I have been studying the properties of the differential operator on manifolds. Given differentiable manifolds $M,~N$ and a function $f \in C^{\infty}(M,N)$, we define the differential at a point $p\in M$ as $d_pf: T_pM \to T_{f(p)}N$, the definition takes two forms:

  1. For a path $\gamma(t)$ in $M$ such that $\gamma (0)=p$, we define $d_pf\gamma'(0)= (f\circ \gamma)'(0)$
  2. For an $X_p \in T_pM$, we define for a $g \in C^{\infty}(N)$, the differential operator as $(d_pfX_p)(g)=X_p(g\circ f)$

Now here is my first question: how can I go on to prove that the two definitions are equivalent?

During lectures, we made quick example computation:

Find the differential of the function $f: \mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y)=R_\theta (x,y)$, where $R_\theta$ is the rotation by an angle $\theta$.

We know that for $f$, we have the vector field $X = - y \partial_x +x \partial_y$.

So we calculate $(dfX)_p= d_pfX_p=(f\circ \gamma)'(0)$ using the 1st definition.

This gives us $(R_\theta \circ \gamma)'(0)= \begin{bmatrix} cos(\theta) \gamma_1' - sin(\theta) \gamma'_2 \\ sin(\theta) \gamma'_1+cos(\theta)\gamma'_2\end{bmatrix}(0)$

Finally for a point $p=(p_1,p_2)$, we have that $(dfX)_p= \begin{bmatrix} - p_2 cos(\theta)\partial_x - p_1 sin(\theta) \partial_y \\ -p_2sin(\theta) \partial_x+p_2cos(\theta)\partial_y\end{bmatrix}$

So using the results from the example, for a function $g:N\to \mathbb{R}$, we should have that: $$ (d_pfX_p)(g)=\begin{bmatrix} - p_2 cos(\theta)\partial_x(g) - p_1 sin(\theta) \partial_y(g) \\ -p_2sin(\theta) \partial_x(g)+p_2cos(\theta)\partial_y(g)\end{bmatrix}$$

This should be the same result that I should get by applying the second definition. But this is not the case as I get, using $g_i=g\circ f_i$, the following result: $$ (d_pfX_p)(g)=X_p(g\circ f)= \begin{bmatrix} g_1(p)\partial_x(g)\\ g_2(p)\partial_y(g)\end{bmatrix}$$

My second question is then: how can I get the same results using both definitions in the computation?

I feel the two matrices represent the same transformation but they are on different basis so that is why they look different. But I don't know how to change the basis.

  • $\begingroup$ I think you are confusing two different maps. The first definition is the gradient $d_pf:T_pM\to\mathbb{R}$ of a smooth function $f:M\to\mathbb{R}$. The second definition is the differential $d_pf:T_pM\to T_{f(p)}N$ of a function $f:M\to N$. Of course they are related but they are not the same thing. That's why the second one (differential) is also known as the tangent map $T_pf$, or the push forward $f_{*,p}$ $\endgroup$ – Jackozee Hakkiuz Oct 13 '18 at 19:58
  • $\begingroup$ That said, I don't understand what your function $R_\theta:\mathbb{R}^{2}\to\mathbb{R}^{2}$ does, and how you obtained $X=-y\partial_x+x\partial_y$ from it. Could you clarify? $\endgroup$ – Jackozee Hakkiuz Oct 13 '18 at 20:02
  • $\begingroup$ The formula for $X$ was obtained from $R_\theta$ taking the derivative with respect to $\theta$. Such that $\gamma' (0)= X_p$ $\endgroup$ – Amphiaraos Oct 13 '18 at 20:42
  • $\begingroup$ We have $d_pf: T_pM\to T_{f(p)}N$. In the first definition by using the curve we get $(d_pf\gamma'(0))(g) \in T_{f(p)}N$ in the second we use the vector field to make the calculation $(d_pfX_p)= X_p(g\circ f)$. Changing $g$, the expression $X_p(g\circ f): C^{\infty}(N) \to \mathbb{R}$. $\endgroup$ – Amphiaraos Oct 13 '18 at 20:55

They are not equivalent.

For a path $\gamma:\mathbb{R}\to M$ and a smooth function $f:M\to\mathbb{R}$ we define the tangent vector $X_{\gamma,p}$ (where $p=\gamma(0)$) to be the map \begin{align*} X_{\gamma,\gamma(0)}:&\,C^{\infty}(M)\to\mathbb{R}\\ &f\mapsto X_{\gamma,\gamma(0)}f:=(f\circ\gamma)'(0) \end{align*} And we call $T_pM$ the space of all such maps at $p$.

Your first definition corresponds to

For an smooth manifold $M$, a smooth map $f:M\to\mathbb{R}$ and a point $p\in M$ we define the map \begin{align*}d_pf:&\,T_pM\to\mathbb{R}\\ &X_p\mapsto d_pf(X_p):=X_pf \end{align*} called the gradient of $f$ at $p$:

And your second definition corresponds to

For smooth manifolds $M$ and $N$, a smooth function $f:M\to N$ and a point $p\in M$ we define the map \begin{align*} T_pf:&\,T_pM\to T_{f(p)}N\\ &X_p\mapsto T_pf(X_p) \end{align*} defined for any smooth $g:N\to\mathbb{R}$ by $$[T_pf(X_p)]g:=X_p(g\circ f)$$ called the tangent map of a $f$ at $p$.

Remark: the tangent map $T_pf$ is also called the pushforward $f_{*,p}$, or the differential $d_pf$.

Please let me know if it helped you, and as I mentioned in the comments, it would be helpful if you clarified what your map $R_\theta$ is so I can help if you have problems.

  • $\begingroup$ I had written the problem wrong. The $X$ vector field had nothing to do with the first part of the exercise. The definitions you gave me helped to solve, not only this problem but other problems. $\endgroup$ – Amphiaraos Oct 14 '18 at 18:17
  • $\begingroup$ @Amphiaraos Hey! I'm glad to help. :) $\endgroup$ – Jackozee Hakkiuz Oct 15 '18 at 1:00

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.