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What are interesting / non-trivial examples of smooth connected closed manifolds that are direct products or involve direct products? I am especially interested in orientable manifolds.

Say, an $n$-torus $T^n$ is a direct product of $n$ copies of a circumference $S^1$. One can build a 3-manifold from a surface of genus $g$ as $M=M^2_g\times S^1$, and use somehow the connected sums of such manifolds.

Typically an "important" manifold would have a name or standard notation. For example, the Kodaira-Thurston manifold (important if it has a proper name!) decomposes into a product of the Heisenberg nil manifold and $S^1$.

I am looking for other important / interesting / non-trivial manifolds that happen to be direct products, preferably having important / interesting applications.

MattAllegro
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Alexander Gelbukh
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  • $\mathbb R^n{}$. – Mariano Suárez-Álvarez Mar 06 '15 at 18:51
  • @MarianoSuárez-Alvarez Thank you! Unfortunately, $\mathbb R^n$ is not closed. – Alexander Gelbukh Mar 06 '15 at 18:52
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    Ah. I missed the closed part. Why are you interested in such a list, by the way? It seems quite random really... – Mariano Suárez-Álvarez Mar 06 '15 at 19:00
  • @MarianoSuárez-Alvarez We are working on a paper where the main result is a theorem about direct product of closed manifolds, and I need to explain why this is important and where the theorem can be applied. – Alexander Gelbukh Mar 06 '15 at 19:09
  • Can you find a product of closed manifolds that's not interesting? –  Mar 06 '15 at 19:23
  • @MikeMiller (Joke: all natural numbers are interesting. Proof by induction: consider the smallest non-interesting number.) Yes, when we proved the theorem, this was our logic. But I think $M^2_7\times M^2_{12}\times S^1$ would not be a convincing example for the reviewers. In the question I added a "definition": an important manifold has a proper name. – Alexander Gelbukh Mar 06 '15 at 19:30

3 Answers3

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The Lie groups $U(n)$ are diffeomorphic to products $SU(n)\times S^1$ (but are not Lie isomorphic, unless $n=1$).

The Lie group SO(8) is diffeomorphic (but not Lie isomorphic) to $SO(7)\times S^7$. (In fact, $S^7$ doesn't have a Lie group structure at all.)

The Lie group $SO(4)$ is diffeomorphic (but not Lie isomorphic) to $S^3\times \mathbb{R}P^3\cong S^3\times SO(3)$.

(To my knowledge, Lie groups with Lie algebra $\mathfrak{so}(8)$ are the only simple Lie group which are diffeomorphic to nontrivial products).

Relatedly, the unit tangent bundles to $S^1$, $S^3$, and $S^7$ are diffeomorphic to products $S^{k-1}\times S^k$, and no other unit tangent bundles of spheres are.

Jason DeVito - on hiatus
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  • Thanks to all, all answers are excellent and useful (I upvoted all of them), but the system requires me to choose only one, so I had to choose one. This answer contains three slightly different examples. – Alexander Gelbukh Mar 07 '15 at 01:08
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A non degenerate (=rank $4$) complex quadratic form $q(x,y,z,t)$ defines a smooth quadric $Q=V(q)=\{q=0\}\subset \mathbb P^3(\mathbb C)$.
This quadric is a complex manifold of dimension $2$ and thus a compact differential manifold of real dimension $4$.
An elementary but extremely beautiful theorem in algebraic geometry states that every such quadric $Q$ is algebraically isomorphic to $\mathbb P^1(\mathbb C)\times \mathbb P^1(\mathbb C)$ and so a fortiori diffeomorphic to the product $S^2\times S^2$ of two spheres.

A completely explicit example is given by the quadratic form $q(x,y,z,t)=xt-yz$.
An isomorphism of algebraic varieties between $\mathbb P^1(\mathbb C)\times \mathbb P^1(\mathbb C)$ and the quadric $Q$ given by the equation $xt-yz=0$ is supplied by the so-called Segre map:

$$ \mathbb P^1(\mathbb C)\times \mathbb P^1(\mathbb C)\stackrel {\cong}{\to} Q:([t:u],[v:w]) \mapsto [x:y:z:t]=[tv:tw:uv:uw] $$

Georges Elencwajg
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Let $M$ and $N$ be homeomorphic, simply connected, smooth 4-manifolds. (They need not be diffeomorphic - that's why this is interesting.) Then there is some $k$ such that $M \; \sharp \; k(S^2 \times S^2) \cong N \; \sharp \; k(S^2 \times S^2);$ it's said that $M$ and $N$ are stably diffeomorphic. This theorem is due to Wall (it's the result of that paper + the work of Freedman: h-cobordant 4-manifolds are homeomorphic).

This tells us that if we want to find an invariant that can tell apart non-diffeomorphic smooth structures on a 4-manifold, it had better not play nicely with connected sum!

  • "it had better not play nicely with connected sum" -- sorry, I don't quite understand the wording, isn't there a typo here? I do understand what you mean, but you may want to edit the answer for a clearer wording. – Alexander Gelbukh Mar 07 '15 at 01:12
  • @Alexander No typo. Call your invariant $D$. Because of the above, if you have some formula relating $D(M\sharp N)$ to $D(M)$ and $D(N)$ - say, $D(M \sharp N) = D(M) +D(N)$ - then $D$ could not distinguish between homeomorphic, but not diffeomorphic, smooth manifolds. –  Mar 07 '15 at 01:14
  • Thank you! I do understand the meaning, I only meant grammar/style: "you'd better not play with connected sum" or "it does not play nicely with connected sum" or so. OK, I guess it's my non-native English... – Alexander Gelbukh Mar 07 '15 at 01:17