# Computing differentials on manifolds

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.

• 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 $f_{*,p}$ – Jackozee Hakkiuz Oct 13 '18 at 19:58
• 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? – Jackozee Hakkiuz Oct 13 '18 at 20:02
• The formula for $X$ was obtained from $R_\theta$ taking the derivative with respect to $\theta$. Such that $\gamma' (0)= X_p$ – Amphiaraos Oct 13 '18 at 20:42
• 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 $X_p(g\circ f): C^{\infty}(N) \to \mathbb{R}$. – 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 $$p$$.

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 $$p$$:
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.
• 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. – Amphiaraos Oct 14 '18 at 18:17