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What are the known examples of closed, connected and simply connected manifolds of positive Euler characteristic in dimension $n$?

Is there any complete list? I think they are rare. The examples that I know are $\Bbb S^{2n}$, $\mathbb{CP}^n$, $\mathbb{HP}^n$ and $\mathbb{CaP}^2$.

Qiaochu Yuan
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C.F.G
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2 Answers2

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Some points.

  1. By Poincare duality the Euler characteristic of a closed odd-dimensional manifold vanishes so we restrict our attention to even-dimensional manifolds. By the classification of surfaces $S^2$ is the only example in dimension $2$.

  2. Euler characteristic is multiplicative with respect to products (e.g. by the Kunneth theorem). So we can find examples of manifolds with positive Euler characteristic by taking products of an even number of manifolds of negative Euler characteristic together with any number of manifolds of positive Euler characteristic. The Euler characteristic is also multiplicative with respect to (nice) fiber bundles so we can consider nontrivial fiber bundles with suitable bases and fibers also.

  3. The Euler characteristic of a connected sum of closed $n$-manifolds satisfies the "inclusion-exclusion" formula $\chi(M\#N) = \chi(M) + \chi(N) - \chi(S^n)$. This means when $n$ is even, connected sum with $N$ increases the Euler characteristic iff $\chi(N) \ge 3$. Also the connected sum of simply connected manifolds is simply connected.

  4. Any manifold whose cohomology is concentrated in even degrees (say, over $\mathbb{Q}$) has positive Euler characteristic, and there's a large source of examples of such manifolds coming from algebraic geometry: every generalized flag variety $G/P$ over $\mathbb{C}$ has this property. $\mathbb{CP}^n$ is a special case of this construction but we also have Grassmannians and complete flag varieties, for example, which are also simply connected (probably generalized flag varieties over $\mathbb{C}$ are always simply connected but I don't know how to prove it).

  5. By Poincare duality, any closed simply connected $4$-manifold has cohomology concentrated in even degrees and hence has Euler characteristic at least $2$. This blog post on hypersurfaces in $\mathbb{CP}^3$ discusses their topology; in particular they are completely classified up to homotopy (Milnor, Whitehead) and even up to homeomorphism (Freedman). Their Euler characteristics can be arbitrarily large: the linked post shows that the Euler characteristic of a smooth degree $d$ hypersurface in $\mathbb{CP}^3$ is $d^3 - 4d^2 + 6d$. When $d = 1$ we get $\mathbb{CP}^2$ which has $\chi = 1 - 4 + 6 = 3$, when $d = 2$ we get $\mathbb{CP}^1 \times \mathbb{CP}^1$ which has $\chi = 8 - 16 + 12 = 4$, and when $d = 4$ we get a K3 surface which has $\chi = 64 - 64 + 24 = 24$.

  6. For the classification of closed simply connected $6$-manifolds see here. Wall showed that any such manifold splits as a connected sum of a number of copies of $S^3 \times S^3$ ($\chi(S^3 \times S^3) = 0$ so this connected sum lowers the Euler characteristic by $2$ and removing it increases the Euler characteristic by $2$) and a manifold $M$ with $b_3 = 0$, hence $M$ has rational cohomology concentrated in even degrees and so has positive Euler characteristic. Incidentally, this makes a connected sum of two copies of $S^3 \times S^3$ the simplest example of a closed simply connected manifold with negative Euler characteristic, so now we can actually carry out the construction I suggested in point 2: the product of two such sums is a closed simply connected $12$-manifold with positive Euler characteristic whose rational cohomology is not concentrated in even degree.

  7. The Hopf conjecture claims in part that a closed even-dimensional manifold admitting a metric with positive sectional curvature has positive Euler characteristic. In dimension $2$ this of course follows from the Gauss-Bonnet theorem, and as Wikipedia discusses this also holds in dimension $4$.

