Let $(M,g)$ be a (say closed) Riemannian manifold. One can try to understand the geometry/topology of $(M,g)$ by studying the eigenvalues of the Laplacian (this I guess has two versions: when considering the Laplacian on functions only, or on differential forms. feel free to answer about whichever). 


Sometimes (for example for nice bounded planar domains) the topology of the manifold is completely determined by the spectrum of the Laplacian (on functions). 


Also, by the Hodge theorem, for an arbitrary closed Riemannian manifold all of the Betti numbers can be read off the spectrum of the Laplacian (on forms). 
Sadly, it is well known that there are examples of isospectral but not isometric Riemannian manifolds.


Q1: Are there examples of pairs of Riemannian manifolds which are isospectral but not diffeomorphic/homeomorphic/homotopy equivalent? If so, what is the simplest one? 


assuming the answer to Q1 is "Yes" or "It is not known":


Q2: What are the best positive results in this direction? How much can I know about the topology of $M$ only from my knowledge of the spectrum of the Laplacian?


I'm sorry if the question is too vague, but still any information will be nice.



Can one hear the (topological) shape of a drum?
There are examples due to Ikeda of [isospectral Lens spaces which are not homotopy equivalent](http://www.numdam.org/item/ASENS_1980_4_13_3_303_0). 


Likeliest the simplest examples are the compact connected 3-dimensional flat manifolds which are [a *tetracosm* and *didicosm*](https://msp.org/gt/2004/8-3/p07.xhtml). These are isospectral but not homotopy equivalent. I'm not sure if they are known to have the same spectrum for the [Laplace-de Rham operator](https://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator#Laplace%E2%80%93de_Rham_operator) (i.e. on forms). Presumably this could be computed explicitly. 


There are many other examples of non-isometric manifolds which are strongly isospectral (having the same spectrum on forms) using Sunada's method, and arithmetic techniques (originally due to [Vignéras](https://mathscinet.ams.org/mathscinet-getitem?mr=584073)) for locally symmetric spaces. Coupled with the Mostow rigidity theorem for locally symmetric spaces, this implies that these manifolds are not homotopy equivalent. 


In 1-dimension, the spectrum determines the manifold. 


For connected two-dimensional surfaces, the Laplace-de Rham spectrum determines the betti numbers, and hence the topological type (the surface $S$ is orientable iff $b\_2(S)=1$, and then $b\_1(S)$ determines the topological type if it is orientable or non-orientable). (For surfaces with boundary, an example of [Bérard-Webb](https://mathscinet.ams.org/mathscinet-getitem?mr=1322332) shows that the Neumann spectrum does not determine orientability.) Thus, the 3-dimensional examples coming from Sunada's construction or Vignéras' are minimal dimensional examples which are strongly isospectral. I don't know if the Laplace spectrum determines the topological type of closed surfaces. 


See section 5 of the [survey paper by Carolyn Gordon](https://mathscinet.ams.org/mathscinet-getitem?mr=1500137) for more information on how topology cannot be detected by the spectrum and references. Note also that [Lubotzky-Samuels-Vishne](https://mathscinet.ams.org/mathscinet-getitem?mr=2267287) showed that there are isospectral arithmetic lattices which are not commensurable (so not obtained by Sunada's method or Vigneras' criterion). 


