Usually one regards manifolds up to dimension 4 as a part of low dimensional topology. There are plenty of various results which work only in low dimensional topology; especially in dimension 4. However still there are phenomena which occur only up from certain dimensions above 4: For example the famous result of Milnor, which states that each $PL$ manifold of dimension $n$ is in fact smooth provided that $n \leq 7$. My question is the following: Could you give an example of the (reasonable) theorem of the type "each manifold of dimension $n$ have some property $P$ provided that $n \leq K$ (and for $n>K$ there are counterexamples)," where $K$ is some large number?
[Math] (Very) High dimensional manifolds
at.algebraic-topologydg.differential-geometrymanifoldssoft-question
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$\mathbb RP^3$ 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.
Consider complex conjugation $\mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$. Perturb this map by taking an affine chart $\mathbb{C}^2$ and setting $f(z) = \rho(|z|)\cdot \bar{z}$, where $\rho$ is a smooth map to $\mathbb{R}_{\geq 0}$ that is 0 for $|z| < 1$, and 1 for $|z| > 2$, say. The map $f\colon \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2$ is orientation-preserving and its effect on $H^2(\mathbb{C}\mathbb{P}^2)$ is that of negative the identity. Now, take two copies of $\mathbb{C}\mathbb{P}^2$, cut out small neighborhoods of the balls in the corresponding affine charts where $\rho$ makes the map $f$ constant, and glue the resulting manifolds together to obtain $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ along with an orientation-preserving diffeomorphism $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2 \to \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ whose effect on second cohomology is given by $(\begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix})$. Repeat this procedure on $\mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ and $\mathbb{C}\mathbb{P}^2$ to obtain an orientation-preserving diffeomorphism $g \colon (\mathbb{C}\mathbb{P}^2)^{\#3} \to (\mathbb{C}\mathbb{P}^2)^{\#3}$ whose induced map on second cohomology is $$\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$ Now consider the mapping torus $M$ of $g$. As can be seen from the long exact sequence in homotopy for the fibration $(\mathbb{C}\mathbb{P}^2)^{\#3} \to M \to S^1$, the rational homotopy $\pi_*(M) \otimes \mathbb{Q}$ has infinite rank. From the long exact sequence in relative cohomology for the inclusion $(\mathbb{C}\mathbb{P}^2)^{\#3} \hookrightarrow M$ we see that $H^2(M, \mathbb{Q}) = 0$. Therefore $M$ has Betti numbers $b_1 = 1, b_2 = 0, b_3 = 0, b_4 = 1$, and so by Theorem 8 in Kotschick's ``On products of harmonic forms'' (https://arxiv.org/abs/math/0004009) it admits a formal metric.
Note that the manifold $M$ has the rational cohomology ring of $S^1 \times S^4$ and so, since geometric formality implies formality, the minimal model of $M$ is given by $$\Lambda(x_1, x_4, y_7; \ \ dx_1 = dx_4 = 0, \ dy_7 = x_4^2),$$ where subscripts denote degrees. It would then seem from this model that $M$ is in fact rationally elliptic. However, $M$ is not a nilpotent space (consider the action of $\pi_1$ on $\pi_2$), and so the ranks of the rational homotopy groups cannot be read off immediately from the amount of generators in the model. Also note that "rationally elliptic" is a concept that, at least in Félix, Halperin, Thomas, is defined only for simply connected spaces. So the above example doesn't truly answer your question.
Best Answer
The smallest example of a manifold that is homotopy equivalent to a topological group, but not rationally equivalent to a Lie group has dimension 1254.