It is known that for a topological space with different metrics, the Hausdorff dimensions may not be equal in general. 


For the case of manifolds, suppose $M$ is a $n$-manifold with a metric(distance), from <https://math.stackexchange.com/questions/931628/hausdorf-dimension-of-a-manifold-of-dimension-n>, we know that $dim\_H M$ and $n$ may not be the same. But whether we can get a relation between $\dim\_H M$ and $n$? I mean, whether we can prove that $\dim\_H M\leq n$ or $\dim\_H M\geq n$? 


Unfortunately, I have no idea how we can get it directly from the definition of Hausdorff dimension.



The relation between Hausdorff dimension of an $n$-manifold and $n$
In a metrizable topological space, Hausdorff dimension is always larger or equal than the topological (covering) dimension. See Theorem 6.3.10 in Edgar's book "Measure, topology and fractal geometry". In particular, for an $n$-dimensional manifold $M$, if $\rho$ is any metric compatible with the Euclidean topology, then $(M,\rho)$ has Hausdorff dimension 
at least $n$.


Because Hausdorff dimension is always at least the topological dimension, and both agree for "nice" spaces such as manifolds, Mandelbrot tentatively defined a fractal to be a metric space whose Hausdorff dimension is strictly larger than its similarity dimension, although this definition is not widely accepted today (the consensus is that you can't really define fractal).


Finally, for any (at least separable) metric space $X$, the topological dimension equals the infimum (which is in fact a minimum) of the Hausdorff dimensions of $(X,\rho)$ where $\rho$ varies among the metrics compatible with the topology of $X$. This is a classical result due to Edward Marczewski. 


