# L-S category versus Betti numbers

Is there a sequence of topological spaces $X_n$ (manifolds ideally), where the sum of the Betti numbers of $X_n$ remains bounded but the Lusternik–Schnirelmann category is unbounded, as $n \to \infty$? What about vice versa?

One might think of both of these numbers as very rough measures of the "complexity" of a space, and it is well known in particular that both quantities are lower bounds on the number of critical points of a Morse function. But it would be nice to hear about any facts governing the relations between them.

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I don't have here an example of X with large SB and small category, but I strongly suspect that the sum of Betti numbers is not sufficient to bound the LS category from below: I think homology is not sufficient an one does need to go and look at the cohomology algebra. Indeed, the classic bound from below is $cat(X)\geq cupl(X)+1$. Also, if I remember well, the free loop space of a sphere has infinite LS category and finite cuplength. –  Pietro Majer Jun 24 '10 at 23:07
actually the example I meant it's quite trivial, for the category is bounded by the dimension, while the BS is unbounded at any dimension (even 1 if you allow disjoint sum) –  Pietro Majer Jun 25 '10 at 21:22

The 2d surfaces have unbounded Betti numbers, but bounded category.

A matrix in $SL_n(\mathbb Z)$ describes a diffeomorphism of the $n$-torus. We may form the mapping torus, a bundle of tori over the circle, with monodromy the matrix. If the matrix is generic, so that none of its exterior powers have eigenvalues that are roots of unity, then the homology of the manifold will be the same as that of $S^1\times S^n$, hence total Betti number $4$. But its universal cover has cup length about $n$, hence large category.

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The dumb answer is infinite projective space, which has zero Betti numbers as classically computed (over $\mathbb{Q}$). But what you really want is very small cohomology and very large L-S category.
Let $X$ be a noncontractible acyclic space. Then all Betti numbers are zero (no matter what coefficients you use), and the same is true for the $n$-fold product $X^n$. But a theorem of Hilton says (as I recall) that an $n$-fold product of noncontractible spaces must have category at least $n$.
On the other hand, if $X$ is simply-connected and $X$ has category $n$, then $X$ must have cohomology in at least $n$ different dimensions.