# The Unapologetic Mathematician

## The Dimension of a Manifold

One thing that may seem clear at first blush about manifolds actually takes a little thought to be sure about. Specifically, our definition says that each point $p\in M$ is contained in some coordinate patch $U$, which comes with a coordinate map $\phi:U\to\mathbb{R}^n$. But does the $n$ here have to be the same for all points $p$?

Well, let’s start by considering what happens if two coordinate patches intersect each other. Let $U_1\subseteq M$ and $U_2\subseteq M$ be open subsets with $U_1\cap U_2$ not empty, and with coordinate maps $\phi_1:U_1\to\mathbb{R}^{n_1}$ and $\phi_2:U_2\to\mathbb{R}^{n_2}$. Now we can restrict our coordinate patches to the intersection, getting $\left(U_1\cap U_2,\phi_1\vert_{U_1\cap U_2}\right)$ and $\left(U_1\cap U_2,\phi_2\vert_{U_1\cap U_2}\right)$. These are two different coordinate patches with two different coordinate maps on the same open subset $U_1\cap U_2\subseteq M$.

So what? So now we can use $\phi_2$ in reverse to lift up from $\phi_2(U_1\cap U_2)\subseteq\mathbb{R}^{n_2}$ into $U_1\cap U_2\subseteq M$, and then use $\phi_1$ forward to drop down into $\phi_1(U_1\cap U_2)\subseteq\mathbb{R}^{n_1}$. This composition $\phi_1\circ\phi_2^{-1}$ is thus a homeomorphism from an open region in $\mathbb{R}^{n_2}$ to an open region in $\mathbb{R}^{n_1}$. And this is absolutely impossible unless $n_1=n_2$.

So now we know that any two coordinate patches that intersect must use the same value of $n$. Does this mean that we always have to use the same value of $n$? Well, not quite.

Take two distinct natural numbers $m$ and $n$. For each one, we can come up with all the coordinate patches with that dimension, and take their unions $U_m$ and $U_n$. Since the union of any collection of open sets is open, each of these sets must be open. But they can’t intersect, or else some coordinate patch with dimension $m$ and some other with dimension $n$ would have to intersect, and we just saw that they can’t.

The only way this is possible is for $U_m$ and $U_n$ to live in different connected components of $M$. And, indeed, our definition so far doesn’t rule out this possibility. We could have a two-dimensional sphere and a one-dimensional circle floating next to each other, never touching, and they would count as a manifold according to what we’ve said so far.

There are two ways around this. One is to only ever talk about connected manifolds, which automatically have a unique dimension since they only have one connected component. However, this imposes restrictions, like making it difficult to take intersections of manifolds and have the result still be a manifold, since it may suddenly be disconnected. The other alternative, which we will use, is simply to assert that all connected components of a manifold must have the same dimension.

Either way, we can specify this dimension by saying that $M$ is an $n$-dimensional manifold, or an $n$-manifold for short.

February 24, 2011 - Posted by | Differential Topology, Topology

1. […] of Manifolds As we discussed the dimension of a manifold yesterday, we passed by an interesting construction that we want to look at in more […]

Pingback by Classes of Manifolds « The Unapologetic Mathematician | February 25, 2011 | Reply

2. “…simply to assert that all connected components of a manifold must have the same dimension.”
should this be “…disconnected components of of a manifold must have the same dimension.” since connected components will have the same
dimension as argued above.

Comment by EE | February 27, 2011 | Reply

3. Sorry if that’s not clear: as the post on connected spaces discusses, we can separate any topological space into its “connected components”, which are maximal connected subsets. As you said, each one of those connected components must have the same dimension in each chart, but in principle each connected component could have a different dimension.

Really it’s just a terminological convention: the “connected” in “connected components” means that each component is a (maximal) connected subspace, not that they are connected to each other.

Comment by John Armstrong | February 27, 2011 | Reply

4. How can we prove the dimension of a manifold is unique using the definition and implicit function theorem?

Comment by Jemal | December 22, 2017 | Reply

• That sounds a bit homeworky, but okay. What does the definition tell you, and what does the implicit function theorem state?

Comment by John Armstrong | December 22, 2017 | Reply