What manifolds are bounded by RP^odd?

Question

Real projective spaces $\mathbb{R}P^n$ have $\mathbb{Z}/2$ cohomology rings $\mathbb{Z}/2[x]/(x^{n+1})$ and total Stiefel-Whitney class $(1+x)^{n+1}$ which is $1$ when $n$ is odd, so it follows that odd dimensional ones are boundaries of compact $(n+1)$-manifolds. My question is: are there any especially nice constructions of these $(n+1)$-manifolds?

I'm especially interested in the case $n=3$. I believe we can get an explicit example of a 4-manifold bound by $\mathbb{R}P^3$ using Rokhlin/Lickorish-Wallace but it doesn't look like that would generalize to higher dimensions at all easily. Are there a lot of different 4-manifolds with this property?

Answer 1

$\mathbb RP^3$ double-covers the lens space $L_{4,1}$, so it's the boundary of the mapping cylinder of that covering map.

In general $\mathbb RP^n$ for $n$ odd double-covers such a lens space. So in general $\mathbb RP^n$ is the boundary of a pretty standard $I$-bundle over the appropriate lens space. To be specific, define the general $L_{4,1}$ as $S^{2n-1} / \mathbb Z_4$ where $Z_4 \subset S^1$ are the 4-th roots of unity, and we're using the standard action of the unit complex numbers on an odd dimensional sphere $S^{2n-1} \subset \mathbb C^n$.

Edit: generalizing Tim's construction, you have the fiber bundle $S^1 \to S^{2n-1} \to \mathbb CP^{n-1}$. This allows you to think of $S^{2n-1}$ as the boundary of the tautological $D^2$-bundle over $\mathbb CP^{n-1}$. You can mod out the whole bundle by the antipodal map and you get $\mathbb RP^{2n-1}$ as the boundary of the disc bundle over $\mathbb CP^{n-1}$ with Euler class $2$. So this gives you an orientable manifold bounding $\mathbb RP^{2n-1}$ while my previous example was non-orientable.

Answer 2

$\mathbb{RP}^3$ is the unit (co)tangent bundle to $S^2$. Thus it bounds the disc bundle in $TS^2$. Alternatively, any time you have a 2-sphere of self-intersection $\pm 2$ in a closed 4-manifold - for example, any Lagrangian sphere in a symplectic 4-manifold - you get a splitting into two pieces along an $\mathbb{RP}^3$. So, you could consider the diagonal $S^2$ in $S^2\times S^2$, for instance.

Answer 3

Take the complex surface $Z_1 ^2 + Z_2 ^2 + Z_3^2 =1$ in complex 3-space, $\mathbb{C}^3$ and intersect it with the ball $|Z_1|^2 + |Z_2|^2 + |Z_3|^2 \le 1$ to get an explicit 4-manifold with boundary embedded in $\mathbb{C}^3$ whose boundary is $\mathbb{R}P^3$, realized by intersecting the complex surface with the 5-sphere $|Z_1|^2 + |Z_2|^2 + |Z_3|^2 = 1$.

To verify, split $Z_a = x_a + i y_a$ into its real and imaginary parts $x_a, y_a$, do a bit of algebra and see that the hypersurface is diffeomorphic to the tangent bundle of the standard two-sphere $S^2$, that two-sphere being realized within $(x_1, x_2, x_3)$ space by setting the $y_a = 0$. (I've heard this complex surface called the 'hypersphere' or 'complex sphere' or some such.)

Intersecting the hypersurface with the 5-ball $|Z_1|^2 + |Z_2|^2 + |Z_3|^2 \le 1$ is the same as taking the disc bundle of Tim Perutz's construction.

Setting $|Z_1|^2 + |Z_2|^2 + |Z_3|^2 = 1$ within the hypersurface is the $RP^3$ which bounds the 4-manifold.

So, you can take your 4-manifold to be a ``Stein domain'' in standard $\mathbb{C}^3$.

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Even more explicitly, with a CR-aside:

After some rescalings, this embedding of $RP^3$ is essentially Rossi's example of an analytic CR structure on the three-sphere which admits no CR embedding into any $C^n$.

Consider the map from $\mathbb{C}^2$ to $\mathbb{C}^3$ given by $$Z_1 = i[ (z^2 + w^2) + t (\bar z ^2 + \bar w ^2)]$$ $$Z_2 = [ (z^2 - w^2) - t (\bar z ^2 - \bar w ^2)]$$ $$Z_3 = 2 [ zw - t \bar z \bar w]$$ where $t$ is real and $i = \sqrt{-1}$.

A computation shows that $Z_1 ^2 + Z_2 ^2 + Z_3^2 =-4 t (|z|^2 + |w|^2)^2$ while $|Z_1|^2 + |Z_2|^2 + |Z_3|^2 = 2(1 + t^2)(|z|^2 + |w|^2)^2.$ It follows that the image of the standard $S^3$ in $C^2$ is the complex hypersurface $Z_1 ^2 + Z_2 ^2 + Z_3^2 =-4 t $ intersected with the 5-ball
$|Z_1|^2 + |Z_2|^2 + |Z_3|^2 = 2(1 + t^2)$.

Since the map $(z, w) \to (Z_1, Z_2, Z_3)$ is $2:1$ restricted to $S^3$, its image is $\mathbb{R}P^3$.

Rossi's 'no CR embedding' assertion for $t \ne 0$ relates to the induced CR structure on $S^3$ (via the pull-back of its image's CR structure). When $t \ne 0$ every CR function for this CR structure on $S^3$ is an analytic function of these $Z_i (z,w)$'s, and hence is invariant under the antipodal map $(z,w) \to (-z, -w)$. Thus the CR functions for these `twisted' CR structures on $S^3$ cannot separate (antipodal) points and hence the $S^3$ cannot be CR-embedded.

It is a fun fact that all left-invariant CR structures on $SU(2) = S^3$ arise this way, with the standard CR structure corresponding to $t = 0$.

some references:

H. Rossi, Attaching analytic spaces to an analytic space along a pseudoconcave boundary. 1965 Proc. Conf. Complex Analysis (Minneapolis, 1964) pp. 242–256, Springer, Berlin.

D. Burns, Global behavior of some tangential Cauchy-Riemann equations in “Partial Differential Equations and Geometry” (Proc. Conf., Park City, Utah, 1977); Dekker, New York, 1979, p. 51.

E. Falbel, Non-embeddable CR-manifolds and Surface Singularities. Invent. Math. 108 (1992), No. 1, 49-65.

Answer 4

Here is a simple general fact: if a manifold $M$ admits a free involution $T$, then $M$ is a boundary. Proof: $M$ will be the boundary of $W =(M \times [-1,1])/((Tx,t) \sim (x,-t))$. In your particular case: when $n$ is odd, the standard free action of $C_4$ on $S^n$ induces a free action of $C_2$ on $\mathbb RP^n$. (Note that this $W$ will tend to be nonorientable, even if $M$ is. Exercise: let $M = S^1$.)