How to construct closed, orientable, smooth, simply-connected $6$-manifolds such that $H^{*}(M,\mathbb{Z}) \cong \mathbb{Z}[a]/(a^{4})$ (Where $a$ is a generator of degree 2) satisfying $p_{1}(M) = n a^{2}$? ($p_{1}(M) \in H^{4}(M,\mathbb{Z})$ denotes the first Pontryagin class).

By Wall's classification of $6$-manifolds for each $n \in \mathbb{Z}$ exists a unique such manifold up to diffeomorphism. The case $n = 4$ is furnished by $\mathbb{C}\mathbb{P}^{3}$, are there at other familiar examples - perhaps the $n=0$ case? How about constructing these manifolds with surgery techniques?


1 Answer 1


I looked at Wall's paper Classification problems in differential topology. V On certain 6-manifolds. In theorem 3 of that paper Wall describes some invariants of 6-mainfolds, and the relation between them. These invariants, in the case you are concerned with, are given by:

  • The abelian group $H = \mathbb{Z} = H^2(M;\mathbb{Z})$, and the zero group $G = 0 = H^3(M; \mathbb{Z})$.
  • The symmetric trilinear form $\mu: \mathbb{Z}^3 \to \mathbb{Z}$ given by $\mu(x,y,z) = \langle xyz, [M]\rangle$, i.e. take cup product and pair with the fundamental class. For the cohomology ring you specify this is just the 3-fold multiplication map "xyz".
  • $p_1: H = \mathbb{Z} \to \mathbb{Z}$, which is your integer $n$, corresponding to the 1st Pontryagin class.
  • An element $w_2 \in H / 2H \cong \mathbb{Z}/2\mathbb{Z}$ (2nd Steifel-Whitney class), which has an integral lift $W_2 \in \mathbb{Z}$.

These invariants necessarily satisfy three relations for arbitrary $x, y \in H$: $$ \mu(x,y,x+y + W_2) \equiv 0 \; \; \; \textrm{mod 2} \\ p_1(x) \equiv \mu(x,W_2, W_2) \; \; \; \textrm{mod 4} \\ p_1(x) \equiv \mu(x,x,x) \; \; \; \textrm{mod 3}$$

The first equation, with the given $\mu$, implies that $W_2$ is even. So $w_2=0$ and the manifold is spin (rather "spin-able"), which is what Wall calls "condition (H)". The other equations then impose further conditions on the allowed values of $p_1 \leftrightarrow n$. In particular we see that $n$ has to be a multiple of 4 and is congruent to 1 mod 3.

All of this comes from some observations about necessarly relations between these invariants. We haven't touched at all on the realization problem, i.e. which of the possible $n$ actually come from 6-manifolds?

Wall helps us there too. He gives a more careful analysis in the case of manifolds satisfying "condition (H)". In that case (his thm 5) diffeomorphism classes of manifolds correspond bijectively to systems of invariants $(H,G, p_1, \mu)$ as before subject to two relations $$ \mu(x,x,y) \equiv \mu(x,y,y) \; \; \; \textrm{mod 2} \\ p_1(x) = 4 \mu(x,x,x) \; \; \; \textrm{mod 24}$$ Note that the final condition is stronger than the previous conditions. Some values of $n$ do not occur.

In the case at hand where $H = \mathbb{Z}$, $G=0$, and $\mu$ is the standard "3-fold product" trilinear map, the first equation is satisfied, and the second condition tells us that $$ n = 4 + 24k $$ for some integer $k$.

Finally, Wall also tells us how to construct this manifold. It can be realized as surgery on a framed knot $S^3 \times D^3 \hookrightarrow S^6$. Specifically it is surgery on the framed knot with invariants $\varphi = -k$ and $\beta' = 1 + 6k$. These knot invariants are discussed at length in sections 4 and 5 of Wall's paper, which has other references.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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