  8. If I were looking for more examples I might look through lists of symmetric spaces.

Qiaochu Yuan
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  • Thanks for a lot of information. I like that part "The Euler characteristic is also multiplicative with respect to (nice) fiber bundles". How much nice? and do you know any example of $\chi(N) \ge 3$ with given conditions above except product of spheres? – C.F.G Sep 07 '20 at 08:19
  • I think you posted your answer before my second edit? I've added simply connectedness condition. – C.F.G Sep 07 '20 at 08:24
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    Yes, I saw, I've edited. It should suffice to have a fiber bundle where the base, fiber, and total space are all closed manifolds. I also edited with a comment about simply connected closed $4$-manifolds with large Euler characteristic. – Qiaochu Yuan Sep 07 '20 at 08:27
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    It is worth to say, in dim 4, in regard to the part 7 of your list, is also correct due to Gauss-Bonnet-Chern theorem. (has been done by Milnor in a conversation and published by Chern.) – C.F.G Sep 07 '20 at 08:45
  • All of symmetric spaces (even-dim) have positive Euler characteristics? – C.F.G Sep 07 '20 at 09:11
  • Is "multiplicative with respect to (nice) fiber bundles" a consequence of multiplicativity of Euler characteristic? – C.F.G Sep 07 '20 at 09:14
  • @C.F.G: 1) I don't think so, but a lot of them seem to. 2) You mean is it a consequence of multiplicativity with respect to products? Maybe if you break up the base carefully enough (and use inclusion-exclusion a lot), but I think it might be cleaner to see it as a consequence of the existence of the Serre spectral sequence. – Qiaochu Yuan Sep 07 '20 at 09:17
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    Thanks again. Your comprehensive answer changed and corrected many of my misunderstandings. – C.F.G Sep 07 '20 at 09:28
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    (Re: Is "multiplicative with respect to (nice) fiber bundles" a consequence…) Yes, just take any CW-structure of the base; the bundle is trivial on each cell so everthing can be lifted to a CW-structure of the total space etc. That's an elementary fact, much easier than spectral sequences. – Grigory M Sep 07 '20 at 12:02
  • @Grigory: thanks, that’s very straightforward. Is it entirely clear that all manifolds are CW complexes? I seem to recall there’s some delicacy around this. – Qiaochu Yuan Sep 07 '20 at 12:15
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    All smooth manifolds (at least compact) are, but it is tricky. You use a handle decomposition which gives you a homotopy equivalent CW complex and argue that this can be refined to actually be homeomorphic to your manifold. In dimensions greater than 4, you can take a topological handle decomposition and at least for compact manifolds you should be able to use topological analogues to argue the same. I believe it is unknown if 4 dimensional topological manifolds are CW complexes. – Connor Malin Sep 07 '20 at 15:08
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    To amplify what @Connor said, see https://math.stackexchange.com/questions/593041/cw-complexes-and-manifolds and the following links. – Jason DeVito - on hiatus Sep 07 '20 at 15:11
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    @QiaochuYuan Well, proving that all manifolds are CW isn't easy, indeed. But any space is CW up to (weak) homotopy equivalence — and that's enough for the argument, no? – Grigory M Sep 13 '20 at 22:24
  • @Grigory: so we replace the base, fiber, and bundle with CW complexes weakly homotopy equivalent to them, compatibly with the fiber map up to (weak?) homotopy? I definitely believe that this is possible but I don't know how to prove this either :P – Qiaochu Yuan Sep 13 '20 at 22:25
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    @QiaochuYuan OK, you win ) I mean, I think I can explain how to find this CW-model — but the argument no longer looks that simple. So let me instead quote my favourite AT textbook (tbc) – Grigory M Sep 13 '20 at 22:32
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    …«Among the natural computational tools used by homology theory, the most efficient ones are delivered by the topology of smooth manifolds, and we cannot help considering this subject. However, the foundations of the theory of manifolds, rooted in geometry and analysis, require a thick volume by themselves. (…) Our approach is based on the idea that the most effective way of treating these difficulties is ignoring them» // the last part is omitted in the English version — so it's given in my clumsy translation – Grigory M Sep 13 '20 at 22:41
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Qiaochu has given a very nice answer, but I wanted to add another large class of examples.

