# Chapter 0, Proposition 2.7 Do Carmo Reimannian Geometry, bit of confusing notation

Going through the proof of the proposition in the title.

Let $$M_1^n, M_2^m$$ be differentiable manifolds and let $$\varphi : M_1 \to M_2$$ be a differentiable mapping. For every $$p \in M_1$$ and for each $$v \in T_p M_1$$, choose a differentiable curve $$\alpha : (-\epsilon,\epsilon) \to M_1$$ with $$\alpha(0) = p$$, $$\alpha'(0) = v$$. Take $$\beta = \varphi \circ \alpha$$. The mapping $$d \varphi_p : T_p M_1 \to T_{\varphi(p)} M_2$$ given by $$d \varphi_p (v) = \beta'(0)$$ is a linear mapping that does not depend on the choice of $$\alpha$$.

There's a very specific bit I can't figure, which I'd like you'd help me to elaborate.

Proof: Let $$x : U \to M_1$$ and $$y : V \to M_2$$ be a parameterizations at $$p$$ and $$\varphi(p)$$, respectively. Expressing $$\varphi$$ in these parameterization, we can write $$y^{-1} \circ \varphi \circ x (q) = (y_1(x_1,\ldots,x_n),\ldots,y_m(x_1,\ldots x_n))$$ where $$\begin{array}{l} q = (x_1,\ldots,x_n) \in U \\ (y_1,\ldots,y_m) \in V \end{array}.$$ On the other hand, expressing $$\alpha$$ in the parameterization $$x$$, we obtain $$x^{-1}\circ \alpha(t) = (x_1(t),\ldots,x_n(t)).$$ Therefore, $$y^{-1} \circ \beta(t) = (y_1(x_1(t),\ldots,x_n(t)),\ldots,y_m(x_1(t),\ldots x_n(t)))$$

Here now the very bit I don't understand

It follows the expression for $$\beta'(0)$$ with respect to the basis $$\left\{ \left( \frac{\partial}{\partial y_i}\right) _0\right\}$$ of $$T_{\varphi(p)}M_2$$, associated to the parameterization $$y$$, is given by $$\beta'(0) = \left(\sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0), \ldots, \sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0) \right).$$

The rest of the proof is clear

How exactly is the basis $$\left\{ \left( \frac{\partial}{\partial y_i}\right)_0 \right\}$$ used to derive an expression for $$\beta'(0)$$?

## 1 Answer

Note that $$\beta:(-\epsilon,\epsilon)\to M_2$$. So $$\beta(0)\in M_2$$ and since $$M_2$$ is of dimension $$m$$ so $$\beta(t) = (y_1(x_1(t),\ldots,x_n(t)),\ldots,y_m(x_1(t),\ldots x_n(t)))....(1)$$

Also $$\{(\frac{\partial}{\partial y_i})|_t:1\le i\le m\}$$ is a basis of the tangent space $$T_{\beta(t)}M_2$$ at $$\varphi(p)=\beta(t)$$ for any $$t\in(-\epsilon,\epsilon)$$.

Now differentiating $$(1)$$ at $$t=0$$ we obtain,

$$\beta'(0) = \left(\sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0), \ldots, \sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0) \right).$$ With respect to the basis means, $$\beta'(0) = \sum_{i=1}^{n} \frac{\partial y_1}{\partial x_i} x'_i(0)\frac{\partial}{\partial y_1}+ \ldots+\sum_{i=1}^{n} \frac{\partial y_m}{\partial x_i} x'_i(0)\frac{\partial}{\partial y_m} .$$

Hope this works.

• Need some time to digest your answer. – user8469759 Mar 21 at 12:06