mathbb{R}^{n+1}$ locally equal to the $\pm f_i$, and you are done.

Now you have a long exercise to solve your technical problem ;) I'm sorry I couldn't just give you a reference, but the only one I have (and from which I took the above sketch of proof) is a french book: Thèmes d'analyse pour l'agrégation, Calcul différentiel by Stéphane Gonnord and Nicolas Tosel (p. 100).

EDIT: actually they give a reference: Elon L. Lima, The Jordan-Brouwer separation theorem for smooth hypersurfaces, American Mathematical Monthly, Volume 95 Issue 1, Jan. 1988

  • 0
    "...together with the simple connectedness of M". Simply connected manifolds are orientable, so isn't this a circular argument, assuming the result you want to prove?2010-11-13
  • 0
    sorry, of $\mathbb{R}^{n+1}$! corrected, thank you2010-11-13
  • 0
    +1. I see now. That works, and a very neat proof! I don't think it proves Jordan-Brouwer's theorem though, which says that M cuts $\mathbb{R}^{n+1}$ into exactly two connected components. How do we know that $\{f > 0\}$ and $\{f < 0\}$ are connected?2010-11-13
  • 0
    Actually, I can see why $S=\{f > 0\}$ is connected (if M is connected). If P,Q are in S, take a curve from P to a point in M and a curve from Q to a point in M. The points in M are connected by a path in M. Shifting this path by a small distance $\epsilon/2$ away from M gives a curve joining P,Q in S. Similarly for $\{f < 0\}$. Very nice!2010-11-13
  • 0
    You now have a normal vector $n_x$ defined at each point $x \in M$. For some $\epsilon \gt 0$, the set comprising the $x + t n_x$ with $0 \lt t \lt \epsilon$ is connected. Same goes for the set comprising the $x - t n_x$. From any $y \in \mathbb{R}^{n+1} \setminus M$, take the shortest path from $y$ to $M$: you will cut one of this two sets before reaching $M$.2010-11-13
  • 0
    Plop - yes, that's what my comment just before yours was getting at. Thanks!2010-11-13
  • 0
    I think this proof would also work (and prove the Jordan-Brouwer theorem) for non-smooth embeddings as long as you know that it "locally cuts" $\mathbb{R}^{n+1}$ into two (which is true, but I don't know a simple proof).2010-11-13
  • 0
    I'm not sure about that: you need the tangent space, and more importantly you use the inverse function theorem2010-11-13
  • 0
    You need the tangent space and inverse function theorem if you want to extend to an embedding of Mx[-1,1] (which is not possible for the Alexander horned sphere http://en.wikipedia.org/wiki/Alexander_horned_sphere). That's not necessary to construct the function f and prove Jordan-Brouwer though. You can still define the $f_i$'s locally, although they won't match up on $A_i\cap A_j$. You don't need them to match. As long as their signs agree, you can join them using a partition of unity.2010-11-14
  • 0
    This is interesting, thanks. I'm a little confused in parts though. For starters, how can you continuously define your $f_i$'s to equal 1 outside of the $\epsilon/2$-neighborhood of $M$, when you modeled them on a function $\lambda$ which is $-1$ below $-\epsilon/2$?2010-11-14
  • 0
    The covering $(A_i)_i$ is given by the $V_x \times ]-\epsilon,\epsilon[$ and $\{ y \in \mathbb{R}^{n+1} | d(y,M)>\epsilon/2 \}$. We define $f_i$ to be $1$ on the latter. Then we have $f_i= \pm f_j$ locally for any $i,j$.2010-11-14
  • 0
    Oh, okay. Of course. And then how does the simple connectedness of $\mathbb{R}^{n+1}$ imply that we can paste together the $\pm f_i$ to obtain such an $f$?2010-11-14
  • 0
    Choose a base point $x_0 \in A_i$ s.t. $f_i(x_0) \neq 0$ (for some $i$). For any $x$ and any path from $x_0$ to $x$, there is a connected open containing this path and a function $f$ on this open equal to $\pm f_j$ locally, and agreeing with $f_i$ at $x_0$. Moreover such a function is unique (using the fact that $M$ has empty interior). Finally, $f(x)$ does not depend on the path chosen (simple connectedness of $\mathbb{R}^{n+1}$.2010-11-14
  • 0
    So the same is true in every simply connected manifold, not just $\mathbb{R}^{n+1}$: every codimension-1 submanifold is orientable.2010-11-14
  • 0
    Ohhhhh, okay. Very nice! I like the generalization, too.2010-11-15
  • 0
    *compact* submanifold, of course2010-11-15
4

Here's a nice solution I just thought of, which may in fact be logically equivalent to Jim's (if it's even correct!). I welcome comments addressing that.

Compactify $\mathbb{R}^{n+1}$ to $S^{n+1}$, so we consider $M \subset S^{n+1}$. By Alexander duality, $$\tilde{H}_0(S^{n+1} \backslash M ; \mathbb{Z}/2) \cong \tilde{H}^{(n+1)-0-1}(M;\mathbb{Z}/2) = \tilde{H}^n(M;\mathbb{Z}/2)=\mathbb{Z}/2,$$ so $H_0(S^{n+1}\backslash M;\mathbb{Z}/2)=\mathbb{Z}/2 \oplus \mathbb{Z}/2$. Hence $M$ separates $S^{n+1}$.

  • 0
    Yes, you're supplying the algebraic details for Jim's argument.2010-11-14
  • 0
    Okay, thanks. Yeah, I only recently saw Alexander duality for the first time (and in the context of spectra, no less!) and so I don't yet have a good handle on what the main point really is.2010-11-15
  • 0
    You might be interested in Corollary 3.45 in Hatcher's Algebraic Topology.2011-04-11
  • 0
    I see. Thanks, that's a nice result. I guess it probably just generalizes my observation, anyways, seeing as it's a corollary of Alexander duality.2011-04-13