The Unapologetic Mathematician

Mathematics for the interested outsider

Partitions of Unity (proof)

Finally we can prove what we’ve asserted: given any open cover \{U_\alpha\} of a smooth manifold M we can find a countable smooth partition of unity \{\phi_k\} subordinate to it.

So, as we’ve seen we can find a countable atlas (V_k,x_k); we use x_k for the coordinate maps since we’ll want the \phi_k free. We’ve also seen that we have a smooth bump function \phi between the two cubes C_1(0) and C_2(0) in \mathbb{R}^n. So let’s define

\displaystyle\theta_k(p)=\left\{\begin{array}{lc}\phi\left(x_k(p)\right)&p\in V_k\\{0}&p\notin V_k\end{array}\right.

Now, it’s easily verified that the furthest points from the origin in C_2(0) are the corners, each of which is 2\sqrt{n} away. Thus we can tell that C_2(0)\subseteq B_{3n}(0), and the support of \phi is contained entirely within x_k(V_k)=B_{3n}(0). This means that the support of \theta_k is entirely contained within V_k — by the time we get to the edge it’s already smoothly tailed off to zero, and so even though we define it piecewise, \theta_k is a smooth function defined on all of M.

Now set up the sum


Since V_k is locally finite and \theta_k is supported within V_k, this sum is guaranteed to be finite at each point, which makes \theta a smooth function on all of M. The ball B_1(0) is contained within the cube C_1(0), so \phi takes the constant value 1 on this ball. Since the preimages x_k^{-1}(B_1(0)) form an open cover of M, there is always at least one k for which \theta_k=1. In particular, for which it’s not zero, and thus the whole sum is nonzero at p.

Since \theta is never zero, we can divide by it. We define \phi_k(p)=\theta_k(p)/\theta(p). These are smooth functions on all of M, and their sum is everywhere exactly 1. Thus the \phi_k form a partition of unity subordinate to \{V_k\}. And since \{V_k\} refines \{U_\alpha\}, the partition is subordinate to this cover as well.

March 14, 2011 Posted by | Differential Topology, Topology | 1 Comment