The Unapologetic Mathematician

Mathematics for the interested outsider

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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April 26, 2011 - Posted by | 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 The Tangent Space of a Product « The Unapologetic Mathematician | 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 Integrable Distributions Have Integral Submanifolds « The Unapologetic Mathematician | June 30, 2011 | Reply


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