Suppose $G$ is a simply connected closed Lie group, and $H$ is a connected closed Lie group. Suppose $H$ acts on $G$ freely via some action, and suppose further that the rank of $H$ is equal to the rank of $G$. Then the orbit space $G/H$ is a closed, simply connected manifold of positive Euler characteristic.

One way to construct such actions is as follows. Beginning with any $G$ as above, let $H$ be a connected subgroup of $G$ which contains a maximal torus of $G$. Then $H$ acts on $G$ by left multiplication, and this meets all the hypothesis above. This gives rise to the so-called homogeneous spaces.

Another way to construct such actions is to allow $H\subseteq G\times G$. Then $H$ acts on $G$ via $(h_1,h_2)\ast g = h_1 g h_2^{-1}$. When this action is free, this gives rise to the so-called biquotients.

The theorem above doesn't require that $H$ acts on $G$ using the multiplicative structure on $G$, but I don't know of any examples that are not of this type.

The proof of the theorem is as follows. First, the fact that the orbit space under a free action by a compact Lie group is a manifold is well known. So let me focus on showing that the quotient is simply connected and that it has positive Euler characteristic.

Since $H$ acts on $G$ freely, there is a principal $H$-bundle $H\rightarrow G\rightarrow G/H$. The long exact sequence in homotopy groups associated to this ends with $$...\rightarrow\pi_1(H)\rightarrow \pi_1(G)\rightarrow \pi_1(G/H)\rightarrow \pi_0(H)\rightarrow ...$$

By assumption, $\pi_1(G)\cong\pi_0(H)\cong 0$, so it follows that $\pi_1(G/H) = 0$.

Lastly, the fun part. Why does $G/H$ have positive Euler characteristic? Well, all Lie groups have the rational homotopy groups of a product of odd spheres. Specifically, $\pi_{even}(G)\otimes \mathbb{Q} = 0$ and $\dim \pi_{odd}(G)\otimes \mathbb{Q} = \operatorname{rank}(G)$.

From the long exact sequence in rational homotopy groups associated to the bundle $H\rightarrow G\rightarrow G/H$, it follows that $G/H$ is rationally elliptic. Further, from the same exact sequence together it follows that $\dim \pi_{even}(G/H)\otimes \mathbb{Q} = \dim \pi_{odd}(G/H)\otimes\mathbb{Q}$.

For rationally elliptic spaces, this condition on rational homotopy groups forces $\chi(G/H) > 0$. See, for example, Felix, Halperin, and Thomas's book "Rational Homotopy Theory", specifically in Part VI (section 32).

Jason DeVito - on hiatus
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    By the way - in the boxed Theorem, $G$ can replaced by any 1-connected closed smooth manifold $M$ for which $\dim \oplus_k \pi_k(M)\otimes \mathbb{Q} < \infty$ and for which $\dim \pi_{odd}(M)\otimes \mathbb{Q} - \pi_{even}(M)\otimes \mathbb{Q} = \operatorname{rank}(H).$ – Jason DeVito - on hiatus Sep 07 '20 at 14:19
  • +1. Thanks. BTW is it true that most of biquotients don't admit metric of positive curvature? You have two papers about classification of biquotients of dim 4 , 5 and 6 , 7. Are they admit metric of positive curvature? As far as I know the only successful biquotients are Bazaikin spaces and Eschenburg of dim 7 and 13. Yes? – C.F.G Sep 07 '20 at 14:36
  • I would say "don't admit metrics of positive curvature" is too strong of a statement - rather, we don't know if they do, and strongly suspect they don't. Other than Bazaikin spaces and Eschenburg spaces, there is just one other known inhomogeneous biquotient of positive curvature - Eschenburg's inhomogeneous $6$-dimensional flag manifold. (It's a circle quotient of some of the Eschenburg spaces.) – Jason DeVito - on hiatus Sep 07 '20 at 15:06