# Are $n$-dimensional manifolds really $n$-dimensional?

I was studying physics, and became curious about coordinate systems of curved space.

The question started from whether it is possible to describe each point on any surface with two parameters.

So I started studying topology and I found that the definition of dimensions for a manifold is different from the intuitive definition.

Intuitively, an $$n$$-dimensional object is an object that requires $$n$$ parameters to specify any point within it.

An $$n$$-dimensional manifold is a topological space that is locally homeomorphic to an open subset of $$\mathbb{R}^n$$.

So I wonder if those dimensions are the same:

Can every $$n$$-manifold have a coordinate system of $$n$$ parameters?

Mathematically, for any $$n$$-manifold $$M$$, does there exist a bijective map $$f: U \to M$$ where $$U \subset \mathbb{R}^n$$?

I know that we can patch the whole manifold with individual charts of $$n$$ parameters, but this doesn't form an $$n$$-dimensional coordinate system because we need to know the chart to which the point of interest belongs. This would make an $$n+1$$-dimensional(or higher) coordinate system.

Some examples where the proposition holds true:

• 2-Sphere: Use $$(0,100)$$ and $$(0,-100)$$ for polar points and use longitudes and latitudes for other points.
• 2-Torus: Poloidal and toroidal angles.
• Mobius strip: Keep the usual coordination system for the rectangle before twisting, but make one boundary open so that it can be connected.

I am very new to this subject; thanks for any help!

• First you should specify that $U$ is (non-empty and) open. Now regarding whether such a map exists: Bijective? Yes for simple cardinality reasons. Homeomorphism? Not necessarily, for example there is no single homeomorphism between a sphere $S^n$, and an open subset of $\Bbb{R}^n$, though you can have multiple local homeomorphisms which cover the sphere. Commented Jan 8 at 10:42
• Also, if you want to go the intuitive route, there is no need to specify bijective. No one said points needed only a single coordinate to them. Only that the coordinate system you choose, if it is to be reasonable, must have $n$ entries to specify a point. It can certainly overlap itself, and there is nothing mathematically wrong with specifying coordinates on the Earth with longitudes way above $360^\circ$. Commented Jan 8 at 10:44
• Arguably, the whole reason mathematicians arrived at the concept of a manifold is because substantial amounts of algebra, analysis, and geometry can be developed "independently of coordinates" but in most examples of interest there is not a single (global) coordinate system. In other words, "most $n$-dimensional manifolds are not open sets in $n$-dimensional Cartesian space." <> Is that what you're asking? Commented Jan 8 at 15:46

For any $$n$$-manifold $$M$$, does there exist a bijective map $$f: U \to M$$ where $$U \subset \mathbb{R}^n$$?

Yes if manifolds are as usual required to be second countable.

Using second countability, it is easy to see that $$M$$ can be covered by countably many open $$U_n \subset M$$ which are homeomorphic to open subsets $$V_n \subset \mathbb R^n$$. This shows that the cardinality of the set $$M$$ agrees with the cardinality of $$\mathbb R^n$$ (and thus of $$\mathbb R$$).

This shows that there exists a bijection $$f : \mathbb R^n \to M$$. This $$f$$ can be compeletely erratic and far from being continuuous, so it will not be what you want.

However, we can assume w.lo.g. that each $$V_n$$ is contained in $$U_1(x_n) =$$ open ball with center $$x_n = (3n, 0, \ldots,0)$$ and radius $$1$$ (take a homeomorphism $$\phi_n : \mathbb R^n \to U_1(x_n)$$ and replace $$V_n$$ by $$V'_n = \phi_n(V_n) \subset U_1(x_n)$$).

Define recursively subsets $$W_n \subset M$$ by $$W_1 = U_1$$ and $$W_{n+1} = U_{n+1} \setminus W_n$$. In general they will not be open in $$M$$. Then choose homeomorphisms $$h_n : U_n \to V_n$$ and define $$V'_n = h_n(W_n)$$. These sets are pairwise disjoint (with distance $$\ge 1$$) and we can define $$f : U = \bigcup_n V'_n \to M, f \mid_{V'_n}(x) = h_n^{-1}(x) .$$ This produces a continuous bijection. However, $$U$$ is not open and not connected, and $$f$$ is far from being a homeormorphism (except in trivial cases like $$M \subset \mathbb R^n$$ open).

The maps in your question arise by similar constructions. For example, for $$S^2$$ you have the continuous bijection

$$f : [0,2\pi) \times (-\frac \pi 2, \frac \pi 2) \cup \{(0,-\frac \pi 2) \} \cup \{(0,-\frac \pi 2) \} \to S^2$$ which maps $$(longitude, latitude) \in [0,2\pi) \times (-\frac \pi 2, \frac \pi 2)$$ in the obvious way and maps $$(0,\pm\frac \pi 2)$$ to north and south pole.

Whether you like to use this map is a matter of taste. Yes, each point of $$S^2$$ is specified uniquely by two parameter, but the disadvantage is that "distant parameters" are mapped tom "close" points of $$S^2$$. Consider for example $$(0,0)$$ and $$(2\pi - 1/n,0)$$.