I'm currently reading about submersions and immersion's Lee's Introduction to Smooth Manifolds (p.77), and I'm slightly confused about what is meant when he says that the rank of $F$ and $p$ is "the rank of the Jacobian matrix of $F$ in any smooth chart."

Letting $(U,\varphi)$ and $(V,\psi)$ be the local charts at $p$ and $F(p)$, respectively, I was wondering if the Jacobian matrix that was mentioned referred to the Jacobian matrix of $\psi\circ f\circ\varphi^{-1}$. This matrix seems to make sense because $\psi\circ f\circ\varphi^{-1}$ maps between Euclidean spaces, but I was wondering if the "Jacobian matrix of $F$" could be defined in a way that is independent of charts (if this makes any sense).


Given your map $f: M \rightarrow N$ (where $\dim M = m$ and $\dim N = n$), the rank of the Jacobian matrix at a point $p\in M$ is independent of your choice of charts, which is why Lee defines it to be the rank of the differential $df_p : T_pM \rightarrow T_{f(p)}N$, a linear map that does not exist as a matrix until you assign a basis on the tangent spaces. In practice, actually computing the rank of $df_p$ requires you to assign a coordinate chart (over the sets $U \subset M$ and $V \subset N$ let's say) as you've indicated.

The reason rank is independent of the chart chosen is because $\phi : U \rightarrow \mathbb{R}^m$ and $\psi : V \rightarrow \mathbb{R}^n$ are diffeomorphisms, and hence their differentials are isomorphisms on the level of tangent spaces. You should convince yourself then that for purely linear algebraic reasons: $$\text{rank }df_p = \text{rank } d\psi\circ df_p \circ d\phi^{-1} = \text{rank } d(\psi\circ f \circ \phi^{-1})_p.$$ I hope that makes it a little clearer for you!


He defined the rank of $F$ at $p$ to be the rank of the linear transformation $dF_p:T_pM \to T_{F(p)}N$ (where the rank of a linear transformation is say defined as the dimension of the image). Next, what he's claiming is that for every chart $(U,\phi)$ about $p$ and $(V,\psi)$ about $F(p)$, we can consider the linear transformation $D(\psi\circ F \circ \phi^{-1})_{\phi(p)}:\Bbb{R}^{\dim M} \to \Bbb{R}^{\dim N}$ (this is the usual derivative at a point for a map between open subsets of a finite-dimensional normed vector space). THe claim is then that \begin{align} \text{rank} \left(dF_p\right) &= \text{rank} \left( D(\psi\circ F \circ \phi^{-1})_{\phi(p)}\right) \\ &= \text{rank} \left( (\psi\circ F \circ \phi^{-1})'(\phi(p))\right) \end{align} where $(\psi\circ F \circ \phi^{-1})'(\phi(p))$ is the matrix representation of the linear transformation $D(\psi\circ F \circ \phi^{-1})_{\phi(p)}$ with respect to the standard ordered bases for $\Bbb{R}^{\dim M}$ and $\Bbb{R}^{\dim N}$, i.e the Jacobian matrix in a chart (the second equality about rank of linear transformation vs rank of matrix representation being equal is a standard linear algebra fact).

The notion of the differential/push-forward/tangent-mapping $dF_p:T_pM \to T_{F(p)}N$ is exactly meant to generalize the concept of "Jacobian matrix" (actually even for maps between open subsets of Euclidean spaces, we can consider the derivative at a point as a linear transformation, and then the Jacobian matrix is just the matrix representation of this linear map relative to the standard bases). See the remarks made on page 62-63.

  • $\begingroup$ Thanks! I was also wondering whether the matrix representation of the linear map $dF_p$ would involve a basis consisting of derivations from the tangent space $T_p M$, which would be similar to the directional/partial derivatives in a typical Jacobian matrix. Is this also true? $\endgroup$ – Chen Aug 1 '20 at 2:08
  • $\begingroup$ @Chen To calculate a matrix representation for $dF_p$, you can choose any basis on the domain and target vector spaces. Matrix representations are not unique and will depend on which basis you choose. But, if you choose the chart-induced basis vectors: i.e if $\phi=(x^1,\dots,x^m)$ and you choose the basis $\frac{\partial}{\partial x^i}(p)$ on $T_pM$ and $\psi=(y^1,\dots,y^n)$ with the basis $\frac{\partial}{\partial y^j}(F(p))$ on $T_{F(p)}N$, then the matrix representation of $dF_p$ will be exactly the matrix representation of $D(\psi\circ F\circ\phi^{-1})_{\phi(p)}$ (cont) $\endgroup$ – peek-a-boo Aug 1 '20 at 2:18
  • $\begingroup$ (cont.) relative to the standard ordered bases on $\Bbb{R}^m$ and $\Bbb{R}^n$, which is simply the Jacobian matrix of the map $\psi\circ F \circ \phi^{-1}$ evaluated at the point $\phi(p)$. $\endgroup$ – peek-a-boo Aug 1 '20 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.