If we have a smooth map 
and a regular value 
of 
, we know that the preimage 
is a smooth 
-dimensional submanifold. It turns out that we also have a nice decomposition of the tangent space 
for every point 
.
The key observation is that the inclusion 
induces an inclusion of each tangent space by using the derivative 
. The directions in this subspace are those “tangent to” the submanifold 
, and so these are the directions in which doesn’t change, “to first order”. Heuristically, in any direction 
tangent to we can set up a curve 
with that tangent vector which lies entirely within . Along this curve, the value of is constantly , and so the derivative of 
is zero. Since the derivative of in the direction only depends on and not the specific choice of curve , we conclude that 
should be zero.
This still feels a little handwavy. To be more precise, if 
and 
is a smooth function on a neighborhood of , then we calculate
&=\left[\left[f\circ\iota\right]_{*p}(v)\right](\phi)\\&=v\left(\phi\circ f\circ\iota\right)\\&=v\left(\phi(q)\right)\\&=0\end{aligned}](https://s-ssl.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5Bf_%7B%2Ap%7D%28%5Ciota_%7B%2Ap%7D%28v%29%29%5Cright%5D%28%5Cphi%29%26%3D%5Cleft%5B%5Cleft%5Bf%5Ccirc%5Ciota%5Cright%5D_%7B%2Ap%7D%28v%29%5Cright%5D%28%5Cphi%29%5C%5C%26%3Dv%5Cleft%28%5Cphi%5Ccirc+f%5Ccirc%5Ciota%5Cright%29%5C%5C%26%3Dv%5Cleft%28%5Cphi%28q%29%5Cright%29%5C%5C%26%3D0%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\left[f_{*p}(\iota_{*p}(v))\right](\phi)&=\left[\left[f\circ\iota\right]_{*p}(v)\right](\phi)\\&=v\left(\phi\circ f\circ\iota\right)\\&=v\left(\phi(q)\right)\\&=0\end{aligned}")
since any tangent vector applied to a constant function is automatically zero. Thus we conclude that 
. In fact, we can say more. The rank-nullity theorem tells us that the dimension of 
and the dimension of 
add up to the dimension of , which of course is 
. But the assumption that 
is a regular point means that the rank of 
is 
, so the dimension of the kernel is . And this is exactly the dimension of , and thus of its tangent space 
! Since the subspace has the same dimesion as , we conclude that they are in fact equal.
What does this mean? It tells us that not only are the tangent directions to contained in the kernel of the derivative 
, every vector in the kernel is tangent to . Thus we can break down any tangent vector in into a part that goes “along” and a part that goes across it. Unfortunately, this isn’t really canonical, since we don’t have a specific complementary subspace to in mind. Still, it’s a useful framework to keep in mind, reinforcing the idea that near the subspace the manifold 
“looks like” the product of 
(from ) and 
, and we can even pick coordinates that reflect this “decomposition”.
April 26, 2011
Posted by John Armstrong |
Differential Topology, Topology |
2 Comments
