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$f$ is a mooth mapping from differential manifold $M$ to $N$ and $T(f)$ is the induced mapping on their tangent bundles. For $x\in M$, does ${\rm{ran}}{{\rm{k}}_{{h_x}}}(T(f)) = 2{\rm{ran}}{{\rm{k}}_x}(f)$ hold? ($h_x$ is a tangent vector on $x$)

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How do you define $\text{rank}_x(f)?$ – Ehsan M. Kermani May 4 '12 at 4:16
up vote 2 down vote accepted


I'm guessing by $\text{rank}_x(f)$ you mean the rank of the linear map $T_x(f):T_xM\to T_{f(x)}N$? I believe it is clearer to denote this by $\text{rank }T_x(f)$, so I'll do that from now on.

We have a map $Tf:TM\to TN$, which we can consider as a map between two differentiable manifolds. Picking a point of $TM$ (let's call it $h_x$, to emphasize that it's a tangent vector of $M$ at the point $x$, even though we will now consider it as a point in $TM$), we can consider the map induced by $Tf$ at the point $h_x$. This is a map $$ T_{h_x}(Tf) : T_{h_x}TM \to T_{(Tf)(h_x)}TN $$ and it is linear. Now I think your question is whether the rank of this map is twice the rank of the linear map $T_x(f)$.

Your question is then:

Is $\text{rank } T_{h_x}(Tf) = 2\cdot\text{rank } T_x(f)$?


If this is your question, then I believe the answer is no. Consider $M=N=\mathbb{R}$ and take $$ f : \mathbb{R}\to\mathbb{R} : x\mapsto x^2.$$ Using the standard coordinates, the matrix of $T_0f$ is just the 1-by-1 zero matrix, so $$\text{rank }T_0f=0.$$ The coordinates on $TM$ are $(x,y)$, say, where $x$ is the point in $M$ where the vector is attached and $y$ is its length. Then we have $$Tf:\mathbb{R}^2\to\mathbb{R}^2:(x,y)\mapsto (x^2,2xy).$$ Taking $h_x=(0,1)$, we have that $T_{h_f}(Tf)$ is given by the matrix $$\begin{pmatrix}0&0\\2&0\end{pmatrix}$$ or maybe its transpose (I'm not sure, and I'm lazy). This means that $$\text{rank } T_{(0,1)}(Tf) = 1,$$ thereby providing a counterexample.


Oh, I believe that the inequality $$\text{rank } T_{h_x}(Tf) \geq 2\cdot\text{rank }T_x(f)$$ does hold, though I have not verified it in detail.

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WLOG take local coordinates for $M,N$ as $\mathbf{R}^m$ and $\mathbf{R}^n$, and consider a neighborhood of zero. A direct computation shows you that at $x = 0$, the map $T_{h_x}Tf$ is given by $$\begin{pmatrix} J_f & 0 \\ H_f\cdot h_x & J_f\end{pmatrix}$$ where $J_f$ is the Jacobian matrix of $f$ and $H_f$ is the Hessian matrix. That is, if $h_x = (0,y)$ in induced local coordinates on $TM$, $(H_f\cdot h_x)_j^k = y^i\partial_i\partial_j f^k$. Hence by the diagonal form of $TTf$ one sees that your remark is true. – Willie Wong May 4 '12 at 11:28
Yes, this is the argument I had in mind. – Daan Michiels May 4 '12 at 12:22
Dear @Willie, what kind of object is $H_f\cdot h_x$ exactly ? I agree with the right hand side, but I actually only have heard of the hessian of one function, not of a sequence of $n$ of them, like here. It would be great, if you could develop your comment into an answer so that we could refer to it, since I have never seen that kind of calculation. But of course I realize that you might not have the time for that and I hope you don't consider me rude for asking... – Georges Elencwajg May 4 '12 at 16:02
@GeorgesElencwajg: component wise Hessian. We are just taking a vector valued function and computing. I expressed the term in index notation anyhow later in the comment. Also, the answer is exactly the same as what Olivier wrote in another answer. – Willie Wong May 4 '12 at 16:09
Thanks, @Willie. – Georges Elencwajg May 4 '12 at 17:15

If you do an explicit calculation in local charts for $M$ and $N$, and with the induced local charts for $TM$ and $TN$ you get explicitely that at some vector $v=v_p\in T_p M$, the matrix of $T_v(Tf)$ (which has $2\cdot\mathrm{dim}M=2m$ columns and $2\cdot\mathrm{dim}N=2n$ lines) decomposes as four $m\times n$ matrices, the top left one and the bottom right one are the matrix you had for $T_pf$ in the original charts, the bottom left one is $0$, and the top right one depends bilinearly on the coordinates of $v$ and the second derivatives of the coordinates of $f$ read in our local charts. From this it follows that the rank of the linear map $T_{v_p}(Tf): T_{v_p} TM\rightarrow T_{T_pf(v_p)}N$ is always at least twice as large as the rank of $T_pf:T_pM\rightarrow T_{f(p)}N$, with equality at all points of the zero section of $TM\rightarrow M$, but may be larger elsewhere (unless $f$ is a submersion).

EDIT I confused the bottom left and top right entries. Usually, the induced charts on $TM|_U$ is defined as $ TX(v)=(x^1(p),\dots,x^m(p),v^1,\dots,v^m)$ where $v=v_p\in T_p M$ and $v=v^1\frac{\partial}{\partial x^1}+\cdots+v^m\frac{\partial}{\partial x^m}$.

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