The Unapologetic Mathematician

Tangent Spaces and Regular Values

If we have a smooth map $f:M^m\to N^n$ and a regular value $q\in N$ of $f$, we know that the preimage $f^{-1}(q)=A\subseteq M$ is a smooth $m-n$-dimensional submanifold. It turns out that we also have a nice decomposition of the tangent space $\mathcal{T}_pM$ for every point $p\in A$.

The key observation is that the inclusion $\iota:A\to M$ induces an inclusion of each tangent space by using the derivative $\iota_{*p}(\mathcal{T}_pA)\subseteq\mathcal{T}_pM$. The directions in this subspace are those “tangent to” the submanifold $A$, and so these are the directions in which $f$ doesn’t change, “to first order”. Heuristically, in any direction $v$ tangent to $A$ we can set up a curve $\gamma$ with that tangent vector which lies entirely within $A$. Along this curve, the value of $f$ is constantly $q\in N$, and so the derivative of $f\circ\gamma$ is zero. Since the derivative of $f$ in the direction $v$ only depends on $v$ and not the specific choice of curve $\gamma$, we conclude that $f_{*p}(v)$ should be zero.

This still feels a little handwavy. To be more precise, if $v\in\mathcal{T}_pA$ and $\phi$ is a smooth function on a neighborhood of $q\in N$, then we calculate

$\displaystyle\begin{aligned}\left[f_{*p}(\iota_{*p}(v))\right](\phi)&=\left[\left[f\circ\iota\right]_{*p}(v)\right](\phi)\\&=v\left(\phi\circ f\circ\iota\right)\\&=v\left(\phi(q)\right)\\&=0\end{aligned}$

since any tangent vector applied to a constant function is automatically zero. Thus we conclude that $\iota_{*p}(\mathcal{T}_pA)\subseteq\mathrm{Ker}(f_{*p})$. In fact, we can say more. The rank-nullity theorem tells us that the dimension of $\mathrm{Ker}(f_{*p})$ and the dimension of $\mathrm{Im}(f_{*p})$ add up to the dimension of $\mathcal{T}_pM$, which of course is $m$. But the assumption that $p$ is a regular point means that the rank of $f_{*p}$ is $n=\dim(N)$, so the dimension of the kernel is $m-n$. And this is exactly the dimension of $A$, and thus of its tangent space $\mathcal{T}_pA$! Since the subspace $\mathcal{T}_pA$ has the same dimesion as $\mathrm{Ker}(f_{*p})$, we conclude that they are in fact equal.

What does this mean? It tells us that not only are the tangent directions to $A$ contained in the kernel of the derivative $f_*$, every vector in the kernel is tangent to $A$. Thus we can break down any tangent vector in $\mathcal{T}_pM$ into a part that goes “along” $A$ and a part that goes across it. Unfortunately, this isn’t really canonical, since we don’t have a specific complementary subspace to $\mathcal{T}_pA$ in mind. Still, it’s a useful framework to keep in mind, reinforcing the idea that near the subspace $A$ the manifold $M$ “looks like” the product of $\mathbb{R}^{m-n}$ (from $A$) and $\mathbb{R}^n$, and we can even pick coordinates that reflect this “decomposition”.

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Posted by John Armstrong | Differential Topology, Topology

2 Comments	»

1. [...] diffeomorphism is . The tangent space at a point on the submanifold is mapped by to , and the kernel of this map is exactly the image of the inclusion . The same statements hold with and swapped appropriately, [...]

Pingback by | April 27, 2011 | Reply

2. [...] further! Let , where is the complementary projection to . This map has maximal rank everywhere, so we know that for each the preimage is an -dimensional submanifold . The tangent space to consists of [...]

Pingback by | June 30, 2011 | Reply

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