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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?

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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. – Josh Guffin Nov 13 '10 at 22:14
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. – George Lowther Nov 13 '10 at 22:19
to me it looks like you're missing one or two tools. Look up the Jordan-Brouwer Separation Theorem and the Tubular Neighbourhood theorem. – Ryan Budney Nov 13 '10 at 23:02
@ Josh: My question was about closed manifolds in general, it just happened to be motivated by stuff about projective space. Thanks for the catch. – Aaron Mazel-Gee Nov 14 '10 at 5:45
@ 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...? – Aaron Mazel-Gee Dec 27 '10 at 9:16
up vote 10 down vote accepted

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.

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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? – Jason DeVito Nov 14 '10 at 14:57
This is beautifully clean. George mentioned something like it as a comment, but I didn't fully understand what he meant. – Aaron Mazel-Gee Nov 14 '10 at 19:11
@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. – George Lowther Nov 15 '10 at 0:32
@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. – George Lowther Nov 15 '10 at 0:35
@ George: I <3 algebraic topology, so this is my favorite answer. I'm glad there's a geometric proof for the smooth case, though. – Aaron Mazel-Gee Nov 15 '10 at 4:30

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

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"...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? – George Lowther Nov 13 '10 at 23:21
sorry, of $\mathbb{R}^{n+1}$! corrected, thank you – Plop Nov 13 '10 at 23:22
+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? – George Lowther Nov 13 '10 at 23:35
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! – George Lowther Nov 13 '10 at 23:43
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 $M$. – Plop Nov 13 '10 at 23:43

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 $S^{n+1}$.

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Yes, you're supplying the algebraic details for Jim's argument. – Ryan Budney Nov 14 '10 at 20:29
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. – Aaron Mazel-Gee Nov 15 '10 at 2:15
You might be interested in Corollary 3.45 in Hatcher's Algebraic Topology. – user641 Apr 11 '11 at 19:19
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. – Aaron Mazel-Gee Apr 13 '11 at 7:21

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