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Given a differentiable manifold $M$ and some chart $(U, \psi)$ near $p$, we can consider the curve $\tilde{\beta}_i: t \mapsto \psi(p)+t e_i$, where $e_i$ denoted the standard basis in $\mathbb{R}^n$, $i \in \{1, \dots, n\}$. Now we can set $\beta_i := \psi^{-1} \circ \tilde{\beta}_i$ to get the corresponding curve on $M$ and we can define the corresponding tangent vector by

$$\left(\frac{\partial}{\partial x^i}\right)_{p,\psi} u := \left.\frac{d}{dt}\right|_{t=0} u(\beta_i(t))$$

for all $u \in C^{\infty}(M)$. We can quickly verify that this is indeed well-defined. A quick computation also shows that any linear combination of these vectors still lies in $T_pM$ and that they span $T_pM$. To show that they form a basis, it is left to show that they are linearly independent.

We have done a proof in class where at some point I must have made a typo or I simply fail to understand what is happening.

There exists a cutoff function $\rho: \mathbb{R}\to\mathbb{R}$ that is smooth and satisfies $$\rho(x) = \begin{cases}1 & x \in (-\frac{1}{2},\frac{1}{2})\\ 0 & x \in (-\infty, -\frac{3}{4}] \cup [\frac{3}{4},\infty)\end{cases}.$$ For $j \in \{1,\dots,n\}$ define $\varphi_j: M \to \mathbb{R}$ by $$\varphi_j(q) := \begin{cases} 0 & q \notin U\\ \left(\prod_{i=1}^n \rho \left(\frac{\psi^i(q)-\psi^i(p)}{\varepsilon}\right)\right)\psi^j(q) & q \in U \end{cases}.$$ Then, $\varphi_j \in C^{\infty}(M)$ and $$\left(\frac{\partial}{\partial x^i}\right)_{p,\psi} \varphi_j = \delta_{ij},$$ which implies that they are linearly independent.

Question 1: What is $\varepsilon$? In an earlier class we defined tangent vectors as linear maps arising as a directional derivative along some smooth curve $\gamma: (-\varepsilon,\varepsilon)\to M$ with $\gamma(0)=p$. This doesn't make any sense in the proof because we can choose $\varepsilon$ rather freely. I am pretty sure this must be a typo, so the real question is: What should the definition of $\varphi_j$ actually be?

Question 2: What does the function $\varphi_j$ do?

Question 3: Why does $$\left(\frac{\partial}{\partial x^i}\right)_{p,\psi} \varphi_j = \delta_{ij}$$ imply that the coordinate basis vectors are linearly independent?

Huy
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1 Answers1

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$\varepsilon$ is a parameter that ensures that $\varphi_j$ is smooth. The smoothness of $\varphi_j$ follows if the factor

$$\prod_{i = 1}^n \rho\biggl(\frac{\psi^i(q) - \psi^i(p)}{\varepsilon}\biggr)\tag{1}$$

is non-zero only on a relatively compact subset of the coordinate patch $U$. By the specification of $\rho$, that factor vanishes outside the subset $V$ of the coordinate patch $U$ corresponding to the cube with sidelength $\frac{3}{2}\varepsilon$ and centre $\psi(p)$, provided that cube is contained in $\psi(U)$. One chooses an arbitrary $\varepsilon > 0$ such that the cube $\prod \bigl[\psi^i(p)-\frac{3}{4}\varepsilon, \psi^i(p) + \frac{3}{4}\varepsilon\bigr]$ is contained in $\psi(U)$, then the function in $(1)$ has compact support in $U$, and $\varphi_j$ is smooth on the whole manifold.

The function $\varphi_j$ is a smooth extension of the coordinate function $\psi^j$ from a small neighbourhood - corresponding to $\prod \bigl( \psi^i(p) - \frac{\varepsilon}{2},\psi^i(p) + \frac{\varepsilon}{2}\bigr)$ - of $p$ to all of $M$. Since apparently the operation of tangent vectors is defined in terms of global smooth functions and not locally, we cannot directly use the coordinate functions $\psi^j$, since those are not necessarily extensible to global smooth functions. Therefore we multiply with a cutoff function to obtain a smooth extension of a restriction of $\psi^j$ to a smaller neighbourhood of $p$.

The relation

$$\biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi} \varphi_j = \delta_{ij}\tag{2}$$

implies linear independence by the linearity of differentiation in the $\frac{\partial}{\partial x^i}$. If we have a linear relation

$$\sum_{i = 1}^n c_i \biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi} = 0,$$

then applying that tangent vector to $\varphi_j$ yields

$$0 = \Biggl(\sum_{i = 1}^n c_i \biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi}\Biggr)\varphi_j = \sum_{i = 1}^n c_i \Biggl(\biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi}\varphi_j\Biggl) = \sum_{i = 1}^n c_i \delta_{ij} = c_j.$$

Doing that for all $1\leqslant j\leqslant n$ shows all coefficients $c_i$ are $0$, i.e. the linear independence.

Daniel Fischer
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