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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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